Mole Concept-1 (a)

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LECTURE # 1 Mole concept-1

Introduction : There are a large number of objects around us which we can see and feel. Anything that occupies space and has mass is called matter. Ancient Indian and Greek Philospher’s beleived that the wide variety of object around us are made from combination of five basic elements : Earth, Fire, Water, Air and Sky. Ancient Greek Philosphers also believed that all matter was composed of tiny building blocks which were hard and indivisible.The Greek philosphere Democritus named these building blocks as atoms, meaning indivisible.All these people have their philosphical view about matter, they were never put to experimental tests, nor ever explain any scientific truth.It was John Dalton who firstly developed a theory on the structure of matter, later on which is known as Dalton’s atomic theory.

I

DALTON’S ATOMIC THEORY :

1.

Matter is made up of very small indivisible particle called atoms.

2.

All the atoms of a given element is idenctical in all respect i.e. mass, shape, size, etc.

3.

Atoms cannot be created or destroyed by any chemical process or physical process.

4.

Atoms of different elements are different in nature. Classification of matter

on the basis of physical behaviour

Solids

Liquids

Gases

on basis the of chemical behaviour

Pure substances Element

Mixtures

Compound

III Some Definitions I. RELATIVE ATOMIC MASS : It is the ratio of the mass of 1 atom of a substance and 1/12 of mass of 1 atom of C12 isotope. For atoms this is done by expressing mass of one atom with respect to a fixed standard. Dalton used hydrogen as the standard (H = 1). Later on oxygen (O = 16) replaced hydrogen as the reference.



C-12 ISOTOPE OF CARBON IS LATEST CHOSEN STANDARD SINCE 1961 Therefore relative atomic mass is given as

Relative atomic mass (R.A.M) =

=

mass of one atom of the element 1  mass of one C12 atom 12

total number of nucleons  mass of 1 nucleon = Total Number of nucleons 1 12  mass of 1 nucleon 12 Page # 1

On Hydrogen scale : Relative atomic mass (R.A.M) =

mass of one atom of the element mass of one H2 atom

Oxygen scale : Relative atomic mass (R.A.M) =

II.

mass of one atom of the element 1  mass of one O  16 atom 16

ATOMIC MASS UNIT (OR AMU)  1 The atomic mass unit (amu) is equal to one twelfth   of the mass of one atom of carbon-12 isotope.  12 



1 × mass of one C-12 atom 12 ~ mass of one neucleon in C-12 atom. = 1.66 × 10–24 gm or 1.66 × 10–27 kg

1 amu =



one amu is also called one Dalton (Da).TODAY , AMU HAS BEEN REPLACED BY ‘u’ WHICH IS KNOWN AS UNIFIED MASS

III.

ATOMIC MASS It is the mass of 1 atom of a substance it is expressed in AMU. Atomic mass = R.A.M × 1 amu

Note : Relative atomic mass is nothing but the number of nucleons present in the atom. Example : Find the relative atomic mass of ‘O’ atom and its atomic mass. Sol. The number of neucleons present in ‘O’ atom is 16.  relative atomic mass of ‘O’ atom = 16. Atomic mass = R.A.M × 1 amu = 16 × 1 amu = 16 amu Q. Ans.

Find the relative atomic mass, atomic mass of the following elements. (i) Na (ii) F (iii) H (iv) Ca (v) Ag (i) 23, 23 amu (ii) 19, 19 amu (iii) 1, 1.008 amu

Q. Ans.

How many neucleons are present in 5 atoms of an element which has atomic mass 14 amu = 70

Note

:

NCERT Reading

NCERT Exercise

DPP No.

Text. 1.6,1.7,1.7.1,1.7.2 Page 13





Exercise–2 Part–I – Part–II –

Exercise–3 Part–I – Part–II –

Sheet Exercise–1 Part–I – Part–II –

Page # 2

LECTURE # 2 IV.

MOLE :THE MASS–NUMBER RELATIONSHIP Mole is a chemical counting S unit and defined as follows : A mole is the amount of a substance that contains as many entities (atoms, molecules or other particles) as there are atoms in exactly 0.012 kg (or 12 gm) of the carbon-12 isotope. From mass spectrometer we found that there are 6.023 × 1023 atoms are present in 12 gm of C-12 isotope. The number of entities in 1 mol is so important that it is given a separate name and symbol known as Avogadro constant denoted by NA. i.e. on the whole we can say that 1 mole is the collection of 6.02 × 1023 entities. Here entities may represent atoms, ions, molecules or even pens, chair, paper etc also include in this but as this number (NA) is very large therefore it is used only for very small things. 1 mole x mass of 1 atom of C12 isotope 1 mole x 12 x mass of one nucleon

= 12g = 12 g

  

Alternatively value of NA can be found in this fashion

1

 1 mole =

= 6.023 x 1023

Note : In modern practice gram-atom and gram-molecule termed as mole.



V.

1.66 x 10  24

GRAM ATOMIC MASS : The atomic mass of an element expressed in gram is called gram atomic mass of the element. For example for oxygen atom : Atomic mass of ‘O’ atom = mass of one ‘O’ atom = 16 amu gram atomic mass = mass of 6.02 × 1023 ‘O’ atoms = 16 amu × 6.02 × 1023 = 16 × 1.66 × 10–24 g × 6.02 ×1023 = 16 g –24 ( 1.66 × 10 × 6.02 × 1023 ~ 1 )

Q.

How many atoms of oxygen are their in 16 g oxygen.

x  1.66 x 10 24 = 16 g 1

x=

1.66 x 10  24

= NA

or It is also defined as mass of 6.02 × 1023 atoms. or It is also defined as the mass of one mole atoms.



Now see the table given below and understand the definition given before. Element

R.A.M. (Relative Atomic Mass)

Atomic mass (mass of one atom)

Gram Atomic mass/weight

N

14

14 amu

14 gm

He

4

4 amu

4 gm

C

12

12 amu

12 gm

Example : What is the weight of 3-g atoms of sulphur R.A.M. of s = 32. Ans. 96 g Example : How many g atoms are present in 144 g of sulphur Ans. 4.5 g atoms Page # 3

Example : The ratio of mass of a silver atom to the mass of a carbon atom is 9 : 1. Find the mass of 1 mole of C atom if molar mass of Ag is 108. Ans. 12 Example : Calculate mass of sodium which contains same number of atoms as are present in 4g of calcium. Atomic masses of sodium and calcium are 23 and 40 respectively. Ans. 2.3 g

VI.

MOLECULES : It is the smallest particle of matter which has free existence. Molecules can be further divided into its constituents atoms by physical & chemical process. Number of atoms presents in molecule is called its atomicity. Element : H2, O2, O3 etc. Compound : KCl, H2SO4, KClO4 etc. Molecule KCl H2SO4 O3 H2

VII.

-

Atomicity 2 7 3 2

MOLECULAR MASS : It is the mass of one molecule Ex. Molecule H2 KCl H2SO4

Molecular mass 2 amu (39 + 35.5) = 74.50 amo (2 + 32 + 64) = 98 amu.

VIII. GRAM MOLECULAR MASS : The molecular mass of a substance expressed in gram is called the gram-molecular mass of the substance. or It is also defined as mass of 6.02 × 1023 molecules or It is also defined as the mass of one mole molecules. (molar mass) For example for ‘O2’ molecule : Molecular mass of ‘O2’ molecule = mass of one ‘O2’ molecule = 2 × mass of one ‘O’ atom = 2 × 16 amu = 32 amu gram molecular mass = mass of 6.02 × 1023 ‘O2’ molecules = 32 amu × 6.02 × 1023 = 32 × 1.66 × 10–24 gm × 6.02 × 1023 = 32 gm

Use the Y-map in the following example.

