IJSO Plus

FORCE AND NEWTON’S LAWS OF MOTION EFFECTS OF FORCE

ILLUSTRATIONS

To define force first of all one has to see the effects of force. By ‘effects of force’ we mean what force can do or what changes a force can bring about.

1.

Effects of Force : A force can produce the following effects : (i)

A force can move a stationary body.

(ii) A force can stop a moving body. (iii) A force can change the speed of a moving body. (iv) A force can change the direction of a moving body. (v) A force can change the shape (and size) of a body.

A force produces an acceleration of 5.0 cm/s2 in a body of mass 20g. Then find out the force acting on the body in Newton.

Sol. Acceleration of the body, a = 5 cm/s = 0.05 m/s Mass of the body, m = 20 g = 0.02 kg  F = ma F = 0.05 × 0.02 = 10–3 N 2.

A force of 15 N acts on a body of mass 5 kg for 2s. What is the change in velocity of body ?

Sol. Given : F = 15 N , t = 2s , m = 5 kg F = ma Based on the effects of force, it may be defined as : a= Force is a pull or push, which changes or tends to change the state of rest or of uniform motion of a body or changes its direction or shape. (a) Mathematical Representation of Force : Mathematically, force F is equal to the product of mass, m of a body and acceleration a, produced in the body due to that force. i.e. F = ma Where a = final velocity – initial velocity/time (b) Units of Force : (i) In C.G.S. system :

a =

F 15 = = 3 m/s2 m 5

v u v – u = at = 3 × 2 = 6m/s t

RESULTANT FORCE Many forces may be simultaneously applied on a body, for example- several persons may jointly make an effort to move a heavy body, each person pushes it i.e. each person applies a force on it. t is also possible that a stronger man pushes that body hard enough and produces same acceleration in it. f a single force acting on a body produces the same acceleration as produced by a number of forces, then that single force is called the resultant force of these individual forces .

 F = ma  gram × cm/s2 = dyne If m = 1 gram, a = 1 cm/s2, then F = 1 dyne When a force is applied on a 1 gram body and the acceleration produced in the body is 1 cm/s2 then the force acting on the body will be one dyne. (ii) In S.I. system : F = ma  kg × m/s2 = Newton If m = 1 kg and a = 1 m/s2 then by F = ma, 2

F = 1 × 1 = 1 kg × m/s = 1 Newton. If a force is applied on a body of mass 1 kg and acceleration produced in the body is 1 m/s2 then the force acting on the body will be one Newton. 

Relationship between the newton and dyne 1 N = 1 kg × 1 m s–2 = 1000 g × 100 cm s–2 = 100000 g cm s–2 = 105 dyne Thus 1 N = 105 dyne

FUNDAMENTAL FORCES All forces observed in nature such as muscular force, tension, reaction, friction, weight, electric, magnetic, nuclear, etc., can be explained in terms of only following four basic interactions. (a) Gravitational Force : The force of interaction which exists between two particles of masses m1 and m2, due to their masses is called gravitational force. The gravitational force acts over long distances and does not need, any intervening medium. Gravitational force is the weakest force of nature. (b) Electromagnetic Force : Force exerted by one particle on the other because of the electric charge on the particles is called electromagnetic force. Following are the main characteristics of electromagnetic force PHYSICS_IJSO_PAGE #11

(i)

These can be attractive or repulsive.

DETAILED ANALYSIS OF CONTACT FORCE

(ii) These are long range forces. (a) Normal force (N) : (iii) These depend on the nature of medium between the charged particles. (iv) All macroscopic forces (except gravitational) which we experience as push or pull or by contact are electromagnetic, i.e., tension in a rope, the force of friction, normal reaction, muscular force, and force experienced by a deformed spring are electromagnetic forces. These are manifestations of the electromagnetic attractions and repulsions between atoms/molecules.

It is the component of contact force perpendicular to the surface. It measures how strongly the surfaces in contact are pressed against each other. It is the electromagnetic force. e.g.1 A table is placed on Earth as shown in figure

(c) Nuclear Force : It is the strongest force. It keeps nucleons (neutrons and protons) together inside the nucleus inspite of large electric repulsion between protons. Radioactivity, fission, and fusion, etc. results because of unbalancing of nuclear forces. It acts within the nucleus that too upto a very small distance. It does not depends on charge and acts equally between a proton and proton, a neutron and neutron, and proton and neutron, electrons does not experience this force. It acts for very short distance order of 10–15 m.

Here table presses the earth so normal force exerted by four legs of table on earth are as shown in figure.

e.g.2 A boy pushes a block kept on a frictionless surface.

(d) Weak Force : It acts between any two elementary particles. Under its action a neutron can change into a proton emitting an electron and a particle called antineutrino. The range of weak force is very small, in fact much smaller than size of a proton or a neutron. It has been found that for two protons at a distance of 1 fermi :

Here, force exerted by boy on block is electromagnetic interaction which arises due to similar charges appearing on finger and contact surface of block, it is normal force.

FN:FEM:FW:FG::1:10–2:10–7:10–38

On the basis of contact forces are classified into two categories (i)

A block is kept on inclined surface. Component of its weight presses the surface perpendicularly due to which contact force acts between surface and block.

Contact forces

(ii) Non contact or field forces (a) Contact force : Forces which are transmitted between bodies by short range atomic molecular interactions are called contact forces. When two objects come in contact they exert contact forces on each other. e.g. Normal, Tension etc.

Normal force exerted by block on the surface of inclined plane is shown in figure. Here normal force is a component of weight of the body perpendicular to the inclined surface i.e. N = mgsin

( b ) Field force : Force which acts on an object at a distance by the interaction of the object with the field produced by other object is called field force. e.g. Gravitational force, Electro magnetic force etc.

Force acts perpendicular to the surface

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3.

Two blocks are kept in contact on a smooth surface as shown in figure. Draw normal force exerted by A on B.

F = 0 spring in natural length does not exerts any force on its ends

x F

F

F = – kx ;k = spring constant or stiffness constant (unit = N/m) x = extension in spring

Sol. In above problem, block A does not push block B, so there is no molecular interaction between A and B. Hence normal force exerted by A on B is zero.

x

 Note : •

Fext

F

F

Normal is a dependent force it comes in role when one surface presses the other.

Fext

F = – kx x = compression in spring

 Note : Spring force is also electromagnetic in nature :

(b) Tension : Tension is the magnitude of pulling force exerted by a string, cable, chain, rope etc. W hen a string is connected to a body and pulled out, the string said to be under tension. It pulls the body with a force T, whose direction is away from the body and along the length of the string. Usually strings are regarded to be massless and unstretchable, known as ideal string.

(d) Friction force : When a body is moving on a rough surface resistance to the motion occurs because of the interaction between the body and its surroundings. We call such resistance as force of friction. Friction is also considered as component of contact force which acts parallel to the surfaces in contact. (i) Origin of friction : The frictional force arises due to molecular interactions between the surfaces at the points of actual contact. When two bodies are placed one over other, the actual area of contact is much

 Note : (i) Tension in a string is an electromagnetic

smaller then the total surface areas of bodies. The

force and it arises only when string is pulled. If a massless string is not pulled, tension in it is zero. (ii) String can not push a body in direct contact.

molecular forces starts operating at the actual points

(c) Force Exerted by spring :

the other, these bonds are broken, and the material

A spring is made of a coiled metallic wire having a definite length. When it is neither pushed nor pulled then its length is called natural length. At natural length the spring does not exert any force on the objects attached to its ends.f the spring is pulled at the ends, its length becomes larger than its natural length, it is known as stretched or extended spring. Extended spring pulls objects attached to its ends.

get deformed and new bonds are formed. The local

A

of contact of the surfaces. Molecular bonds are formed at these contact points. When one body is pulled over

deformation sends vibrations into the bodies. These Vibrations ultimately dumps out and energy of vibrations appears as heat. Hence to start or carry on the motion, there is a need of force.

Body 1

B Normal spring

Spring force on A

Body 2 Spring force on B

A

B

Actual area of contact

Stretched spring Spring force on A A

(ii) Statics and Kinetic Frictions :

Spring force on B B

Compressed spring

If the spring is pushed at the ends, its length becomes less than natural length. It is known as compressed spring. A compressed spring pushes the objects attached to its ends.

•

Experiment : (A) Consider a block placed on a table, and a small force F1 is acted on it. The block does not move. It indicates that the frictional force fs starts acting in opposite direction of applied force and its magnitude is equal of F1(figure b). That is for the equilibrium of PHYSICS_IJSO_PAGE #33

(B) About kinetic friction :

the block, we have F1 – fs = 0 or F1 = fs

1.

The kinetic friction depends on the materials of the surface in contact.

the surface is called static friction (fs).

2.

(B) As the applied force increases the frictional force also increases. When the applied force is increased up to a certain limit (F2) such that the block is on the verge of motion. The value of frictional force at this stage is called limiting friction flim (figure c).

It is also independent of apparent area of contact as long as the magnitude of normal reaction remains the same.

3.

Kinetic friction is almost independent of the velocity, provided the velocity is not too large not too small. The kinetic friction is directly proportional to

The force of friction when body is in state of rest over

4.

the magnitude of the normal reaction between the surfaces. fk = k N. Here k is coefficient of kinetic friction.

 We can write, k = •

fk N

There are two types of kinetic frictions: (i)

Sliding friction : The force of friction when one body slides over the surface of the another body is called sliding friction.

(C) Once the motion started, the smaller force is now necessary to continue the motion (F3) and thus frictional force decreases. The force of friction when body is in state of motion over the surface is called kinetic or dynamic friction fk (figure d).

(ii) Rolling friction : When a wheel rolls without slipping over a horizontal surface, there is no relative motion of the point of contact of the wheel with respect to the plane. Theoretically for a rolling wheel the frictional force is zero. This can only possible when bodies in contact are perfectly rigid and contact of wheel with the surface is made only at a point. But in practice no material body is perfectly rigid and therefore bodies get deformed when they pressed each other. The actual area of their contact no longer remains a point, and thus a small amount of friction starts acting between the body and the surface. Here frictional force is

(iii) More about frictional force : (A) About static friction 1.

The limiting friction depends on the materials of the surfaces in contact and their state of polish.

2.

The magnitude of static friction is independent of the apparent area of contact so long as the normal reaction remains the same.

3.

The limiting friction is directly proportional to the magnitude of the normal reaction between the two surfaces i.e. flim= SN. Here s is coefficient of static friction.

f  We can write, s = lim N

called rolling friction. It is clear from above discussion that rolling friction is very much smaller than sliding friction.

flim > fkinetic > frolling.

 Note : s and k are dimensionless quantities and independent of shape and area of contact . It is a property of two contact surfaces. s will always be greater than k .Theoretical value of  can be o to  but practical value is 0 <   1.6

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GALILEO’S EXPERIMENTS (a) Conservative Force :

Experiment 1 : It was observed by Galileo that when a ball is rolled

A force is said to be conservative if the amount of work done in moving an object against that force is independent on the path. One important example of conservative force is the gravitational force. It means that amount of work done in moving a body against gravity from location A to location B is the same whichever path we may follow in going from A to B. This is illustrated in figure.

down on an inclined frictionless plane its speed increases, whereas if it is rolled up an inclined frictionless plane its speed decreases .If it is rolled on a horizontal frictionless plane the result must be between the cases describe above i.e. the speed should remain constant. It can be explain as :

A force is conservative if the total work done by the force on an object in one complete round is zero, i.e. when the object moves around any closed path (returning to its initial position).

v’

v

A force is conservative if there is no change in kinetic energy in one complete round. KE = 0

v’ = v

moving down : speed increases moving up : speed decreases moving horizontal : speed remains constant

This definition illuminates an important aspect of a conservative force viz. Work done by a conservative force is recoverable. Thus in figure, we shall have to do mgh amount of work in taking the body from A to B. However, when body is released from B, we recover mgh of work.

Experiments 2 : When a ball is released on the inner surface of a smooth hemisphere, it will move to the other side and reach the same height before coming to rest momentarily. f the hemisphere is replaced by a surface shown in figure(b) in order to reach the same height the ball will have to move a larger distance.

Other examples of conservative forces are spring force, electrostatic force etc. (b) Non-Conservative Force : A force is non-conservative if the work done by that force on a particle moving between two points depends on the path taken between the points.

h

The force of friction is an example of non-conservative force. Let us illustrate this with an instructive example.

h (a)

(b)

Suppose we were to displace a book between two points on a rough horizontal surface (such as a table). If the book is displaced in a straight line between the two points, the work done by friction is simply FS where :

v

F = force of friction ; S = distance between the points.

v

(c)

However, if the book is moved along any other path between the two points (such as a semicircular path),

If the other side is made horizontal, the ball will never stop because it will never be able to reach the same height, it means its speed will not decrease. It will have uniform velocity on the horizontal surface. Thus, if unbalanced forces do not act on a body, the body will either remain at rest or will move with a uniform velocity. It will remain unaccelerated.

the work done by friction would be greater than FS. Finally, if the book is moved through any closed path, the work done by friction is never zero, it is always negative. Thus the work done by a non-conservative force is not recoverable, as it is for a conservative force. 

Newton concluded the idea suggested by Galileo and was formulated in the laws by Newton.

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NEWTON’S FIRST LAW OF MOTION Every body remain in its state of rest or uniform motion in a straight line unless it is compelled by some external force. It means a body remain unaccelerated if and only if, the resultant force on it is zero. In such a case the body is said to be in equilibrium.

INERTIA (a) Definition of Inertia : The tendency of the body to oppose the change its states of rest or uniform motion in a straight line is called inertia. Newton’s first law of motion is also called law of inertia. (b) Description : It follows from first law of motion that in absence of any external force, a body continues to be in its state of rest or in uniform motion along a straight line. In other words, the body cannot change by itself its position of rest or of uniform motion. (c) Inertia Depends upon Mass : We know that it is difficult to move a heavier body than the lighter one. Similarly it is difficult to stop a moving heavier body than a lighter body moving with the same velocity. Thus, we conclude that mass of the body is the measure of inertia, more the mass, more the inertia.

TYPES OF INERTIA

(c) Inertia of Direction : The tendency of a body to oppose any change in its direction of motion is known as inertia of direction. Example based on Inertia of direction : Tie a stone to one end of a string and holding other end of the string in hand, rotate the stone in a horizontal circle. If during rotation, the string breaks at certain stage, the stone is found to fly off tangentially at that point of the circle.

String Breaks

String breaks, stone goes away tangentially

There are three types of Inertia which are : (a) Inertia of Rest : The tendency of the body to oppose the change in its state of rest when some external unbalance force is applied on it, is called the inertia of rest. Example based on Inertia of rest : A person sitting in a bus falls backwards when the bus suddenly starts. The reason is that lower part of his body begins to move along with the bus but the upper part of his body tends to remain at rest due to inertia of rest. (b) Inertia of Motion : The tendency of the body to oppose its state of motion when some unbalance forces are applied on it, is called the inertia of motion. Example based on Inertia of motion : A man carelessly getting down a moving bus falls forward, the reason being that his feet come to rest suddenly, whereas the upper part of his body retains the forward motion.

Definition of force from first law of motion : According to first law of motion, if there is no force, there is no change in state of rest or of uniform motion. In other words, if a force is applied, it may change the state of rest or of uniform motion. If the force is not sufficient, it may not produce a change but only try to do so. Hence force is that which changes or tries to change the state of rest or of uniform motion of a body in straight line.

PHYSICS_IJSO_PAGE #66

So the magnitude of the resultant force acting on a body is equal to the product of mass of the body and the acceleration produced. Direction of the force is same as that of the acceleration.

MOMENTUM Definition : Momentum of a particle may be defined as the quantity of motion possessed by it and it is measured by the product of mass of the particle and its velocity.

Momentum is a vector quantity and it is represented

UNITS OF FORCE



by p

(a) In C.G.S. System : 

 F = ma  gm × cm/s2 = Dyne



p mv



Definition of one dyne : Unit of momentum : If m = 1 gm, a = 1 cm/s2, then F = 1 dyne.

(In C.G.S. system)  p = mv  gram × cm/s = dyne × s

When a force is applied on a body of mass 1 gram and the acceleration produced in the body is 1 cm/s2, then the force acting on the body will be one dyne.

(In M.K.S. system)  p = mv  kg × m/s = Newton × s 4.

A ball of mass 100 gm. is moving with a velocity of 15 m/s. Calculate the momentum associated with the ball.

(b) In S.I. System : F = ma  kg × m/s2 = Newton

Sol . Definition of one Newton : Mass of the ball

Velocity of the ball So, momentum

100 = 100 gm. = kg. 1000

If m = 1 kg and a = 1 m/s2 then by, F = ma F = 1 × 1 = 1 kg × m/s2 = 1 N. If a force is applied on a body of mass 1 kg and acceleration produced in the body is 1 m/s2, then the force acting on the body will be one Newton.

= 0.1 kg. = 15 m/s = mass of the ball × velocity of the ball = 0.1 kg. × 15 m/s = 1.5 kg. m/s

(c) Kilogram Force (kgf) : Kilogram force (kgf) or Kilogram weight (kg. wt.) is force with which a mass of 1 kg is attracted by the earth towards its centre. 1kgwt = 1kgf = 9.8 N

NEWTON’S SECOND LAW OF MOTION The rate of change of momentum of a body is directly proportional to the applied unbalanced forces i.e. Rate of change of momentum  Force applied

(d) Gram Force (gf) : Gram force or gram weight is the force with which a mass of 1 gram is attracted by the earth towards its centre. 1gwt = 1gf = 981 dyne

Let a body is moving with initial velocity u and after applying a force F on it, its velocity becomes v in time t. Initial momentum of the body p1 = mu Final momentum of the body p2 = mv Change in momentum in time t is mv – mu So rate of change of momentum =

F

m( v – u) t

Here,

Relation between Newton and dyne. We know : 1 N = 1kg × 1ms-2 or 1 N = 1000 g × 100 cms-2 or 1 N = 105 g cms-2 = 105 dyne  1 N = 105 dyne

mv – mu t

But according to Newton’s second law,

or

Abou both the units are called gravitational unit of force.

mv – mu F t

v–u = a (acceleration) t

So Fma or F = kma (Here k is proportionality constant. If 1N force is applied on a body of mass 1 kg and the acceleration produced in the body is 1 m/s 2, then 1 = k × 1 × 1 or k = 1 Hence, F = ma

5.

A force of 20N acting on a mass m1 produces an acceleration of 4 ms–2. The same force is applied on mass m2 then the acceleration produced is 0.5 ms–2. What acceleration would the same force produce, when both masses are tied together ?

Sol. For mass m1: F = 20N, a = 4 ms–2 then

m1 =

20 F = = 5 kg a 4

For mass m2 : F = 20N, a = 0.5 ms–2

PHYSICS_IJSO_PAGE #77

then

m2 =

F 20 = = 40 kg a 0 .5

When m1 and m2 are tied together : Total mass = m1 + m2 = 45 kg, F = 20N

20 F a= = = 0.44 ms–2 45 (m1  m 2 )

then

IMPULSE OF FORCE A large force acting for a short time to produce a finite change in momentum is called impulsive force. The product of force and time is called impulse of force. i.e., Impulse = Force × Time

Demonstration- Newton’s third law of motion The other end of the spring balance A is pulled out to the left. Both balances show the same reading (20 N) for the force. The pulled balance A exerts a force of 20N on the balance B. It acts as action, B pulls the balance A in opposite direction with a force of 20 N. This force is known as reaction. We conclude that action-reaction forces are equal and opposite and act on two different bodies.

NO ACTION IS POSSIBLE WITHOUT REACTION Examples : (i) A nail cannot be fixed on a suspended wooden ball.

or

I = Ft

(ii) A paper cannot be cut by scissors of single blade.

The S.I. unit of impulse is Newton-second (N-s) and the C.G.S unit is dyne- second (dyne-s)

(iii) A hanging piece of paper cannot be cut by blade. (iv) Writing on a hanging page is impossible.

Impulse and Momentum : From Newton’s second law of motion Force, F =

p 2  p1 t

or Ft = p2 – p1

i.e., Impulse = Change in momentum This relation is called impulse equation or momentumimpulse theorem. It has an important application in our everyday life.

IMPULSE DURING AN IMPACT OR COLLISION The impulsive force acting on the body produces a change in momentum of the body on which it acts. We know, Ft = mv – mu, therefore the maximum force needed to produce a given impulse depends upon time. If time is short, the force required in a given impulse or the change in momentum is large and viceversa.

NEWTON’S THIRD LAW (a) Statement : The law states that “ To every action there is an equal and opposite reaction“. Moreover, action and reaction act on different bodies.

(v) Hitting on a piece of sponge does not produce reaction. You do not enjoy hitting.

ACTION AND REACTION ARE NOT BALANCED Action and reaction, though equal and opposite are not balanced because they act on two different bodies. In case when they act on two different bodies forming a single system, they become balanced. ANY PAIR OF EQUAL AND OPPOSITE FORCES IS NOT AN ACTION–REACTION PAIR Consider a book kept on a table. We have seen that the table pushes the book in the upward direction. Then why does not the book fly up? It does not fly up because there is another force on the book pulling it down. This is the force exerted by the earth on the book, which we call the weight of the book. So, there are two forces on the book– the normal force, N acting upwards, applied by the table and the force, W acting downwards, applied by the earth. As the book does not accelerate, we conclude that these two forces are balanced. In other words, they have equal magnitudes but opposite directions.

N

(b) Demonstration : Two similar spring balances A and B joined by hook as shown in the figure. The other end of the spring balance B is attached to a hook rigidly fixed in a rigid wall.

N=W V

V

V

Can we call N the action and W the reaction ? We cannot. This is because, although they are equal and PHYSICS_IJSO_PAGE #88

By Newtons third law, F1 = –F2

opposite, they are not forces applied by two bodies on each other. The force N is applied by the table on the book, its reaction will be the force applied by the book on the table. Weight W is the force applied by the earth on the book, its reaction will be the force applied by the book on the earth. So, although N and W are equal and opposite, they do not form an action–reaction pair.

m1v 1 – m1u1  m v – m 2u 2  = – 2 2   m1v1 – m1u1 t t  

= –m2v2 + m2u2 or

m1u1 + m2u2 = m1v1 + m2v2

or

Initial momentum = Final momentum

6. PRINCIPLE OF CONSERVATION OF LINEAR MOMENTUM

By Newton’s second law, the rate of change of momentum is equal to the applied force. Change in momentum = Force time

Change in momentum = F × t If F = 0 then,

Two ice hockey players, suitably padded collide directly with each other and immediately become entangled. One has a mass of 120 kg and is travelling at 2 m/s, while the other has a mass of 80 kg and is travelling at 4 m/s towards the first player at what speed do they travel after they become entangled? Sol. m1u1 + m2u2 = m1v1 + m2v2 120 × 2 – 80 × 4 = (120 + 80) 240 – 320 = 200 V – 80 = 200 V

Change in momentum = 0 If the force applied on the body is zero then its momentum will be conserved, this law is also applicable on the system. If in a system the momentum of the objects present in the system are P1, P2, P3........... and external force on the system is zero, then– P1 + P2 + P3 +................. = Constant 

NOTE : If only internal forces are acting on the system then its linear momentum will be conserved. (a) The Law of Conservation of Linear Momentum by Third Law of Motion : Suppose A and B are two objects of masses m1 and m2 are moving in the same direction with velocity u1 and u2 respectively (u1 > u2). Object A collides with object B and after time t both move in their original direction with velocity v1 and v2 respectively. The change in momentum of object A = m1v1 – m1u1

m1

m2 u1 u2 before collision (u1 > u2)

The force on B by A is F1 =

F1 =

Change in momentum time

m1v 1 – m1u1 t

.............(1)

The change in momentum of object B = m2v2 – m2u2

V=–

2 m/s 5

SOME ILLUSTRATION ON CONSERVATION OF MOMENTUM (a) Recoil of Gun : A loaded gun (rifle) having bullet inside it forming one system is initially at rest. The system has zero initial momentum.

When the trigger (T) is pressed, the bullet is fired due to internal force of explosion of powder in cartidge inside. The bullet moves forward with a high velocity and the gun move behind (recoils) with a lesser velocity. Let the bullet and the gun have masses m and M respectively. Let the bullet move forward with velocity v and the gun recoils with velocity V. Then final momentum of the gun and bullet is MV + mv By the law of conservation of momentum– Initial momentum of the system = Final momentum of the system. 0 = MV + mv or V = –

The force on A by B is F2 =

Change in momentum time

v

V

mv M

Hence the recoil velocity of gun = =

m 2 v 2 – m 2u 2 t m1

mv M

.............(2)

m2 v1 v2 after collision

and the velocity of the gun is = –

mv M

PHYSICS_IJSO_PAGE #99

(ii) Are forces acting on block forms action- reaction pair.

(b) The Working of a Rocket :

7.

the momentum of a rocket before it is fired is zero. When the rocket is fired, gases are produced. These gases come out of the rear of the rocket with high speed. The direction of the momentum of the gases coming out of the rocket is in the downward direction. Thus, to conserve the momentum of the system i.e., (rocket + gases), the rocket moves upward with a momentum equal to the momentum of the gases. So, the rocket continues to move upward as long as the gases are ejected out of the rocket. Thus a rocket works on the basis of the law of conservation of momentum. A bullet of mass 0.01 kg is fired from a gun weighing

(iii) If answer is no, draw action reaction pair. Sol.(i)

F.B.D. of block

(ii) ‘N’ and mg are not action -reaction pair. Since pair act on different bodies, and they are of same nature.

5.0 kg. If the speed of the bullet is 250 m/s, calculate the speed with which the gun recoils.

(iii) Pair of ‘mg’ of block acts on earth in opposite direction.

1 250 Sol. V = = – 50 m/s 5

SYSTEM

earth

mg

Two or more than two objects which interact with each other form a system. Classification of forces on the basis of boundary of system :

and pair of ‘N’ acts on surface as shown in figure. N

(a) Internal Forces : Forces acting with in a system among its constituents. (b) External Forces : Forces exerted on the constituents of a system by the outside surroundings are called as external forces.

9.

Two sphere A and B are placed between two vertical walls as shown in figure. Draw the free body diagrams of both the spheres. B

FREE BODY DIAGRAM A free body diagram consists of a diagrammatic representations of single body or a subsystem of bodies isolated from surroundings showing all the forces acting on it.

A

Sol.F.B.D. of sphere ‘A’ :

 Steps for F.B.D. Step 1 : Identify the object or system and isolate it from other objects, clearly specify its boundary. Step 2 : First draw non-contact external force in the diagram, generally it is weight. Step 3 : Draw contact forces which acts at the boundary of the object of system. Contact forces are normal ,

F.B.D. of sphere ‘B’ : (exerted by A)

friction, tension and applied force. In F.B.D, internal forces are not drawn only external are drawn. 8.

A block of mass ‘m’ is kept on the ground as shown in figure.

 Note : Here NAB and NBA are the action - reaction pair (Newton’s third law).

(i)

Draw F.B.D. of block. PHYSICS_IJSO_PAGE #10 10

10. Draw F.B.D. for systems shown in figure below.

O (b) Unstable equilibrium : If on slight displacement from equilibrium position a body moves in the direction of displacement, the equilibrium is said to be unstable. In this situation potential energy of body is maximum and so center of gravity is highest. Sol.

O

(c) Neutral equilibrium : If on slight displacement from equilibrium position a body has no tendency to come back to its original position or to move in the direction of displacement, it is said to be in neutral equilibrium. In this situation potential energy of body is constant and so center of gravity remains at constant height.

. (a) Newtons 2 nd law of motion :

TRANSLATORY EQUILIBRIUM The rate of change of linear momentum of a body is directly proportional to the applied force and the change takes place in the direction of the applied force.

When several forces acts on a body simultaneously in such a way that resultant force on the body is zero, i.e.,

  F = 0 with F =











In relation F = ma the force F stands for the net

Fi the body is said to be in translatory

equilibrium. Here it is worthy to note that :

external force. Any internal force in the system is not to

(i)

be included in F .



