Mole Concept and Stoichiometry (arihant)

CHAPTER 1

... le Concept and oichiometry Chapter Contents Concept of atoms and molecules, Dalton's atomic theory; �lolc concept; Chemical formulae; Chemical equations; Calcul.llions (based on mole concept) involving common oxidation, reduction, neutralization and displacement reactions; Concenlmt1on in terms of mole fraction, molarity, molality and nol'lnali t) and Various le'els ,

of multiple-choice questions.

DALTON'S ATOMIC THEORY

•

An atom is the smallest particle of an element which is neutral in nature, retains all the proper­ ties of the element and takes part in a chemical reaction. The word atom was introduced by Dalton

and properties while atoms of different element differ in these aspects. •

Modern view about the atom According •

Main Assumptions • •

Maner (of any type) is composed of atoms

.

An atom is the smallest, fundamental, undivided particle.

• An

atom can neither be created nor destroyed.

Atoms combine in whole number ratios to form a molecule. therefore. a molecule is the smallest

identity that exists individually.

(alamos means undivided).

The Dalton's atomic theory was proposed by Dalton on the basis of laws of chemical combi­ nation.

Atoms of an element have similar size, energy

•

to

the modem view:

An atom is divisible into other smaller par­ ticles which are known as subatomic particles. It can also combine in non-whole number ratio as in the case of non-stoichiometric compounds (Berthollide compounds) like Fe09p. Atoms of same element also difef r in mass and mass related propenies as in the case of isotopes.

1.2

•

Mole Concep t and Stoichiomet1ry

Molecule The term molecule was inr:rocluced by Avo­ gadro. It is the smallest panicle (identity) of matter

that can exist independently and retains

all che prop­

erries of the substance. Normally, the diameter of the

molecules is in the range of4-20

A and the molecular

2-1000. ln case ofmacromolecules, the diameter is in the range of50-250 A and che molecular

mass is between

weight may be in Wchs.

Berzelius Hypothesis According ro the Berzelius hypothesis, equal volumes ofall the gases concain same number of aroms under the similar conditions of tem­

per.icure and pressure." This hypothesis on application to

law ofcombining volume confirms chat atoms are divis­ ible which

is

in contrary ro Dalton's theory.

AVOGADRO'S LAW • Avogadro's

law

volumes.

explains

law

of

combining

• According to chis law, "Under similar coodicions of remperarure and pressure, equal volume of gases con­ tain equal number ofmolecules."

;_ •

Ir means LO ml of H2, 02, N2 or a mixture of gases lmve same number of molecules .

Ir is

usw in:

jM,. 2

x

x

x

1

cm� of a gas ac STP

is equal to Loschmidr number, thar is, 2.68

• Reciprocal

Avogram.

of Avogadro number

Gram molecular mass

Volume ofsubstance in litre Mole = 22.4 lirre Volume ofone mole ofany gas is

equal co 22.4 litres of

dmJ ac STP. le is known as molar volume. Mole =

Number ofidencities Avogadro's number

Mole=

RY RT

Here P= Pressure in atmosphere V .. Volume in litre T =Temperature in Kelvin R = Un.iversal gas cons•anc

Relationship ofMole:

x

1ou acorns ofnitrogen

To Find Total Number of Identities

102;.

• Avogadro number of gas molecules occupy 22.4 litre or 22400 ml or cm' volume at STP. The number ofmolecules in

•

• 22.4 liue ofN2 ac STP.

v.o.I

(iv) Deriving che gram molecular volume

•

Here G.m.m.

• 28 gm ofnitrogen

vapour deosicy

• Avogadro number (N0 or N11). 6.023

Wt. of substance in gm. Molar mass ofsubstance (G.m.m)

• 2 x 6.023

(iii) Deriving a relacion

2

w

Mole .. M

• 6.023 x 10" molecules ofN2

(ii) Dececmining aromicicy ofa gas =

related to the mass ofsubstance, the volwne of

gaseous substance and clY number ofparticles

A mole ofany substance (like Ni> scands for:

(i) Deriving molecular formula ofa gas

molecular mass

• Mole is

x 1019

is known

as

• Tocal number ofMolecule

•

mole(n)

x

N,.