Page # 4

Example: Find the mass in grams of 3 mol of zinc. (GMM = 65) Sol. Mass = mol × At. wt. = 3 × 65 gm = 195 gm Example : How many atoms of copper are present in 0.5 mol of pure copper metal? Sol. No. of atoms = no. of moles × NA = 0.5 × 6.02 × 1023 = 3.01 × 1023 Example : The molecular mass of H2SO4 is 98 amu. Calculate the number of moles of each element in 294 g of H2SO4. Solution Gram molecular mass of H2SO4 = 98 gm moles of H2SO4 = H2SO4 One molecule 1 × NA  one mole  3 mole

294 = 3 moles 98

H 2 atom 2 × NA atoms 2 mole 6 mole

S one atom 1 × NA atoms one mole 3 mole

O 4 atom 4 × NA atoms 4 mole 12 mole

Example : A sample of (C2H6) ethane has the same mass as 107 molecules of methane. How many C2H6 molecules does the sample contain ? Ans. n = 5.34 × 106 Example : How many molecules of water are present in 252 mg of (H2C2O4.2H2O) Ans. 2.4 × 1021 Example : From 48 g of the He sample ,6.023 x 1023 atoms of He are removed. Find out the moles of He left.Also Calculate the mass of carbon which contains same number of atoms as left over in this sample. Ans. 11 mole, 132 g of C. Note

:

NCERT Reading Text. Formula Mass Page 14 Sheet Exercise–1 Part–I –1,2,3,4,5,6,7. Part–II –6,7,8,9,10,11,12

NCERT Exercise

DPP No.

Q.No.-1.1,1.10,1.28,1.30,1.33

1

Exercise–2 Part–I – Part–II –

Exercise–3 Part–I – Part–II –

Page # 5

LECTURE # 3 GAY-LUSSAC’S LAW OF COMBINING VOLUME : Gases combine in a simple ratio of their volumes provided all measurements should be done at the same temperature and pressure H2 (g)

+

1 vol

Cl2 (g)

 2HCl

1 vol

2 vol

AVOGADRO’S HYPOTHESIS : Equal volume of all gases have equal number of molecules (not atoms) at same temperature and pressure condition. mathematically, for ideal gases, V  n (CONSTANT T & P) S.T.P. (Standard Temperature and Pressure): At S.T.P. / N.T.P. condition : temperature = 0°C or 273 K pressure = 1 atm = 760 mm of Hg volume of one mole of an ideal gas = 22.4 litres (experimentally determined) NOTE FOR FACULTY : The gas equation PV = nRT should never be used in this chapter. Ex. Sol.

Calculate the volume in litres of 20 g hydrogen gas at STP. 20 gm No. of moles of hydrogen gas = mass  atomic weight = 2 gm = 10 mol volume of hydrogen gas at STP = 10 × 22.4 lt. Number

×N



A

N A

Mole

 mol. wt.  At. wt.

Ex. Ans. Ex. Ans. Ex.

lt 2.4 2 × lt 2.4 2

Volume at STP

× mol. wt. × At. wt.

Mass Calculate the volume in litres of 142 g chlorine gas at STP. 44.8 lt.

Find the volume at STP occupied by 16 g of ozone at STP. 22 . 4 = 7.5  3 From 160 g of SO2 (g) sample, 1.2046 x 1024 molecules of SO2 are removed then find out the volume of left over SO2 (g) at STP. Ans. 11.2 Ltr.

Ex.

14 g of Nitrogen gas and 22 g of CO2 gas are mixed together. Find the volume of gaseous mixture at STP. Ans. 22.4 Ltr.

Ex.

672 ml of ozonized oxygen (mix of O2 and O3) at N.T.P. were found to weight one gram. Calculate the volume of ozone in the ozonized oxygen. 56 ml

Ans.

Note :

NCERT Reading – Sheet Exercise–1 Part–I – 1 to 8 Part–II – 1 to 12

NCERT Exercise –

DPP No. 2

Exercise–2 Part–I – Part–II –

Exercise–3 Part–I – Part–II – Page # 6

LECTURE # 4 II

THE LAWS OF CHEMICAL COMBINATION Atoine Lavoisier, John Dalton and other scientists formulate certain law concerning the composition of matter and chemical reactions.These laws are known as the laws of chemical combination.

1.

THE LAW OF CONSERVATION OF MASS :( ANTOINE LAVOISIER) It states that matter can neither be created nor destroyed in chemical reaction i.e . In a chemical change , total mass remains conserved.i.e. mass of all reactants = mass of products after reaction. ( In a closed system ) 1 Example : H2 (g) + O2 (g)  H2O (l) 2 1 Before reaction 1 mole mole 2 After the reaction 0 0 1 mole 1 mass before reaction = mass of 1 mole H2 (g) + mole O2 (g) 2 = 2 + 16 = 18 gm mass after reaction = mass of 1 mole water = 18 gm

2.

LAW OF CONSTANT OR DEFINITE PROPORTION :{JOSEPH PROUST} A given compound always contains exactly the same proportion of elements by weight irrespective of their source or method of preparation .

Ex.

In water (H2O), Hydrogen and Oxygen combine in 2 : 1 molar ratio, this ratio remains constant whether it is tap water, river water or sea water or produced by any chemical reaction.

Ex.

1.80 g of a certain metal burnt in oxygen gave 3.0 g of its oxide. 1.50 g of the same metal heated in steam gave 2.50 g of its oxide. Show that these results illustrate the law of constant proportion. In the first sample of the oxide, Wt. of metal = 1.80 g, Wt. of oxygen = (3.0 – 1.80) g = 1.2 g

Sol.



wt. of metal 1.80g   1.5 wt. of oxygen 1.2g

In the second sample of the oxide, Wt. of metal = 1.50 g, Wt. of oxygen = (2.50 – 1.50) g = 1 g. 

wt. of metal 1.50 g   1.5 wt. of oxygen 1g

Thus, in both samples of the oxide the proportions of the weights of the metal and oxygen are fixed. Hence, the results follow the law of constant proportion.

3.

THE LAW OF MULTIPLE PROPORTION : {JOHN DALTON} When one element combines with the other element to form two or more different compounds, the mass of

one elements, which combines with a constant mass of the other, bear a simple ratio to one another. Note : Simple ratio here means the ratio between small natural numbers, such as 1 : 1, 1 : 2, 1 : 3, later on this simple ratio becomes the valency and then oxidation state of the element. Ex.

Carbon and oxygen when combine, can form two oxides viz CO (carbonmonoxide), CO2 (Carbondioxides) In CO, 12 gm carbon combined with 16 gm of oxygen. In CO2, 12 gm carbon combined with 32 gm of oxygen. Thus, we can see the mass of oxygen which combine with a constant mass of carbon (12 gm) bear simple ratio of 16 : 32 or 1 : 2 Note : See oxidation number of carbon also have same ratio 1 : 2 in both the oxide.

Page # 7

CONCEPTS RELATED TO DENSITY : It is of two type. 1. Absolute density 2. Relative density For liquid and solids mass volume density of the subs tance specific gravity = density of water at 4C For gases : Molar mass of the gas Absolute density (mass/volume) = Molar volume of the gas For simplification, we can conclude that the density and specific gravity of any substance is numerically same, but density has a definite unit, but specific gravity has no unit. (dimension less) Specific gravity of a solution is 1.8 then find the mass of 100 ml of solution. 180 gm.

Absolute density =

* Ex. Ans.

RELATIVE DENSITY : Ex.

It is the density of a substance with respect to any other substance. What is the V.D. of SO2 with respect to CH4

M.W . SO 2 V.D. = M.W . CH 4 64 =4 16

V.D = Ex.