As if a vector is zero all its components must vanish i.e. in equilibrium as -

  F = 0 with F =

F

x

=0;





F

Fi = 0

y

 =0;

F

z

=0

So in equilibrium forces along x axes must balance each other and the same is true for other directions. If a body is in translatory equilibrium it will be either at rest or in uniform motion. If it is at rest, equilibrium is called static, otherwise dynamic. Static equilibrium can be divided into following three types : (a) Stable equilibrium : If on slight displacement from equilibrium position a body has a tendency to regain its original position it is said to be in stable equilibrium. In case of stable equilibrium potential energy is minimum and so center of gravity is lowest.

In S.I. the absolute unit of force is newton (N) and gravitational unit of force is kilogram weight or kilogram force (kgf.) Note : The absolute unit of force remains the same everywhere, but the gravitational unit of force varies from place to place because it depends on the value of g. ( b ) Applications of Newton’s 2 nd Law (i)

When objects are in equilibrium : Steps to solve problem involving objects in equilibrium :

Step 1 : Make a sketch of the problem. Step 2 : Isolate a single object and then draw the freebody diagram for the object. Label all external forces acting on it. Step 3 : Choose a convenient coordinate system and resolve all forces into rectangular components along x and Y direction. Step 4 : Apply the equations

F

x

 0 and

F

y

 0.

Step 5 : Step 4 will give you two equations with several unknown quantities. If you have only two unknown quantities at this point, you can solve the two equations

PHYSICS_IJSO_PAGE #11 11

y

for those unknown quantities. Step 6 : If step 5 produces two equations with more than two unknowns, go back to step 2 and select

60º T3

another object and repeat these steps. Eventually at step 5 you will have enough equations to solve for all

30º

F

y

11. A ‘block’ of mass 10 kg is suspended with string as shown in figure. Find tension in the string. (g = 10 m/s2).

= 0 T4 cos 60º = T2 cos 30º

 T4 = 200 N and

F

x

= 0 T3 + T2 sin30º = T4 sin 60º

200

Sol.F.B.D. of block For equilibrium of block along Y axis y

x

T2

unknown quantities.

F

B

T4

T3 =

3

N

13. Two blocks are kept in contact as shown in figure. Find :-

0

(a) forces exerted by surfaces (floor and wall) on blocks.

T – 10 g = 0 T = 100 N 12. The system shown in figure is in equilibrium. Find the magnitude of tension in each string ; T1 , T2, T3 and T4. (g = 10 m/s2).

(b) contact force between two blocks.

Sol. A : F.B.D. of 10 kg block

Sol.F.B.D. of 10 kg block For equilibrium of block along Y axis.

F

y

N1 = 10 g = 100 N .......(1) N2 = 100 N .........(2) F.B.D. of 20 kg block

T0

0

T0 = 10 g T0 = 100 N F.B.D. of point ‘A’

10g y

F

y

T2

0 30º

T2 cos 30º = T0 = 100 N

T1

x

A

200  T2 =



3

N T0

N2 = 50 sin 30º + N3 N3 = 100 – 25 = 75 N & N4 = 50 cos 30º + 20 g N4 = 243.30 N 14. Find magnitude of force exerted by string on pulley.

Fx  0

T1 = T2 . sin 30º

200 =

3

.

1 100 = N. 3 2

Sol B.

F.B.D. of 10 kg block :

F.B.D. of point of ‘B’

T = 10 g = 100 N F.B.D. of pulley :

PHYSICS_IJSO_PAGE #12 12

F

x

Since string is massless, so tension in both sides of string is same. So magnitude of force exerted by string on pulley

N = m2 . a N=

=

100 2  100 2

= 100 2 N

 Note : Since pulley is in equilibrium position, so net forces on it is zero. (ii) Accelerating Objects : Steps to solve problems involving objects that are in accelerated motion : Step 1 : Make a sketch of the problem. Step 2 : Isolate a single object and then draw the free - body diagram for that object. Label all external forces acting on it. Be sure to include all the forces acting on the chosen body, but be equally careful not to include any force exerted by the body on some other body. Some of the forces may be unknown , label them with algebraic symbols. Step 3 : Choose a convenient coordinate system, show location of coordinate axis explicitly in the free - body diagram, and then determine components of forces with reference to these axis and resolve all forces into x and y components.

Fx = max & Fy = may. Step 4 : Apply the equations Step 5 : Step 4 will give two equations with several unknown quantities. If you have only two unknown quantities at this point, you can solve the two equations for those unknown quantities. Step 6 : If step 5 produces two equations with more than two unknowns, go back to step 2 and select another object and repeat these steps. Eventually at step 5 you will have enough equations to solve for all unknown quantities.





15. A force F is applied horizontally on mass m1 as shown in figure. Find the contact force between m1 and m2.

= max

m2F m1  m2 

  F sin ce a   m1  m 2  

16. A 5 kg block has a rope of mass 2 kg attached to its underside and a 3 kg block is suspended from the other end of the rope. The whole system is accelerated upward at 2 m/s2 by an external force F0. (a) What is F0 ? (b) What is the net force on rope ? (c) What is the tension at middle point of the rope ? (g = 10 m/s2)

Sol.For calculating the value of F0. F.B.D of whole system F0 2 (a) 2m/s

10 g = 100 N

F0 –100 = 10 × 2 F0 = 120 N

........(1)

(b) According to Newton’s second law, net force on rope. F = ma = 2 × 2 = 4N ............(2) (c) For calculating tension at the middle point we draw F.B.D. of 3 kg block with half of the rope (mass 1 kg) as shown.

Sol.Considering both blocks as a system to find the common acceleration. Common acceleration

F a= m1  m2  .......(1)

F

m1

T–4g=4.2

T = 48 N 17. A block of mass 50 kg is kept on another block of mass

m2

a

1 kg as shown in figure. A horizontal force of 10 N is applied on the 1Kg block. (All surface are smooth).

To find the contact force between ‘A’ and ‘B’ we draw F.B.D. of mass m2. F.B.D. of mass m2

Find : (g = 10 m/s2) (a) Acceleration of blocks A and B.

PHYSICS_IJSO_PAGE #13 13

(b) Force exerted by B on A. B A

Sol.(a)

50 kg 1 kg

F.B.D. of 50 kg

Sol. for (man + platform) system : 2mg – 4T = 2m(a)

N2 = 50 g = 500 N along horizontal direction, there is no force aB = 0 (b) F.B.D. of 1 kg block :

 mg  mg  = 2m (a) [ T = ]  2  2



2mg – 4 



a=0

N1 N2 10 N 1g

along horizontal direction 10 = 1 aA. aA = 10 m/s2 along vertical direction

N1 = N2 + 1g = 500 + 10 = 510 N 18. One end of string which passes through pulley and connected to 10 kg mass at other end is pulled by 100 N force. Find out the acceleration of 10 kg mass. (g

WEIGHING MACHINE

=9.8 m/s2) A weighing machine does not measure the weight but measures the force exerted by object on its upper surface. 20. A man of mass 60 Kg is standing on a weighing Sol.Since string is pulled by 100 N force. So tension in the string is 100 N F.B.D. of 10 kg block

machine placed on ground. Calculate the reading of

weighing machine

machine (g = 10 m/s2).

Sol.For calculating the reading of weighing machine, we draw F.B.D. of man and machine separately. F.B.D of man 100 – 10 g = 10 a 100 – 10 × 9.8 = 10 a a = 0.2 m/s2. 19. A man of mass m stands on a platform of equal mass m and pulls himself by two ropes passing over pulleys as shown. If he pulls each rope with a force equal to half his weight, find his upward acceleration ?

F.B.D of man taking mass of man as M

F.B.D. of weighing machine

N

N weighing machine

N = Mg

N1

Mg

Mg

PHYSICS_IJSO_PAGE #14 14

Here force exerted by object on upper surface is N Reading of weighing machine N = Mg

T

= 60 × 10 N = 600 N. 20 g

SPRING BALANCE It does not measure the weight. It measures the force exerted by the object at the hook. Symbolically, it is represented as shown in figure. A block of mass ‘m’ is suspended at hook. When spring balance is in equilibrium, we draw the F.B.D. of mass m for calculating the reading of balance.

spring balance hook m

F.B.D. of ‘m’.

mg – T = 0 T = 20 g = 200 N Since both the balances are light so, both the scales will read 200 N. 22. (i)

A 10 kg block is supported by a cord that runs to a spring scale, which is supported by another cord from the ceiling figure (a). What is the reading on the scale ?

(ii) In figure (b) the block is supported by a cord that runs around a pulley and to a scale. The opposite end of the scale is attached by cord to a wall. What is the reading of the scale. (iii) In figure (c) the wall has been replaced with a second 10 kg block on the left, and the assembly is stationary. What is the reading on the scale now ?

mg – T = 0 T = mg Magnitude of T gives the reading of spring balance.

T

21. A block of mass 20 kg is suspended through two light spring balances as shown in figure . Calculate the : spring balance

T

hook

10 kg

(a) T

T T 10kg

(b) T

T

(1) reading of spring balance (1).

Sol.For calculating the reading, first we draw F.B.D.of 20 kg block. F.B.D. 20 kg

T

T

(2) reading of spring balance (2).

10kg

10kg

(c) Sol. In all the three cases the spring balance reads 10 kg. To understand this let us cut a section inside the spring as shown;

PHYSICS_IJSO_PAGE #15 15

or As each part of the spring is at rest, so F= T. As the block is stationary, so T= 10g = 100N.

F=

mg .......(i) cos  –  sin 

 Pull : Along y-axis we have ;

23. A block of mass m is placed on a weighing machine and it is also attached with a spring balance. The whole system is placed in a lift as shown in figure. Lift is moving with constant acceleration 10 m/s2 in upward direction. The reading of the weighing machine is 500 N and the reading of spring balance is 300 N. Find the actual mass of the block in kg. (Take g = 10 m/s2)

F

y

 0;

 N = mg – F sin  To just move the block along x-axis, we have F cos = N =  (mg – F sin )

 mg  . .......(ii)  cos    sin   

Sol. m = mass of the block Applying Newton's Law on the block in vertical direction T + N – mg = ma ...............(1) T = 300 N (given reading of spring balance)

or F = 

It is clear from above discussion that pull force is smaller than push force. 25. Discuss the direction of friction in the following cases : (i)

A man walks slowly, without change in speed.

(ii) A man is going with increasing speed. (iii) When cycle is gaining speed. N = 500 N (given reading of W.M.) a = 10 m/s2

T N 500  300 m= ga = 20 m = 40 kg Ans. 24. Pull is easier than push

 Push : Consider a block of mass m placed on rough horizontal surface. The coefficient of static friction between the block and surface is . Let a push force F is applied at an angle  with the horizontal.

(iv) When cycle is slowing down . Sol. (i)

Consider a man walks slowly without acceleration, and both the legs are touching the ground as shown in figure (a). The frictional force on rear leg is in forward direction and on front leg will be on backward direction of motion. As a = 0,

 Fnet = 0 or

f1 – f 2 = 0

 f1 = f2

N1 = N2.

&

N1

N2

f1

f1

f2

f2

Ground N1

F

y

N2

(b)

As the block is in equilibrium along y-axis, so we have

 0;

or N = mg + F sin  To just move the block along x-axis, we have

(ii) When man is gaining the speed : The frictional force on rear leg f1 will be greater than frictional force on front leg f2 (fig. b).

F cos = N = (mg + F sin )

 acceleration of the man, a =

f1  f2 . m

PHYSICS_IJSO_PAGE #16 16

(iii) When cycle is gaining speed : In this case torque

In IInd case, let the force exerted by the man on the floor

is applied on the rear wheel of the cycle by the

in N2 . Consider the forces inside the dotted box, we

chain-gear system. Because of this the slipping

have

tendency of the point of contact of the rear wheel is

N2 = 50 g – T

backward and so friction acts in forward direction.

and T = 25 g

The slipping tendency of point of contact of front

N2 = 50 g – 25 g

wheel is forward and so friction acts in backward

= 25 g = 25 × 9.8 = 245 N.

direction. If f1 and f2 are the frictional forces on rear

As the floor yields to a downward force of 700 N, so the

and front wheel, then acceleration of the cycle a =

man should adopt mode .

f1 – f2 , where M is the mass of the cycle together M with rider (fig. a).

N1

N2

f1

f2 (a)

N1

N2

f1

27. Figure shows a weighing machine kept in a lift is moving upwards with acceleration of 5 m/s2. A block is kept on the weighing machine. Upper surface of block is attached with a spring balance. Reading shown by weighing machine and spring balance is 15 kg and 45 kg respectively.

f2 (b)

(iv) When cycle is slowing down : When torque is not applied (cycle stops pedaling), the slipping tendency of points of contact of both the wheels are forward, and so friction acts in backward direction (fig. b). If f1 and f2 are the frictional forces on rear and front wheel, then retardation a=

f1  f2 M

26. A block of mass 25 kg is raised by a 50 kg man in two

Answer the following questions. Assume that the weighing machine can measure weight by having negligible deformation due to block, while the spring balance requires larger expansion. (take g = 10 m/s2)

different ways as shown in fig.. What is the action on the floor by the man in the two cases ? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding.

(i) Find the mass of the object in kg and the normal force acting on the block due to weighing machine? (ii) Find the acceleration of the lift such that weighing machine shows its true weight ?

Sol. (i) 50g 50g

T + N – Mg = Ma Sol. The FBD for the two cases are shown in figure. In Ist case, let the force exerted by the man on the floor is N1. Consider the forces inside the dotted box, we have N1 = T + 50 g.

45 g + 15 g = M(g + a) 450 + 150 = M(10 + 5) M = 40 kg Normal force is the reaction applied by weighing machine i.e. 15 × 10 = 150 N.

Block is to be raised without acceleration, so T = 25 g.

 N1 = 25 g + 50 g =

75 g = 75 × 9.8 = 735 N

PHYSICS_IJSO_PAGE #17 17

6.

What is the reading of the spring balance shown in the figure below? (IJSO/Stage-I/2012) T

T

(ii) T 0.2kg

T + N – Mg = Ma 45 g + 40 g = 40(g + a) 450 + 400 = 400 +40 a a=

450 45 = m/s2 40 4

(A) 0 (C) 4N 7.

2.

3.

4.

5.

Two blocks of masses 2 kg and 1 kg are placed on a smooth horizontal table in contact with each other. A horizontal force of 3 newton is applied on the first so that the blocks move with a constant acceleration. The force between the blocks would be : (A) 3 Newton (B) 2 Newton (C) 1 Newton (D) Zero An object of mass 10 kg. moves at a constant speed of 10ms–1. A constant force acts for 4 s on the object and gives it a speed of 2 ms–1 in opposite direction. The force acting on the object is : (A) – 3N (B) –30 N (C) 3 N (D) 30 N A machine gun has a mass 5kg. It fires 50 gram bullets at the rate of 30 bullets per minute at a speed of 400 ms–1. What force is required to keep the gun in position : (A) 10 N (B) 5 N (C) 15 N (D) 30N

Two blocks are in contact on a frictionless table. One has mass m and the other 2m.A force F is applied on 2m as shown in the figure. Now the same force F is applied from the right on m. In the

EXERCISE 1.

two cases respectively, the ratio of force of contact between the two blocks will be :

(A) Same (C) 2 : 1 8.

(B) 1 : 2 (D) 1 : 3

Two forces of 6N and 3N are acting on the two blocks of 2kg and 1kg kept on frictionless floor. What is the force exerted on 2kg block by 1kg block ?:

6N 2kg 1kg

9.

3N

(A)1N

(B) 2N

(C) 4N

(D) 5N

There are two forces on the 2.0 kg box in the overhead view of figure but only one is shown. The second force is nearly : y

A particle of mass m1 moving with velocity v collides with a mass m2 at rest, then they get embedded. Just after collision, velocity of the system : (A) Increases (B) Decreases (C) remains constant (D) becomes zero When a car turns on a curved road, you are pushed against one of the doors of the car because of : (IJSO/Stage-I/2012) (A) inertia (B) the centripetal force (C) the centrifugal force (D) the frictionaI force

(B) 2N (D) 6N

F1 = 20 N x 30º a = 12 m/s

(A) –20 ˆj N

2

(B) – 20 ˆi + 20 ˆj N

(C) –32 ˆi – 12 3 ˆj N (D) –21 ˆi – 16 ˆj N 10. A dish of mass 10 g is kept horizontally in air by firing bullets of mass 5 g each at the rate of 100 per second. If the bullets rebound with the same speed, what is the velocity with which the bullets are fired : (A) 0.49 m/s (B) 0.098 m/s (C) 1.47 m/s (D) 1.96 m/s PHYSICS_IJSO_PAGE #18 18

11.

A block of metal weighing 2 kg is resting on a frictionless plank. If struck by a jet releasing water at a rate of 1 kg/s and at a speed of 5 m/s. The initial acceleration of the block will be : (A) 2.5 m/s2 (B) 5.0 m/s2 2 (C) 10 m /s (D) none of the above

16. A mass M is suspended by a rope from a rigid support at A as shown in figure. Another rope is tied at the end B, and it is pulled horizontally with a force F. If the rope AB makes an angle  with the vertical in equilibrium,then the tension in the string AB is :

12. A constant force F is applied in horizontal direction as shown. Contact force between M and m is N and between m and M’ is N’ then

(A) F sin  (C) F cos  (A) N= N’ (C) N’> N (D) cannot be determined

(B) F /sin  (D) F / cos 

17. In the system shown in the figure, the acceleration of the 1kg mass and the tension in the string connecting between A and B is :

(B) N > N’

13. STATEMENT-1 : Block A is moving on horizontal surface towards right under action of force. All surface are smooth. At the instant shown the force exerted by block A on block B is equal to net force on block B.

STATEMENT-2 : From Newtons’s third law, the force exerted by block A on B is equal in magnitude to force exerted block B on A (A) statement-1 is true, Statement 2 is true, statement-2 is correct explanation for statement-1. (B) statement-1 is true, Statement 2 is true, statement-2 is NOT a correct explanation for statement-1. (C) statement-1 is true, Statement 2 is false (D) statement-1 is False, Statement 2 is True

(A)

g 8g downward, 4 7

(B)

g g upward, 4 7

(C)

g 6 downward, g 7 7

(D)

g upward, g 2

18. A body of mass 8 kg is hanging from another body of mass 12 kg. The combination is being pulled by a string with an acceleration of 2.2 m s–2. The tension T1 and T2 will be respectively :(Use g =9.8 m/s2)

14. A certain force applied to a body A gives it an acceleration of 10 ms–2 . The same force applied to body B gives it an acceleration of 15 ms–2 . If the two bodies are joined together and same force is applied to the combination, the acceleration will be : (IJSO/Stage-I/2011) (B) 25 ms–2

(A) 6 ms–2

(C) 12.5 ms–2 (D) 9 ms–2  15. Four blocks are kept in a row on a smooth horizontal table with their centres of mass collinear as shown in the figure. An external force of 60 N is applied from left on the 7 kg block to push all of them along the table. The forces exerted by them are :(IAO/Sr./Stage-I/2008) 60N

P

Q

R

S

7 kg

5 kg

2 kg

1 kg

(A) 32 N by P on Q (C) 12 N by Q on R

(B) 28 N by Q on P (D) 4 N by S on R

(A) 200 N, 80 N (C) 240 N, 96 N

(B) 220 N, 90 N (D) 260 N, 96 N

19. Two masses M1 and M2 are attached to the ends of a light string which passes over a massless pulley attached to the top of a double inclined smooth plane of angles of inclination  and . If M2 > M1 then the acceleration of block M2 down the inclined will be :

M2 (sin ) g M1  M2  M2 sin   M1 sin    g (C)  M1  M2   (A)

(B)

M1g(sin  ) M1  M2

(D) Zero

PHYSICS_IJSO_PAGE #19 19

20. Three masses of 1 kg, 6 kg and 3 kg are connected to

23. Figure shows four blocks that are being pulled along a

each other by threads and are placed on table as shown in figure. What is the acceleration with which

smooth horizontal surface. The mssses of the blocks and tension in one cord are given. The pulling force F is :

the system is moving ? Take g = 10 m s–2: F 30N 60º

4kg

3kg

2kg

1kg

(A) 50 N

(B) 100 N

(C) 125 N

(D) 200 N

24. A10 kg monkey climbs up a massless rope that runs (A) Zero

(B) 1 ms–2

(C) 2 m s–2

(D) 3 m s–2

21. The pulley arrangements shown in figure are identical

over a frictionless tree limb and back down to a 15 kg package on the ground. The magnitude of the least acceleration the monkey must have if it is to lift the package off the ground is :

the mass of the rope being negligible. In case I, the mass m is lifted by attaching a mass 2m to the other end of the rope. In case II, the mass m is lifted by pulling the other end of the rope with a constant downward force F= 2 mg, where g is acceleration due to gravity. The acceleration of mass in case I is :

(A) 4.9 m/s2

(B) 5.5 m/s2

(C) 9.8 m/s2

(D) none of these

25. Two blocks, each of mass M, are connected by a massless string, which passes over a smooth



massless pulley. Forces F act on the blocks as shown. The tension in the string is :

(A) Zero (B) More than that in case II (C) Less than that in case II (D) Equal to that in case II 22. A 50 kg person stands on a 25 kg platform. He pulls massless rope which is attached to the platform via the frictionless, massless pulleys as shown in the figure. The platform moves upwards at a steady velocity if the force with which the person pulls the rope is :

(A) Mg

(B) 2 Mg

(C) Mg + F

(D) none of these

26. Two blocks of mass m each is connected with the string which passes over fixed pulley, as shown in figure. The force exerted by the string on the pulley P is :

(A) 500 N (C) 25 N

(B) 250 N (D) 50 N

(A) mg

(B) 2 mg

(C) 2 mg

(D) 4 mg PHYSICS_IJSO_PAGE #20 20

27. One end of a massless rope, which passes over a massless and frictionless pulley P is tied to a hook C while the other end is free. Maximum tension that rope can bear is 360 N, with what minimum safe acceleration (in m/s2) can a monkey of 60 kg move down on the rope :

31. A weight is supported by two strings 1.3 and 2.0 m long fastened to two points on a horizontal beam 2.0 m apart. The depth of this weight below the beam is : (IAO/Jr./Stage-I/2007) (A) 1.0 m

(B) 1.23 m

(C) 0.77 m

(D) 0.89 m

P

32. A fully loaded elevator has a mass of 6000 kg. The tension in the cable as the elevator is accelerated downward with an acceleration of 2ms–2 is (Take g = I0 ms –2 ) C

(A) 16

(B) 6

(C) 4

(D) 8

28. Which figure represents the correct F.B.D. of rod of mass m as shown in figure :

(A)

(B)

(C)

(D) None of these

29. Two persons are holding a rope of negligible weight tightly at its ends so that it is horizontal. A 15 kg weight is attached to the rope at the mid point which now no longer remains horizontal. The minimum tension required to completely straighten the rope is : (A) 15 kg (B)

15 kg 2

(C) 5 kg (D) Infinitely large (or not possible) 30. In the figure, the blocks A, B and C of mass each have acceleration a1 , a2 and a3 respectively . F1 and F2 are external forces of magnitudes 2 mg and mg respectively then which of the following relations is correct :

(A) a1 = a2 = a3 (C) a1 = a2 , a2 > a3

(B) a1 > a2 > a3 (D) a1 > a2 , a2= a3

(A) 7·2 × 104 N (C) 6 × 104 N

(B) 4.8 × 104 N (D) 1.2 × 104 N

33. A light string goes over a frictionless pulley. At its one end hangs a mass of 2 kg and at the other end hangs a mass of 6 kg. Both the masses are supported by hands to keep them at rest. When the masses are released, they being to move and the string gets taut. (Take g = 10 ms–2) The tension in the string during the motion of the masses is : (A) 60 N (B) 30 N (C) 20 N (D) 40 N 34. In the given figure. What is the reading of the spring balance:

(A) 10 N (C) 5 N

(B) 20 N (D) Zero

35. Two bodies of masses M1 and M2 are connected to each other through a light spring as shown in figure. If we push mass M1 with force F and cause acceleration a1 in mass M1 what will be the acceleration in M2 ?

(A) F/M2 (C) a1

(B) F/(M1 + M2) (D) (F–M1a1)/M2

36. A spring balance is attached to 2 kg trolley and is used to pull the trolly along a flat surface as shown in the fig. The reading on the spring balance remains at 10 kg during the motion. The acceleration of the trolly is (Use g= 9.8 m–2) :

(A) 4.9 ms–2 (C) 49 ms–2

(B) 9.8 ms–2 (D) 98 ms–2 PHYSICS_IJSO_PAGE #21 21

37. A body of mass 32 kg is suspended by a spring balance from the roof of a vertically operating lift and going downward from rest. At the instants the lift has covered 20 m and 50 m, the spring balance showed 30 kg & 36 kg respectively. The velocity of the lift is : (A) Decreasing at 20 m & increasing at 50 m (B) Increasing at 20 m & decreasing at 50 m (C) Continuously decreasing at a constant rate throughout the journey (D) Continuously increasing at constant rate throughout the journey 38. A ship of mass 3 × 107 kg initially at rest is pulled by a force of 5 × 104 N through a distance of 3m. Assume that the resistance due to water is negligible, the speed of the ship is : (A) 1.5 m/s (B) 60 m/s (C) 0.1 m/s (D) 5 m/s 39. When a horse pulls a cart, the force needed to move the horse in forward direction is the force exerted by : (A) The cart on the horse (B) The ground on the horse (C) The ground on the cart (D) The horse on the ground 40. A 2.5 kg block is initially at rest on a horizontal surface.