• Total number of Atoms "' mole (n) x N11 aroms p�nc in one molecule

x

• Total number ofElectrons mole (n) x N11 ofelectron present in one electron •

• Toca! charge on any ion one ion )( l.6 )( l o-•9c

=

No. of x

No.

mole (n) x N11 x charge on

MOLE •

Mole

is

a

unit which represents

6.023

x

1013 par­

ticles, aroms, molecules or ions etc., irrespective of

their nature.

Illustrations l. If a piece ofcopper weights atoms does ic concafo?

0.635 g, how many

Mole Concept and Stoichiometry 6.023 x 1ou Rs I 06 x 60 x 60 x 24 x 365 Rs/year

Solutlon Number ofmoles ofCu in 0.635 g 0.635 g =

= 1o-2 mol

63.5 g mot·•

=

As I mole Cu contains 6.023 x 1on atoms ofCu

6.023 x I 021 atoms ofCu

2. Oxygen is present in a one litre flask at a pres­

·

S. Calculate the number ofatoms ofoxygen pres­

ent in 88 g ofC02• What would be the mass of CO having the same number ofoxygen atoms?

Solution Number of moles ofC02 = 44

Solution PV = nRT

=

CO

112 g

Solution Mass of diamond (C) = 0.5 x 200 mg = I 00 mg = I 00 x 1O-l g Number of mole ofC

So 1.6 g CH4 contains 6.02 x I 022 molecules ofCH, As one molecule of CH 4 contains (6 + 4)

=

10

6.02 x 1022 molecules of CH, contain

10 x 6.02 )( 1022•

6.02 x 1ou electrons.

4. How many years would it take to spend Avo­ gadro number of rupees at the rate of I 0 lac

rupees per second?

As Avogadro number= 6.023 x Jon =-

6.023 x Ion Rs

=

I 0 lac rupees,'s ""' I 06 Rs/s

Number of years to spend all the rupees

=

0.1 g

0.1 g 12 g mol·' .

=

contains 6.02 x 10" molecules ofCH,

Rate ofspending

=

200 mg)

Solution Molecular mass ofmethane"" J 6 g mot"'

So total rupees

4 moles oxygen atoms arc present in 4 moles

ring contains 0.5 carat diamond? (1 carat =

in 1.6 gram of methane.

Solution

so

given to his bride-to-be if the engagement

2.68 x 1010

so

2.5092 x 102'

6. How many atoms of carbon has a young man

(6.02 x I on mol-1) (I 0-12/22.41 mol)

electrons,

x l OU

As I mole oxygen atom is present in I mole of

ofCO. Its mass is 4 (12 + 16)

10-n mol 22.41

As 16 g CH,

= 4 >< 6.023 =

3. Calculate the total number of electrons present

=

As one mole of C02 contains two moles of oxy­

n x 0.0821 L atm K-1 moJ-1 x 273 K

Number ofmolecules =

-' .

gen atoms, so two moles ofco2 contain 4 moles of

Number ofoxygen atoms

7 6 x I Q-lll L atm

=

g mot

oxygen atoms.

7.6 x 10-10 atm x I L 760 =

88 g

.. 2 moles

sure of 7.6x I 0-10 mm of Hg. Calculate the

number oxygen molecules in the flask at O"C.

1.3

1.90988 x l 010 years

So I 0'2 mole Cu contains 6.023 x I 021 x 10'2 =

•

= 1/120 mole

io x 6.023

Number of C atoms =

1

=

x

I 023

5.02 x I 021

7. lf the Faraday were to be 60230 coulombs instead of 96500 coulombs, what will be the charge on an electron?

Solution

One mole electron carries 1 Faraday charge.

As 6.023 x I on electrons carry

.

So I electron cames

=

,..

60230 C

60230 C 6.023 x 1OU =

I x J0-19 C.

1.4 • Mole Concept and Stoichiometry

Mass can be expressed in terms of atoms or mol­

molecules.

of the

litres

'V'

Suppose

MASS ecules as follows:

Vapour density of gas

Atomic Mass

_Weight of 'n' molecules of the

It is the relative mass ofan atom which shows how many

times an atom is heavier than 1112 mass ofC-12. •

The atomic mass of any element expressed in grams is called G.A.M. (gram atomic mass). Atomic mass - E

><

V

V = Valency

t4eat .

Spe c1.ti1 c

m

Weight of I molecule of H1

-

REMEMBER Gram atomic \\eight ts atomic weight expressed

in grami; but ii hls a special significance with refc.:rencc to a mole.

·

=

Molecular mass represents the total mass of a mol­

ecule, that is, number of times a molecule is heavier than that of l 112 weight ofC- I 2. • It is non-variable.