Find the density of CO2(g) with respect to N2O(g).

VAPOUR DENSITY : Vapour density is defined as the density of the gas with respect to hydrogen gas at the same temperature and pressure. dgas Molar mass of gas / molar volume Vapour density = d = Molar mass of H / molar volume H2

Mgas M gas V.D. = M = ; H2 2 Ex.

Mgas = 2 V.D.

What is V.D. of oxygen gas V.D =

Ex.

2

32 = 16 unitss 2

7.5 litre of the particular gas as S.T.P. as weight 16 gram. What is the V.D. of gas 7.5 litre = 16 gram moles =

7.5 16  22.4 M M = 48 gram

48 = 24. 2 Note : V.D. can be expressed with respect to some other gas other than hydrogen. V.D.

Note

:

NCERT Reading Text. 1.5.1 to 1.5.5 Page 11 to 30. Sheet Exercise–1 Part–I –12,13,14. Part–II –17,18,19,20,21.

NCERT Exercise

DPP No.

Q.No.1.21

3

Exercise–2 Part–I –2. Part–II –33,35.

Exercise–3 Part–I – Part–II – Page # 8

LECTURE # 5 % PERCENTAGE COMPOSITION : Here we are going to find out the percentage of each element in the compound by knowing the molecular formula of compound. We know that according to law of definite proportions any sample of a pure compound always possess constant ratio with their combining elements. Ex.

Every molecule of ammonia always has formula NH3 irrespective of method of preparation or sources. i.e. 1 mole of ammonia always contains 1 mol of N and 3 mole of H. In other wards 17 gm of NH3 always contains 14 gm of N and 3 gm of H. Now find out % of each element in the compound. Mass % of N in NH3 =

Mass of N in 1 mol NH3  100 = 14 × 100 = 82.35 % Mass of 1 mol of NH3 17

Mass of H in 1 mol NH3 3 Mass % of H in NH3 = Mass of 1 mol e of NH  100 = × 100 = 17.65 % 3 17

Ex. Ans.

What is the percentage of calcium and oxygen in calcium carbonate (CaCO3) ? 40%, 48%.

Ex. Ans.

A compound of sodium contains 11.5% sodium then find the minimum molar mass of the compound. 200 gm/mole.

EMPIRICAL AND MOLECULAR FORMULA : We have just seen that knowing the molecular formula of the compound we can calculate percentage composition of the elements. Conversely if we know the percentage composition of the elements initially, we can calculate the relative number of atoms of each element in the molecules of the compound. This gives us the empirical formula of the compound. Further if the molecular mass is known then the molecular formula can easily be determined. Thus, the empirical formula of a compound is a chemical formula showing the relative number of atoms in the simplest ratio, the molecular formula gives the actual number of atoms of each element in a molecule. The molecular formula is generally an integral multiple of the empirical formula. i.e. molecular formula = empirical formula × n where Ex. Sol.

molecular formula mass

n = empirical formula mass

Acetylene and benzene both have the empirical formula CH. The molecular masses of acetylene and benzene are 26 and 78 respectively. Deduce their molecular formulae.  Empirical Formula is CH Step-1 The empirical formula of the compound is CH  Empirical formula mass = (1 × 12) + 1 = 13. Molecular mass = 26 Step-2 To calculate the value of ‘n’

Molecular mass 26 n = Empirical formula mass = =2 13 Step-3 To calculate the molecular formula of the compound. Molecular formula = n × (Empirical formula of the compound) = 2 × CH = C2 H2 Thus the molecular formula is C2 H2 Similarly for benzene To calculate the value of ‘n’ Molecular mass 78 n = Empirical formula mass = =6 13 thus the molecular formula is 6 × CH = C6H6 Page # 9

Ex.

Sol.

An organic substance containing carbon, hydrogen and oxygen gave the following percentage composition. C = 40.684% ; H = 5.085% and O = 54.228% The molecular weight of the compound is 118. Calculate the molecular formula of the compound. Step-1 To calculate the empirical formula of the compound.

Percentage At. mass Relative no. Percentage of element of element of atoms = At. mass

Element

Symbol

Carbon

C

40.687

12

Hydrogen

H

5.085

1

Oxygen

O

54.228

16

Simplest atomic ratio

Simplest whole no. atomic ratio

40.687 = 3.390 12 5.085 = 5.085 1

3.390 3.389

=1

2

5.085 3.389

=1.5

3

54.228 = 3.389 16

3.389 3.389

=1

2

Empirical Formula is C2 H3 O2

 Step-2

To calculate the empirical formula mass. The empirical formula of the compound is C2 H3 O2 .  Empirical formula mass = (2 × 12) + (3 × 1) + (2 × 16) = 59. Step-3 To calculate the value of ‘n’

Molecular mass 118 n = Empirical formula mass = =2 59 Step-4 To calculate the molecular formula of the salt. Molecular formula = n × (Empirical formula) = 2 × C2 H3 O2 = C4 H6 O4 Thus the molecular formula is C4 H6 O4. Ex.

Ans. Ex.

An oxide of nitrogen gave the following precentage composition : N = 25.94 and O = 74.06 Calculate the empirical formula of the compound. N2O5 Hydroquinone, used as a photographic developer, is 65.4%C, 5.5% H, and 29.1%O, by mass. What is the empirical formula of hydroquinone ? Ans. C3H3O

Note :

NCERT Reading Text. 1.9,1.9.1 Page 15 Sheet Exercise–1 Part–I –9,10,11. Part–II –13,14,15,16.

NCERT Exercise

DPP No.

Q.No.1.2,1.3,1.8.

4

Exercise–2 Part–I – 3,11,13. Part–II – 3,4,7,8,9,14,28,29.

Exercise–3 Part–I – Part–II –

Page # 10

LECTURE # 6 CHEMICAL REACTION : It is the process in which two or more than two substances interact with each other where old bonds are broken and new bonds are formed.

VI CHEMICAL EQUATION : All chemical reaction are represented by chemical equations by using chemical formule of reactants and products. Qualitatively a chemical equation simply describes what the reactants and products are. However, a balanced chemical equation gives us a lot of quantitative information mainly the molar ratio in which reactants combine and the molar ratio in which products are formed. Example : When potassium chlorate (KClO3) is heated it gives potassium chloride (KCl) and oxygen (O2).

 KClO3  KCl + O2 (unbalanced chemical equation )  2KClO3  2 KCl + 3 O2 (balanced chemical equation) Attributes of a balanced chemical equation: (From NCERT PAGE - 17) (a) It contains an equal number of atoms of each element on both sides of equation.(POAC) (b) It should follow law of charge conservation on either side. (c) Physical states of all the reagents should be included in brackets. (d) All reagents should be written in their standard molecular forms (not as atoms ) (e) The coefficients give the relative molar ratios of each reagent. Balancing a chemical equation According to the law of conservation of mass, a balanced chemical equation has the same number of atoms of each element on both sides of the equation. Many chemical equations can be balanced by trial and error. Let us take the reactions of a few metals and non-metals with oxygen to give oxides 4 Fe(s) + 3O2(g)  2Fe2O3(S) (a) balanced equation 2 Mg(s) + O2(g)  2MgO(S) (b) balanced equation P4(s) + O2(g)  P4O10(S) (c) unbalanced equation Equations (a) and (b) are balanced since there are same number of metal and oxygen atoms on each side of equations. However equation (c) is not balanced. In this equation. phosphorus atoms are balanced but not the oxygen atoms. To balance it, we must place the coefficient 5 on the left of oxygen on the left side of the equation to balance the oxygen atoms appearing on the right side of the equation. P4(S) + 5O2(g)  P4O10(S) balanced equation Now let us take combustion of propane, C3H8, This equation can be balanced in steps. Step 1. Write down the correct formulas of reactants and products. Here propane and oxygen are reactants, and carbon dioxide and water are products. C3H8(g) + O2(g)  CO2 (g) + H2O (l) unbalanced equation Step 2. Balance the number of C atoms : Since 3 carbon atoms are in the reactant, therefore, three CO2 molecules are required on the right side. C3H8(g) + O2 (g)  3CO2 (g) + H2O (l) Step 3. Balance the number of H atoms : on the left there are 8 hydrogen atoms in the reactants however, each molecule of water has two hydrogen atoms , so four molecules of water will be required for eight hydrogen atoms on the right side. C3H8 (g) + O2 (g)  3CO2 (g) + 4H2O (l) Step 4. Balance the number of O atoms : There are ten oxygen on the right side (3 × 2 = 6 in CO2 and 4 × 1 = 4 in water). Therefore, five O2 molecules are needed to supply to supply the required ten oxygen atoms. C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (l) Step 5. Verify that the number of atoms of each element is balanced in the final equation. Always remember that subscripts in formula of reactants and products cannot be changed to balance an equation.