A 6.0 N horizontal force and a vertical force P are applied to the block as shown in figure. The coefficient of static friction for the block and surface is 0.4. The magnitude of friction force when P = 9N : (g = 10 m/s2)

43. N bullets each of mass m are fired with a velocity v m/ s at the rate of n bullets per sec., upon a wall. If the bullets are completely stopped by the wall, the reaction offered by the wall to the bullets is : (A) N m v / n (B) n m v (C) n N v / m (D) n v m / N 44. A vehicle of mass m is moving on a rough horizontal road with momentum P. If the coefficient of friction between the tyres and the road be , then the stopping distance is : (A)

P 2 mg

2 m2g

41. The upper half of an inclined plane with inclination  is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom, if the coefficient of friction for the lower half is : (A) 2 tan  (B) tan  (C) 2 sin  (D) 2 cos  42. Minimum force required to pull the lower block is (take g = 10 m/s2) :

(A) 1 N (C) 7 N

(B) 5 N (D) 10 N

(D)

P2 2 m2g

45. What is the maximum value of the force F such that the block shown in the arrangement, does not move : F 60º

1 2 3 m = 3kg

(A) 20 N (C) 12N

(B) 10 N (D) 15 N

46. A bock of mass 5 kg is held against wall by applying a horizontal force of 100N. If the coefficient of friction between the block and the wall is 0.5, the frictional force acting on the block is : (g =9.8 m/s2)

5kg

(A)100 N (C) 49 N (B) 6.4 N (D) zero

P2 2 mg

P (C)

100N

(A) 6.0 N (C) 9.0 N

(B)

(B) 50 N (D) 24.9 N

47. A heavy roller is being pulled along a rough road as shown in the figure. The frictional force at the point of contact is : (IAO/Jr./Stage-I/2007)

F

(A) parallel to F (C) perpendicular to F

(B) opposite to F (D) zero

48. When a motor car of mass 1500 kg is pushed on a road by two persons, it moves with a small uniform velocity. On the other hand if this car is pushed on the same road by three persons, it moves with an acceleration of 0.2 m/s2. Assume that each person is producing the same muscular force. Then, the force of friction between the tyres of the car and the surface of the road is : (IAO/Jr./Stage-I/2009) (A) 300 N (C) 900 N

(B) 600 N (D) 100 N

PHYSICS_IJSO_PAGE #22 22

49. A block of mass M is at rest on a plane surface inclined at an angle  to the horizontal The magnitude of force exerted by the plane on the block is : (A) Mg cos (B) Mg sin  (C) Mg tan (D) Mg 50. A block of mass M rests on a rough horizontal table. A steadily increasing horizontal force is applied such that the block starts to slide on the table without toppling. The force is continued even after sliding has started. Assume the coefficients of static and kinetic friction between the table and the block to be equal. The correct representation of the variation of the frictional forces, ƒ, exerted by the table on the block with time t is

52. On a horizontal frictional frozen lake, a girl (36 kg) and a box (9kg) are connected to each other by means of a rope. Initially they are 20 m apart. The girl exerts a horizontal force on the box, pulling it towards her. How far has the girl travelled when she meets the box ? (A) 10 m (B) Since there is no friction, the girl will not move (C) 16 m (D) 4m 53. Which of the following does NOT involve friction ? (IJSO/Stage-I/2011) (A) Writing on a paper using a pencil (B) Turning a car to the left on a horizontal road. (C) A car at rest parked on a sloping ground (D) Motion of a satellite around the earth.

given by :

 54. In the two cases shown below, the coefficient of kinetic friction between the block and the surface is the same, and both the blocks are moving with the same uniform speed. Then, (IAO/Sr./Stage-I/2008) (A)

(C)

(B)

(D)

51. A small child tries to move a large rubber toy placed on the ground. The toy does not move but gets deformed  under her pushing force (F) which is obliquely upward as shown . Then

 (A) The resultant of the pushing force (F ) , weight of

F1

F2

(A) F1 = F2 (B) F1 < F2 (C) F1 > F2 (D) F1 = 2F2 if sin = Mg/4F2 55. The ratio of the weight of a man in a stationary lift and when it is moving downward with uniform acceleration ‘a’ 3:2. The value of ‘a’ is : (g = acceleration, due to gravity) (A) (3/2)g (B) g (C) (2/3) g (D) g/3 56. A person standing on the floor of an elevator drops a coin. The coin reaches the floor of the elevator in time t1 when elevator is stationary and in time t2 if it is moving uniformly. Then (A) t1 = t2 (B) t1 > t2 (C) t1 < t2 (D) t1 < t2 or t1 > t2 depending 57. STATEMENT-1 : A man standing in a lift which is moving

the toy, normal force by the ground on the toy and the

upward, will feel his weight to be greater than when

frictional force is zero. (B) The normal force by the ground is equal and oppo-

the lift was at rest.

site to the weight of the toy.  (C) The pushing force (F) of the child is balanced by

then the man of mass m shall feel his weight to be

the equal and opposite frictional force  (D) The pushing force (F) of the child is balanced by the total internal force in the toy generated due to deformation

STATEMENT-2 : If the acceleration of the lift is ‘a’ upward equal to normal reaction (N) exerted by the lift given N = m(g+a) (where g is acceleration due to gravity (A) statement-1 is true, Statement 2 is true, statement2 is correct explanation for statement 1. (B) statement-1 is true, Statement 2 is true, statement2 is NOT a correct explanation for statement-1. (C) statement-1 is true, Statement 2 is false (D) statement-1 is False, Statement 2 is True PHYSICS_IJSO_PAGE #23 23

58. A beaker containing water is placed on the platform of a digital weighing machine. It reads 900 g. A wooden

 63. Figure shows the displacement of a particle going along the x-axis as a function of time :

block of mass 300 g is now made to float in water in the beaker (without touching walls of the beaker). Half the wooden block is submerged inside water. Now, the reading of weighing machine will be : (IAO/Jr./Stage-I/2009) (A) 750 g

(B) 900 g

(C) 1050 g

(D) 1200 g

(A) The force acting on the particle is zero in the region AB (B) The force acting on the particle is zero in the region BC (C) The force acting on the particle is zero in the region CD (D) The force is zero no where

59. An object will continue accelerating until : (A) Resultant force on it begins to decreases (B) Its velocity changes direction (C) The resultant force on it is zero (D) The resultant force is at right angles to its direction of motion 60. In which of the following cases the net force is not zero ? (A) A kite skillfully held stationary in the sky (B) A ball freely falling from a height (C) An aeroplane rising upward at an angle of 45° with the horizontal with a constant speed (D) A cork floating on the surface of water.

 61. Figure shows the displacement of a particle going along the X-axis as a function of time. The force acting on the particle is zero in the region.

(A) AB (C) CD

(B) BC (D) DE

62. A 2 kg toy car can move along x axis. Graph shows force Fx, acting on the car which begins to rest at time t = 0. The velocity of the car at t = 10 s is :

64. A force of magnitude F1 acts on a particle so as to accelerate if from rest to velocity v. The force F1 is then replaced by another force of magnitude F2 which decelerates it to rest. (A) F1 must be the equal to F2 (B) F1 may be equal to F2 (C) F1 must be unequal to F2 (D) None of these 65. In a imaginary atmosphere, the air exerts a small force F on any particle in the direction of the particle’s motion. A particle of mass m projected upward takes a time t1 in reaching the maximum height and t2 in the return journey to the original point. Then (A) t1 < t2 (B) t1 > t2 (C) t1 = t2 (D) The relation between t1 and t2 depends on the mass of the particle 66. A single force F of constant magnitude begins to act on a stone that is moving along x axis. The stone continues to move along that axis. W hich of the following represents the stone’s position ? (A) x = 5t – 3 (B) x = 5t2 + 8t – 3 2 (C) x = –5t + 5t – 3 (D) x = 5t3 + 4t2 – 3 67. Three forces act on a particle that moves with



unchanging velocity v = (3 ˆi – 4 ˆj ) m/s. Two of the   forces are F1 = (3 ˆi + 2 ˆj – 4 kˆ ) N and F2 = (–5 ˆi + 8 ˆj + 3 kˆ ) N. The third force is : (A) (–2 ˆi + 10 ˆj – 7 kˆ ) N (B) (2 ˆi – 10 ˆj + kˆ ) N (C) (7 ˆi – 2 kˆ + 10 ˆj ) N (D) none of these 68. An 80 kg person is parachuting and experiencing a downward acceleration of 2.5 m/s2 . The mass of the (A) – ˆi m/s

(B) – 1.5 ˆi m/s

(C) 6.5 ˆi m/s

(D) 13 ˆi m/s

parachute is 5.0 kg. The upward force on the open parachute from the air is : (A) 620 N (C) 800 N

(B) 740 N (D) 920 N

PHYSICS_IJSO_PAGE #24 24

69. A block of mass m is pulled on the smooth horizontal surface with the help of two ropes, each of mass m, connected to the opposite faces of the block. The forces on the ropes are F and 2F. The pulling force on the block is :

(B) 2F (D) 3F/2

70. A body of mass 5 kg starts from the origin with an initial  velocity u = 30 ˆi + 40 ˆj ms–1 . If a constant force  F = –( ˆi + 5 ˆj ) N acts on the body, the time in which the y-component of the velocity becomes zero is : (A) 5 s (B) 20 s (C) 40 s (D) 80 s 71. STATEMENT-1 :According to the newton’s third law of motion, the magnitude of the action and reaction force is an action reaction pair is same only in an inertial frame of reference. STATEMENT-2 : Newton’s laws of motion are applicable in every inertial reference frame. (A) statement-1 is true, Statement 2 is true, statement2 is correct explanation for statement 1. (B) statement-1 is true, Statement 2 is true, statement2 is NOT a correct explanation for statement-1. (C) statement-1 is true, Statement 2 is false (D) statement-1 is False, Statement 2 is True 72. A body of mass 10 g moves with constant speed 2 m/ s along a regular hexagon. The magnitude of change in momentum when the body crosses a corner is : (IAO/Sr./Stage-I/2007) (A) 0.04 kg-m/s (B) zero (C) 0.02 kg-m / s (D) 0.4 kg-m/s 73. An object with uniform density  is attached to a spring that is known to stretch linearly with applied force as shown below

When the spring object system is immersed in a liquid of density 1 as shown in the figure, the spring stretches by an amount x1 ( > 1). When the experiment is repeated in a liquid of density 2 < 1 . the spring is stretched by an amount x2. Neglecting any buoyant force on the spring, the density of the object is: 1x1  2 x 2 1x 2  2 x1 (A)   x  x (B)   x  x 1 2 2 1 1x 2  2 x1 (C)   x  x 1 2

the influence of a varying force F (in Newtons) as shown below :

3

F(N)

(A) F (C) F/3

74. A body of 0.5 kg moves along the positive x - axis under

3

1

0,0

2

4

6

8

10

x(m)

If the speed of the object at x = 4m is 3.16 ms–1 then its speed at x = 8 m is : (A) 3.16 ms–1

(B) 9.3 ms–1

(C) 8 ms

(D) 6.8 ms–1

–1

75. A soldier with a machine gun, falling from an airplane gets detached from his parachute. He is able to resist the downward acceleration if he shoots 40 bullets a second at the speed of 500 m/s. If the weight of a bullet is 49 gm, what is the weight of the man with the gun ? Ignore resistance due to air and assume the acceleration due to gravity g = 9.8 m/s2 . (A) 50 kg (B) 75 kg (C) 100 kg (D) 125 kg 76. Blocks of mass M1 and M2 are connected by a cord which passes over the pulleys P1 and P2 as shown in the figure. If there is no friction, the acceleration of the block of mass M2 will be:

M2 g

2M2 g

(A)

(4M1M2 )

(B)

(C)

2M1g (M1  4M2 )

(D)

(4M1M2 ) 2M1g

(M1M2 )

1x1  2 x 2 (D)   x  x 1 2

PHYSICS_IJSO_PAGE #25 25

MOLE CONCEPT ATOMS

Symbol Derived from English Names

All the matter is made up of atoms. An atom is the smallest particle of an element that can take part in a chemical reaction. Atoms of most of the elements are very reactive and do not exist in the free state (as single atom).They exist in combination with the atoms of the same element or another element. Atoms are very, very small in size. The size of an atom is indicated by its radius which is called "atomic radius" (radius of an atom). Atomic radius is measured in "nanometres"(nm). 1 metre = 109 nanometre or 1nm = 10-9 m. Atoms are so small that we cannot see them under the most powerful optical microscope. 

Note : Hydrogen atom is the smallest atom of all , having an atomic radius 0.037nm. (a) Symbols of Elements : A symbol is a short hand notation of an element which can be represented by a sketch or letter etc. Dalton was the first to use symbols to represent elements in a short way but Dalton's symbols for element were difficult to draw and inconvenient to use, so Dalton's symbols are only of historical importance. They are not used at all.

It was J.J. Berzelius who proposed the modern system of representing an element. The symbol of an element is the "first letter" or the "first letter and another letter" of the English name or the Latin name of the element. e.g. The symbol of Hydrogen is H. The symbol of Oxygen is O. There are some elements whose names begin with the same letter. For example, the names of elements Carbon, Calcium, Chlorine and Copper all begin with the letter C. In such cases, one of the elements is given a "one letter "symbol but all other elements are given a "first letter and another letter" symbol of the English or Latin name of the element. This is to be noted that "another letter" may or may not be the "second letter" of the name. Thus, The symbol of Carbon is C. The symbol of Calcium is Ca. The symbol of Chlorine is Cl. The symbol of Copper is Cu (from its Latin name Cuprum) It should be noted that in a "two letter" symbol, the first letter is the "capital letter" but the second letter is the small letter



English Name of the Element Hydrogen Helium Lithium Boron Carbon Nitrogen Oxygen Fluorine Neon Magnesium Aluminium Silicon Phosphorous Sulphur Chlorine Argon Calcium

Symbol H He Li B C N O F Ne Mg Al Si P S Cl Ar Ca

Symbols Derived from Latin Names English Name of the Element

Symbol

Latin Name of the Element

Sodium

Na

Natrium

Potassium

K

Kalium

(b) Significance of The Symbol of an Element : (i) Symbol represents name of the element. (ii) Symbol represents one atom of the element. (iii) Symbol also represents one mole of the element. That is, symbol also represent 6.023 × 1023 atoms of the element. (iv) Symbol represent a definite mass of the element i.e. atomic mass of the element. Example : (i) Symbol H represents hydrogen element. (ii) Symbol H also represents one atom of hydrogen element. (iii) Symbol H also represents one mole of hydrogen atom. (iv) Symbol H also represents one gram hydrogen atom.

IONS An ion is a positively or negatively charged atom or group of atoms. Every atom contains equal number of electrons PAGE # 26

(negatively charged) and protons (positively charged). Both charges balance each other, hence atom is electrically neutral.

Note : Size of a cation is always smaller and anion is always greater than that of the corresponding neutral atom.



(c) Monoatomic ions and polyatomic ions : (i) Monoatomic ions : Those ions which are formed from single atoms are called monoatomic ions or simple ions. e.g. Na+, Mg2+ etc.

(a) Cation : If an atom has less electrons than a neutral atom, then it gets positively charged and a positively charged ion is known as cation. e.g. Sodium ion (Na+), Magnesium ion (Mg2+) etc. A cation bears that much units of positive charge as there are the number of electrons lost by the neutral atom to form that cation.

(ii) Polyatomic ions : Those ions which are formed from group of atoms joined together are called polyatomic ions or compound ions. e.g. Ammonium ion (NH4+) , hydroxide ion (OH–) etc. which are formed by the joining of two types of atoms, nitrogen and hydrogen in the first case and oxygen and hydrogen in the second.

e.g. An aluminium atom loses 3 electrons to form aluminium ion, so aluminium ion bears 3 units of positive charge and it is represented as Al3+.

(d) Valency of ions : The valency of an ion is same as the charge present on the ion. If an ion has 1 unit of positive charge, its valency is 1 and it is known as a monovalent cation. If an ion has 2 units of negative charge, its valency is 2 and it is known as a divalent anion.

(b) Anion : If an atom has more number of electrons than that of neutral atom, then it gets negatively charged and a negatively charged ion is known as anion. – e.g. Chloride ion (Cl ), oxide ion (O2-) etc. An anion bears that much units of negative charge as there are the number of electrons gained by the neutral atom to form that anion. e.g. A nitrogen atom gains 3 electrons to form nitride ion, so nitride ion bears 3 units of negative charge and it is represented as N3-.

LIST OF COMMON ELECTROVALENT POSITIVE RADICALS

LIST OF COMMON ELECTROVALENT NEGATIVE RADICALS Bivalent Electronegative

Monovalent Electronegative 1. Fluoride

F–

1. Sulphate 2. Sulphite

SO 3

3. Sulphide

S

4. Thiosulphate

S 2O3

Br I

5. Hydride

H

–

ZnO2

–

7. Peroxide

O2

–

8. Dichromate

Cr 2O7

–

9. Carbonate

–

10. Silicate

2CO 3 2SiO 3

NO2

NO3

HSO3 – HS

1. Nitride

N

3-

Tetravalent Electronegative 1. Carbide

C

4-

3-

2. Phosphide

P

3. Phosphite

PO 3

4. Phosphate

PO 4

33-

2-

O

9. Bicarbonate or Hydrogen carbonate HCO3



2-

6. Oxide

8.Nitrate

12. Bisulphate or Hydrogen sulphate 13. Acetate

2-

5. Zincate

–

OH

7. Nitrite

10. Bisulphite or Hydrogen sulphite 11. Bisulphide or Hydrogen sulphide

2-

–

Cl

3. Bromide

6. Hydroxide

2-

SO 4

–

2. Chloride 4. Iodide

Trivalent Electronegative

222-

–

HSO4 – CH3COO

Note : Cation contains less no. of electrons and anion contains more no. of electrons than the no. of protons present in them.

PAGE # 27

Example :

LAWS OF CHEMICAL COMBINATION

Water is a compound of hydrogen and oxygen. It can be obtained from various sources (like river, sea, well etc.) or even synthesized in the laboratory. From whatever source we may get it, 9 parts by weight of water is always found to contain 1 part by weight of hydrogen and 8 parts by weight of oxygen. Thus, in water, this proportion of hydrogen and oxygen always remains constant.

The laws of chemical combination are the experimental laws which led to the idea of atoms being the smallest unit of matter. The laws of chemical combination played a significant role in the development of Dalton’s atomic theory of matter. There are two important laws of chemical combination. These are:  (i) Law of conservation of mass (ii) Law of constant proportions (a) Law of Conservation of Mass or Matter :

(c) Law of Multiple Proportions :

This law was given by Lavoisier in 1774 . According to the law of conservation of mass, matter can neither be created nor be destroyed in a chemical reaction. Or The law of conservation of mass means that in a chemical reaction, the total mass of products is equal to the total mass of the reactants. There is no change in mass during a chemical reaction. Suppose we carry out a chemical reaction between A and B and if the products formed are C and D then, A + B  C + D Suppose 'a' g of A and 'b' g of B react to produce 'c' g of C and 'd' g of D. Then, according to the law of conservation of mass, we have, a+b = c+d Example : When Calcium Carbonate (CaCO 3) is heated, a chemical reaction takes place to form Calcium Oxide (CaO) and Carbon dioxide (CO2). It has been found by experiments that if 100 grams of calcium carbonate is decomposed completely, then 56 grams of Calcium Oxide and 44 grams of Carbon dioxide are formed.

Note : The converse of Law of definite proportions that when same elements combine in the same proportion, the same compound will be formed, is not always true. According to it, when one element combines with the other element to form two or more different compounds, the mass of one element, which combines with a constant mass of the other, bears a simple ratio to one another. Simple ratio means the ratio between small natural numbers, such as 1 : 1, 1 : 2, 1 : 3 e.g. Carbon and oxygen when combine, can form two oxides that are CO (carbon monoxide), CO2 (carbon dioxide). In CO,12 g carbon combined with 16 g of oxygen. In CO2,12 g carbon combined with 32 g of oxygen. Thus, we can see the mass of oxygen which combine with a constant mass of carbon (12 g) bear simple ratio of 16 : 32 or 1 : 2



Note : The law of multiple proportion was given by Dalton in 1808.

Sample Problem : 1.

Since the total mass of products (100g ) is equal to the total mass of the reactants (100g), there is no change in the mass during this chemical reaction. The mass remains same or conserved. (b) Law of Constant Proportions / Law of Definite Proportions : Proust, in 1779, analysed the chemical composition (types of elements present and percentage of elements present ) of a large number of compounds and came to the conclusion that the proportion of each element in a compound is constant (or fixed). According to the law of constant proportions: A chemical compound always consists of the same elements combined together in the same proportion by mass. 

Note : The chemical composition of a pure substance is not dependent on the source from which it is obtained.

Carbon is found to form two oxides, which contain 42.8% and 27.27% of carbon respectively. Show that these figures illustrate the law of multiple proportions. Sol. % of carbon in first oxide = 42.8 % of oxygen in first oxide = 100 - 42.8 = 57.2 % of carbon in second oxide = 27.27  % of oxygen in second oxide = 100 - 27.27 = 72.73 For the first oxide Mass of oxygen in grams that combines with 42.8 g of carbon = 57.2  Mass of oxygen that combines with 1 g of carbon = 57.2  1.34 g 42.8 For the second oxide Mass of oxygen in grams that combines with 27.27 g of carbon = 72.73  Mass of oxygen that combines with 1 g of carbon = 72.73  2.68 g 27.27

Ratio between the masses of oxygen that combine with a fixed mass (1 g) of carbon in the two oxides = 1.34 : 2.68 or 1 : 2 which is a simple ratio. Hence, this illustrates the law of multiple proportion. PAGE # 28

(d) Law of Reciprocal Proportions :

(e) Gay Lussac’s Law of Gaseous Volumes : Gay Lussac found that there exists a definite relationship among the volumes of the gaseous reactants and their products. In 1808, he put forward a generalization known as the Gay Lussac’s Law of combining volumes. This may be stated as follows :

According to this law the ratio of the weights of two element A and B which combine separately with a fixed weight of the third element C is either the same or some simple multiple of the ratio of the weights in which A and B combine directly with each other. e.g.

When gases react together, they always do so in volumes which bear a simple ratio to one another and to the volumes of the product, if these are also gases, provided all measurements of volumes are done under similar conditions of temperature and pressure. e.g. Combination between hydrogen and chlorine to form hydrogen chloride gas. One volume of hydrogen and one volume of chlorine always combine to form two volumes of hydrochloric acid gas.

The elements C and O combine separately with the third element H to form CH4 and H2O and they combine directly with each other to form CO2. 4 H2 CH4

12 C 12

H2O

CO2

16 O 32

In CH4, 12 parts by weight of carbon combine with 4 parts by weight of hydrogen. In H2O, 2 parts by weight of hydrogen combine with 16 parts by weight of oxygen. Thus the weight of C and O which combine with fixed weight of hydrogen (say 4 parts by weight) are 12 and 32 i.e. they are in the ratio 12 : 32 or 3 : 8. Now in CO2, 12 parts by weight of carbon combine directly with 32 parts by weight of oxygen i.e. they combine directly in the ratio 12 : 32 or 3 : 8 which is the same as the first ratio. 

H2 (g) + Cl2 (g) 2HCl (g) 1vol. 1 vol. 2 vol. The ratio between the volume of the reactants and the product in this reaction is simple, i.e., 1 : 1 : 2. Hence it illustrates the Law of combining volumes. (f) Avogadro’s Hypothesis : This states that equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules. This hypothesis has been found to explain elegantly all the gaseous reactions and is now widely recognized as a law or a principle known as Avogadro’s Law or Avogadro’s principle. The reaction between hydrogen and chlorine can be explained on the basis of Avogadro’s Law as follows :

Note : The law of reciprocal proportion was put forward by Ritcher in 1794.

2.

Sample Problem : Ammonia contains 82.35% of nitrogen and 17.65% of hydrogen. Water contains 88.90% of oxygen and 11.10% of hydrogen. Nitrogen trioxide contains 63.15% of oxygen and 36.85% of nitrogen. Show that these data illustrate the law of reciprocal proportions.

Hydrogen + Chlorine

Sol

In NH3, 17.65 g of H combine with N = 82.35 g

 1g of H combine with N =

82.35 17.65

88.90 11.10

1 vol.

2 vol.

n molecules.

n molecules.

2n molecules.(By Avogadro's Law)

1 molecules. 2

1 molecules. 2

1 molecules. (By dividing throughout by 2n)

1 Atom

1 Atom

1 Molecule

(By experiment)

(Applying Avogadro's hypothesis)

g = 4.67 g

In H2O, 11.10 g of H combine with O = 88.90 g

 1 g H combine with O =

Hydrogen chloride gas

1 vol.

g = 8.01 g

 Ratio of the weights of N and O which combine

It implies that one molecule of hydrogen chloride gas is made up of 1 atom of hydrogen and 1 atom of chlorine. (i) Applications of Avogadro’s hypothesis :

with fixed weight (=1g) of H = 4.67 : 8.01 = 1 : 1.72

(A) In the calculation of atomicity of elementary gases.

In N2O3, ratio of weights of N and O which combine with each other = 36.85 : 63.15 = 1 : 1.71

e.g.

Thus the two ratio are the same. Hence it illustrates the law of reciprocal proportions.

2 volumes of hydrogen combine with 1 volume of oxygen to form two volumes of water vapours. Hydrogen + Oxygen  Water vapours 2 vol. 1 vol. 2 vol. PAGE # 29

Applying Avogadro’s hypothesis

GRAM-ATOMIC MASS

Hydrogen + Oxygen  Water vapours 2 n molecules n molecules 2 n molecules

1 or 1 molecule molecule 2

1 molecule

1 Thus1 molecule of water contains molecule of 2

The atomic mass of an element expressed in grams is called the Gram Atomic Mass of the element. The number of gram -atoms =

oxygen. But 1 molecule of water contains 1 atom of oxygen. Hence.

1 molecule of oxygen = 1 atom of 2

e.g. Calculate the gram atoms present in (i) 16g of oxygen

oxygen or 1 molecules of oxygen = 2 atoms of oxygen i.e. atomicity of oxygen = 2.

and (ii) 64g of sulphur.

(B) To find the relationship between molecular mass and vapour density of a gas.

(i) The atomic mass of oxygen = 16.

Density of gas Vapour density (V.D.) = Density of hydrogen =

Mass of a certain volume of the gas Mass of the same volume of hydrogen at the same temp. and pressure

 Gram-Atomic Mass of oxygen (O) = 16 g. No. of Gram-Atom =

V.D. =

64 64 = Gram Atomic Mass of sulphur = =2 32

Atomic Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Mass of n molecules of the gas Mass of n molecules of hydrogen

=

Mass of 1 molecule of the gas Mass of 1 molecule of hydrogen

=

Molecular mass of the gas Molecular mass of hydrogen

Molecular mass 2 (since molecular mass of hydrogen is 2) Hence, Molecular mass = 2 × Vapour density =

ATOMIC MASS UNIT The atomic mass unit (amu) is equal to one-twelfth (1/12) of the mass of an atom of carbon-12.The mass of an atom of carbon-12 isotope was given the atomic mass of 12 units, i.e. 12 amu or 12 u. The atomic masses of all other elements are now expressed in atomic mass units.

RELATIVE ATOMIC MASS The atomic mass of an element is a relative quantity and it is the mass of one atom of the element relative to one -twelfth (1/12) of the mass of one carbon-12 atom. Thus, Relative atomic mass =

Mass of one atom of the element 1 12

 mass of one C  12 atom

[1/12 the mass of one C-12 atom = 1 amu, 1 amu = 1.66 × 10–24 g = 1.66 × 10–27 kg.] Note : One amu is also called one dalton (Da).

16 =1 16

(ii) The gram-atom present in 64 gram of sulphur.

If n molecules are present in the given volume of a gas and hydrogen under similar conditions of temperature and pressure.



Mass of the element in gram Gram Atomic mass of the element

Element Symbol Hydrogen H Helium He Lithium Li Beryllium Be Boron B Carbon C Nitrogen N Oxygen O Fluorine F Neon Ne Sodium Na Magnesium Mg Aluminium Al Silicon Si Phosphorus P Sulphur S Chlorine Cl Argon Ar Potassium K Calcium Ca

Atomic mass 1 4 7 9 11 12 14 16 19 20 23 24 27 28 31 32 35.5 40 39 40

RELATIVE MOLECULAR MASS The relative molecular mass of a substance is the mass of a molecule of the substance as compared to one-twelfth of the mass of one carbon -12 atom i.e., Relative molecular mass =

Mass of one molecule of the substance 1 12

 mass of one C  12 atom

The molecular mass of a molecule, thus, represents the number of times it is heavier than 1/12 of the mass of an atom of carbon-12 isotope. PAGE # 30

GRAM MOLECULAR MASS

The molecular mass of a substance expressed in grams is called the Gram Molecular Mass of the substance . The number of gram molecules Mass of the subs tance in grams

=

Gram molecular mass of the subs t ance

e.g. (i) Molecular mass of hydrogen (H2) = 2u.

 Gram Molecular Mass of hydrogen (H2) = 2 g . (ii) Molecular mass of methane (CH4) = 16u

 Gram Molecular Mass of methane (CH4) = 16 g.

EQUIVALENT MASS (a) Definition : Equivalent mass of an element is the mass of the element which combine with or displaces 1.008 parts by mass of hydrogen or 8 parts by mass of oxygen or 35.5 parts by mass of chlorine. (b) Formulae of Equivalent Masses of different substances : (i) Equivalent mass of an element = Atomic wt. of the element Valency of the element (ii) Eq. mass an acid =

Mol. wt. of the acid Basicity of the acid

e.g. the number of gram molecules present in 64 g of methane (CH4).

Basicity is the number of replaceable H+ ions from one molecule of the acid.

64 64 = Gram molecular mass of CH = = 4. 4 16

(iii) Eq. Mass of a base =

(a) Calculation of Molecular Mass : The molecular mass of a substance is the sum of the atomic masses of its constituent atoms present in a molecule. Ex.1 Calculate the molecular mass of water. (Atomic masses : H = 1u, O = 16u). Sol. The molecular formula of water is H2O.

Molecular mass of water = ( 2 × atomic mass of H) + (1 × atomic mass of O) = 2 × 1 + 1 × 16 = 18 i.e., molecular mass of water = 18 amu. Ex.2 Find out the molecular mass of sulphuric acid. (Atomic mass : H = 1u, O = 16u, S = 32u). Sol. The molecular formula of sulphuric acid is H2SO4.

 Molecular mass of H2SO4 = (2 × atomic mass of H) + ( 1 × atomic mass of S) + ( 4 × atomic mass of O) = (2 × 1) + (1× 32) + (4×16) = 2 + 32 + 64 = 98 i.e., Molecular mass of H2SO4= 98 amu.