Page # 11

INTERPRETATION OF BALANCED CHEMICAL EQUATIONS : (STOICHIOMETRY AND STOICHIOMETRIC CALCULATIONS, NCERT, PAGE - 17) 3.

Mole-mole analysis : This analysis is very much important for quantitative analysis point of view. Students are advised to clearly understand this analysis. Now consider again the decomposition of KClO3 . 2KClO3  2KCl + 3O2 In very first step of mole-mole analysis you should read the balanced chemical equation like 2 moles KClO3 on decomposition gives you 2 moles KCl and 3 moles O2. and from the stoichiometry of reaction we can write Moles of O 2 Moles of KClO 3 Moles of KCl = = 3 2 2 Now for any general balance chemical equation like a A + b B  c C + d D

you can write. Mole of A reacted moles of B reacted moles of C reacted moles of D reacted = = = a b c d

Ex. Sol.

3 moles (367.5 gm) of KClO3 when heated how many moles KCl and O2 is produced. The reaction is 2KClO3  2KCl + 3O2

Moles of KClO 3 Moles of KCl = 2 2 Now,



Moles of KCl produced = 3

Moles of O 2 Moles of KClO 3 = 3 2

33 = 4.5 moles 2 CH4 + 2O2  CO2 + 2H2O (from NCERT Page - 18) following conclusions can be drawn from above reaction by observing its stoichiometry mole of O2 produced =

Ex.

 One mole of CH4 (g) reacts with two moles of O2 (g) to give one mole of CO2 (g) and two moles of H2O (g) One molecule of CH4 (g) reacts with 2 molecues of O2 (g) to give one molecule of CO2 (g) and 2 molecules of H2O (g)  22.4 L of CH4 (g) reacts with 44.8 L of O2 (g) to give 22.4L of CO2 (g) and 44.8 L of H2O (g)  16 g CH4 (g) reacts with 2×32 g of O2 (g) to give 44 g of CO2 (g) and 2 × 18 g of H2O (g). Note : In fact mass-mass and mass-vol analysis are also interpreted in terms of mole-mole analysis you can use following chart also.

Mass

At. wt. / Mol. Wt.

Mole

Mole-mole relationship of equation

×

Mass

Ex.

Sol.

t. t. w ./A t l. w mo

Mole × 22.4 lt

Volume at STP

367.5 gm KClO3 (M = 122.5) when heated (a) How many grams O2 is produced (b) How many litre of O2 is produced at STP Now consider the balanced chemical equation 2KClO3  2KCl + 3O2 Now go with the above chart

Page # 12

367.5 gm KClO3

122.5 gm

Mole of O2 Mole of KClO3 = 3 2

3 mole KClO3

(Mole-mole relationship of equation)

(a) 144 gm

(b) 100.8 lt

Ex.

Ans. Ex.

× 32 g (mol. m wt.)

9/2 mole of O2

lt × 22.4 at STP) e m (volu

Iron in the form of fine wire burns in oxygen to form iron (III) oxide 4Fe(s) + 3O2(g)  2Fe2O3(s) How many moles of O2 are needed to produce 5 mol Fe2O3 ? 7.5 mol O2 Nitric acid, HNO3, is manufactured by the Ostwald process, in which nitrogen dioxide, NO2, reacts with water. 3NO2(g) + H2O(l)  2HNO3(aq) + NO(g)

Ans.

How many grams of nitrogen dioxide are required in this reaction to produced 6.3 g HNO3 ? 6.9g NO2

Ex.

How many grams of Fe2 O3 is formed by heating 18 gm FeO with Oxygen. 4FeO + O2  2Fe2 O3

Ans.

20. gm

Ex. Ans.

How many litre O2 at N.T.P. is required for complete combustion of 1 mole C5 H10. 168 lt.

Ex.

Calculate the weight of residue obtained when CaCO3 is strongly heated and 5.6 litre CO2 is produced at N.T.P. 14 gm

Ans. Ex.

When sodium bicarbonate is heated 1.806 x 1024 molecules of water is obtained. Then find the volume of CO2(g) obtained at STP and amount of NaHCO3 needed for this reaction.

Sol.

2NaHCO3  Na2CO3 + H2O + CO2 So volume of CO2 = 3 × 22.4 = 67.2 Lt. Mass of NaHCO3 needed = 6 × 84 = 504 gm.

Problem 1.3 Solution :

(NCERT, Page 18) Calculate the amount of water (g) produced by the combustion of 16 g of methane. The balanced equation for combustion of methane is : CH4 (g) + 202 (g)  CO2 (g) + 2H2O (g) (i) 16 g of CH4 corresponds to one mole. (ii) From the above equation, 1 mol of CH4 (g) gives 2 mol of H2O (g) 2 mol of water (H2O) = 2 × (2 + 16) = 2 × 18 = 36 g

Page # 13

Problem 1.4 Solution:

(NCERT, Page 18) How many moles of methane are required to produce 22 g CO2 (g) after combustion? According to the chemical equation, CH4 (g) + 202 (g)  CO2 (g) + 2H2O (g) 44g CO2 (g) is obtained from 16 g CH4 (g). [  ] 1mol CO2 (g) is obtained from 1 mol of CH4 (g) mole of CO2 (g)

1molCO 2 (g) = 22g CO2 (g) × 44gCO (g) 2

= 0.5 mol CO2 (g) Hence, 0.5 mol CO2 (g) would be obtained from 0.5 mol CH4 (g) or 0.5 mol of CH4 (g) would be required to produce 22 g CO2 (g). Note :

NCERT Reading Text. 1.10 Page 17 Sheet Exercise–1 Part–I –15,16,17. Part–II – 22,23,24.

NCERT Exercise

DPP No.

Q.No. 1.3,1.4.

5

Exercise–2 Part–I – Part–II –16,17,18,26,27.

Exercise–3 Part–I – Part–II –

Page # 14

LECTURE # 7 PRINCIPLE OF ATOM CONSERVATION (POAC) : POAC is based on law of mass conservation if atoms are conserved, moles of atoms shall also be conserved hence mass of atoms is also conserved. This principle is fruitful for the students when they don’t get the idea of balanced chemical equation in the problem. This principle can be under stand by the following example. Consider the decomposition of KClO3 (s)  KCl (s) + O2 (g) (unbalanced chemical reaction) Apply the principle of atom conservation (POAC) for K atoms. Moles of K atoms in reactant = moles of K atoms in products or moles of K atoms in KClO3 = moles of K atoms in KCl. Now, since 1 molecule of KClO3 contains 1 atom of K or 1 mole of KClO3 contains 1 mole of K, similarly,1 mole of KCl contains 1 mole of K. Thus, moles of K atoms in KClO3 = 1 × moles of KClO3 and moles of K atoms in KCl = 1 × moles of KCl.  moles of KClO3 = moles of KCl or



wt. of KClO3 in g wt. of KCl in g = mol. wt. of KCl mol. wt. of KClO3

The above equation gives the mass-mass relationship between KClO3 and KCl which is important in stoichiometric calculations. Again, applying the principle of atom conservation for O atoms, moles of O in KClO3 = 3 × moles of KClO3 moles of O in O2 = 2 × moles of O2  3 × moles of KClO3 = 2 × moles of O2 or

wt. of KClO 3 vol. of O 2 at NTP 3 × mol. wt. of KClO = 2 × s tan dard molar vol. (22.4 lt.) 3



The above equations thus gives the mass-volume relationship of reactants and products.