FORMULA MASS The term ‘formula mass’ is used for ionic compounds and others where discrete molecules do not exist, e.g., sodium chloride, which is best represented as (Na+Cl–)n, but for reasons of simplicity as NaCl or Na+Cl–. Here, formula mass means the sum of the masses of all the species in the formula. Thus, the formula mass of sodium chloride = (atomic mass of sodium) + (atomic mass of chlorine) = 23 + 35.5 = 58.5 amu

Mol. wt. of the base Acidity of the base

Acidity is the number of replaceable OH– ions from one molecule of the base (iv) Eq. mass of a salt =

Mol. wt. of the salt Number of metal atoms  valency of metal

(v) Eq. mass of an ion =

Formula wt. of the ion Charge on the ion

(vi) Eq. mass of an oxidizing/reducing agent =

Mol wt. or At. wt No. of electronslost or gainedby one molecule/ atomof the substance

Equivalent weight of some compounds are given in the table :

S.No.

Compound

Equivalent weight

1

HCl

36.5

2

H2SO4

49

3

HNO3

63

4

45 COOH

5 COOH

.2H2O

63

6

NaOH

40

7

KOH

56

8

CaCO3

50

9

NaCl

58.5

10

Na2CO3

53

PAGE # 31

SOME IMPORTANT RELATIONS AND FORMULAE In Latin, mole means heap or collection or pile. A mole of atoms is a collection of atoms whose total mass is the number of grams equal to the atomic mass in magnitude. Since an equal number of moles of different elements contain an equal number of atoms, it becomes convenient to express the amounts of the elements in terms of moles. A mole represents a definite number of particles, viz, atoms, molecules, ions or electrons. This definite number is called the Avogadro Number (now called the Avogadro constant) which is equal to 6.023 × 1023.

(i) 1 mole of atoms = Gram Atomic mass = mass of 6.023 × 1023 atoms (ii) 1 mole of molecules = Gram Molecular Mass = 6.023 x 1023 molecules (iii) Number of moles of atoms

=

(iv) Number of moles of molecules

A mole is defined as the amount of a substance that contains as many atoms, molecules, ions, electrons or other elementary particles as there are atoms in exactly 12 g of carbon -12 (12C).

=

(a) Moles of Atoms :

=

(i) 1 mole atoms of any element occupy a mass which is equal to the Gram Atomic Mass of that element. e.g. 1 Mole of oxygen atoms weigh equal to Gram Atomic Mass of oxygen, i.e. 16 grams. (ii) The symbol of an element represents 6.023 x 1023 atoms (1 mole of atoms) of that element. e.g : Symbol N represents 1 mole of nitrogen atoms and 2N represents 2 moles of nitrogen atoms. (b) Moles of Molecules : (i) 1 mole molecules of any substance occupy a mass which is equal to the Gram Molecular Mass of that substance.

Mass of element in grams Gram Atomic Mass of element

Mass of substance in grams Gram Molecular Mass of substance

(v) Number of moles of molecules No. of molecules of element

N 

Avogadro number

NA

Ex.3 To calculate the number of moles in 16 grams of Sulphur (Atomic mass of Sulphur = 32 u). Sol. 1 mole of atoms = Gram Atomic Mass. So, 1 mole of Sulphur atoms = Gram Atomic Mass of Sulphur = 32 grams. Now, 32 grams of Sulphur = 1 mole of Sulphur So, 16 grams of Sulphur = (1/32) x 16 = 0.5 moles Thus, 16 grams of Sulphur constitute 0.5 mole of Sulphur. 22.4 litre

e.g. : 1 mole of water (H2O) molecules weigh equal to Gram Molecular Mass of water (H2O), i.e. 18 grams.

In term of volume 23

(ii) The symbol of a compound represents 6.023 x 10 23 molecules (1 mole of molecules) of that compound.

6.023 × 10 (NA) Atoms

23

1 Mole

e.g. : Symbol H 2O represents 1 mole of water molecules and 2 H2O represents 2 moles of water molecules. 

Note : The symbol H2O does not represent 1 mole of H2 molecules and 1 mole of O atoms. Instead, it represents 2 moles of hydrogen atoms and 1 mole of oxygen atoms.



Note : The SI unit of the amount of a substance is Mole. (c) Mole in Terms of Volume : Volume occupied by 1 Gram Molecular Mass or 1 mole of a gas under standard conditions of temperature and pressure (273 K and 1atm. pressure) is called Gram Molecular Volume. Its value is 22.4 litres for each gas. Volume of 1 mole = 22.4 litre (at STP)



Note : The term mole was introduced by Ostwald in 1896.

6.023 × 10 (NA) molecules

In terms of particles

In terms of mass

1 gram atom of element

1 gram molecule of substance

1 gram formula mass of substance

PROBLEMS BASED ON THE MOLE CONCEPT Ex.4 Calculate the number of moles in 5.75 g of sodium. (Atomic mass of sodium = 23 u) Sol. Number of moles Mass of the element in grams

=

Gram Atomic Mass of element

=

5.75 23

= 0.25 mole

or, 1 mole of sodium atoms = Gram Atomic mass of sodium = 23g. 23 g of sodium = 1 mole of sodium. 5.75 g of sodium =

5.75 23

mole of sodium = 0.25 mole

PAGE # 32

Ex.5 What is the mass in grams of a single atom of chlorine ? (Atomic mass of chlorine = 35.5u) Sol. Mass of 6.022 × 1023 atoms of Cl = Gram Atomic Mass of Cl = 35.5 g.

 Mass of 1 atom of Cl =

35.5 g 6.022  10 23

= 5.9 × 10–23 g.

Ex.6 The density of mercury is 13.6 g cm . How many moles of mercury are there in 1 litre of the metal ? (Atomic mass of Hg = 200 u). –3

Sol. Mass of mercury (Hg) in grams = Density (g cm–3)× Volume (cm3) = 13.6 g cm–3 × 1000 cm3 = 13600 g.  Number of moles of mercury Mass of mercury in grams 13600 = Gram Atomic Mass of mercury = = 68 200

Ex.7 The mass of a single atom of an element M is 3.15× 10–23 g . What is its atomic mass ? What could the element be ? Sol. Gram Atomic Mass = mass of 6.022 × 1023 atoms = mass of 1 atom × 6.022 × 1023 = (3.15 × 10–23g) × 6.022 × 1023 = 3.15 × 6.022 g = 18.97 g.  Atomic Mass of the element = 18.97u Thus, the element is most likely to be fluorine. Ex.8 An atom of neon has a mass of 3.35 × 10–23 g. How many atoms of neon are there in 20 g of the gas ? Sol. Number of atoms =

20 Total mass = = 5.97 × 1023 Mass of 1 atom 3.35  10 – 23

Ex.9 How many grams of sodium will have the same number of atoms as atoms present in 6 g of magnesium ? (Atomic masses : Na = 23u ; Mg =24u) Sol. Number of gram -atom of Mg Mass of Mg in grams

1

6

= Gram Atomic Mass = = 24 4

 Gram Atoms of sodium should be =

1 4

 1 Gram Atom of sodium = 23 g 1 4

gram atoms of sodium = 23 ×

1

= 5.75 g

4

Ex.10 How many moles of Cr are there in 85g of Cr2S3 ? (Atomic masses : Cr = 52 u , S =32 u) Sol. Molecular mass of Cr2S3=2 × 52 + 3 × 32 = 200 u. 200g of Cr2S3 contains = 104 g of Cr.

 85 g of Cr2S3 contains =

104  85 200

Thus, number of moles of Cr =

g of Cr = 44.2g

44.2 52

= 0.85 .

Ex.11 What mass in grams is represented by (a) 0.40 mol of CO2, (b) 3.00 mol of NH3, (c) 5.14 mol of H5IO6 (Atomic masses : C=12 u, O=16 u, N=14 u, H=1 u and I = 127 u) Sol. Weight in grams = number of moles × molecular mass. Hence, (a) mass of CO2 = 0.40 × 44 = 17.6 g (b) mass of NH3 = 3.00 × 17 = 51.0 g (c) mass of H5IO6 = 5.14 × 228 = 1171.92g Ex.12 Calculate the volume in litres of 20 g of hydrogen gas at STP. Sol. Number of moles of hydrogen Mass of hydrogen in grams

=

Gram Molecular Mass of hydrogen

=

20 = 10 2

 Volume of hydrogen = number of moles × Gram Molecular Volume. = 10 ×22.4 = 224 litres. Ex.13 The molecular mass of H 2 SO 4 is 98 amu. Calculate the number of moles of each element in 294 g of H2SO4. Sol. Number of moles of H2SO4 =

294 98

=3.

The formula H 2SO 4 indicates that 1 molecule of H2SO4 contains 2 atoms of H, 1 atom of S and 4 atoms of O. Thus, 1 mole of H2SO4 will contain 2 moles of H,1 mole of S and 4 moles of O atoms Therefore, in 3 moles of H2SO4 : Number of moles of H = 2 × 3 = 6 Number of moles of S = 1 × 3 = 3 Number of moles of O = 4 × 3 = 12 Ex.14 Find the mass of oxygen contained in 1 kg of potassium nitrate (KNO3). Sol. Since 1 molecule of KNO 3 contains 3 atoms of oxygen, 1 mol of KNO 3 contains 3 moles of oxygen atoms.  Moles of oxygen atoms = 3 × moles of KNO3 =3×

1000

= 29.7 101 (Gram Molecular Mass of KNO3 = 101 g)  Mass of oxygen = Number of moles × Atomic mass = 29.7 × 16 = 475.2 g Ex.15 You are asked by your teacher to buy 10 moles of distilled water from a shop where small bottles each containing 20 g of such water are available. How many bottles will you buy ? Sol. Gram Molecular Mass of water (H2O) = 18 g  10 mol of distilled water = 18 × 10 = 180 g. Because 20 g distilled water is contained in 1 bottle, 180 g

of distilled water is contained in =

bottles = 9 bottles.

 Number of bottles to be bought = 9 PAGE # 33

180 20

Ex.16 6.022 × 1023 molecules of oxygen (O2) is equal to how many moles ? Sol. No. of moles = No. of molecules of oxygen Avogadro' s no. of molecules

N 

N

=

A

6.023  10 23 6.023  10 23

(iii) If the simplest ratio is fractional, then values of simplest ratio of each element is multiplied by smallest integer to get the simplest whole number for each of the element.

=1

(iv) To get the empirical formula, symbols of various elements present are written side by side with their respective whole number ratio as a subscript to the lower right hand corner of the symbol.

PERCENTAGE COMPOSITION The percentage composition of elements in a compound is calculated from the molecular formula of the compound. The molecular mass of the compound is calculated from the atomic masses of the various elements present in the compound. The percentage by mass of each element is then computed with the help of the following relations. Percentage mass of the element in the compound

(v) The molecular formula of a substance may be determined from the empirical formula if the molecular mass of the substance is known. The molecular formula is always a simple multiple of empirical formula and the value of simple multiple (n) is obtained by dividing molecular mass with empirical formula mass. n

=

Total mass of the element

=

× 100

Molecular mass

Ex.17 What is the percentage of calcium in calcium carbonate (CaCO3) ? Sol. Molecular mass of CaCO 3 = 40 + 12 + 3 × 16 = 100 amu. Mass of calcium in 1 mol of CaCO3 = 40g.

Percentage of calcium =

40  100

Ex-20 A compound of carbon, hydrogen and nitrogen contains these elements in the ratio of 9:1:3.5 respectively. Calculate the empirical formula. If its molecular mass is 108, what is the molecular formula ? Sol. Ele m e nt M as Atom i

= 40 %

100 Ex.18 What is the percentage of sulphur in sulphuric acid (H2SO4) ? Sol. Molecular mass of H2SO4 = 1 × 2 + 32 + 16 × 4 = 98 amu.

 Percentage of sulphur =

32  100 98

Sol. Molecular mass of water, H2O = 2 + 16 = 18 amu. H2O has two atoms of hydrogen. So, total mass of hydrogen in H2O = 2 amu. Percentage of H =

18

= 11.11 %

Similarly, percentage of oxygen =

16  100 18

Re lative Num be r

s

c

Carbon

9

12

9  0.75 12

Hy droge n

1

1

1  1 1

Nitrogen

3.5

14

Sim ple s t Ratio 0. 7 5 3 0. 2 5 1 4 0. 25

3.5  0.25 14

0 .2 5 1 0 .2 5

= 32.65 % Empirical ratio = C3H4N Empirical formula mass = (3 × 12) + (4× 1) + 14 = 54

Ex.19 W hat are the percentage compositions of hydrogen and oxygen in water (H2O) ? (Atomic masses : H = 1 u, O = 16 u)

2  100

Molecular Mass Empirical Formula Mass

= 88.88 %

The following steps are involved in determining the empirical formula of a compound : (i) The percentage composition of each element is divided by its atomic mass. It gives atomic ratio of the elements present in the compound. (ii) The atomic ratio of each element is divided by the minimum value of atomic ratio as to get the simplest ratio of the atoms of elements present in the compound.

n=

Molecular Mass 108 2 = Empirical Formula Mass 54

Thus, molecular formula of the compound = (Empirical formula)2 = (C3H4N)2 = C6H8N2 Ex.21 A compound on analysis, was found to have the following composition : (i) Sodium = 14.31%, (ii) Sulphur = 9.97%, (iii) Oxygen = 69.50%, (iv) Hydrogen = 6.22%. Calculate the molecular formula of the compound assuming that whole hydrogen in the compound is present as water of crystallisation. Molecular mass of the compound is 322. Sol.

Element

Percentage

Atomic mass

Relative Number of atoms

Sodium

14.31

23

14 .31  23

Sulphur

9.97

32

9.97  32

Hydrogen

6.22

1

Oxygen

69.50

16

Simplest ratio

0.622

0.622 2 0.311

0.311

0.311 1 0.311



6.22

6.22  20 0.311

69 .50  16

4.34

4.34  14 0.311

6 . 22 1

PAGE # 34

The empirical formula = Na2SH20O14 Empirical formula mass = (2 × 23) + 32 + (20 × 1) + (14 × 16) = 322 Molecular mass = 322 Molecular formula = Na2SH20O14 Whole of the hydrogen is present in the form of water of crystallisation. Thus, 10 water molecules are present in the molecule. So, molecular formula = Na2SO4. 10H2O

Normality of a solution is defined as the number of gram equivalents of the solute dissolved per litre (dm3) of given solution. It is denoted by ‘N’. Mathematically, Number of gram equivalent s of solute

N =

N=

Weight of solute in gram / equivalent weight of solute Volume of the solution in litre

Strength in grams per litre

(a) Strength in g/L :

N=

The strength of a solution is defined as the amount of the solute in grams present in one litre (or dm3) of the solution, and hence is expressed in g/litre or g/dm3.

Equivalent mass of solute

=

S E

If ‘w’ gram of the solute is present in V cm3 of a given solution. 1000

w

Weight of solute in gram

N = Equivalent mass of the solute ×

Volume of solution in litre

(b) Molarity : Molarity of a solution is defined as the number of moles of the solute dissolved per litre (or dm3) of solution. It is denoted by ‘M’. Mathematically,

w

N = Equivalent mass of the solute × N=

Volume of the solution in litre

V

e.g. A solution of sulphuric acid having 0.49 gram of it dissolved in 250 cm 3 of solution will have its normality,

Number of moles of solute

M=

Volume of the solution in litre

N can be calculated from the strength as given below :

CONCENTRATION OF SOLUTIONS

Strength in g/L =

(d) Normality :

0.49 49

1000

×

250

1000 V

= 0.04

(Eq. mass of sulphuric acid = 49). Mass of solute in gram/Gram Molecular Mass of solute Volume of solution in litre

M can be calculated from the strength as given below : M=

1000

w Molecular

×

mass

V e.g. a solution of sulphuric acid having 4.9 grams of it dissolved in 500 cm3 of solution will have its molarity, w

M=

Normality

Deci normal

1 2

1000

Molecular mass

4.9 98

1000

×

500

×

Centi normal

1 10

1 100

Some Important Formulae :

Molecular mass of solute

If ‘w’ gram of the solute is present in V cm of a given solution , then

M=

Semi normal

Strength in grams per litre 3

M=

Solution

V

= 0.1

(c) Formality : In case of ionic compounds like NaCl, Na2CO3 etc., formality is used in place of molarity. The formality of a solution is defined as the number of gram formula masses of the solute dissolved per litre of the solution. It is represented by the symbol ‘F’. The term formula mass is used in place of molecular mass because ionic compounds exist as ions and not as molecules. Formula mass is the sum of the atomic masses of the atoms in the formula of the compound. Mass of solute in gram/Formula Mass of solute Volume of solution in litre

(i) Milli equivalent of substance = N × V where , N  normality of solution V  Volume of solution in mL (ii) If weight of substance is given, milli equivalent (NV) =

w  1000 E

Where, W  Weight of substance in gram E  Equivalent weight of substance (iii) S = N × E S  Strength in g/L N  Normality of solution E  Equivalent weight (iv) Calculation of normality of mixture : N N HCl is mixed with 50 ml of H SO . 10 5 2 4 Find out the normality of the mixture. Sol. Milli equivalent of HCl + milli equivalent H 2SO 4 = milli equivalent of mixture { where, V3 =V1 + V2 ) N1 V1 + N2 V2 = N3 V3

Ex.22 100 ml of

 1  1   100     50   N3 × 150  10 5    

N3 =

20 150

=

2 15

= 0.133

PAGE # 35

Ex.23 100 ml of

N

HCl is mixed with 25 ml of

10 Find out the normality of the mixture.

N 5

NaOH.

Sol. Milli equivalent of HCl – milli equivalent of NaOH = milli equivalent of mixture N1 V1 – N2 V2 = N3 V3 { where, V3 =V1 + V2 )  1  1   100  –   25  = N3 × 125  10 5    

N3 = 

1 25

Note : 1 milli equivalent of an acid neutralizes 1 milli equivalent of a base. (e) Molality : Molality of a solution is defined as the number of moles of the solute dissolved in 1000 grams of the solvent. It is denoted by ‘m’. Mathematically,

Number of moles of the solute × 1000 Weight of the solvent in gram ‘m’ can be calculated from the strength as given below : m=

Strength per 1000 gram of solvent m= Molecular mass of solute If ‘w’ gram of the solute is dissolved in ‘W’ gram of the solvent then m=

w 1000 × Mol. mass of the solute W

e.g. A solution of anhydrous sodium carbonate (molecular mass = 106) having 1.325 grams of it, dissolved in 250 gram of water will have its molality m= 

1.325 1000  = 0.05 106 250

Ex.25 Find the molarity and molality of a 15% solution of H2SO4 (density of H2SO4 solution = 1.02 g/cm3) (Atomic mass : H = 1u, O = 16u , S = 32 u) Sol. 15% solution of H2SO4 means 15g of H2SO4 are present in 100g of the solution i.e. Wt. of H2SO4 dissolved = 15 g Weight of the solution = 100 g Density of the solution = 1.02 g/cm3 (Given) Calculation of molality : Weight of solution = 100 g Weight of H2SO4 = 15 g Wt. of water (solvent) = 100 – 15 = 85 g Molecular weight of H2SO4 = 98

 15 g H2SO4 =

15 98

= 0.153 moles

Thus, 85 g of the solvent contain 0.153 moles . 1000 g of the solvent contain=

0.153 85

× 1000 = 1.8 mole

Hence ,the molality of H2SO4 solution = 1.8 m Calculation of molarity : 15 g of H2SO4 = 0.153 moles Wt. of solution

Vol. of solution = =

100 1.02

Density of solution

= 98.04 cm3

This 98.04 cm3 of solution contain H2SO4 = 0.153 moles 1000 cm3 of solution contain H2SO4

0.153 =

98.04

× 1000 = 1.56 moles

Note :

Hence the molarity of H2SO4 solution = 1.56 M

Relationship Between Normality and Molarity of a Solution : Normality of an acid = Molarity × Basicity Normality of base = Molarity × Acidity

(f) Mole Fraction :

Ex.24 Calculate the molarity and normality of a solution containing 0.5 g of NaOH dissolved in 500 cm 3 of solvent. Sol. Weight of NaOH dissolved = 0.5 g Volume of the solution = 500 cm3 (i) Calculation of molarity : Molecular weight of NaOH = 23 + 16 + 1 = 40 Weight of solute/ molecular weight of solute Volume of solution in litre 0.5/40

Molarity = =

500/1000

= 0.025

(ii) Calculation of normality : Normality Weight of solute/ equivalent weight of solute = Volume of solution in litre 0.5/40 = = 0.025 500/1000

The ratio between the moles of solute or solvent to the total moles of solution is called mole fraction.

Moles of solute mole fraction of solute = =

Moles of solution



n nN

w/m w/m  W/M

Moles of solvent Mole fraction of solvent =

Moles of solution



N nN =

W/M w/m  W/M where,

n  number of moles of solute N  number of moles of solvent m  molecular weight of solute M  molecular weight of solvent w  weight of solute W  weight of solvent PAGE # 36

Ex.26 Find out the mole fraction of solute in 10% (by weight) urea solution. weight of solute (urea) = 10 g weight of solution = 100 g weight of solvent (water) = 100 – 10 = 90g

Moles of solution 10 / 60

w/m w/m  W/M 

=

10 / 60  90 / 18

2KClO3(s)  2KCl(s) + 3O2(g) 2 mole KClO3  3 mole O2

 3 mole O2 formed by 2 mole KClO3 2

Moles of solute mole fraction of solute =

Sol. Decomposition of KClO3 takes place as,

=

= 0.032

Note : Sum of mole fraction of solute and solvent is always equal to one.



 2.4 mole O 2 will be formed by  3  2.4    moles of KClO3 = 1.6 mole KClO3 Mass of KClO3 = Number of moles × molar mass = 1.6 × 122.5 = 196 g (ii) Calculations based on mass-mass relationship: In making necessary calculation, following steps are followed (a) Write down the balanced chemical equation.

STOICHIOMETRY (a) Quantitative Relations in Chemical Reactions : Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction. It is based on the chemical equation and on the relationship between mass and moles. N2(g) + 3H2(g)  2NH3(g)

(b) Write down theoretical amount of reactants and products involved in the reaction. (c) The unknown amount of substance is calculated using unitary method. Ex.28 Calculate the mass of CaO that can be prepared by heating 200 kg of limestone CaCO3 which is 95% pure. Sol. Amount of pure CaCO 3 =

95  200 = 190 kg 100

A chemical equation can be interpreted as follows -

= 190000 g

1 molecule N2 + 3 molecules H2  2 molecules NH 3 (Molecular interpretation)

CaCO3(s)  CaO(s) + CO2(g)

1 mol N2 + 3 mol H2  2 mol NH3 (Molar interpretation)

100 g CaCO3  56 g CaO

28 g N2 + 6 g H2  34 g NH3 (Mass interpretation) 1 volume N2 + 3 volume H2  2 volume NH3 (Volume interpretation) Thus, calculations based on chemical equations are divided into four types (i) Calculations based on mole-mole relationship. (ii) Calculations based on mass-mass relationship. (iii) Calculations based on mass-volume relationship. (iv) Calculations based on volume -volume relationship. (i) Calculations based on mole-mole relationship : In such calculations, number of moles of reactants are given and those of products are required. Conversely, if number of moles of products are given, then number of moles of reactants are required. Ex.27 Oxygen is prepared by catalytic decomposition of potassium chlorate (KClO 3). Decomposition of potassium chlorate gives potassium chloride (KCl) and oxygen (O2). How many moles and how many grams of KClO 3 are required to produce 2.4 mole O2.

1 mole CaCO3  1 mole CaO

100 g CaCO3 give 56 g CaO 56  190000 g CaCO3 will give= × 190000 g CaO 100 = 106400 g = 106.4 kg Ex.29 Chlorine is prepared in the laboratory by treating manganese dioxide (MnO 2 ) with aqueous hydrochloric acid according to the reaction MnO2 + 4HCl  MnCl2 + Cl2 + 2H2O How many grams of HCl will react with 5 g MnO2 ? Sol. 1 mole MnO2 reacts with 4 mole HCl or 87 g MnO2 react with 146 g HCl

 5 g MnO2 will react with =

146 × 5 g HCl = 8.39 g HCl 87

Ex.30 How many grams of oxygen are required to burn completely 570 g of octane ? Sol. Balanced equation

2C8H18 + 25O2 2 mole 2 × 114

16CO2 + 18H2O

25 mole 25 × 32

First method : For burning 2 × 114 g of the octane, oxygen required = 25 × 32 g For burning 1 g of octane, oxygen required =

25  32 g 2  114

Thus, for burning 570 g of octane, oxygen required =

25  32 × 570 g = 2000 g 2  114 PAGE # 37

Mole Method : Number of moles of octane in 570 grams

Ex-33 What quantity of copper (II) oxide will react with 2.80 litre of hydrogen at STP ?

570 = 5.0 114

Sol. CuO + 1 mol 79.5 g

For burning 2.0 moles of octane, oxygen required = 25 mol = 25 × 32 g For burning 5 moles of octane, oxygen required =

25  32 × 5.0 g = 2000 g 2 .0

Proportion Method : Let x g of oxygen be required for burning 570 g of octane. It is known that 2 × 114 g of the octane requires 25 × 32 g of oxygen; then, the proportion. x 25  32 g oxygen = 570 g oc tane 2  114 g oc tan e

2Fe2O3 + 8H2SO4

4 mole 4 × 120 g

8 mole 8 × 98 g

4 × 120 g of FeS2 yield H2SO4 = 8 × 98 g 1000 g of FeS2 will yield H2SO4 =

8  98 × 1000 4  120

= 1633.3 g (iii) Calculations involving mass-volume relationship : In such calculations masses of reactants are given and volume of the product is required and vice-versa. 1 mole of a gas occupies 22.4 litre volume at STP. Mass of a gas can be related to volume according to the following gas equation PV = nRT PV =

w RT m

NH4Cl(s)

=

79.5 × 2.80 g = 9.93 g 22.4

Ex-34 Calculate the volume of carbon dioxide at STP evolved by strong heating of 20 g calcium carbonate. Sol. The balanced equation is -

CaCO3

NH3(g) + HCl(g) 1 mol

 53.5 g NH4Cl give 1 mole NH3 1  26.75 g NH4Cl will give × 26.75 mole NH3 53.5 = 0.5 mole PV = nRT 1 ×V = 0.5 × 0.0821 × 300 V = 12.315 litre

CaO +

1 mol 100 g

CO2

1 mol = 22.4 litre at STP

22.4 × 20 = 4.48 litre 100

Ex.35 Calculate the volume of hydrogen liberated at 27ºC and 760 mm pressure by heating 1.2 g of magnesium with excess of hydrochloric acid. Sol. The balanced equation is

Mg + 2HCl

MgCl2

+

24 g

H2 1 mol

24 g of Mg liberate hydrogen = 1 mole 1.2 g Mg will liberate hydrogen = 0.05 mole PV = nRT 1 × V = 0.05 × 0.0821 × 300 V = 1.2315 litre (iv) Calculations based on volume volume relationship : These calculations are based on two laws : (i) Avogadro’s law (ii) Gay-Lussac’s Law e.g. N 2(g)

+

1 mol 1 × 22.4 L

3H 2(g) 3 mol 3 × 22.4 L

2NH 3(g) (Avogadro's law) 2 mol 2 × 22.4 L

(under similar conditions of temperature and pressure, equal moles of gases occupy equal volumes) N2(g)

+

1 vol

Ex-32. What volume of NH3 can be obtained from 26.75 g of NH4Cl at 27ºC and 1 atmosphere pressure. Sol. The balanced equation is 1 mol 53.5 g

22.4 litre of hydrogen at STP reduce CuO = 79.5 g 2.80 litre of hydrogen at STP will reduce CuO

=

Ex.31 How many kilograms of pure H 2SO 4 could be obtained from 1 kg of iron pyrites (FeS2) according to the following reactions ? 4FeS2 + 11O2  2Fe2O3 + 8SO2 2SO2 + O2  2SO3 SO3 + H2O  H2SO4 Sol. Final balanced equation, 4FeS2 + 15O2 + 8H2O

+ H2O

100 g of CaCO3 evolve carbon dioxide = 22.4 litre 20 g CaCO3 will evolve carbon dioxide

25  32  570 = 2000 g 2  114

x=

Cu

H2

1 mol 22.4 litre at NTP

3H2(g) 3 vol

2NH3(g) (Gay- Lussac's Law) 2 vol

(under similar conditions of temperature and pressure, ratio of coefficients by mole is equal to ratio of coefficient by volume). Ex-36 One litre mixture of CO and CO2 is taken. This is passed through a tube containing red hot charcoal. The volume now becomes 1.6 litre. The volume are measured under the same conditions. Find the composition of mixture by volume. Sol. Let there be x mL CO in the mixture , hence, there will be (1000 – x) mL CO2. The reaction of CO2 with red hot charcoal may be given as -

CO2(g) + C(s) 1 vol. (1000 –x)

2CO(g) 2 vol. 2(1000 – x)

Total volume of the gas becomes = x + 2(1000 – x) x + 2000 – 2x = 1600 x = 400 mL

 volume of CO = 400 mL and volume of CO2 = 600 mL PAGE # 38

Ex-37 What volume of air containing 21% oxygen by volume is required to completely burn 1kg of carbon containing 100% combustible substance ? Sol. Combustion of carbon may be given as,

C(s) + O2(g) 1 mol 12 g

CO2(g)

1 mol 32 g

USEFUL FORMULAE FOR VOLUMETRIC CALCULATIONS (i) milliequivalents = normality × volume in millilitres. (ii) At the end point of titration, the two titrants, say 1 and 2, have the same number of milliequivalents, i.e., N1V1 = N2V2, volume being in mL.

 12 g carbon requires 1 mole O2 for complete combustion 1  1000 mole O2 for 12 combustion, i.e. , 83.33 mole O2

 1000 g carbon will require

m.e. . 1000 (iv) No. of equivalents for a gas =

(iii) No. of equivalents =

Volumeat STP equivalent volume( vol. of 1eq. at STP)

Volume of O2 at STP = 83.33 × 22.4 litre

(v) Strength in grams per litre = normality × equivalent

= 1866.66 litre

weight.

 21 litre O2 is present in 100 litre air 100 1866.66 litre O2 will be present in × 1866.66 litre air 21 = 8888.88 litre or 8.89 × 103 litre

Ex-38 An impure sample of calcium carbonate contains 80% pure calcium carbonate 25 g of the impure sample reacted with excess of hydrochloric acid. Calculate the volume of carbon dioxide at STP obtained from this sample. Sol. 100 g of impure calcium carbonate contains = 80 g pure calcium carbonate 25 g of impure calcium carbonate sample will contain =

80 × 25 = 20 g pure calcium carbonate 100

The desired equation is CaCO3 + 2HCl

(vi) (a) Normality = molarity × factor relating mol. wt. and eq. wt. (b) No. of equivalents = no. of moles × factor relat ing mol. wt. and eq. wt. Ex.39 Calculate the number of milli equivalent of H2SO4 present in 10 mL of N/2 H2SO4 solution. 1 × 10 = 5. 2 Ex.40 Calculate the number of m.e. and equivalents of

Sol. Number of m.e. = normality × volume in mL =

NaOH present in 1 litre of N/10 NaOH solution. Sol. Number of m.e. = normality × volume in mL =

1 × 1000 = 100 10

Number of equivalents =

CaCl2 + CO2 + H2O 22.4 litre at STP

1 mol 100 g

100 g pure CaCO3 liberate = 22.4 litre CO2. 20 g pure CaCO3 liberate =

22.4  20 100

= 4.48 litre CO2

VOLUMETRIC CALCULATIONS

no. of m.e. 100 = = 0.10 1000 1000

Ex.41 Calculate number of m.e. of the acids present in (i) 100 mL of 0.5 M oxalic acid solution. (ii) 50 mL of 0.1 M sulphuric acid solution. Sol. Normality = molarity × basicity of acid (i) Normality of oxalic acid = 0.5 × 2 = 1 N m.e. of oxalic acid = normality × vol. in mL = 1 × 100 = 100. (ii) Normality of sulphuric acid = 0.1 × 2 = 0.2 N m.e. of sulphuric acid = 0.2 × 50 = 10

The quantitative analysis in chemistry is primarily carried out by two methods, viz, volumetric analysis and gravimetric analysis.In the first method the mass of a chemical species is measured by measurement of volume, whereas in the second method it is determined by taking the weight.

Ex.42 A 100 mL solution of KOH contains 10 milliequiva

The strength of a solution in volumetric analysis is generally expressed in terms of normality, i.e., number of equivalents per litre but since the volume in the volumetric analysis is generally taken in millilitres (mL), the normality is expressed by milliequivalents per millilitre.

Again, strength in grams/litre = normality × eq. wt.

lents of KOH. Calculate its strength in normality and grams/litre. no. of m.e. Sol. Normality = volume in mL

=

10  0 .1 100

 strength of the solution = N/10

=

1  56 = 5.6 gram/litre. 10

  molecular wt. 56  eq. wt. of KOH    56  acidity 1  

PAGE # 39

Ex.43 What is strength in gram/litre of a solution of H2SO4, N 12 cc of which neutralises 15 cc of NaOH 10 solution ?

Sol. m.e. of NaOH solution =

1 × 15 = 1.5 10

m.e. of 12 cc of H2SO4 = 1.5 1.5 12 Strength in grams/litre = normality × eq. wt.

 normality of H2SO4 =

1.5 × 49 grams/litre 12 = 6.125 grams/litre.

=

  molecular wt. 98  eq. wt. of H2 SO 4    49  basicity 2   Ex.44 What weight of KMnO4 will be required to prepare N solution if eq. wt. of KMnO4 is 31.6 ? 10 Sol. Equivalent weight of KMnO4 = 31.6

250 mL of its

1 10 Volume of solution (V) = 250 ml

Normality of solution (N) =

W

NEV ; W = 1  31 .6  250 31 .6  0.79 g 10 1000 1000 40

Ex.45 100 mL of 0.6 N H2SO4 and 200 mL of 0.3 N HCl were mixed together. What will be the normality of the resulting solution ? Sol. m.e. of H2SO4 solution = 0.6 × 100 = 60 m.e. of HCl solution = 0.3 × 200 = 60  m.e. of 300 mL (100 + 200) of acidic mixture = 60 + 60 = 120.

m.e. total vol. 120 2 = = N. 300 5

Normality of the resulting solution =

Ex.46 A sample of Na2CO3. H2O weighing 0.62 g is added to 100 mL of 0.1 N H2SO4. Will the resulting solution be acidic, basic or neutral ?

0.62 Sol. Equivalents of Na2CO3. H2O = = 0.01 62 124    62   eq. wt. of Na 2 CO 3 .H2 O  2   m.e. of Na2CO3. H2O = 0.01 × 1000 = 10 m.e. of H2SO4 = 0.1 × 100 = 10 Since the m.e. of Na2CO3. H2O is equal to that of H2SO4, the resulting solution will be neutral.

(a) Introduction : Volumetric analysis is a method of quantitative analysis. It involves the measurement of the volume of a known solution required to bring about the completion of the reaction with a measured volume of the unknown solution whose concentration or strength is to be determined. By knowing the volume of the known solution, the concentration of the solution under investigation can be calculated. Volumetric analysis is also termed as titrimetric analysis.

(b) Important terms used in volumetric analysis : (i) Titration : The process of addition of the known solution from the burette to the measured volume of solution of the substance to be estimated until the reaction between the two is just complete, is termed as titration. Thus, a titration involves two solutions: (a) Unknown solution and (b) Known solution or standard solution. (ii) Titrant : The reagent or substance whose solution is employed to estimate the concentration of unknown solution is termed titrant. There are two types of reagents or titrants: (A) Primary titrants : These reagents can be accurately weighed and their solutions are not to be standardised before use. Oxalic acid, potassium dichromate, silver nitrate, copper sulphate, ferrous ammonium sulphate, sodium thiosulphates etc., are the examples of primary titrants. (B) Secondary titrants : These reagents cannot accurately weighed and their solutions are to be standardised before use. Sodium hydroxide, potassium hydroxide, hydrochloric acid, sulphuric acid, iodine, potassium permanganate etc. are the examples of secondary titrants. (iii) Standard solution : The solution of exactly known concentration of the titrant is called the standard solution. (iv) Titrate : The solution consisting the substance to be estimated is termed unknown solution. The substance is termed titrate. (v) Equivalence point : The point at which the reagent (titrant) and the substance (titrate) under investigation are chemically equivalent is termed equivalence point or end point. (vi) Indicator : It is the auxiliary substance used for physical (visual) detection of the completion of titration or detection of end point is termed as indicator. Indicators show change in colour or turbidity at the stage of completion of titration. (c) Concentraion representation of solution (A) Strength of solution : Grams of solute dissolved per litre of solution is called strength of solution' (B) Parts Per Million (ppm) : Grams of solute dissolved per 10 6 grams of solvent is called concentration of solution in the unit of Parts Per Million (ppm). This unit is used to represent hardness of water and concentration of very dilute solutions. (C) Percentage by mass : Grams of solute dissolved per 100 grams of solution is called percentage by mass. (D) Percentage by volume : Millilitres of solute per 100 mL of solution is called percentage by volume. For example, if 25 mL ethyl alcohol is diluted with water to make 100 mL solution then the solution thus obtained is 25% ethyl alcohol by volume. (E) Mass by volume percentage :Grams of solute present per 100 mL of solution is called percentage mass by volume. For example, let 25 g glucose is dissolved in water to make 100 mL solution then the solution is 25% mass by volume glucose. PAGE # 40

(d) Classification of reactions involved in volumetric analysis (A) Neutralisation reactions The reaction in which acids and bases react to form salt called neutralisation. e.g., HCI + NaOH  NaCI + H2O H+(acid) + OH–(base)  H2O (feebly ionised) The titration based on neutralisation is called acidimetry or alkalimetry. (B) Oxidation-reduction reactions The reactions involving simultaneous loss and gain of electrons among the reacting species are called oxidation reduction or redox reactions, e.g., let us consider oxidation of ferrous sulphate (Fe2+ ion) by potassium permanganate (MnO 4– ion) in acidic medium. MnO4– + 8H+ + 5e–  Mn2+ + 4H2O (Gain of electrons or reduction) 5  [Fe2+  Fe3+ + e–] (Loss of electrons or oxidation) MnO4– + 5Fe2+ + 8H+  Mn2+ + 5Fe3+ + 4H2O ________________________________________________________________ In the above reaction, MnO4– acts as oxidising agent and Fe2+ acts as reducing agent. The titrations involving redox reactions are called redox titrations. These titrations are also called according to the reagent used in the titration, e.g., iodometric, cerimetric, permanganometric and dichromometric titrations (C) Precipitation reactions : A chemical reaction in which cations and anions combine to form a compound of very low solubility (in the form of residue or precipitate), is called precipitation. BaCl2 + Na2SO4  BaSO4+ 2NaCl (white precipitate) The titrations involving precipitation reactions are called precipitation titrations. (D) Complex formation reactions : These are ion combination reactions in which a soluble slightly dissociated complex ion or compound is formed. Complex compounds retain their identity in the solution and have the properties of the constituent ions and molecules. e.g. CuSO4 + 4NH3  [Cu(NH3)4]SO4 (complex compound) AgNO3 + 2KCN  K[Ag(CN)2] + KNO3 (complex compound) 2CuSO4 + K4[Fe(CN)6]  Cu2[Fe(CN)6] + 2K2SO4 (complex compound) The titrations involving complex formation reactions are called complexometric titrations.

The determination of concentration of bases by titration with a standard acid is called acidimetry and the determination of concentration of acid by titration with a standard base is called alkalimetry. The substances which give different colours with acids and base are called acid base indicators. These indicators are used in the visual detection of the equivalence point in acid-base titrations. The acid-base indicators are also called pH indicators because their colour change according to the pH of the solution. In the selection of indicator for a titration, following two informations are taken into consideration : (i) pH range of indicator (ii) pH change near the equivalence point in the titration. The indicator whose pH range is included in the pH change of the solution near the equivalence point, is taken as suitable indicator for the titration. (i) Strong acid-strong base titration : In the titration of HCl with NaOH, the equivalence point lies in the pH change of 4–10. Thus, methyl orange, methyl red and phenolphthalein will be suitable indicators. (ii) Weak acid-strong base titration : In the titration of CH3COOH with NaOH the equivalence point lies between 7.5 and 10. Hence, phenolphthalein (8.3– 10) will be the suitable indicator. (iii) Weak base-strong acid titration : In the titration of NH4OH (weak base) against HCl (strong acid) the pH at equivalence point is about 6.5 and 4. Thus, methyl orange (3.1–4.4) or methyl red (4.2–6.3) will be suitable indicators. (iv) Weak acid-weak base titration : In the titration of a weak acid (CH3COOH) with weak base (NH4OH) the pH at the equivalence point is about 7, i.e., lies between 6.5 and 7.5 but no sharp change in pH is observed in these titrations. Thus, no simple indicator can be employed for the detection of the equivalence point. (v) Titration of a salt of a weak acid and a strong base with strong acid: H2CO3 + 2NaOH  Na2CO3 + 2H2O Weak acid Strong base Na 2CO3 when titrated with HCl, the following two stages are involved : Na2CO3 + HCl  NaHCO3 + NaCl (First stage) pH = 8.3, near equivalence point NaHCO3 + HCl  NaCl + H2CO3 (Second stage) pH = 4, near equivalence point

PAGE # 41

(i) Phenolphthalein (weak organic acid) : It shows colour change in the pH range (8 – 10) (ii) Methyl orange (weak organic base) : It shows

For first stage, phenolphthalein and for second stage, methyl orange will be the suitable indicator. Titration of mixture of NaOH, Na2CO3 and NaHCO3 by strong acid like HCl

colour change in the pH range (3.1 – 4.4). Due to lower pH range, it indicates complete neutralisation

In this titration the following indicators are mainly used : of whole of the base.

Let for complete neutralisation of Na2CO3, NaHCO3 and NaOH, x,y and z mL of standard HCl are required. The titration of the mixture may be carried by two methods as summarised below :

Volume of HCl used with Mixture

Phenlphthalein from beginning

Methyl orange from beginning

Methyl orange after first end point

1. NaOH + Na2CO3

z + (x/2)

(x + z)

x/2 (for remaining 50% Na2CO3)

2. NaOH + NaHCO3

z+0

(z + y)

y (for remaining 100% NaHCO3

3. Na2CO3 + NaHCO3

(x/2) + 0

(x + y)

x/2 + y (for remaining 50% of Na2CO3 and 100% NaHCO3 are indicated)

An indicator is a substance which is used to determine the end point in a titration. In acid-base titrations organic substances (weak acids or weak bases) are generally used as indicators. They change their colour within a certain pH range. The colour change and the pH range of some common indicators are tabulated below:

PAGE # 42

________________________________________ Indicator pH range Colour change ________________________________________ Methyl orange 3.2 – 4.5 Orange to red Methyl red 4.4 – 6.5 Red to yellow Litmus 5.5 – 7.5 Red to blue Phenol red 6.8 – 8.4 Yellow to red Phenolphthalein 8.3 – 10.5 Colourless to pink ________________________________________ Theory of acid-base indicators : Two theories have been proposed to explain the change of colour of acid-base indicators with change in pH.

[HIn] [H3+O] = KIn 

pH = –log10 [H3+O] = –log10 [KIn] – log10

(b) The ionisation of the indicator is largely affected in acids and bases as it is either a weak acid or a weak base. In case, the indicator is a weak acid, its ionisation is very much low in acids due to common ions while it is fairly ionised in alkalies. Similarly if the indicator is a weak base, its ionisation is large in acids and low in alkalies due to common ions. Considering two important indicators phenolphthalein (a weak acid) and methyl orange (a weak base), Ostwald's theory can be illustrated as follows: Phenolphthalein: It can be represented as HPh. It ionises in solution to a small extent as: H+ + Ph–

HPh Colourless

Pink

Applying law of mass action, K=

[H ][Ph  ] [HPh]

The undissociated molecules of phenolphthalein are colourless while ph– ions are pink in colour. In presence of an acid, the ionisation of HPh is practically negligible as the equilibrium shifts to left hand side due to high concentration of H+ ions. Thus, the solution would remain colourless. On addition of alkali, hydrogen ions are removed by OH– ions in the form of water molecules and the equilibrium shifts to right hand side. Thus, the concentration of ph– ions increases in solution and they impart pink colour to the solution. Let us derive Henderson's equation for an indicator HIn + H2O

H3+O + In–

'Acid form'

'Base form'

pH = pKIn + log10

KIn =

[In  ][H3 O] [HIn]

KIn = Ionization constant of indicator

[In  ]

[In ] (Henderson's equation for [HIn]

indicator) At equivalence point ; [In–] = [HIn] and pH = pKIn Methyl orange : It is a very weak base and can be represented as MeOH. It is ionised in solution to give Me+ and OH– ions. MeOH

Me+ +OH–

Orange

Red

Applying law of mass action, K=

[Me  ][OH  ] [MeOH]

In presence of an acid, OH– ions are removed in the form of water molecules and the above equilibrium shifts to right hand side. Thus, sufficient Me+ ions are produced which impart red colour to the solution. On addition of alkali, the concentration of OH– ions increases in the solution and the equilibrium shifts to left hand side, i.e., the ionisation of MeOH is practically negligible. Thus, the solution acquires the colour of unionised methyl orange molecules, i.e. orange. This theory also explains the reason why phenolphthalein is not a suitable indicator for titrating a weak – base against strong acid. The OH ions furnished by a weak base are not sufficient to shift the equilibrium towards right hand side considerably, i.e., pH is not reached to 8.3. Thus, the solution does not attain pink colour. Similarly, it can be explained why methyl orange is not a suitable indicator for the titration of weak acid with strong base.

SOLUBILITY The solubility of a solute in a solution is always expressed with respect to the saturated solution. (a) Definition : The maximum amount of the solute which can be dissolved in 100g (0.1kg) of the solvent to form a saturated solution at a given temperature. Suppose w gram of a solute is dissolved in W gram of a solvent to make a saturated solution at a fixed temperature and pressure. The solubility of the solute will be given by Mass of the solute

w Conjugate acid-base pair

[HIn]



1. Ostwald's theory: According to this theory (a) The colour change is due to ionisation of the acidbase indicator. The unionised form has different colour than the ionised form.

[In  ]

W

× 100 =

Mass of the solvent

× 100

For example, the solubility of potassium chloride in water at 20ºC and 1 atm. is 34.7 g per 100g of water. This means that under normal conditions 100 g of water at 20ºC and 1 atm. cannot dissolve more than 34.7g of KCl. PAGE # 43

(b) Effect of Temperature and Pressure on Solubility of a Solids : The solubility of a substance in liquids generally increases with rise in temperature but hardly changes with the change in pressure. The effect of temperature depends upon the heat energy changes which accompany the process. 

Note : If heat energy is needed or absorbed in the process, it is of endothermic nature. If heat energy is evolved or released in the process, it is of exothermic nature. (i) Effect of temperature on endothermic dissolution process : Most of the salts like sodium chloride, potassium chloride, sodium nitrate, ammonium chloride etc. dissolve in water with the absorption of heat. In all these salts the solubility increases with rise in temperature. This means that sodium chloride becomes more soluble in water upon heating. (ii) Effect of temperature on exothermic dissolution process : Few salts like lithium carbonate, sodium carbonate monohydrate, cerium sulphate etc. dissolve in water with the evolution of heat. This means that the process is of exothermic nature. In these salts the solubility in water decreases with rise in temperature.



Ex.48 4 g of a solute are dissolved in 40 g of water to form a saturated solution at 25ºC. Calculate the solubility of the solute. Mass of solute

Sol. Solubility =

Mass of solvent

× 100

Mass of solute = 4 g Mass of solvent = 40 g Solubility =

4 × 100 = 10 g 40

Ex.49 (a) What mass of potassium chloride would be needed to form a saturated solution in 50 g of water at 298 K ? Given that solubility of the salt is 46g per 100g at this temperature. (b) What will happen if this solution is cooled ? Sol. (a) Mass of potassium chloride in 100 g of water in saturated solution = 46 g Mass of potassium chloride in 50 g of water in saturated solution. =

46 × 50 = 23 g 100

(b) When the solution is cooled, the solubility of salt in water will decrease. This means, that upon cooling, it will start separating from the solution in crystalline form.

Note : 1. While expressing the solubility, the solution must be saturated but for expressing concentration (mass percent or volume percent), the solution need not to be saturated in nature. 2. While expressing solubility, mass of solvent is considered but for expressing concentration the mass or volume of the solution may be taken into consideration. (c) Ef fect of Temperature on the Solubility of a Gas (i) The solubility of a gas in a liquid decreases with the rise in temperature. (ii) The solubility of gases in liquids increases on increasing the pressure and decreases on decreasing the pressure.

SAMPLE PROBLEMS Ex.47 12 grams of potassium sulphate dissolves in 75 grams of water at 60ºC. What is the solubility of potassium sulphate in water at that temperature ? Sol. Solubility =

mass of solute  100 mass of solvent 12

=

75

×100 = 16 g

Thus, the solubility of potassium sulphate in water is 16 g at 60ºC.

It is defined as the charge (real or imaginary ) which an atom appears to have when it is in combination. In the case of electrovalent compounds, the oxidation number of an element or radical is the same as the charge on the ion. This is the real charge and is developed by the loss and gain of electron or electrons. For example, in the electrovalent compound , sodium chloride (NaCl), the charge on sodium and chlorine is +1 and –1, respectively. The oxidation numbers of atoms in covalent compounds can be derived by assigning the electrons of each bond to the more electronegative atom of the bonded atoms. For a molecule of HCl both the electrons of the covalent bond are assigned to the chlorine atom since it is more electronegative than hydrogen.

H × Cl Thus, chlorine atom has one more electron than the neutral chlorine atom which brings one unit negative charge on chlorine. The oxidation number of chlorine in the compound is –1. The hydrogen atom has lost the only electron possessed by it, thus acquiring one unit positive charge. The oxidation number of hydrogen is , therefore , +1 in this compound. PAGE # 44

Counting of electrons in this fashion is not convenient

Ex.51 What is the oxidation number of Cr in K2Cr2O7 ?

in many molecules and therefore the following operational rules are followed which are helpful and

Sol. Let the Ox.no. of Cr in K2Cr2O7 be x. We known that Ox. no. of K = + 1

convenient in determining the oxidation numbers : (i) The oxidation number (Ox. no.) of an atom in free

Ox. no. of O = 2 2(ox. no. K) + 2(ox. no. Cr) + 7 (ox. no. O) = 0

So,

elements is zero, no matter how complicated the molecule is, hydrogen in H2, or O3,all have zero value

or

2(+1) +2

of oxidation numbers.

or

2x = + 14 –2 = + 12

or

x=+

(ii) The fluorine, which is the most electronegative element, has oxidation number –1 in all of its compounds. (iii) Oxidation number of oxygen is –2 in all compounds except in peroxides, superoxides and oxygen fluorides. In peroxides (O22–) oxygen has oxidation number –1,

+ 2(x) + 2x

+ 7(–2) – 14

=0 =0

12 =+6 2

Hence, oxidation number of Cr in K2Cr2O7 is + 6 Ex.52 Find the oxidation number of (a) S in SO42– (b) S in HSO3– (c) Pt in [Pt Cl6]2– (d) Mn in (MnO4)– ion Sol. (a) Let the oxidation number of S be x.

in superoxide (O2–) oxygen has oxidation number –½ and in OF2, the oxygen has an oxidation number +2.

we known that ox. no. of O = – 2 s o ox. no. S + 4 (ox. no. O) =–2

(iv) The oxidation number of hydrogen is +1 in all of its compounds except in metallic hydrides like NaH,

or or

BaH2, etc. hydrogen is in –1 oxidation state in these hydrides.

or x=+8–2=+6 The oxidation number of S in SO42– ions is +6.

(v) The oxidation number of an ion is equal to the

(b) Let the oxidation number of S be x in HSO3– ion.

electrical charge present on it.

we known that

(vi) The oxidation number of IA elements (Li, Na, K,

ox. no. of H = + 1

Rb, Cs and Fr) is +1 and the oxidation number of IIA elements (Be, Mg, Ca, Sr, Ba and Ra) is +2.

ox. no. of O = – 2

(vii) For complex ions, the algebraic sum of oxidation numbers of all the atoms is equal to the net charge

x + x –

sum of the oxidation numbers of all the atoms present in the molecule is zero.

=–2 =–2

s o ox. no. H + ox. no. S + 3 (ox. no. O) = – 1 or

+1

+x

+ 3(–2)

=–1

or

+1

+x

–

6

=–1

x

–

5

=–1

on the ion. (viii) In the case of neutral molecules, the algebraic

4 (–2) 8

or

x=+5–1=+4

The oxidation number of S in HSO3– ion is + 4. (c) Let oxidation number of Pt be x.

Ex.50 What is the oxidation number of Mn in KMnO4 and of S in Na2S2O3 ?

We know that Ox. no. of Cl = – 1 So ox. no. Pt + 6(ox. no. Cl)

Sol. Let the ox. no. of Mn in KMnO4 be x We known that ox. no. of K = +1

=–2

or

x

+ 6(–1)

=–2

or

x

–

= –2

or

x=+6–2=+4

ox. no. of O = –2 s o ox. no. K + ox. no. Mn + 4 (ox. no. O) = 0 or or

+1 +1

+ +

x x

or

x=8–1=+7

+ 4(–2) – 8

=0 =0

6

The oxidation number of Pt in [PtCl6]2– ions is + 4. (d) Let oxidation number of Mn be x.

Hence, ox. no. of Mn in KMnO4 is + 7.

We known that ox. no. of O = – 2.

Similarly for S in Na2S2O3, 2 (ox. no. Na) +2 (ox. no. S) +3(ox. no. O)= 0

So Ox. no. Mn + 4 (ox. no. O) = – 1

x = +2 Hence ox. no. of S in Na2S2O3 = +2

or

x

+

4(–2)

=–1

or

x

–

8

=–1

or

x = +8 – 1 = + 7

The oxidation number of Mn in [MnO4–] ion is +7. PAGE # 45

Note : (i) Oxidation state of chromium in CrO5. CrO5 has butterfly structure having two peroxo bonds. Peroxo oxygen has (–1) oxidation state

Balancing oxidation reduction reactions by oxidation number method : In a balanced redox reaction, total increase in oxidation number must be equal to the decrease in oxidation number. This equivalence provides the basis for balancing redox reaction. This method is applicable to both molecular and ionic equation. The general procedure involves the following steps: (i) Write the skeleton equation (if not given, frame it) representing the chemical change.

Let oxidation state of chromium be ‘x’ x + 4(–1) + (–2) = 0 x=+6 (ii) Oxidation state of chlorine in bleaching powder : Bleaching powder has two chlorine atoms having different oxidation states.

(iii) Fractional values of oxidation numbers are possible as in Na2S4O6, Fe3O4, N3H, etc.

(ii) Assign oxidation number to the atoms in the equation and find out which atoms are undergoing oxidation and reduction. Write separate equations for the atoms undergoing oxidation and reduction. (iii) Find the change in oxidation number in each equation. Make the change equal in both the equations by multiplying with suitable integers. Add both the equations. (iv) Complete the balancing by inspection. First balance those substance which have undergone change in oxidation number and then other atoms except hydrogen and oxygen. Finally balance hydrogen and oxygen by putting H2O molecules wherever needed. The final balanced equation should be checked to ensure that there are as many atoms of each element on the right as there are on the left. (v) In ionic equations the net charges on both sides of the equations must be exactly the same. Use H+ ion/ions in acidic reactions and OH– ion/ions in basic reactions to balance the charge and number of hydrogen and oxygen atoms. The following examples illustrate the above rules. Ex.53 Balance the following equation by oxidation number method : Cu + HNO3  Cu(NO3)2 + NO2 + H2O Sol. Writing the oxidation number of all the atoms.

Change in ox. no. has occurred in copper and nitrogen. (iv) Fe0.94 O (Oxidation state of iron is to be determined) : 0.94x – 2 = 0 x = 2/0.94 = 200/94 (v) Na2[Fe(CN)5NO] : In iron complex NO lies in NO+ state ; thus oxidation state of ‘Fe’ may be determined as : +2 + x – 5 + 1 = 0 x=+2 (vi) [Fe(NO)(H2O)5]SO4 : x + 1 + 5(0) – 2 = 0 x=+1

Increasing in ox. no. of copper = 2 unit per molecule Cu decrease in ox. no. of nitrogen = 1 unit per molecule HNO3 To make increase and decrease equal eq. (ii) is multiplied by 2. Cu + 2HNO3  Cu(NO3)2 + 2NO2 + H2O Balancing nitrate ions, hydrogen and oxygen, the following equation is obtained. Cu + 4HNO3  Cu(NO3)2 + 2NO2 + 2H2O This is the balanced equation.