Q.

Write POAC equation for all the atoms in the following reaction. (i) N2O + P4  P4O10 + N2 (ii) P4 + HNO3  H3PO4 + NO2 + H2O

Example : 27.6 g K2CO3 was treated by a series of reagents so as to convert all of its carbon to K2 Zn3 [Fe(CN)6]2. Calculate the weight of the product. [mol. wt. of K2CO3 = 138 and mol. wt. of K2Zn3 [Fe(CN)6]2 = 698] Sol. Here we have not knowledge about series of chemical reactions but we know about initial reactant and final product accordingly Several K2CO3    K2Zn3 [Fe(CN)6]2 Steps

Since C atoms are conserved, applying POAC for C atoms, moles of C in K2CO3 = moles of C in K2Zn3 [Fe(CN)6]2 1 × moles of K2CO3 = 12 × moles of K2Zn3 [Fe(CN)6]2 ( 1 mole of K2CO3 contains 1 moles of C)

wt. of K 2CO 3 wt. of the product = 12 × mol. wt. of K 2CO 3 mol. wt. of product wt. of K2Zn3 [Fe(CN)6]2 =

27.6 698 × = 11.6 g 138 12

Page # 15

Q.1

Ans. Note :

0.32 mole of LiAlH4 in ether solution was placed in a flask and 74 g (1 moles) of t-butyl alcohol was added. The product is LiAlHC12H27O3 . Find the weight of the product if lithium atoms are conserved. [Li = 7, Al = 27, H = 1, C = 12, O = 16] 81.28 g

NCERT Reading

NCERT Exercise

DPP No.



Q.No. 1.7,1.34,1.36.

6

Sheet Exercise–1 Part–I – 26,27. Part–II – 31 to 34.

Exercise–2 Part–I – Part–II – 1,2,5,6,11,13,15,20,22

Exercise–3 Part–I – Part–II –

Page # 16

LECTURE # 8 LIMITING REAGENT : The reactant which is consumed first and limits the amount of product formed into the reaction, and is therefore called limiting reagent. Limiting reagent is present in least stoichiometric amount and therefore controls amount of product. The remaining or leftout reactant is called the excess reagent. When you are dealing with balanced chemical equation then if number of moles of reactants are not in the ratio of stoichiometric coefficient of balanced chemical equation, then there should be one reactant which should be limiting reactant. Example : Three mole of Na2 CO3 is reacted with 6 moles of HCl solution. Find the volume of CO2 gas produced at STP. The reaction is Na2 CO3 + 2HCl  2 NaCl + CO2 + H2O Sol.



From the reaction :

Na2 CO3 + 2HCl  2 NaCl + CO2 + H2O

given moles given mole ratio Stoichiometric coefficient ratio

3 mol 1 : 1 :

6 mol 2 2

See here given moles of reactant are in stoichiometric coefficient ratio therefore none reactant left over. Now use Mole-mole analysis to calculate volume of CO2 prdouced at STP

Moles of Na 2 CO3 Mole of CO 2 Pr oduced = 1 1 Moles of CO2 produced = 3 volume of CO2 produced at STP = 3 × 22.4 L = 67.2 L Example : 6 moles of Na2 CO3 is reacted with 4 moles of HCl solution. Find the volume of CO2 gas produced at STP. The reaction is Na2 CO3 + 2HCl  2 NaCl + CO2 + H2O Sol.



From the reaction :

Na2 CO3 + 2HCl  2 NaCl + CO2 + H2O

given mole of reactant give molar ratio Stoichiometric coefficient ratio

6 3 1

: : :

4 2 2

See here given number of moles of reactants are not in stoichiometric coefficient ratio. Therefore there should be one reactant which consumed first and becomes limiting reagent. But the question is how to find which reactant is limiting, it is not very difficult you can easily find it according to the following method.

HOW TO FIND LIMITING REAGENT : Step : Divide the given moles of reactant by the respective stoichiometric coefficient of that reactant. Step :  See for which reactant this division come out to be minimum. The reactant having minimum value is limiting reagent for you.

Page # 17

Step :  Now once you find limiting reagent then your focus should be on limiting reagent From Step  &  Na2 CO3 HCl

6 =6 1  HCl is limiting reagent From Step  From  

4 = 2 (division is minimum) 2

Moles of CO 2 produced Mole of HCl = 2 1 mole of CO2 produced = 2 moles volume of CO2 produced at S.T.P. = 2 × 22.4 = 44.8 lt.

Examples on limiting reagent : Ex. The reaction 2C + O2  2CO is carried out by taking 24g of carbon and 96g O2, find out : (a) Which reactant is left in excess ? (b) How much of it is left ? (c) How many mole of CO are formed ? Ex. For the reaction 2P + Q  R, 8 mol of P and 5 mol of Q will produce (A) 8 mol of R (B) 5 mol of R (C*) 4 mol of R (D) 13 mol of R Ex.

Ans. Sol.

X + Y  X3 Y4 Above reaction is carried out by taking 6 moles each of X and Y respectively then (A) X is the limiting reagent (B) 1.5 moles of X3 Y4 is formed (C) 1.5 moles of excess reagent is left behind (D) 75% of excess reagent reacted X + Y  X3 Y4 B, C, D 3X + 4Y  X3 Y4 6 mole 6 mole 6 – 4.5 0 1.5 mole left

Ex.

Ans. Sol.

1.5 mole formed

A + B  A3B2 (unbalanced) A3B2 + C  A3B2C2 (unbalanced) Above two reactions are carried out by taking 3 moles each of A and B and one mole of C. Then (A) 1 mole of A3B2C2 is formed (B*) 1 2 mole of A3B2C2 is formed (D*) 1 2 mole of A3B2 is left finally

(C*) 1 mole of A3B2 is formed B, C, D 3A + 2B  A3 B2 3 mole

3 mole 1 mole formed A3B2 + 2C  A3 B2 C2 1 mole 1 mole 0.5 mole 0 0.5 mole Ex.

CS2 and Cl2 in the weight ratio 1 : 2 are allowed to react according to equation, find the fraction of excess reagent left behind. CS2 + 3Cl2  CCl4 + S2Cl2 mole

w 76

2w 71

;

w 76

2w 71  3



remaining =

L.R. = Cl2 . Moles of CS2 =

2w 71  3

w 2w  76 213 x 100 fraction of Cl2 left = w 76

w 2w – 76 213

= 28.6%.

Page # 18

Ex. Ans. Ex.

Ans. Ex.

Ans.

27 gm Al is heated with 49 ml of H2SO4 (sp. gr = 2) produces H2 gas. Calculate the volume of H2 gas at N.T.P. and % of Al reacted with H2SO4 Vol of H2 = 22.4 litre Al reacted = 66.66% Equal weights of carbon and oxygen are heated in a closed vessel producing CO and CO2 in a 1 : 1 mol ratio. Find which component is limiting and which component left and what is its percentage out of total weight taken. Limiting reagent is O2, carbon will left behind, 50% Three moles of Phosphorus is reacted with 2 moles of iodine to form P3 according to the reaction P + 2  P3 P3 formed in the above reaction is further reacted with 27 g of water. According to the reaction. P3 + H2O  H3PO3 + H HI formed in the above reaction is collected in the gaseous form. At higher temperature HI dissociated 50% then find the molecules of H2 gas liberated H  H2 + 2 3/8 NA.