PAGE # 46

Ion - electron method for balancing redox reactions: The method for balancing redox reactions by ion electron method was developed by Jette and LaMev involves the following steps : (i) Write down the redox reaction in ionic form. (ii) Split the redox reaction into two half reactios, one for oxidation and the other for reduction. (iii) Balance each half reaction for the number of atoms of each element . For this purpose. (a) Balance the atoms other than H and O for each half reaction using simple multiples. (b) Add water molecules to the side deficient in oxygen and H+ to the side deficient in hydrogen. This is done in acidic or neutral solutions. (c) In alkaline solution, for each excess of oxygen, add one water molecule to the same side and two OH – ions to the other side. If hydrogen is still unbalanced, add one OH – ion for each excess hydrogen on the same side and one water molecule to the other side. (iv) Add electrons to the side deficient in electrons as to equalise the charge on both sides. (v) Multiply one or both the half reactions by a suitable number so that the number of electrons become equal in both the equations. (vi) Add the two balanced half reaction and cancel any term common to both sides. The following solved problems illustrate the various steps of ion electron method : Ex.54 Balance the following equations by ion electrons method : (a) MnO4– + Fe2+ + H+  Mn2+ + Fe3+ + H2O (b) Cl2 + lO3– + OH–  IO4– + Cl– + H2O Sol. (a) MnO4– + Fe2+ + H+  Mn2+ + Fe3+ + H2O Ist step : splitting into two half reactions. MnO4–  Mn2+ ; Fe2+  Fe3+ (Reduction half reaction) (oxidation half reaction) 2nd step : In first reaction add 4H2O in R.H.S. to balance oxygen. MnO4–  Mn2+ + 4H2O

Sol. (b) Cl2 + IO3– + OH–  IO4– + Cl– + H2O Ist step : Splitting into two half reactions, IO3–  IO–4 ; Cl2  Cl– (oxidation half reaction (Reduction half reaction) 2nd step : Adding 2OH– ions and H2O to balance H and O IO3– + 2OH–  IO4– + H2O 3rd step : Adding electrons to the sides deficient in electrons, IO3– + 2OH–  IO4– + H2O + 2e– Cl2 + 2e–  2Cl– 4th step : Adding both the half reaction. IO3– + 2OH– + Cl2  IO4– + 2Cl– + H2O

EXERCISE 1.

2.

3.

5th step : Balancing electrons in both half reaction. MnO4– + 8H+ + 5e–  Mn2+ + 4H2O 5Fe2+  5Fe3+ + 5e– 6th steps : Adding both the half reaction. MnO4– + 5Fe2+ + 8H+  Mn2+ + 5Fe3+ + 4H2O

K2SO4 is (A) 10 g

(B) 25 g

(C) 50 g

(D) 75 g

Molarity of H2SO4 (density 1.8g/mL) is 18M. The molality of this solution is (A)36

(B) 200

(C) 500

(D) 18

8g of sulphur are burnt to form SO2, which is oxidised by Cl 2 water. The solution is treated with BaCl 2 solution. The amount of BaSO4 precipitated is (A) 1.0 mole (C) 0.75 mole

4.

(B) 0.5 mole (D) 0.25 mole

In a compound AxBy (A) Mole of A = Mole of B = mole of AxBy (B) Eq. of A = Eq. of B = Eq. of AxBy (C) X × mole of A = y × mole of B = (x + y) × mole of AxBy (D) X × mole of A = y × mole of B

5.

The percentage of sodium in a breakfast cereal labelled as 110 mg of sodium per 100 g of cereal is (A) 11% (B) 1.10%

3rd Step : Adding hydrogen ions to the side deficient in hydrogen. MnO4– + 8H+  Mn2+ + 4H2O 4th steps : Adding electrons to the side deficient in electrons. MnO4– + 8H+ + 5e–  Mn2+ + 4H2O Fe2+  Fe3+ + e–

The solubility of K2SO4 in water is 16 g at 50ºC. The minimum amount of water required to dissolve 4 g

(C) 0.110% 6.

(D) 110%

Two elements A (at. wt. 75) and B (at. wt. 16) combine to yield a compound. The % by weight of A in the compound was found to be 75.08. The empirical formula of the compound is (A) A2B (B) A2B3 (C) AB

7.

(D) AB2

No. of oxalic acid molecules in 100 mL of 0.02 N oxalic acid are (A) 6.023 × 1020 (C) 6.023 × 10

22

(B) 6.023 × 1021 (D) 6.023 × 1023

PAGE # 47

8.

Which of the following sample contains the maximum number of atoms (A) 1 mg of C4H10 (B) 1 mg of N2 (C) 1 mg of Na (D) 1 mL of water

19. The mole fraction of NaCl, in a solution containing 1 mole of NaCl in 1000 g of water is (A) 0.0177 (B) 0.001 (C) 0.5 (D) 0.244

9.

The total number of protons, electrons and neutrons

20. 3.0 molal NaOH solution has a density of 1.110 g/ mL. The molarity of the solution is (A) 2.9732 (B) 3.05 (C) 3.64 (D) 3.0504

in 12 g of

12 6 C is

(A) 1.084 × 1025 (C) 6.022 × 1022

(B) 6.022 × 1023 (D) 18

10. 4.4 g of CO2 and 2.24 litre of H2 at STP are mixed in a container. The total number of molecules present in the container will be (A) 6.022 × 1023 (B) 1.2044 × 1023 (C) 2 mole (D) 6.023 × 1024

21. How many atoms are contained in a mole of Ca(OH)2 (A) 30 × 6.02 × 1023 atoms/mol (B) 5 × 6.02 × 1023 atoms/mol (C) 6 × 6.02 × 1023 atoms/mol (D) None of these

11. Which is not a molecular formula ? (A) C6H12O6 (B) Ca(NO3)2 (C) C2H4O2 (D) N2O

22. Insulin contains 3.4% sulphur. The minimum molecular weight of insulin is (A) 941.176 u (B) 944 u (C) 945.27 u (D) None of these

12. The hydrated salt, Na2SO4. nH2O undergoes 55.9% loss in weight on heating and becomes anhydrous. The value of n will be (A) 5 (B) 3 (C) 7 (D) 10 13. W hich of the following mode of expressing concentration is independent of temperature (A) Molarity (B) Molality (C) Formality (D) Normality 14. The haemoglobin of the red blood corpuscles of most of the mammals contains approximately 0.33% of iron by weight. The molecular weight of haemoglobin is 67,200. The number of iron atoms in each molecule of haemoglobin is (Atomic weight of iron = 56) (A) 2 (B) 3 (C) 4 (D) 5 15. An oxide of metal have 20% oxygen, the eq. wt. of metal oxide is (A) 32 (B) 40 (C) 48 (D) 52 16. How much water is to be added to dilute 10 mL of 10 N HCl to make it decinormal ? (A) 990 mL (B) 1010 mL (C) 100 mL (D) 1000 mL 17. The pair of compounds which cannot exist in solution is (A) NaHCO3 and NaOH (B) Na2SO3 and NaHCO3 (C) Na2CO3 and NaOH (D) NaHCO3 and NaCl 18. If 250 mL of a solution contains 24.5 g H2SO4 the molarity and normality respectively are (A) 1 M, 2 N (B) 1M,0.5 N (C) 0.5 M, 1N (D) 2M, 1N

23. Number of moles present in 1 m3 of a gas at NTP are (A) 44.6 (B) 40.6 (C) 42.6 (D) 48.6 24. Weight of oxygen in Fe2O3 and FeO is in the simple ratio of (A) 3 : 2 (B) 1 : 2 (C) 2 : 1 (D) 3 : 1 25. 2.76 g of silver carbonate on being strongly heated yield a residue weighing (A) 2.16g (B) 2.48 g (C) 2.32 g (D) 2.64 g 26. How many gram of KCl would have to be dissolved in 60 g of H2O to give 20% by weight of solution (A) 15 g (B) 1.5 g (C) 11.5 g (D) 31.5 g 27. When the same amount of zinc is treated separately with excess of H2SO4 and excess of NaOH, the ratio of volumes of H2 evolved is (A) 1 : 1 (B) 1 : 2 (C) 2 : 1 (D) 9 : 4 28. Amount of oxygen required for combustion of 1 kg of a mixture of butane and isobutane is (A) 1.8 kg (B) 2.7 kg (C) 4.5 kg (D) 3.58 kg 29. Rakesh needs 1.71 g of sugar (C12H22O11) to sweeten his tea. What would be the number of carbon atoms present in his tea ? (A) 3.6 × 1022 (B) 7.2 × 1021 23 (C) 0.05 × 10 (D) 6.6 × 1022 30. The total number of AlF3 molecule in a sample of AlF3 containing 3.01 × 1023 ions of F– is (B) 3.0 × 1024 (A) 9.0 × 1024 (C) 7.5 × 1023 (D)1023 PAGE # 48

31. The volume occupied by one molecule of water (density 1 g/cm3) is (A) 18 cm3 (B) 22400 cm3 –23 (C) 6.023 × 10 (D) 3.0 × 10–23 cm3 32. 224 mL of a triatomic gas weigh 1 g at 273 K and 1 atm. The mass of one atom of this gas is (A) 8.30 × 10–23 g (B) 2.08 × 10–23 g –23 (C) 5.53 × 10 g (D) 6.24 × 10–23 g 33. The percentage of P2O5 in diammonium hydrogen phosphate is (A) 77.58 (B) 46.96 (C) 53.78 (D) 23.48 34. The mole fraction of water in 20% (wt. /wt.) aqueous solution of H2O2 is (A) (C)

77 68 20 80

(B) (D)

43. The number of molecules present in 11.2 litre CO2 at STP is (A) 6.023 × 1032 (C) 3.011 × 10

23

(B) 6.023 × 1023 (D) None of these

44. 250 ml of 0.1 N solution of AgNO3 are added to 250 ml of a 0.1 N solution of NaCl. The concentration of nitrate ion in the resulting solution will be (A) 0.1N

(B) 1.2 N

(C) 0.01 N

(D) 0.05 N

45. Amount of BaSO4 formed on mixing the aqueous solution of 2.08 g BaCl2 and excess of dilute H2SO4 is (A) 2.33 g

(B) 2.08 g

(C) 1.04 g

(D) 1.165 g

68

46. 2g of NaOH and 4.9 g of H2SO4 were mixed and

77

volume is made 1 litre. The normality of the resulting

80

solution will be -

20

(A) 1N

(B) 0.05 N

(C) 0.5 N

(D) 0.1N

35. Which of the following has the maximum mass ? (A) 25 g of Hg (B) 2 moles of H2O (C) 2 moles of CO2 (D) 4 g atom of oxygen 36. Total mass of neutrons in 7mg of 14C is (A) 3 × 1020 kg (B) 4 × 10–6 kg –7 (C) 5 × 10 kg (D) 4 × 10–7 kg 37. Vapour density of a metal chloride is 66. Its oxide contains 53% metal. The atomic weight of metal is (A) 21 (B) 54 (C) 26.74 (D) 2.086 38. The number of atoms in 4.25 g NH3 is approximately (A) 1 × 1023 (B) 1.5 × 1023 23 (C) 2 × 10 (D) 6 × 1023 39. The modern atomic weight scale is based on (A) C12 (B) O16 1 (C) H (D) C13 40. Amount of oxygen in 32.2g of Na2SO4. 10H2O is (A) 20.8 g (B) 22.4 g (C) 2.24 g (D) 2.08 g 41. Which of the followings does not change on dilution ? (A) Molarity of solution (B) Molality of solution (C) Millimoles and milli equivalent of solute (D) Mole fraction of solute

47. 1g of a metal carbonate neutralises completely 200 mL of 0.1N HCl. The equivalent weight of metal carbonate is (A) 25

(B) 50

(C) 100

(D) 75

48. 100 mL of 0.5 N NaOH were added to 20 ml of 1N HCl and 10 mL of 3 N H2SO4. The solution is (A) acidic

(B) basic

(C) neutral

(D) none of these

49. 1M solution of H2SO4 is diluted from 1 litre to 5 litres , the normality of the resulting solution will be (A) 0.2 N

(B) 0.1 N

(C) 0.4 N

(D) 0.5 N

50. The volume of 7g of N2 at S.T.P. is (A) 11.2 L

(B) 22.4 L

(C) 5.6 L

(D) 6.5 L

51. One mole of calcium phosphide on reaction with excess of water gives (A) three moles of phosphine (B) one mole phosphoric acid (C) two moles of phosphine (D) one mole of P2O5 52. Mg (OH)2 in the form of milk of magnesia is used to

42. Equal masses of O2, H2 and CH4 are taken in a

neutralize excess stomach acid. How many moles of

container. The respective mole ratio of these gases

stomach acid can be neutralized by 1 g of Mg(OH)2 ?

in container is -

(Molar mass of Mg(OH)2 = 58.33)

(A) 1 : 16 : 2

(B) 16 : 1 : 2

(A) 0.0171

(B) 0.0343

(C) 1 : 2 : 16

(D) 16 : 2 : 1

(C) 0.686

(D) 1.25

PAGE # 49

53. Calcium carbonate decomposes on heating according to the following equation CaCO3(s) How

CaO(s) + CO2(g)

many moles of CO 2 will be obtained by

decomposition of 50 g CaCO3 ? 3 2 1 (C) 2

(A)

(B)

5 2

(D) 1

54. Sulphur trioxide is prepared by the following two reactions S8(s) + 8O2(g)  8SO2(g) 2SO2(g) + O2(g)  2SO3(g)

60. The volume of CO2 (in litres) liberated at STP when 10 g of 90% pure limestone is heated completely, is(A) 22.4 L (B) 2.24 L (C) 20.16 L (D) 2.016 L 61. A metal oxide has the formula Z2O3. It can be reduced by hydrogen to give free metal and water. 0.1596 g of the metal requires 6 mg of hydrogen for complete reduction. The atomic mass of the metal is (A) 27.9 (B) 159.6 (C) 79.8 (D) 55.8 Question number 62, 63, 64 and 65 are based on the following information : Q.

Dissolved oxygen in water is determined by using a redox reaction. Following equations describe the procedure -

I

2Mn2+(aq) + 4OH–(aq) + O2 (g)

I

MnO2(s)+2I–(aq)+4H+(aq)  Mn2+(aq)+I2(aq) + 2H2O(  )

III

2S 2O32 – (aq) + I2(aq)  2S 4 O 26 – (aq) + 2I–(aq)

How many grams of SO3 are produced from 1 mole of S8 ? (A) 1280

(B) 640

(C) 960

(D) 320

55. PH 3 (g) decomposes on heating to produce phosphorous and hydrogen. The change in volume when 100 mL of such gas decomposed is (A) + 50 mL (C) – 50 mL

(B) + 500 mL (D) – 500 mL

56. What amount of BaSO4 can be obtained on mixing 0.5 mole BaCl2 with 1 mole of H2SO4 ? (A) 0.5 mol (B) 0.15 mol (C) 0.1 mol

(D) 0.2 mol

57. In the reaction , CrO5 + H2SO4  Cr2(SO4)3 + H2O + O2 one mole of CrO5 will liberate how many moles of O2 ? (A) 5/2 (B) 5/4 (C) 9/2

(D) None

58. Calcium carbonate decomposes on heating according to the equation CaCO3(s)  CaO(s) + CO2(g) At STP the volume of CO 2 obtained by thermal decomposition of 50 g of CaCO3 will be (A) 22.4 litre (B) 44 litre (C) 11.2 litre

(D) 1 litre

59. When FeCl 3 is ignited in an atmosphere of pure oxygen, the following reaction takes place4FeCl3(s) + 3O2(g)  2Fe2O3(s) + 6Cl2(g) If 3 moles of FeCl3 are ignited in the presence of 2 moles of O2 gas, how much of which reagent is present in excess and therefore, remains unreacted ? (A) 0.33 mole FeCl3 remains unreacted (B) 0.67 mole FeCl3 remains unreacted (C) 0.25 mole O2 remains unreacted (D) 0.50 mole O2 remains unreacted

 2MnO2(s) + 2H2O(  )

62. How many moles of S 2O 23 – are equivalent to each mole of O2 ? (A) 0.5 B) 1 (C) 2 (D) 4 63. What amount of I2 will be liberated from 8 g dissolved oxygen ? (A) 127 g (B) 254 g (C) 504 g (D) 1008 g 64. 3 × 10–3 moles O2 is dissolved per litre of water, then what will be molarity of I– produced in the given reaction ? (A) 3 × 10–3 M (B) 4 × 3 × 10–3 M 1  3  10 – 3 M 2 65. 8 mg dissolved oxygen will consume (A) 5 × 10–4 mol Mn+2 (B) 2.5 × 10–4 mol Mn2+ (C) 10 mol Mn2+ (D) 2 mol Mn2+

(C) 2 × 3 × 10–3 M

(D)

66. 2 g of a base whose eq. wt. is 40 reacts with 3 g of an acid. The eq. wt. of the acid is(A) 40 (B) 60 (C) 10 (D) 80 67. Normality of 1% H2SO4 solution is nearly (A) 2.5 (B) 0.1 (C) 0.2 (D) 1 68. What volume of 0.1 N HNO3 solution can be prepared from 6.3 g of HNO3 ? (A) 1 litre (B) 2 litres (C) 0.5 litre (D) 4 litres 69. The volume of water to be added to 200 mL of seminormal HCl solution to make it decinormal is (A) 200 mL (B) 400 mL (C) 600 mL

(D) 800 mL PAGE # 50

70. 0.2 g of a sample of H2O2 required 10 mL of 1N KMnO4

80. A 3 N solution of H2SO4 in water is prepared from

in a titration in the presence of H2SO4. Purity of H2O2 is(A) 25% (B) 85%

Conc. H2SO4 (36 N) by diluting

(C) 65%

(B) 10 ml of the conc. H2SO4 to 240 ml

71. Which of the following has the highest normality ? (A) 1 M H2SO4 (C) 1 M H3PO4

(B) 1 M H3PO3 (D) 1 M HNO3

72. The molarity of 98% H2SO4(d = 1.8g/mL) by wt. is (A) 6 M

(B) 18.74 M

(C) 10 M

(D) 4 M

73. 0.7 g of Na2CO3 . xH2O is dissolved in 100 mL. 20 mL of which required to neutralize 19.8 mL of 0.1 N HCl. The value of x is (A) 4

(B) 3

(C) 2

(D) 1

(A) 20 ml of the conc. H2SO4 to 240 ml (C) 1 ml of the conc. H2SO4 to 36 ml (D) 20 ml of the conc. H2SO4 to 36 ml 81. The solubility curve of KNO3 as a function of temperature is given below Solubility (g/100 ml water)

(D) 95%

250 200 150 100 50 0 0

basicity of the acid is (A) 1

(B) 2

(C) 3

(D) 4

75. 1 litre of 18 molar H2SO4 has been diluted to 100 litres. The normality of the resulting solution is (A) 0.09 N

(B) 0.18

(C) 1800 N

(D) 0.36

N HCl is required to react completely with 10 1.0 g of a sample of limestone. The percentage purity

76. 150 ml of

of calcium carbonate is (A) 75%

(B) 50%

(C) 80%

(D) 90%

N N HCl is treated with 70 ml NaOH. 10 10 Resultant solution is neutralized by 100 ml of

77. 50 ml of

sulphuric acid. The normality of H2SO4 (A) N/50

(B) N/25

(C) N/30

(D) N/10

60

80

100

The amount of KNO3 that will crystallize when a saturated solution of KNO3 in 100 ml of water is cooled from 90°C to 30 °C, is (A) 16 g (B) 100 g (C) 56 g (D)160 g 82. The volume of 0.5 M aqueous NaOH solution required to neutralize 10 ml of 2 M aqueous HCl solution is : (A) 20ml (B) 40ml (C) 80ml (D) 120ml 83. 3.01×1023 molecules of elemental Sulphur will react with 0.5 mole of oxygen gas completely to produce (A) 6.02 × 1023 molecules of SO3 (B) 6.02 × 1023 molecules of SO2 (C) 3.01 × 1023 molecules of SO3 (D) 3.01 x 1023 molecules of SO2 84. The solubility of a gas in a solution is measured in three cases as shown in the figure given below where w is the weight of a solid slab placed on the top of the cylinder lid. The solubility will follow the order : w

gas

N HCl were added to 1 g calcium car10 bonate, what would remain after the reaction ?

40

Temperature (°C)

74. 0.45 g of an acid of molecular weight 90 was neutralised by 20 mL of 0.5 N caustic potash. The

20

w

gas

w

w

w

w

gas

78. 200 mL of

(A) CaCO3

(B) HCl

(C) Neither of the two

(D) Part of both

79. Equivalent mass of KMnO4, when it is converted to

solution

(A) a > b > c (C) a = b = c

solution

solution

(B) a < b < c (D) a >b < c

MnSO4 is (A) M/5

(B) M/3

(C) M/6

(D) M/2

PAGE # 51

85. The density of a salt solution is1.13 g cm–3 and it contains 18% of NaCI by weight. The volume of the solution containing 36.0 g of the salt will be : (A) 200 cm3 (B) 217 cm3 3 (C) 177 cm (D) 157cm3 86. One mole of nitrogen gas on reaction with 3.01 x 1023 molecules of hydrogen gas produces (A) one mole of ammonia (B) 2.0 x 1023 molecules of ammonia (C) 2 moles of ammonia (D) 3.01 × 1023 molecules of ammonia 87.

Solubility g/I 250

KNO3

KCl

100 50 20

40

91. 8 Grams of oxygen at NTP contain [IJSO-Stage-I/2012] (A) 1.5 × 1023 molecules (B) 3.0 × 1023 molecules (C) 6.023 × 1023 molecules (D) 1.5 × 1022 molecules 92. When 1g of CaCO3 reacts with 50 ml of 0.1 M HCI, the volume of CO2 produced is - [IJSO-Stage-I/2012] (A) 11.2 mL (B) 22.4 mL (C) 112 mL (D) 56 mL 93. Molality of a solution is the number of [IJSO-Stage-I/2012] (A) moles of the solute per 1000 mL of the solution. (B) moles of the solute per 1000 mL of the solvent. (C) moles of the solute per 1000 g of the solvent. (D) moles of the solute per 100g of the solvent.

200 150

90. One mole of oxalic acid is equivalent to (A) 0.5 mole of NaOH (B) 1 mole of NaOH (C) 1.5 mole of NaOH (D) 2 mole of NaOH

60 80 100 Temperature (ºC)

Given the solubility curves of KNO3 and KCl, which of the following statements is not true ? (A) At room temperature the solubility of KNO3 and KCI are not equal (B) The solubilities of both KNO3 and KCI increase with temperature (C) The solubility of KCI decreases with temperature (D) The solubility of KNO3 increases much more compared to that of KCl with increase in temperature 88. 10 ml of an aqueous solution containing 222 mg of calcium chloride (mol. wt. = 111) is diluted to 100 ml. The concentration of chloride ion in the resulting solution is (A) 0.02 mol/lit. (B) 0.01 mol/lit. (C) 0.04 mol/lit (D) 2.0 mol/lit. 89. Aluminium reduces manganese dioxide to manganese at high temperature. The amount of aluminium required to reduce one gram mole of manganese dioxide is (A) 1/2 gram mole (B) 1 gram mole (C) 3/4 gram mole (D) 4/3 gram mole

94. The oxidation number of chlorine in CaOCI2 is [IJSO-Stage-I/2012] (A) 0 (C) +1

(B) –1 (D) +3

SUBJECTIVE QUESTIONS 95. (a) Sachin was suffering from problem of acidity , so he visited a physician who advised him to take 0.0025 dm3 of milk of magnesia for a fast relief. He exactly followed what the doctor told him to do. Out of curiosity he saw the label on milk of magnesia bottle and he found that there were different ingredients written on it and the concentration of milk of magnesia mentioned was 29 ppm. Assuming, the volume of milk of magnesia required for neutralization of acid is equal to intake of milk of magnesia, help Sachin to find out the following (i) How many moles of acid was produced in Sachin’s stomach ? (ii) Write down the neutralization reaction of this process. (iii)Calculate the concentration of acid produced in mol/dm 3



PAGE # 52

POLYNOMIALS (iii) Quadratic polynomial : A polynomial of degree two is called a quadratic polynomial. The general form of a quadratic polynomial

(a) Definition : An

algebraic

expression

f(x)

of

the

form

is ax2 + bx + c where a  0

f(x) = a0 + a1x + a2x2 + ..........+ anxn, Where a0 ,a1, a2.....an

e.g. x2 + x + 1, 2x2 + 1, 3x2 + 2x + 1 etc.

are real numbers and all the indices of x are non

(iv) Cubic polynomial :

negative integers is called a polynomial in x and the highest index n is called the degree of the polynomial, if an  0. Here a0 , a1x, a2x2 .....,anxn are called the terms of the polynomial and a0, a1, a2, ...... an are called various co-efficients of the polynomial f(x). A polynomial in x is

A polynomial of degree three is called a cubic polynomial. The general form of a cubic polynomial is ax3 + bx2 + cx + d, where a  0 e.g. x3 + x2 + x + 1, x3 + 2x + 1, 2x3 + 1 etc.

said to be in its standard form when the terms are

(v) Biquadratic polynomial :

written either in increasing order or decreasing order

A polynomial of degree four is called a biquadratic or

of the indices of x in various terms.

quartic polynomial. The general form of biquadratic polynomial is ax4 + bx3 + cx2 + dx + e where a  0

EXAMPLES :

e.g. x4 + x3 + x2 + x + 1 , x4 + x2 + 1 etc.

(i) 2x3 + 4x2 + x + 1 is a polynomial of degree 3. 7

5

2

(ii) x + x + x + 1 is a polynomial of degree 7.

NOTE : A polynomial of degree five or more than five does not have any particular name. Such a polynomial is usually

(iii) x3/2 + x2 + 1 is not a polynomial as the indices of x

called a polynomial of degree five or six or ..... etc.

are not all non negative integer (b) Polynomial Based on Terms : (iv) x2 +

2 x + 1 is a polynomial of degree 2.

There are three types of polynomial based on number of terms.

(v) x–2 + x + 1 is not a polynomial as –2 is not non negative.

(i) Monomial : A polynomial is said to be a monomial if it has only one term. For example, x, 9x2, – 5x2 are all monomials (ii) Binomial : A polynomial is said to be a binomial if it contains two terms.

(a) Polynomial Based on Degree : For example 2x2 + 3x, There are five types of polynomials based on degree. (i) Constant polynomial : A polynomial of degree zero is called a zero degree

binomials. (iii) Trinomial : A polynomial is said to be a trinomial if it contains three terms.

polynomial or constant polynomial.

For example 3x3 – 8x +

e.g. f(x) = 4 = 4x0

trinomials.

(ii) Linear polynomial : A polynomial of degree one is called a linear polynomial. The general form of a linear polynomial is ax + b, where a and b are any real numbers and a  0 e.g. 4x + 5, 2x + 3, 5x + 3 etc.

3 x + 5x4, – 8x3 + 3 etc are all

5 , 2

7 x10 + 8x4 – 3x2 etc are all

REMARKS : (i) A polynomial having four or more than four terms does not have any particular name. They are simply called polynomials. (ii) A polynomial whose coefficients are all zero is called a zero polynomial, degree of a zero polynomial is not defined. PAGE # 5353

Ex.2 Show that x + 1 and 2x – 3 are factors of 2x3 – 9x2 + x + 12. (a) Value of a Polynomial : The value of a polynomial f(x) at x =  is obtained by

Sol. To prove that (x + 1) and (2x – 3) are factors of

substituting x =  in the given polynomial and is

2x3 – 9x2 + x + 12 it is sufficient to show that p(–1) and

denoted by f().

3 p   both are equal to zero. 2

Consider the polynomial f(x) = x3 – 6x2 + 11x – 6,

p (– 1) = 2 (– 1)3 – 9 (– 1)2 + (– 1) + 12

If we replace x by – 2 everywhere in f(x), we get

= – 2 – 9 – 1 + 12

f(– 2) = (– 2)3 – 6(– 2)2 + 11(– 2) – 6

= – 12 + 12 = 0.

f(– 2) = – 8 – 24 – 22 – 6 f(– 2) = – 60  0. So, we can say that value of f(x) at x = – 2 is – 60.

3

2

3 3 3 3 and p   = 2   – 9       12 2 2 2 2

(b) Zero or root of a Polynomial :

=

27 81 3 –   12 4 4 2

=

27 – 81  6  48 4

=

– 81  81 4

The real number  is a root or zero of a polynomial f(x), if f( = 0. Consider the polynomial f(x) = 2x3 + x2 – 7x – 6, If we replace x by 2 everywhere in f(x), we get

= 0.

 f(2) = 2(2)3 + (2)2 – 7(2) – 6 = 16 + 4 – 14 – 6 = 0 Hence, x = 2 is a root of f(x).

Hence, (x + 1) and (2x – 3) are the factors 2x3 – 9x2 + x + 12. Ex. 3 Find the values of a and b so that the polynomials

Let ‘p(x)’ be any polynomial of degree greater than or equal to one and a be any real number and If p(x) is divided by (x – a), then the remainder is equal to p(a). Ex.1 Find the remainder, when f(x) = x3 – 6x2 + 2x – 4 is

x3 – ax2 – 13x + b has (x – 1) and (x + 3) as factors. Sol. Let f(x) = x3 – ax2 – 13x + b Because (x – 1) and (x + 3) are the factors of f(x),  f(1) = 0 and f(– 3) = 0 f(1) = 0

divided by g(x) = 1 – 2x.

 (1)3 – a(1)2 – 13(1) + b = 0

Sol. f(x) = x3 – 6x2 + 2x – 4 Let, 1 – 2x = 0

 1 – a – 13 + b = 0

 2x = 1



– a + b = 12

.... (i)

f(–3) = 0

 x= 1 2

 (– 3)3 – a(– 3)2 – 13(– 3) + b = 0

 1 Remainder = f   2 3

 – 27 – 9a + 39 + b = 0 2

 1  1  1  1 f   =   – 6   2  – 4 2 2 2 2 1 3 1 – 12  8 – 32 35 = –  1– 4 = . – 8 2 8 8



– 9a + b = –12

...(ii)

Subtracting equation (ii) from equation (i) (– a + b) – (– 9a + b) = 12 + 12  – a + 9a = 24  8a = 24

FACTOR THEOREM

a = 3.

Let p(x) be a polynomial of degree greater than or equal

Put a = 3 in equation (i)

to 1 and ‘a’ be a real number such that p(a) = 0, then

– 3 + b = 12

(x – a) is a factor of p(x). Conversely, if (x – a) is a factor

b = 15.

of p(x), then p(a) = 0.

Hence, a = 3 and b = 15.

PAGE # 5454

(ii) (3x2 + 5y)2

ALGEBRAIC IDENTITIES

= (3x2)2 + 2(3x2)(5y) + (5y)2 Some important identities are

= 9x4 + 30x2y + 25y2

(i) (a + b)2 = a2 + 2ab + b2 (iii) ( 2 x – 3y)( 2 x + 3y)

(ii) (a – b)2 = a2 – 2ab + b2

( 2 x)2 – (3y)2

= (iii) a2 – b2 = (a + b) (a – b)

2x2 – 9y2

=

(iv) (a + b + c)2 = a2 + b2 + c2 + 2ab + 2 bc + 2ca

1 1  (iv)  a  b  1 4 2  

(v) a3 + b3 = (a + b) (a2 – ab + b2)

2

2

2

1   1  1   1  =  a  +   b  + (1)2 + 2  a    b  4   2  4   2 

(vi) a3 – b3 = (a – b) (a2 + ab + b2) (vii) (a + b)3 = a3 + b3 + 3ab (a + b)

 1  1  + 2   b  (1) + 2(1)  a  2   4 

(viii) (a – b)3 = a3 – b3 – 3ab (a – b) (ix) a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ac) Special case : if a + b + c = 0 then a3 + b3 + c3 = 3abc. Value Form :

=

1 1 2 ab a a2 + b +1– –b+ . 16 4 4 2

Ex. 5 Simplify :

(i) a2 + b2 = (a + b)2 – 2ab, if a + b and ab are given.

1  1  2 1  4 1   (i)  x   x   x  2  x  4  x  x  x  x  

(ii) a2 + b2 = (a – b)2 + 2ab, if a – b and ab are given.

(ii) 2x  y 2x  y  4 x 2  y 2

a  b 2  4ab , if a – b and ab are given.

(iii) a + b =





Sol.  x 

 (iv) a – b = (v) a2 +

(vi) a2 +

1 a2 1 a2

a  b 2  4ab , if a + b and ab are given. 2

1 1  =  a   – 2, if a + is given. a a   2

1  1 =  a   + 2, if a – is given. a  a

3 1  1  1 =  a   + 3  a   , if a – is given. 3 a   a a   a

1

(xi) a4 – b4 = (a2 + b2) (a2 – b2) = [(a + b)2 – 2ab](a + b) (a – b). Ex.4 Expand :

(iii)



2

2 x  3y



 2x  3y 

1    Sol. (i)  2x  3x  

2



1 1  (iv)  a  b  1 4 2  

2

2

 1  1  + = (2x)2 – 2(2x)   3x  3x 2

= 4x2 –

1   2 1  1    x  2   x4  4  x2   x  x 



3

(ii) 3 x 2  5 y



2 = x 

2   1    4 1  2 2   x  ( x )     =   x 2    x4 

1 1 1   (ix) a3  1 =  a   – 3  a   , if a + is given. a a a   a3

1    (i)  2x  3x  

1  1  1  1    x    x2  2   x4  4  x  x  x  x 



(vii) a3 + b3 = (a + b)3 – 3ab (a + b), if (a + b) and ab are given. (viii) a3 – b3 = (a – b)3 + 3ab (a – b), if (a – b) and ab are given.

(x) a 3 



1 4 + . 3 9x 2

4 = x 

 =

(x4)2

= x8 –

1  4 1   x  4  x4   x 

 1  –  4 x  1 x8

2

.

(ii) (2x + y)(2x – y)(4x2 +y2) = [(2x)2 – (y)2](4x2 + y2) = (4x2 – y2)(4x2 + y2) = (4x2)2 – (y2)2 = 16x4 – y4. Ex.6 Find the value of x – y when x + y = 9 & xy = 14: Sol. x + y = 9 On squaring both sides x2 + y2 + 2xy = 81 Putting value of xy = 14 x2 + y2 + 28 = 81 x2 + y2 = 81 – 28 = 53 ...(i) (x – y)2 = x2 + y2 – 2xy Putting xy = 14 and (i) (x – y)2 = 53 – 2 (14) = 53 – 28 (x – y)2 = 25 x–y=±

25 = ±5 PAGE # 5555

Ex.7 If x2 +

Sol.

1

= 23, find the values of

x2

x2 1



x2 +



(x2) + 

x2

= 23

…(i)

+ 2 = 25

[Adding 2 on both sides of (i)]

 1 1  +2x = 25 x x 2

1   x   = (5)2 x  1  x+ =  5 x 



1   x   =  21 . x 

= x3 +

= 23 – 2 = 21

2

2

1    4 1   x  4  =  x2  2  – 2 x    x 



1   4  x  4  = (23)2 – 2 = 529 – 2 x  



1   4  x  4  = 527. x  

 a 2 – 5ab a2 – b2   2 . Ex. 8 Find the value of  2 2 a  ab   a – 6ab  5b

a 2 – 5ab a 2 – 6ab  5b 2

a2 – b2 ×

a 2  ab

a(a – 5b) (a – b)(a  b) × =1 (a – b )(a – 5b ) a(a  b)

7.98  7.98 – 2.02  2.02 Ex. 9 Find the value of 5.96

Sol.

8 x3

= 2x3 +

1   4  2 1  x  2  =x  4  + 2 x   x  

=

3

2 2   (ii)  x   +  x   x x  

2

1  x   x 

3

3

3

2 2  2 + 3 (x)    x   + x 3 –   x x  x

3

2 2  – 3(x)    x   x x 

2



2  2  (ii)  x     x   x  x 

3

Sol. (i)(3x + 4)3 – (3x – 4)3 = [(3x)3 + (4)3 + 3 (3x) (4) (3x + 4)] – [(3x)3 – (4)3 – 3 (3x) (4) (3x – 4)] = [27x3 + 64 + 36x (3x + 4)] – [27x3 – 64 – 36x (3x – 4)] = [27x3 + 64 + 108x2 + 144x] – [27x3 – 64 – 108x2 + 144x] = 27x3 + 64 + 108x2 + 144x – 27x3 + 64 + 108x2 – 144x = 128 + 216x2.

2 = x3 +   x

1  1  x   = x2 + –2 x  x2

Sol.

3

(i) (3x + 4) – (3x – 4)

 . 

2



Ex.10 Simplify : 3

1 1    x   and  x 4  4 x x   1 2 x +

1  x   , x 

7.98  7.98 – 2.02  2.02 5.96

+ 6x +

8 12 12 + x3 – 3 – 6x + x x x

24 . x

Ex.11 Evaluate : (i) (1005)3 (ii) (997)3 3 3 Sol. (i) (1005) = (1000 + 5) = (1000)3 + (5)3 + 3 (1000) (5) (1000 + 5) = 1000000000 + 125 + 15000 (1000 + 5) = 1000000000 + 125 + 15000000 + 75000 = 1015075125. (ii) (997)3 = (1000 – 3)3 = (1000)3 – (3)3 – 3 × 1000 × 3 × (1000 – 3) = 1000000000 – 27 – 9000 × (1000 – 3) = 1000000000 – 27 – 9000000 + 27000 = 991026973. 1 1 = 5, find the value of x3 – 3 . x x 1 Sol. We have, x – =5 ...(i) x 3 1    x   = (5)3 [Cubing both sides of (i)] x 

Ex.12 If x –

1

1 1    x   = 125 x  x



x3 –



x3 –

1  – 3  x   = 125 x  x



x3 –

1  – 3 × 5 = 125 [Substituting  x   = 5] x  x



x3 –



x3 –

(7.98  2.02)(7.98 – 2.02) 10  5.96 = = = 10. 5.96 5.96

x3

– 3x 

1

3

1

3

1 x3

1 x3

– 15 = 125

= (125 + 15) = 140. PAGE # 5656

Ex.13 Find the products of the following expression : (i) (4x + 3y) (16x2 – 12xy + 9y2) (ii) (5x – 2y) (25x2 +10xy + 4y2) Sol. (i) (4x + 3y) (16x2 – 12 xy + 9y2) = (4x + 3y) [(4x)2 – (4x) × (3y) + (3y)2] = (a + b) (a2 – ab + b2) [Where a = 4x, b = 3y ] = a3 + b3 = (4x)3 + (3y)3 = 64x3 + 27y3. (ii) (5x – 2y) (25x2 + 10xy + 4y2) = (5x – 2y) [(5x)2 + (5x) × (2y) + (2y)2] = (a – b) (a2 + ab + b2) [Where a = 5x, b = 2y] = a3 – b3 = (5x)3 – (2y)3 = 125x3 – 8y3. Ex. 14 If a + b + c = 9 and ab + bc + ac = 26, find the value of a3 + b3 + c3 – 3abc. Sol. We have a + b + c = 9 ...(i) 2  (a + b + c) = 81 [On squaring both sides of (i)]  a2 + b2 + c2 + 2(ab + bc + ac) = 81  a2 + b2 + c2 + 2 × 26 = 81 [ ab + bc + ac = 26]



2 3

 

2

2 3

 

2

 b c  c a Ex.15 Simplify : a  b 3 3 a  b  b  c   c  a 3



=

2

2

2 3

2

2

2

2

4.

If

=

3a  b  a  b  b  c b  c c  a c  a  3a  b  b  c  c  a 







= a  bb  c c  a .

3

3

3

So, (28) + (–78) + (50) = 3 × 28 × (–78) × 50 = – 327600.

3x

III. 10 (B) II only (D) I and III only

mean) of a and b ?

1 2 ab (D) 2c (B)

5.

On simplifying (a + b)3 + (a – b)3 + 6a(a2 – b2) we get : (A) 8a2 (B) 8a2b (C) 8a3b (D) 8a3

6.

Find the value of

7.

If (x + y + z) = 1, xy + yz + zx = –1, xyz = –1, then value of x3 + y3 + z3 is : (A) –1 (B) 1 (C) 2 (D) –2

8.

If (x + a) is a factor of x2 + px + q and x2 + mx + n then the value of a is :

9.

a 3  b 3  c 3 – 3abc

ab  bc  ca – a 2 – b 2 – c 2 a = – 5, b = – 6, c = 10. (A) 1 (B) –1 (C) 2 (D) –2

(A)

m–p n–q

(B)

n–q m–p

(C)

nq mp

(D)

mp nq

, when

If x2 – 4 is a factor of 2x3 + ax2 + bx + 12, where a and b are constant. Then the values of a and b are : (A) – 3, 8 (B) 3, 8 (D) 3, – 8 (C) –3, – 8

10. If

Ex.16 Find the value of (28)3 – (78)3 + (50)3. Sol. Let a = 28, b = – 78, c = 50 Then, a + b + c = 28 – 78 + 50 = 0  a3 + b3 + c3 = 3abc.

2 + x

1 1 1   and ab =c, what is the average (arithmetic a b c

 Given expression



II. 5 2

I. –10 (A) I only (C) III only

a  b3  b  c 3  c  a 3 = 3a  b b  c c  a 3 a 2  b2 b2  c 2 c 2  a2 3a  b  b  c  c  a 

(D) 5x2 +

3

If a2 –b2 =21 and a2 + b2 = 29, which of the following could be the value of ab ?

2

=

3

2x +

3.

Also, a  b   b  c   c  a   0 

2 x+

(B) x2 +

The remainder obtained when t6 +3t2 + 10 is divided by t3 + 1 is : (A) t2 – 11 (B) 3t2 + 11 3 (C) t – 1 (D) 1 – t3

2 3

2

3

2.

.

     a  b   b  c   c  a  3 a  b  b  c  c  a  2 3

2 x+

(C) x3/2 +

2 3



2

(A) x2 +

(C) 1

Sol. Here, a 2  b 2  b 2  c 2  c 2  a 2  0 

Which of the following is a polynomial :

(A) 0

 a2 + b2 + c2 = (81 – 52)  a2 + b2 + c2 = 29. Now, we have a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ac) = (a + b + c) [(a2 + b2 + c2) – (ab + bc + ac)] = 9 × [(29 – 26)] = (9 × 3) = 27. 2

1.

1 3 x

1 3 y



1 3 z

= 0 then which one of the following

expression is correct : (A) x3 + y3 + z3 = 0 1

1

1

(B) x + y + z = 3 x 3 y 3 z 3 (C) x + y + z = 3xyz (D) x3 + y3 + z3 = 3xyz PAGE # 5757

11. If x51 + 51 is divided by (x + 1) the remainder is : (A) 0

(B) 1

(C) 49

(D) 50

12. If a4 +

1 a4

= 119, then find the value of a3 –

(A) 11

1 a3

4 4 2  2 , then x + x 4 is : (A) 2(3 – 2 ) (B) 6 2 – 2 (C) 6 – 2 (D) 12

22. If x =

.

(B) 36

(C) 33

23. If 4x – 5z = 16 and xz = 12, 64x3 – 125z3 = (A) 14512 (B) 15676 (C) 25833 (D) 15616

(D) 12 2

(a – b ) (b – c ) 2 (c – a ) 2   13. Evaluate : . (b – c )(c – a ) (a – b)(c – a) (a – b )(b – c )

24.

x 3  y 3 x  3 y 1  ( xy )  2  y 3 x 1

(A) 0

(B) 1

(A) x + y

(C) 2

(D) 3

(C)

14. The polynomials ax3 + 3x2 – 3 and 2x3 – 5x + a when divided by (x – 4) leaves remainders R1 & R2 respectively then value of ‘a’ if 2R1 – R2 = 0. (A) –

18 127

18 127

(B)

17 17 (D) – 127 127 15. A quadratic polynomial is exactly divisible by (x + 1) & (x + 2) and leaves the remainder 4 after division by (x + 3) then that polynomial is : (A) x2 + 6x + 4 (B) 2x2 + 6x + 4 2 (C) 2x + 6x – 4 (D) x2 + 6x – 4

(C)

16. If x2 – 4 is a factor of 2x3 + ax2 + bx + 12, where a and b are constant. Then the values of a and b are : (A) – 3, 8 (B) 3, 8 (C) –3, – 8 (D) 3, – 8 17. The value of

0.76  0.76  0.76  0.24  0.24  0.24 is : 0.76  0.76  0.76  0.24  0.24  0.24

(A) 0.52 (C) 0.01

(B) 1 (D) 0.1

18. If x + y = 3 and xy = 2, then the value of x3 – y3 is equal to (A) 6 (B) 7 (C) 8 (D) 0

19. If x =

1 , then the value of x + 2

1

is :

1

1

1

1 x

(A)

5 4

(B)

(C)

3 4

(D) None of these

20. If (a2 + b2)3 = (a3 + b3)2 then (A)

2 3

(C)

5 6

4 5

a b + = b a

3 2 6 (D) 5

(B)

21. m5 + m4 + m3 + m2 + m + 1 = (m3 + 1) × _______ (A) m5 + m4 + m2 + m (B) m2 + m3 3 3 (C) m + m + m + 1 (D) m2 + m + 1

(B) y – x

1 1 – x y

(D) 2

25. If

 a – b  4

ab

a–b

=

1 1 + x y

5 , then the value of a : b is : 3

(A) 1 : 16 (C) 4 : 1

(B) 1 : 4 (D) 16 : 1

26. If x = 0.50, then the value of the expression

 x 3  2 (1  x  x )   1  x  is :  (A) 4 (C) 1.50 27. If p = 22/3 + 21/3, then : (A) p3 – 6p + 6 = 0 (C) p3 – 6p – 6 = 0

(B) 2 (D) 1 (B) p3 – 3p – 6 = 0 (D) p3 – 3p + 6 = 0

28. The polynomial p(x) = 2x4 – x3 – 7x2 + ax + b is divisible by x2 – 2x – 3 for certain values of a and b. The value of (a + b), is : (A) – 34 (B) – 30 (C) – 26 (D) – 18 29. When the polynomial (6x4 + 8x3 + 17x2 + 21x + 7) is divided by (3x2 + 4x + 1), the remainder is (ax – b). Therefore : [IJSO-2011] (A) a = 1, b = 2 (B) a = 1, b = – 2 (C) a = 2, b = 1 (D) a = –1, b = – 2 30. If 22x–1 + 21–2x = 2, then the value of x is : [IJSO-2011] (A) 0.5 (B) –0.5 (C) 1 (D) 0 31. Given that a (a+b) = 36 and b (a + b) = 64, where a and b are positive, (a – b) equals : [IJSO-2011] (A) 2.8 (B) 3.2 (C) –2.8 (D) –2.5

ac is : bc a (A) always smaller than b a (B) always greater than b a (C) greater than only if a > b. b a (D) greater than only if a < b. b

32. If a, b, c are positive,

[IJSO-2011]

PAGE # 5858

33. Find x2 + y2 + z2 if x2 + xy + xz = 135, y2 + yz + yx = 351 and z2 + zx + zy = 243. [IJSO-2012] (A) 225 (B) 250 (C) 275 (D) 300

34. lf a + b + c = 1, a2 + b2 + c2 = 21 and abc = 8 then find the value of (1– a)(1– b) (1– c). [IJSO-2012] (A) – 10 (B) – 18 (C) – 24 (D) – 30



PAGE # 5959

NATURAL RESOURCES (ii) Based on origin : On the basis of their origin resources may be biotic (organic) or abiotic (inorganic). Biotic resources are obtained from the biosphere. Forest and forest products, crops, birds, animal, fish and other marine life forms are examples of biotic resources. Coal and mineral oil also belong to this category since they originate from organic matter. Some biotic resources like forest and livestock are renewable, whereas coal and oil are non–renewable. Resources composed of non–living inorganic matter are called abiotic resources. Land, water and minerals like iron, copper, lead and gold are abiotic resources.

NATURAL RESOURCES

•

•

•

It indicates the potential wealth of a country. The variety of substances that man gets from earth and nature to meet his basic needs are called natural resources. The word resource means a source of supplying a material generally held in reserve. Natural resources are both living and non – living. Some of these resources are found in abundance, while others are found in limited quantities & that too in some restricted parts of our land. For this reason, the natural resources have to be wisely used. However, in reality it is not so. They are being used indiscriminately.

(iii) Based on utility : Every resource has some utility. For example, some are used as food, some as raw materials and others as sources of energy.

Types of Natural Resources : (i) Based on availability : The natural resources are categorized into two types i.e. (a) nexhaustible natural resources

AIR OR ATMOSPHERE

•

(b) Exhaustible natural resources

Natural Resources

Inexhaustible(i) Resources that are in unlimited quantity. (ii) Resources that are not likely to be exhausted by human activity or their use. Examples : Air, Water & Solar Radiations. Exhaustible (i) Resources that are in limited quantity. (ii) Resources that are likely to be exhausted by human activities. Non–Renewable (i) Cannot replenish themselves by recycling & replacement. (ii) These may be exhausted. (iii) Example : Minerals, Fossil fuels Renewable (i) Can replenish themselves by quick recycling and replacement within a reasonable time. (ii) Not likely to be exhausted. (iii) Examples : Soil, Forests and Wild life.

•

The multilayered, transparent and protective envelope of gases surrounding the planet earth is called atmosphere. In other words atmosphere is the layer of air above the earth’s surface and air is a mixture of several gases. About 95% of total air is present up to the height of 20 km above earth's surface, remaining 5% is up to the height of 280 km. (a) Composition of Air : Gas Nitrogen Oxygen Argon Carbon dioxide He, Ne, Kr, Xe,

•

Relative percentage / volume 78.08 % 20.94 % 0.9 % 0.03 % in trace amounts.

Besides these gaseous components air also possesses water vapour, industrial gases, dust, smoke particles, microorganisms, pollen grains, fungal spores etc. (b) The Different Zones of Atmosphere or Air (i) Troposphere : t is the basal part that extends about 8-16 km above the earth's surface. (upto 8 km on poles). In this layer important climatic events occur like cloud formation, lightning, thundering etc.In this region air temperature gradually decreases with height. Its upper limit is called Tropopause (ii) Stratosphere : t lies next to troposphere and is up to 50 km high. In this layer temperature rises. There is a formation of ozone layer in this region which can absorb the harmful ultra violet rays coming from sun. (iii) Mesosphere : t lies next to stratosphere and is up to 80km in height. Temperature decreases in this region.

BIOLOGY_IJSO_PAGE #60 60

(iv) Ionosphere : t lies upto 400 km above earth's surface. In this layer gaseous components become ionized due to sun's energy and remain there as ions. Different Regions of Atmosphere S.No. 1.

Region Troposphere

Range of distance 11 kms from the surface of earth

Density of air Highest

Importance Most of the atmospheric air is present here. It is a medium for locomotion of flying animals, helps in dispersal of seeds and fruits; region of cloud formation.

2.

Stratosphere

50 kms from the surface of earth Less than troposphere Contains ozone layer that traps most of UV rays and cosmic rays of the Sun.

3.

Mesosphere

80 kms from the surface of earth Low

4.

Thermosphere

100 kms upwards

--

Extremely low

Reflect radio waves back to earth, artificial satellites are present here.

ATMOSPHERIC REGIONS BASED ON TEMPERATURE

HETEROSPHERE

TEMPERATURE INCREASING WITH HEIGHT

IONOSPHERE

THERMOSPHERE

100 TEMPERATURE DECREASING WITH HEIGHT

HOMOSPHERE

HEIGHT IN KILOMETERS

150

MESOSPHERE

50

STRATOSPHERE

OZONOSPHERE

TEMPERATURE INCREASING WITH HEIGHT

TROPOSPHERE

0

0 100 200 300 TYPICAL TEMPERATURE IN DEGREE CELSIUS Different Regions of Atmosphere

(c) Role of Air or Atmosphere :

• • •

It acts as medium for movement of insects, birds etc. It protects the life on earth from harmful ultra violet rays. It is a source of oxygen, carbon dioxide and nitrogen required for various metabolic activities of living beings.

• • • •

It helps in dispersal of spores, pollen grains etc. t maintains temperature on earth required for life. It transmits sound for communication. Ionosphere reflects the radio waves back to earth for long distance communication due to presence of ions and free electrons. BIOLOGY_IJSO_PAGE #61 61

• • •

Burning (combustion) takes place in presence of oxygen and produces carbon dioxide. Specific climatic conditions and water cycle is maintained due to circulation of air. Eukaryotic cells and many prokaryotic cells require O2 for break down of glucose to get energy through respiration and release CO2. (d) The Role of Atmosphere in Climate Control :

•

• • •

Climate is an average weather of an area. Temperature, light and rainfall are important factors that determine climate of an area. Atmosphere plays a crucial role in its control : It acts like a blanket covering the whole earth. It keeps the temperature of earth steady. It acts as bad conductor of heat thus prevents the sudden increase in temperature during the day as well as slows down the escape of heat into the outer space during night.The temperature range on the surface of moon –190ºC to 110ºC. (e) Wind :

•

Air in motion is called wind. Speed of wind can be determined by : Heating of air Formation of water vapour Atmosphere can be heated from below by radiations, such radiations are reflected back. Convection currents appear in air on being heated.

•

Factors controlling movement of air

• • • •

(f) Rain :

• • • • • • •

POLLUTION

• •

Any material or act of man, or nature which leads to pollution is called as pollutants.

•

(A) Rotation of earth.

•

•

The general pattern of winds over earth is known as general circulation and specific winds are named for the direction from which they originate (e.g. wind blowing from west to east is westerly). W ind speeds are often classified according to Beaufort scale.

•

•

Secondary pollutants`

Pollutants which persist in the environment in the form it is produced.

Pollutant formed from a primary one through change or reaction. (No)x and hydrocarbons react photochemically to produce PAN (Peroxyacetyl nitrate) and O3 .

These include particulate matter, CO, CO2, SO2, (NO)x and hydrocarbons.

Photochemical smog, O3 and peroxy acetyl nitrate are secondary pollutants.

These are less toxic.

These are more toxic than the primary pollutants & this phenomenon is called synergism.

(b) Air Pollution :

•

The pollution is usually brought about by the addition to the environment of waste products of human activity. W hen the waste products are not efficiently assimilated, decomposed or other wise removed by natural, biological and physical processes (recycling) and the system is unable to utilize them properly, so that the balance of the system breaks down. Therefore such type of pollutants can stimulate or inhibit the biological reactions or change in their capacity. Therefore changes also take place in the ecosystem. The amount, number and type of pollutants are increasing with the growth of the population.

Difference between primary and secondary pollutants

Primary pollutants

•

Any undesirable change in physical, chemical or biological characteristics in the air, water and land which is harmful to the men directly or indirectly through animals, plants, industrial units or raw materials is called as pollution. Pollution is man made. But it can also be natural. 99.95 % of pollution is natural only 0.05 % pollution is manmade. (a) Pollutants :

•

(B)n the path of wind, mountain ranges may come across.

The warm, moist and rising air cools and forms clouds in the sky. This happens due to heating of water bodies during day time which get mixed with atmosphere. The air rises, it expands and cools. Cool air in the atmosphere sinks towards the ground. Due to cooling water vapours present in air get facilitated. These tiny droplets become bigger and bigger due to condensation. When they become heavy, they fall down in the form of rain.

Air pollution is caused due to the addition of the unwanted substances or gases. The atmospheric pollution is mainly caused by the activities of man and concentrated to the inhabited and the industrial complexes in cities.

•

There are two main categories of air pollutants. (i) Gaseous : The gaseous materials include various gases and vapours of volatile substances or the compound with a boiling point below 200ºC. (ii) Particulate : Dust particles , carbon particles, particles of other metals etc. BIOLOGY_IJSO_PAGE #62 62

(v) Smoke : Many constituents are present in smoke such as Sulphur dioxide (SO 2), Sulphur trioxide (SO3),Sulphuric acid (H2SO4), Ozone (O3), Carbon dioxide (CO2), PAN (peroxyacetyl nitrate), Arsenic and Fluoride etc.

(c) Major air Pollutants and Their Effects : (i) Carbon monoxide (CO) : This is the main air pollutant.