Problem 1.5 (NCERT Page - 19) 50.0 kg of N2 (g) and 10.0 kg of H2 (g) are mixed to produce NH3 (g) Calculate the NH3 (g) formed. Identify the limiting reagent in the production of NH3 in this situation. Take reaction N2 + 3H2  2NH3. Sol. A balanced equation for the above reaction is written as follows : Calculation of moles : N2 (g) + 3H2 (g)  2NH3 (g)

N2 

moles of N2

= 50.0 kg

moles of H2

= 10.00 kg H2 ×

1000 g N2 1 mol H2 1kg N2 × 28.0 g N2

= 17.86 × 102 mol

1000 g H2 1 mol H2 × 1kg H2 2.016 g H2

= 4.96 × 103 mol

According to the above equation, 1 mol N2 (g) requires 3 mol H2 (g), for the reaction, Hence, for 17.86 × 102 mol of N2, the moles of H2 (g) required would be

3 mol H2 (g) 17.86 × 102 mol N2 × 1molN (g) 2

= 5.36 × 103 mol H2

But we have only 4.96×103 mol H2. Hence, dihydrogen is the limiting reagent in this case. So NH3 (g) would be formed only from that amount of available digydrogen i.e., 4.96 × 103 mol Since 3 mol H2 (g) gives 2 mol NH3 (g) 

2 mol NH3 (g) 4.96 × 103 mol H2 (g) × 3 mol H (g) 2

= 3.30 × 103 mol NH3 (g) is obtained. If they are to be converted to grams, it is done as follows :

17.0gNH3 (g) 3.30 × 103 mol NH3 (g) × 1molNH (g) . 3 = 56.1 × 103 g NH3

= 3.30 × 103 × 17 g NH3 (g)

= 56.1 kg NH3

Note :

NCERT Reading Text. 1.10.1. Page 18 Sheet Exercise–1 Part–I – 18 to 25. Part–II – 25 to 30.

NCERT Exercise

DPP No.

Q.No. 1.4,1.23,1.24,1.26.

7

Exercise–2 Part–I – 7. Part–II – 16,23,25,26.

Exercise–3 Part–I – Part–II –

Page # 19

LECTURE # 9 SOLUTIONS : A mixture of two or more substances can be a solution. We can also say that a solution is a homogeneous mixture of two or more substances ‘Homogeneous’ means ‘uniform throughout’. Thus a homogeneous mixture, i.e., a solution, will have uniform composition throughout.

CONCENTRATION TERMS : The following concentration terms are used to express the concentration of a solution. These are : 1. strength of solution 2. Molarity (M) 3. Molality (m) 4. Mole fraction (x) 5. % calculation 6. Normality (N) 7. ppm



Remember that all of these concentration terms are related to one another. By knowing one concentration term you can also find the other concentration terms. Let us discuss all of them one by one.

1.

STRENGTH OF SOLUTION : The concentration of solution in gram/litre is said to be strength of solution.

(a)

A 65% solution has the following meanings 65% by weight i.e. 100 gm solution contain 65 gm solute 65% by volume i.e. 100 ml of solution contain 65 ml solute 65% by strength i.e. 100 ml of solution contain 65 gm solute If, anything is not specified, 65% generally mean 65% by mass

(b)

For concentrated acids, like 98% H2SO4, 65% HNO3 etc, if anything is not specified than percentage by mass/volume is usually considered.

(c)

For the calculation of strength (% w/w, %w/v etc) the solute must be completely dissolved into the solution, otherwise, the given terminologies will be invalid. For example, the specific gravity of gold = 19.3 gm/cm3, if 193 x 100 = 16.17 is appears to be correct, 1000  193 but gold is not dissolvable in water, its % w/w in water cannot be calculated.

we add 193 gm gol powder in 1 litre of water, its % w/w =

2.

MOLARITY (M) : The number of moles of a solute dissolved in 1 L (1000 ml) of the solution is known as the molarity of the solution. number of moles volume of solution in litre Let a solution is prepared by dissolving w gm of solute of mol.wt. M in V ml water.

i.e., Molarity of solution =



Number of moles of solute dissolved =



V ml water have



1000 ml water have M  V ml



Molarity (M) = (Mol. wt of solute)  V ml

w M

w mole of solute M w  1000

w  1000

Page # 20

Problem 1.7 (NCERT Page - 20) Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g in enough water to form 250 mL of the solution. Solution Since molarity (M) =

No. of moles of solute Volume of solution in litr es

=

Mass of NaOH/Molar mass of NaOH 0.250 L

4 g / 40 g 0.1mol  = 0.4 mol L–1 = 0.4 M 0.250L 0.250 L Note that molarity of a solution depends upon temperature because volume of a solution is temperature dependent. Some other relations may also useful.

=

Number of millimoles

=

mass of solute  1000 = (Molarity of solution × V ) inml (Mol. wt. of solute)

Molarity is an unit that depends upon temperature .it varies inversely with temperature . mathematically : molarity decreases as temperature increases. Molarity 



Molarity of solution may also given as :

(i)

(ii)

Number of millimole of solute Total volume of solution in ml If a particulars solution having volume V1 and molarity = M1 is diluted to V2 mL then M1V1 = M2V2 M2 : Resultant molarity

If a solution having volume V1 and molarity M1 is mixed with another solution of same solute having volume V2 mL & molarity M2 then M1V1 + M2V2 = MR (V1 + V2) MR = Resultant molarity =

Ex. Sol.

Ex.

149 gm Moles of KCl = 74 .5 gm = 2

2 = 0.2 M 10 117 gm NaCl is dissolved in 500 ml aqueous solution. Find the molarity of the solution. 0.4 M



Ex.

M1V1  M2 V2 V1  V2

149 gm of potassium chloride (KCl) is dissolved in 10 Lt of an aqueous solution. Determine the molarity of the solution (K = 39, Cl = 35.5) Molecular mass of KCl = 39 + 35.5 = 74.5 gm 

Q. Ans.

1 1  temperature volume

Molarity of the solution =

Calculate molarity of the following : (a) 0.74 g of Ca(OH)2 in 5 mL of solution (b) 3.65 g of HCl in 200 ml of solution (c) 1/10 mole of H2SO4 in 500 mL of solution

[ 2M ] [0.5M ] [0.2 M ]

Calculate the resultant molarity of following : (a) 200 ml 1M HCl + 300 ml water (b) 1500 ml 1M HCl + 18.25 g HCl (c) 200 ml 1M HCl + 100 ml 0.5 M H2SO4 (d) 200 ml 1M HCl + 100 ml 0.5 M HCl

Page # 21

Ex. Sol.

Calculate the molarity of H+ ion in the resulting solution when 200 ml 1M HCl is mixed with 200 ml 1M H2SO4 For HCl M1 = 1 M V1 = 200 mL For

H2SO4 M2 = 1 V2 = 200 mL

nHCl = MHCl × VHCl = 1 × 0.2 HCl  H+ + Cl–

nH = 0.2 (from HCl) nH2SO 4  MH2SO4  VH2SO4 = 1 × 0.2 = 0.2 mole H2SO4  2H+ + SO4–2

nH = 0.2 × 2

(from H2SO4)

Total H+ = nH (from HCl) + nH (from H2SO4) = 0.2 + 0.4 = 0.6 Total volume = 200 + 200 = 400 mL = 0.4 L MR = Resultant molarity = Ex.

nH Vsolution



0 .6 = 1.5 Ans. 0 .4

What are the final concentration of all the ion when following are mixed 50 ml of 0.12 M Fe(NO3)3 + 100 ml of 0.1 M FeCl3 + 100 ml of 0.26 M Mg(NO3)2

[NO3–] =

50  0.12  3  100  0.26  2 250 18  52 70  = 0.28 250 250 [Cl–] = 0.12 M [Mg++] = 0.104 M [Fe3+] = 0.064 M

=

Ex.