• •

Carbon monoxide is highly toxic & it is colourless and odourless in nature. It combines with haemoglobin of the blood and blocks the transportation of oxygen.Thus, it impairs respiration and it causes death. (ii) Unburnt hydrocarbons : Out of them 3, 4 – benzpyrene is the main pollutant. This causes cancer in lungs.

•

• •

(iii) Ethylene : The falling of leaves without particular reason, falling buds etc. effects are seen in plants are due to ethylene.

• • • •

(iv) Oxides of nitrogen : These oxides of nitrogen form photochemical smog in the atmosphere and release ozone. Ozone causes harm to mucilagenous membrane. The oxide pollutants of nitrogen are nitric oxide (NO), and nitrogen di oxide (NO2). These oxides and ozone are very harmful for the plants. The entry of these pollutants causes various diseases in animals like-respiratory trouble such as emphysema, bronchitis, swelling of lungs and lung cancer etc.

• • •

The distribution area of lichen and mosses are the indicators of SO2 pollution because lichen and mosses cannot grow in the industrial regions or the regions containing SO2 pollutants. The higher concentration of ozone produces harmful effects. Ozone layer absorbs U.V. rays which are harmful for the living beings. (vi) Aerosol : The aerosol like C.F.C. (Chloro fluoro carbon) released into the atmosphere from the refrigerators, air conditioners and jet planes, deplete or reduce the ozone layer. This thin layer of ozone is also known as ozone hole results in the increase in temperature of the earth. Secondary effect of Air pollution : Green -House Effect : Usually carbon dioxide is not considered as pollutant, but its higher concentration forms the thick layer above the earth surface which checks the radiation of the heat from the earth surface. Because of this the temperature of the earth surface increases.This is called as"Green house effect".

Sun light

Gre en Hou s

eG ase s

re he p s o Atm

Heat Earth Surface

Heat

Radiant heat trapped by CO2 Green House Effect

•

•

•

•

The various green house gases are CO2 (Warming effect 60% ), CH4 (warming effect 20%) , chlorofluoro carbon or CFCs (14%) and nitrous oxide (N2O 6%) Even 2–3º C rise in temperature will lead to melting of glaciers & ice caps of polar regions and consequently causes floods in rivers, rise in sea level and changes in cycle of rain. Islands may be emerged in sea water. Global Warming : Global warming is the increase in average global temperature due to increase in amount of GHGs in earth’s atmosphere. Consequences of global warming : (i) Increase in the sea level : Global warming will melt polar ice caps.

•

If all the ice on the earth will melt water would be added to surface of all oceans. Thus low lying coastal cities like Shanghai, Kolkata, Bangkok, Dhaka, Venice, etc. will be inundated. (ii) Increase in global temperature : f present input of GHGs will be continued, the earth’s global temperature will rise. (iii) Effect on agriculture : Grain production will be reduced. India’s annual monsoon rains may even cease together. One third of global forest might be swept away. Deserts are likely to increase (iv) Chances of hurricanes, cyclones and floods will be more.

BIOLOGY_IJSO_PAGE #63 63

(v) Increased temperature and humidity caused by global warming will lead to spread of diseases like malaria, filariasis etc. due to spread of vectors. Incidences of respiratory and skin diseases are likely to increase.

•

and soil. About 50% of acidity is passed to earth as dry deposition. Rainfall will wash it down from trees and other articles. Acid rain is caused by large scale emission of acidic gases into the atmosphere from thermal power plants, industries and automobiles. The common ones are sulphur dioxide, nitrogen oxides (NOX), volatile organic carbons (VOCs) and hydrogen chloride. NO X are also produced in atmosphere through lightning. Sulphur dioxide and nitrogen oxides are changed in the atmosphere into sulphuric acid and nitric acid by combining with oxygen and water. 2SO2 + O2  2SO3 ; SO3 + H2O  H2SO4 2NO + [O]  N2O5 ; N2O5 + H2O  2HNO3

Acid rain : The term was coined by Robert August (1872). Acid rain is rainfall and other forms of precipitation with a pH of less than 5. pH of normal rain is 5.6 – 6.5. The most acidic rain has occured over West Virginia U.S.A. with a pH of 1.5. Acids from atmosphere are deposited over earth in two forms, wet and dry. Wet deposition occurs through rain, snow and fog. Dry deposition is settling down of wind blown acidic gases and particles over trees, various articles

Air Pollutants SO2, NOx Acid Precursors

Transport and change, Complex oxidation reactions

H2SO4, HNO3

Clouds

Clouds

Acid rain, snow or fog

Acid rain, snow or fog

Buildings and monuments Surface run off Urban area, power plants, vehicles, etc

Forest ecosystem Lake ecosystem

River ecosystem

Urban areas,

Direction of wind flow and acid rain path Pollution source Acid rain formation Acid rain damages plants by direct effect on foliage • This layer is established due to an equilibrium and growing points (eg – chlorosis, necrosis, between photo dissociation of ozone by UV – defoliation, dieback.) It causes leaching of essential radiations and regeneration of ozone. minerals of soil. Toxic minerals left in the soil further • The thickness of this ozonosphere averages 5 km. kill the plants. 50% of natural forests have been • The ozone layer acts as a shield and absorbs the destroyed by acid rain in Germany, Sweden, north east, harmful UV–radiations of the sunlight so protects the U.S.A., Romania, Poland, etc. Acid rain has also ruined earth’s biota from the harmful effects of strong UV– fresh water reservoirs of most industrialised countries. radiations. So this layer is very important for the survival Acidity dissolves toxic metals like Hg, Pb, Zn, Al. Both and existence of life on earth. acidity and toxic metals kill all types of aquatic life (a) Causes of Thinning of Ozone Layer : except some algae and fungi. Acid rain corrodes metals, marble, painted surfaces, slate, stone, etc. • The decline in spring – layer thickness is called ozone The phenomenon is called stone leprosy. hole. • Ozone hole is largest over Antarctica and was just OZONE DEPLETION short of 27 million sq. km. during Spetember 2003. • Main chemicals responsible for destruction of ozone • Between 20 and 26 km above the sea level it occurs – layer are : chlorofluorocarbons (CFCs), halogens ozone layer and the part of atmosphere containing it (used in fire extinguishers) methane and nitrous oxide. is called ozonosphere (Stratosphere). Out of these, most damaging is the effect of CFCs BIOLOGY_IJSO_PAGE #64 64

•

which are a group of synthetic chemicals and are used as coolants in refrigerators and air conditioners; as cleaning solvents, propellants and sterilants etc. These CFCs produce “active chlorine” (Cl and ClO radicals) in the presence of UV–radiations. These active chlorine radicals catalytically destroy ozone and convert it into oxygen.

Cl + O3  ClO + O2

•

(b) Effects of Ozone Depletion : The thinning of ozone layer results in an increase in the UV radiations (in the range of 290 – 320 nm) reaching the earth’s surface. It is estimated that a 5 per cent loss of ozone results in a 10 per cent increase in UV – radiations. These UV – radiations : (i) Increase incidences of cataract and skin cancer. (ii) Decrease the functioning of immune system : due to killing of melanin – producing cells of the skin. (iii) nhibit photosynthesis in most of phytoplanktons so adversely affect the food chains of aquatic ecosystems. (iv) Damage nucleic acids of the living organisms.

ClO + O3  Cl + 2O2

Nitrous oxide : It is produced by industrial processes, forest fires, solid waste disposal , spraying of insecticides and pesticides, etc. Methane and nitrous oxide also cause ozone destruction.

Ultraviolet radiation from sun Atmosphere Ozone shield

Hole over Antarctica

• • • •

(c) Measures to Control Air Pollution :

(a) Types of Water Resources :

Barium compounds should be mixed with petrol which reduce the smoke. It is also very essential to check the quality of gases released from the factories. Industries should not be established at one place. The smoke should be released into the atmosphere after filtration and purification (by cyclone collector or electrostatic precipitators).

(i) Fresh water resource : t consists of ponds, lakes, large rivers. t can be recycled. It is essential for life on earth as well as for survival. It can be obtained by three different types of natural resources.

WATER (HYDROSPHERE)

• • •

•

It is a renewable resource which is essential for sustainance of life. It covers 3/4th of the earth’s surface. Of the total water present in hydrosphere 97% is present in oceans which is not utilizable by living beings. Only 3% water is fresh water. Among this 3%, 02.0% in Ice caps, 0.68% in Ground Water, 0.009% in Freshwater lakes, 0.009% in Salt lakes and 0.0019% in Atmosphere is present.

(A) Rain water : ndia receives 3 trillion m3 of water from rainfall or precipitation. Its intensity is different in different zones, on this basis zones are classified as : • • • •

Wet zone : with very high rainfall ntermediate zone : with heavy rainfall Semi arid zone : with moderate rainfall Arid zone : with low rainfall. (B) Surface water :

•

These are major river systems with plenty of lakes & ponds etc. (C) Ground water :

•

t is the water which percolate into the ground. There is a certain level below the soil surface where the rocks are saturated with water and this level is known as the zone of saturation. BIOLOGY_IJSO_PAGE #65 65

•

The upper level of the zone of saturation is called the water table. However, the vertical distance from the surface of a region to the water table is called the water level.

•

•

(ii) Salt water resource : t consists of oceans, seas etc. It cannot be used by living beings for drinking. • (b) Role of Water or Hydrosphere : • • • • • • • • • • •

Water is the main constituent of protoplasm. It is the universal solvent. Through which mineral salts are transported from one part of the plant to the other. Various metabolic reactions take place in the medium containing water. It acts as a reactant in numerous metabolic reactions. During photosynthesis, water releases oxygen. Turgidity of the growing cells is maintained with water. Various movements of plant organs like movements in sensitive plant (touch-me-not) are controlled by water. The growth of the cells during elongation phase mainly depends upon absorption of water. Metabolic end product of respiration is water. It acts as a temperature buffer as its specific heat is highest (only exception - liquid ammonia). It shows the properties of cohesion and adhesion which account for the capillary action of water.

•

• • •

•

•

(iii) Industrial wastes : The wastes of industries are discharged into the running water, rivers and canals, Industrial wastes mainly contain inert suspended particles such as dust, coal, toxins like acid, base, phenols, cyanides, mercury, zinc etc., inorganic materials like – ferrous salts, sulphides, oils and other residues of organic material and hot water.

(c) Water Pollution : •

•

The water pollution is the addition of organic and inorganic chemicals as well as the biological materials which change the physical and chemical properties of water. The water pollution is caused by many sources such as sewage matter, industrial wastage, agricultural wastage, domestic wastage, hot water of thermal plants and nuclear reactors etc. Water pollution can be caused by the following man made sources : (i) Household detergents : The household detergents include the compounds of phosphate, nitrate, ammonium and alkylbenzene sulphonate etc. harmful substances which are gathered in water. Alkyl benzene sulphonate (ABS) is not degradable, so that its concentration increases which is harmful for aquatic life.

•

Control measures : For the control of this pollution lime, ferric chloride etc. are used to precipitate the phosphate. Zirconium is considered best for this purpose.

• •

•

•

•

(ii) Sewage : Sewage contains highest amount of organic materials and biological materials. • • •

These organic materials increase the number of decomposers like bacteria and fungus. The acceleration of microbial activity increases the BOD of water. BOD is very less in pure water. The higher BOD (Biological oxygen demand) is the indication of water pollution and the water of polluted reservoir can not be utilized and produces a very bad smell spreading around the locality.

The infection or disease also takes place. Daphnia and some fishes are sensitive to water pollution and show the intensity of water pollution. Control measures : To control the water pollution of sewage water it should be left into reservoir after the primary and secondary treatment. The big particles are mainly separated in primary treatment through floatation and sedimentation. Micro organisms are used for secondary treatment such as oxidation chamber or activated sludge process. Oxidation chamber is a shallow reservoir in which the sewage is stored. Algae and bacteria grow very well because of the higher amount of organic materials in it. Bacteria decomposes the organic materials and produce CO 2 which is utilized by the algae in photosynthesis. Oxygen released by photosynthesis protects the water pollution. Therefore oxidation pond is the good example of symbiosis in between algae and bacteria. The infectious bacteria are destroyed during the activity (reactions) in the oxidation pond. So that the simple substances are left after decomposition of organic matter.

•

•

The water polluted by mercury, lead etc. causes disorganization of nervous system. It means it produces insanity. The minamata disease caused in Japan by eating of mercury polluted fishes. So many people died because of this disease. Control measures : The industrial wastes and toxic components should be made pure before releasing into rivers, lakes, ponds or sea .So that the water pollution of industrial effluents can be controlled by suitable treatment to remove the pollutants. Bioaccumulation of pesticides : Pesticides like DDT are poisonous chemicals sprayed on crops to protect them from pests and diseases. This increase in concentration of harmful nonbiodegradable chemical substances in the body of living organisms at each trophic level of a food chain is called biological magnification. Eutrophication : The discharge of sewage water and detergents in water bodies promotes excessive growth of phytoplanktons (minute aquatic algae). This excessive growth causes reduction in oxygen level of water. The excessive growth of phytoplanktons brings about a reduction in dissolved oxygen which affects other aquatic organisms. Consequently potential sources of food are highly reduced.

BIOLOGY_IJSO_PAGE #66 66

worms appear and die. Organic matter gets accumulated. Roots of some plants grow into the cervices of rocks.

LITHOSPHERE Lithosphere is the main life supporting system. Top layer of earth is called soil. It is the main natural resource essential for survival and development.

(e) Soil Pollution : Soil is also polluted through the polluted water and air.

(a) Structure and Formation of Soil : Soil is formed due to interaction between weathering of rocks, rain, wind, temperature (physical components) and plants, animals and microbes (biological components). •

It is formed by combined action of climatic factors such as temperature, rainfall, light etc. and biotic factors such as plants and microbes on earth crust.

•

•

•

(b) Constituents of Soil : •

•

Soil contains : (a) Inorganic constituents of parent rocks (b) Organic products of living organisms (c) Living organisms including microorganisms (d) Air in the pores. There are four important components of soil. They are (i) Mineral matter (ii) Organic matter (A) Living organisms (B) Decomposed matter (iii) Soil water (iv) Soil air

50–60% 10%

25 – 35 % 15–25 %

(c) Types of Soil :

•

• •

•

•

On the basis of its nature and composition, soil is mainly of six types — (i) Alluvial soil : rich in loam and clay. (ii) Black soil – which has clay. (iii) Red soil : which is sandy to loam. (iv) Mountain soil — which is a stony and sandy soil.

•

•

•

Outer most layer of earth is called crust. Many types of minerals are found in crust. They provide many types of nutrients to living beings. (d) Factors / Processes Responsible for Formation of Soil : (i) Sun : Rocks get expanded due to heat produced by sun during day time. At night, the rocks cool down and contract. Due to this unequal expansion and contraction of rocks, cracks in rocks appear. This leads to formation of smaller pieces of rocks.

Control measures : Soil pollution can be controlled through biological degradation of waste materials. The various carbonic materials are of agricultural waste, cattle dung etc.which can be minimized by the use of biogas plants which can produce energy also. Inspite all measures pesticides and weedicides should be used in limited quantity only when they are required. Bhopal GasTragedy is the best example of human hazard which took the life of many persons the tank of methyl isocyanate burst during the manufacturing of Savin insecticide on 3rd December 1984. (f) Soil erosion :

(v) Desert soil – which is sandy. (vi) Laterite soil — which has porous clay.

These pollutants are mixed into the soil through the rainy water. Such as H2SO4 acid is formed by mixing of SO2 with rainy water in the air. The fertilizers are used to increase yield of the crops.Various types of pesticides and weedicides etc. are sprayed over the crops. All these mixed with soil to produce harmful effects. The growth of plants inhibited or reduced due to this type of pollution and sometimes death also takes place. Excluding to these soil pollution is also caused by the disposal of house hold detergents, sewage, flowing oils, radioactive substances and hot water etc. The main substances of pesticides in soil pollutants are D.D.T. and weedicides 2, 4 – D (2, 4 di– chlorophenoxy acetic acid) 2, 4, 5–T (2, 4, 5, tri– chlorophenoxy acetic acid).

•

•

(ii) Water : Due to continuous movement of rain and fast flowing river water, rock pieces collide and break down in still finer particles due to their abrasive effect.

•

(iii) Wind : Wind has abrasive effect on rocks. Finer rock particles are blown away and get deposited at other distant places.

•

(iv) Living organisms : The step of weathering is brought about by plants and animals. Lichens are first to appear on bare rocks. They produce acids which corrode the rocky surface to produce fine particles. Now plants like mosses can appear on it. In such type of soil, certain microbes, algae, insects and

• •

Fertility of soil depends on (A) Presence of organic matter (humus) and nutrients, (B) Capacity of soil to retain water and air.A loamy soil is the best - suited for plant growth. The fertility of soil is threatened due to various activities of humans. The main threat to the fertility of soil is from soil erosion, which is the loss of soil due to wind or water flow. Methods of Preventing Soil Erosion : Prevention of soil erosion can be brought about by controlling the factors which cause soil erosion. The methods would thus be as follows : Deforestation should be stopped, rather, trees should be planted (afforestation). Afforestation should be undertaken not only in areas already cut , but additional areas should be brought under plantation. To reduce the effect of strong wind in the fields, the boundaries of the fields should be planted with trees in two or three rows. To maintain the soil in its natural condition, it is advisable to grow different crops. Crop rotation helps to maintain the fertility of the soil. The water - holding capacity of the soil is also maintained by this method. Proper drainage and irrigation arrangements should be made in the fields. On the sloping areas in hills, strip cropping (means the planting of crops in rows or strips to check flow of water). should be practised, thereby reducing the steepness of the slopes and checking soil erosion. BIOLOGY_IJSO_PAGE #67 67

Two types of biogeochemical cycles are : 1. Gaseous cycles 2. Sedimentary cycles

•

BIOGEOCHEMICAL CYCLES These are the cyclic pathways through which chemical elements move from environment to organisms and back to the environment. The earth and its environment, with reference to these elements, are considered as closed system and there is no inflow of such elements from outside the earth and their amount is limited.

Differences between gaseous & sedimentary cycles. Characters

Gaseous cycles

Sedimentary Cycles

Reservoir pool

Air or water

Rocks

Speed

Faster

Slower

Examples

Carbon, nitrogen & oxygen cycles

Calcium, phosphorous & sulphur cycles.

(a) Water Cycle : Water is the most abundant (60–90%) component of protoplasm.It acts as a habitat for hydrophytes & many aquatic animals, a good ionizer, good solvent, temperature, buffer and perform transportation of materials. It also helps in digestion of organic compounds & in photosynthesis of plants.

Water Cycle

•

Types of water cycles are : (A) Global water cycle : It does not involve living organisms and involves the interchange of water between the earth’s surface and the atmosphere via the processes of precipitation and evaporation.

• • •

Ocean is the biggest store house of water. Evaporation involves the conversion of liquid and solid forms of water into vapours and later form the clouds. Precipitation involves the rainfall, hail, snow, etc. Energy for global water cycle is provided by sunlight. (B) Biological water cycle : t is the interchange of water between abiotic and biotic components of environment

•

e.g. the plants absorb water from water bodies and soil while loose most of the water by the process of transpiration, animals consume water from water bodies or the food ingested, while release water via the processes of respiration and excretion. (b) Nitrogen Cycle : Nitrogen is an essential component of amino acids, proteins, enzymes and nucleic acids of the protoplasm.Reservoir pool of nitrogen is atmosphere which contains about 78.08% of nitrogen in gaseous state. But it cannot be used directly and is changed into nitrites and nitrates and then utilized.

BIOLOGY_IJSO_PAGE #68 68

•

Steps of nitrogen cycle are : (A) Nitrogen fixation : It involves the conversion of free diatomic nitrogen (N2) into nitrites and nitrates.t occurs in three ways :

• •

• • • • • • •

Physical nitrogen fixation : Atmospheric nitrogen fixation in the presence of photochemical and electrochemical reactions induced by thundering and lightning. Industrial nitrogen fixation in the industries at high temperature and high pressure. Biological nitrogen fixation : Biological nitrogen fixation occurs in the presence of certain living organisms such as. Rhizobium bacterium in the root nodules of legumes. Azotobacter bacterium in the soil. Anabaena (blue green algae) in water in the paddy fields. Azospirillum bacterium in loose association with the roots of maize, sorghum, etc. (B) Ammonification : t involves the decomposition of proteins of dead plants and animals to ammonia in the presence of ammonifying bacteria like Bacillus ramosus.

(C) Nitrification : It involves the oxidation of ammonia to nitrites (NO2–) and nitrates (NO3–) in the presence of nitrifying bacteria like Nitrosomonas (Ammonia to nitrite), Nitrobacter (Nitrite to nitrate), etc.Plants absorb the nitrites and nitrates from the soil through their roots and convert them into organic compounds (e.g. proteins) of protoplasm by the process called nitrogen assimilation. (D) Denitrification : t involves reduction of ammonium compounds, nitrites and nitrates to molecular nitrogen in the presence of denitrifying bacteria like Thiobacillus denitrificans.

•

• • •

Carbon Cycle : Carbon is the basic component of all the organic compounds like carbohydrates, proteins, lipids, enzymes and nucleic acid of the protoplasm. In atmosphere, it is present as carbon dioxide. It involves two types of processes, one involving CO2 utilization and another involving CO2 production. They are expressed as follows :

BIOLOGY_IJSO_PAGE #69 69

(i) CO2 utilization : Carbon dioxide is utilized by the photosynthetic organisms like green plants, photosynthetic bacteria, diatoms and blue green algae in the process of photosynthesis, t occurs in the presence of chlorophyll and radiant energy of sunlight.Glucose synthesized in photosynthesis is used to synthesize other organic compounds.

•

• •

The oxides can be reduced both chemically and biologically to produce oxygen. Microbial oxidation can also occur. Due to burning of materials oxygen form carbon dioxide. When oxygen combines with nitrogen, it forms oxides of nitrogen, amino acids, proteins etc.

(ii) CO2 production :

• • • • •

CO2 is released during respiration of both producers and consumers. During decomposition of organic compounds of dead bodies. During burning of fossil fuels like wood, coal, petro leum, etc. Volcanic eruptions and hot springs. During weathering of rocks by acids produced by microorganisms and roots of higher plants. (d) Oxygen Cycle :

•

•

Oxygen is present in water and forms 20% of air in atmosphere. All living beings need it for respiration. Oxygen content of atmosphere has remained constant for the last several million years. Most of O2 is replenished by photosynthesis. During photosynthesis CO2 is used by plants to form food along with release of oxygen.

BIOLOGY_IJSO_PAGE #70 70

OBJECTIVE DPP - 16.1 1.

Soil is a part of (A) atmosphere (C) hydrosphere

(B) lithosphere (D) ionosphere

2.

Maximum air in which we breath is present at (A) troposphere (B) stratosphere (C) ionosphere (D) mesosphere

3.

Corrosion of statues and monuments occurs due to : (A) Photochemical smog (B) CO (C) Acid rain (D) Methane

4.

Lichens indicate pollution by : (A) O3 (B) SO2 (C) NO2 (D) CO

5.

Pollutant released by jet planes is : (A) Fog (B) Aerosol (C) Smog (D) Colloid

6.

Thickness of ozonosphere is (A) 3km. (B) 7km. (C) 5cm. (D) 5km.

7.

8.

9.

15. Coal is an / a – (A) exhaustible resource (B) inexhaustible resource (C) potential resource (D) none of these 16. Ozonosphere occurs at height of – (A) 8 – 10 km above poles (B) 8 – 10 km above equator (C) 20 – 26 km above the earth surface (D) 11 – 16 km above equator 17. Biosphere is made of – (A) living beings and their remains (B) living beings, lithosphere, hydrosphere and atmosphere (C) living beings and lithosphere (D) living beings, lithosphere and hydrosphere. 18. Free living bacteria involve in nitrogen fixation – (A) Rhizobium (B) Azotobacter (C) Anabaena (D) Azospirillum

Soil erosion can be prevented by – (A) deforestation (B) afforestation (C) overgrazing (D) removal of vegetation

19. Which one of the following is renewable resource ? (A) Water (B) Metals (C) Fossil fuel (D) All of these 20. Which gas is responsible for the global warming – (A) O2 (B) N2 (C) H2 (D) CO2

A renewable source of energy is – (A) petroleum (B) coal (C) nuclear fuel (D) trees Percentage of nitrogen in air is – (A) 77.02 % (B) 78.09 % (C) 76.08% (D) 74.09%

10. Ozone layer is present in atmosphere in – (A) troposphere (B) stratosphere (C) mesosphere (D) thermosphere 11. Nodules in the roots of legume plants contain (A) nitrogen fixing bacteria (B) sulphur fixing bacteria (C) potassium fixing bacteria (D) none of the above 12. Which gas is mainly responsible for the depletion of ozone layer ? (A) Oxygen (B) CFC (C) Nitrogen dioxide (D) All of the above 13. Acid rain mainly contains – (A) nitric acid (C) sulphuric acid

14. Plants and animals are known as – (A) biotic resources (B) abiotic resources (C) machines (D) none of these

(B) hydrochloric acid (D) (A) and (C) both

21. Biogeochemical cycles are also known as (A) sedimentary cycles (B) gaseous cycles (C) material cycles (D) cycles of water 22. Which of the following is a free living nitrogen fixing bacteria present in soil ? (A) Azotobacter (B) Nitrosomonas (C) Rhizobium (D) Pseudomonas 23. CO2 and O2 balance in atmosphere is due to (A) photosynthesis (B) respiration (C) leaf anatomy (D) photorespiration 24. Nitrogen fixation is -

(A) Nitrogen (B) Nitrogen (C) Nitrogen (D) Both A and B

Nitrite nitrates Amino acid

25. Ozone depletion in stratosphere would result in : (A) Forest fires (B) Increased incidence of skin cancer (C) Global warming (D) None of the above



BIOLOGY_IJSO_PAGE #71 71

ANSWER KEY Force and Newton’s law of motion Que.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Ans.

C

D

A

C

A

B

B

C

C

B

A

B

D

A

ACD

Que.

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

Ans.

B

C

C

C

C

C

B

D

A

A

A

C

C

D

B

Que.

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

Ans.

B

B

B

A

D

C

B

C

B

A

A

C

B

D

A

Que.

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

Ans. C A B Que. 61 62 63 Ans. A,C C A,B Que. 76 Ans. A

D

A

A

D

D

CD

D

A

A

D

C

B

64

65

66

67

68

69

70

71

72

73

74

75

B

A

B

B

A

C

C

D

C

B

D

C

MOLE CONCEPT(CHEMISTRY) Ques. Ans. Ques. Ans. Ques. Ans. Ques. Ans. Ques. Ans. Ques. Ans. Ques. Ans.

1 B 16 A 31 D 46 B 61 C 76 A 91 A

2 C 17 A 32 C 47 B 62 D 77 A 92 D

3 D 18 A 33 C 48 C 63 A 78 C 93 C

4 B 19 A 34 B 49 C 64 B 79 A 94 A

5 C 20 A 35 C 50 C 65 A 80 A

POLYNOMIALS

6 B 21 B 36 B 51 C 66 B 81 D

7 A 22 A 37 C 52 B 67 C 82 B

8 D 23 A 38 D 53 C 68 A 83 D

9 A 24 D 39 A 54 B 69 D 84 B

10 B 25 A 40 B 55 A 70 B 85 C

11 B 26 A 41 C 56 A 71 C 86 B

12 D 27 A 42 A 57 D 72 B 87 C

13 B 28 D 43 C 58 C 73 C 88 C

14 C 29 A 44 D 59 A 74 B 89 D

15 B 30 D 45 A 60 D 75 D 90 D

(MATHEMATICS)

Q.

1

2

3

4

5

6

7

8

9

10

Ans.

A

B

D

B

D

A

B

B

C

B

Q.

11

12

13

14

15

16

17

18

19

20

Ans.

D

B

D

B

B

C

B

B

A

A

Q.

21

22

23

24

25

26

27

28

29

30

Ans.

D

D

D

B

D

B

C

A

B

A

Q.

31

32

33

34

Ans.

C

D

C

B

NATURAL RESOURCES(BIOLOGY)

Q. 1 2 3 4 5 A. B A C B B Q. 21 22 23 24 25 A. C A A D B

6 D

7 B

8 D

9 B

10 11 12 13 14 15 16 17 18 19 20 B A B D A A C B B A D

ANSWER KEY IJSO_PAGE #72 72