Calculate the molarity of water H2O  18 gm = 1 mole Volume of water = 1 Litre Mass = 1000 gm mole =

1000 18

1000 = 55.55 M 18 Find the minimum volume of 0.2 M HCl solution for the complete neutralisation of 0.4 M, 40 ml of NaOH solution.

Molarity of water = Ex. Ex.

CaCO3 reacts with aq. HCl to give CaCl2 and CO2 according to reaction CaCO3(s) + 2HCl(aq)  CaCl2 + CO2 + H2O How much mass of CaCO3 is required to react completly with 100 ml of 0.5 m HCl

Page # 22

Sol.

millimole of HCl 100 × 0.5 = 50 2 mole of HCl reacts  1 moles CaCO3

1 1 mole of HCl reacts  2 1 50 mmole of HCl reacts  × 50 = 25 mmole of CaCO3 2 mole of CaCO3 = mass of CaCO3 = Ex.

Sol.

25 1000 25 × 100 = 2.5 gm. 1000

Na2CO3 + 2HCl  2NaCl + CO2 + H2O (a) moles of NaCl formed when 10.6 gm of Na2CO3 is mixed with 100 ml of 0.5 M HCl solution. (b) Calculate the concentration of each ion in the solution after the reaction (c) Volume of CO2 liberated at S.T.P. molar mass of Na2CO3 = 106 gm 10.6 × 1000 = 100 m mole 106 HCl = 100 × 0.5 = 50 milli mole

Na2CO3 =

limiting reagent  HCl  2 HCl reacts = 2 NaCl 50 mmole HCl reacts = 50 mmole of Na2CO3 remaining Na2CO3 = 100 – 25 = 75 mmole (a)

moles of NaCl =

(c)

volume of CO2 =

[Na+] = [Cl–] =

50 = 0.05 1000

25  22.4 = 0.556 L 1000

50  150 200  =2M 100 100 50 = 0.5 M 100

(CO32–] =

75 = 0.75 M 100

Note :

NCERT Reading Text. 1.10.2. Page 19 Sheet Exercise–1 Part–I – 28 to 41 Part–II – 35 to 41.

NCERT Exercise

DPP No.



8

Exercise–2 Part–I – Part–II – 19.

Exercise–3 Part–I – Part–II –

Page # 23

LECTURE # 10 3.

MOLALITY (M) : The number of moles of solute dissolved in1000 gm (1 kg) of a solvent is known as the molality of the solution. i.e., molality = number of moles of solute  1000 mass of solvent in gram Let y gm of a solute is dissolved in x gm of a solvent. The molecular mass of the solute is m. Then Y/m mole of the solute are dissolved in x gm of the solvent. Hence Y  1000 Molality = m x

Ex. Sol.

225 gm of an aqueous solution contains 5 gm of urea. What is the concentration of the solution in terms of molality. (Mol. wt. of urea = 60) Mass of urea = 5 gm Molecular mass of urea = 60 5 = 0.083 60 Mass of solvent = (255 – 5) = 250 gm

Number of moles of urea =

Number of moles of solute  1000 = 0.083  1000 = 0.332 Mass of solvent in gram 250 Note : molality is independent of temperature changes. 

Molality of the solution =

Problem 1.8 (NCERT Page - 21) The density of 3 M solution of NaCl is 1.25 g mL–1. Calculate molality of the solution. Sol. M = 3 mol L–1 Mass of NaCl in 1 L solution = 3 × 58.5 = 175.5 g Mass of 1L solution = 1000 × 1.25 = 1250 g (since density = 1.25 mL–1) Mass of water in solution = 1250 – 175.5 = 1074.5 g

No. of moles of solute Molaity = Mass of solvent in kg

3 mol = 1.0745 kg

= 2.79 m

Often in a chemistry laboratory, a solution of a desired concentration is prepared by diluting a solution of known higher concentration. The solution of higher concentration is also known as stock solution. Note that molality of a solution does not change with temperature since mass remains unaffected with temperature. Q. Ans.

518 gm of an aqueous solution contains 18 gm of glucose (mol.wt. = 180). What is the molality of the solution. 0.2 m

Q.

Molality of an NaOH solution is 4. Find the wt. of solution if solvent is 500 g.

4.

MOLE FRACTION (x) :



The ratio of number of moles of the solute or solvent present in the solution and the total number of moles present in the solution is known as the mole fraction of substances concerned. Let number of moles of solute in solution = n Number of moles of solvent in solution = N n  Mole fraction of solution (x1) = nN N  Mole fraction of solvent (x2) = nN also x1 + x2 = 1 Note : mole fraction is a pure number its also independent of temperature changes.

Page # 24

5.

% CALCULATION : The concentration of a solution may also expressed in terms of percentage in the following way. (i) % weight by weight (w/w) : It is given as mass of solute present in per 100 gm of solution. i.e.

mass of solute in gm % w/w = mass of solution in gm  100

(ii) % weight by volume (w/v) : It is given as mass of solute present in per 100 ml of solution. i.e.,

% w/v =

mass of solute in gm  100 mass of solution in ml

(iii) % volume by volume (V/V) : It is given as volume of solute present in per 100 ml solution. i.e.,

% V/V =

Volume of solute in ml  100 Volume of solution

Example 0.5 g of a substance is dissolved in 25 g of a solvent. Calculate the percentage amount of the substance in the solution. Solution. Mass of substance = 0.5 g Mass of solvent = 25 g 

percentage of the substance (w/w) =

0 .5  100 = 1.96 0.5  25

Problem 1.6 (NCERT Page - 19) A solution is prepared by adding 2 g of a substance A to 18 g of water. Calculate the mass per cent of the solute. Solution Mass per cent of A =



Mass of A × 100 Mass of solution

=

2g 2gofA  18gofwater × 100

2g  100 = 10 % 20g

Example 20 cm3 of an alcohol is dissolved in80 cm3 of water. Calculate the percentage of alcohol in solution. Solution Volume of alcohol = 20 cm3 Volume of water = 80 cm3 

percentage of alcohol =

20  100 = 20. 20  80

Interconversion of Concentration units : Representing solvent and solute in a binary solution by subscripts 1 and 2 respectively, the various conversion expressions are as follows : DERIVE THE FOLLOWING CONVERSION : 1. Sol.

x 2   1000 Mole fraction of solute into molarity of solution M = x M  M x 1 1 2 2 Mole fraction into molarity M mole fraction of solvent and solute are X1 and X2 so X1 + X2 = 1 supposs total mole of solution is = 1 mole of solute and solute and solvent is X2, X1 weight of solute = X2M2 , weight of solvent = X1M1 total wt of solution = X1M1 + X2M2

Page # 25

volume of solution =

volume in L =

X1M1  X 2M2 ml 

X1N1  X 2N2   1000

X 2    1000 molarity (M) = X N  X N 1 1 2 2 MM1  1000   1000  MM2

2.

Molarity into mole fraction x2 =

Sol.

Molarity into mole fraction molarity (M) = moles solute in 1000 ml of solution so moles of solution = M mass of solution =  x 1000 wt. of solute = MM2 wt. of solvent = 1000  – MM2 moles of solvent =

1000  MM2 M1 x 2  1000 x1M1

3.

Mole fraction into molality m =

Sol.

Mole fraction into molarity mole fraction of solute X2 and solvent X1 mole is n2 & n1

 n2 x 2  n2 molality = n M x 1000  n  x  1 1 1  1 x2 = x M x 1000 1 1 4. Sol.

mM1 Molality into mole fraction x2 = 1000  mM 1 Moalrity into mole fraction molality = moles of solute in 1000 gm of solute = m

1000 mole of solvent = M 1 mM m mole fraction X2 = 1000 = 1000  mM 1 m M1

5. Sol.

m   1000 Molality into molarity M = 1000  mM 2 Molality in molarity molality = moles of solute in 1000 gm of solvent mole of solute = m wt. of solute = mM2 wt. of solution = 1000 + mM2 volume of solution =

1000  mM2  Page # 26

volume in (L) =

1000  mM2   1000

m    1000 molarity = 1000  mM 2 6.

Sol.

M  1000 Molarity into Molality m = 1000   MM 2 M1 and M2 are molar masses of solvent and solute.  is density of solution (gm/mL) M = Molarity (mole/lit.), m = Molality (mole/kg), x1 = Mole fraction of solvent, x2 = Mole fraction of solute Molarity (M) into molality (m) molarity = mole of solute in 1000 ml of solution moles of solute = M wt. of solute = MM2 wt. of solution = 1000  mass of solvent = 1000  – MM2 molality =

moles of solute wt of solvent x 1000

M  1000 m = 1000  MM 2 Note :

NCERT Reading

NCERT Exercise

DPP No.



Q.No. 1.5,1.6,1.11,1.12,1.25,1.29,1.35.

9

Sheet Exercise–1 Part–I – 42 to 50. Part–II – 42 to 45.

Exercise–2 Part–I – Part–II – 21,24,30,31,32.

Exercise–3 Part–I – Part–II –

Page # 27

LECTURE # 11 Mix Questions : 1.

Sol.

Density for 2 M CH3COOH solution is (1.2 g/mL) Calculate (1) Molality (2) mole fraction of CH3COOH acid (3) % (w/w) for the solution (4) % (w/v) for the solution Let volume & solution = 1 Litre

n solute Molarity = v =2 solution n solute = 2 mole (CH3COOH)

mass of solute molar mass of solute = 2 mass of solute = 2 × 60 = 120 g. mass of solution = volume of solution × density of solution = 1200 g = mass of solute + mass of solvent  mass of solvent = 1200 – 120 = 1080 g = 1.08 kg (H2O) moles of solvent =

1080 = 60 mole 18

(H2O) (1) Molality = w

n solute



solvent (kg )

2 1.08 = 1.85 m Ans.

nCH3CCOH 2  (2) X CH3COOH = n 2  60 = 0.0322 CH3 CCOH  nH2O

wt. of solute (gm) 120 (3) % w/w = wt. of solution (gm) × 100 = ×100 = 10% 1200 wt. of solute (gm) 120 (4) % w/v = volume. of solution (mL ) × 100 = ×100 = 12% 1000 2.

Sol.

Mole fraction for aqueous glucose soltuion is 0.1 specific gravity is 1.1 Find (1) Molarity (2) Molality (3) % (w/w) for the solution (4) % (w/v) for the solution

nGlu cos e 1  XGlucose = 0.1 = n 10 Glu cos e  nH2O Let 1 mole of glucose is present is solution n Glucose = 1 m Glucose = 1 × 180 = 180 gm

nH2O + nGlucose = So

n Glu cos e 1 = 10 0.1 0.1

nH2 O = 9 mass of H2 O = 9. MH2O

MH2O = 9 × 10 = 162 gm = 0.162 kg mass of solution = mass of solute + mass of solvent = mas of Glucose + mass of H2O = 180 + 162 = 342 gm Page # 28

mass of solution 342 g volume of solution = Density of solution = 1.1 g / mL = 311 mL = 0.311 Litre.

{Sp. gravity = 1.1Density = 1.1 g/mL}

(1)

n Glu cos e 1  molarity = v 0.311 = 3.215 M solution

(2)

nGlu cos e 1  molality = w 0.162 = 6.17 M H2O (kg)

(3)

wt. of solute(gm) 180 % (w/w) = wt. of solution(gm )  100  342 × 100 = 52.63%

(4) 3.

Sol.

wt. of solute(gm) 180 % (w/v) = volume. of solution(mL )  100  311 × 100 = 57.87%

10% w/w urea solution is 1.2 g/mL calculate (1) % w/v (2) molarity (3) molality (4) mole fraction of urea % w/w = 10% It means 10 g urea is present in 100 gm of solution Let 100 gm of solution is taken msolution = 100 gm

m solution 100 gm vsolution = Density  1.1g / mL = 91 mL = 0.91 L Murea = 10 gm nurea =

Mass of urea 10 1   Murea 60 6

msolution = msolute + msolvent= 100 msolvent =

n H2O 90  = 5 mole MM 18

(1)

wt. of urea 10  100 = 10.99% % (w/v) = volume. of solution(mL ) × 100 = 91

(2)

nurea 1/ 6  molarity = v 0 .91 = 1.83 M solution

(3)

nurea 1/ 6  molarity = w 0.9 = 1.85 M solution

(4)

mole fraction of urea

nurea 1/ 6 1/ 6  Xurea = n =  n urea H2 O 1/ 6  5 31/ 6 = 1/31 Ans. Q.

Ans.

Molality of HNO3 soltuion is 2 (spf gr = 1.50) then find : (a) Molarity of the solution (b) Mole fraction of the solute (a) 2.66 M. (b) 0.0347.

Page # 29

4.

AVERAGE/ MEAN ATOMIC MASS : The weighted average of the isotopic masses of the element’s naturally occuring isotopes.

Mathematically, average atomic mass of X (Ax) =

a1x 1  a 2 x 2  .....  an x n 100

Ex.

Naturally occuring chlorine is 75% Cl35 which has an atomic mass of 35 amu and 25% Cl37 which has a mass of 37 amu. Calculate the average atomic mass of chlorine (A) 35.5 amu (B) 36.5 amu (C) 71 amu (D) 72 amu

Sol.

(A) Average atomic mass =

=

% of  isotope x its atoms mass  % of I isotope x its atomic mass 100 75 x 35  25 x 37 100

= 35.5 amu Note : (a) In all calculations we use this mass. (b) In periodic table we report this mass only.

6.

MEAN MOLAR MASS OR MOLECULAR MASS: The average molar mass of the different substance present in the container =

Ex.

n1M1  n 2M2  ......nnMn n1  n 2  ....nn

The molar composition of polluted air is as follows : Gas At. wt. mole percentage composition Oxygen 16 16% Nitrogen 14 80% Carbon dioxide 03% Sulphurdioxide 01% What is the average molecular weight of the given polluted air ? (Given, atomic weights of C and S are 12 and 32 respectively. jn

n M j

Sol.

Mavg =

j

j1 jn

n

j

j1

j n

Here

n

j

= 100

j1

 Mavg =

16 x 32  80 x 28  44 x 3  64 x 1 2948 512  2240  132  64 = = = 29.48 Ans. 100 100 100

Note :

NCERT Reading

NCERT Exercise

DPP No.



Q.No. 1.9,1.32.

10

Sheet Exercise–1 Part–I – 4,5,10. Part–II – 34.

Exercise–2 Part–I – Part–II – 12.

Exercise–3 Part–I – Part–II –

Page # 30

CHEMISTRY LECTURE NOTES

COURSE - VIKAAS (A) (LECTURE No. 1 TO 11)

TOPIC : Mole Concept-1

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