• It is now called relative molecular mass.

Determination of Molar Mass It is possible by these methods

Vapour Density Method Vapour density of a gas is defined as the ratio of the weight of a certain volume of the gas to the

weight of the same volume of hydrogen at the same temperature and pressure.

Vapour density of a gas Weight of 'V' litres ofthe gas

Weight of·v· litres ofH1 at the same temperature and pressure

_ _ _ _

·

W >< 22400 Volume at STP (in ml)'

Here W = Weight of substance in gm. Molecular mass = 2

><

Vapour density

Graham Diffusion Method r/r2 = '1(MjM1) Here r1, r2

while M1, M1

tively.

Molecular Mass

•

e.. �ag le f:..:.t:::. h::... .: s= _m_o_l..: . :e :..: c .:_ o. :::... u.:. Weight of 1 atom of H,

_ _

2

·

It is known as Dulong Petit's law.

gas

g ht o f I e i"! >< W_,

VD ca1. .

'n'

Weight of'n' molecules ofH,

_Weight of I molecule of the gas

=

Here E = Equivalent weight

• Atomic mass=

-

contains

gas

(iii)

are rates of diffusion for two species are their molecular masses respec­

By Colligative Propenies Methods:

w

7tV= m RT Here

1t=

Osmotic pressure in attn.

= Volume in litre

V W

=

Weight in grams

R= Universal gas constant T

=

Given temperature

m "" Molar mass

Illustrations 8. The weight ofone litre sample of ozonized oxy­ gen at NTP was found to be 1.5 g. When l 00 ml of this mixture at NTP were treated with terpen­

tine oil, the volume was reduced to 90 ml. Hence

calculate the molecular weight ofozone.

Solution Volume absorl>cd by terpentine oil= IO ml As volume of ozone = I0 ml (as terpentine oil

absorbed 01)

Mole Concep t and Stoichiometry • 1.5 So Volume of 02

100 - 1 0

=

=

For example, for SO/-, E

90

M ot. wt. of ozonized oirygen l.5

WRT

= -py- "' =

x

0.082 l lxl

x

For PO�.i-, E

273

As volume or mole racio of02 and 0 is 900:

3

100

So mot. wt. of ozonized oxygen x

32

+

100

1000

x

=

E

Molecular weighr

=

Valency of cation or anion

For example, for CaCO,, E

x

AlCl3, E

E

48.2

Hence mol. we. of ozone

=

=

48.2

M/3

Molecular weight

=

Change in oxidation number

(a) [n acidic medium: E

KlS04 + 2Mnso.. + 3Hp + 5 [OJ

35.5 pare by weighr ofC4.

• Equivalent weight is a number and when ic is

5

• It depends upon che oarure of chemical reaction in

(b) ln basic medium: E

denoted in grams, it is calJed gram equivalenr. which substance cakes part.

.7

2KMn04

+

unit change in oxidation number.

(i) For Acids

Mlt

·6

2K2Mn04 �Hp

Procociry or Basicicy of Acid ·

For H2S04, E

=

=

M/2

(c) ln neuual medium: E

=

M/3

..

2KOH + 2Mn 02 + 3[0]

3 unir change in ox.idacion number Molecular weight

Acidity or number of OH- ions

For example, for Ca(OH)2, E For Al(OH) , E ,

=

M/3

=

M/2

(vi) For Acidic Salt E "'

Molecular weight Number of replaceable H-aroms

(iii) For Ions _

-

[OJ

MJ3

(ii) for Bases

E

+

one unit change in oxidation number

Molecular weight

For example, for HlO�, E

•

=

2KOH ......

Methods to Find Equivalent Weight

E

M/5

·2

part by weight of H2 or 8 part by weighr of 01 or

=

=

of an element or a

compound which combines with or displaces 1.008

E

M/2

For example, for KMo04

Equivalent Weight • Equivalent weight is the weight

=

(v) For Redox Reaccions

00 x 32 + 100 x x 33.62 -- 9 1000 x

M/3

(iv) For Compounds

33.62

900

=

M/2

=

Molecular weight Charge on ion

Ca( OH)2 +

H}PO•-+ CaHP04

E: M/2

+

2Hp

1.6 • Mole Concept and Stoichiometry (vii) Some other methods

Weight of metal

(a) Hydrogen displacement method E

=

0.199 - 0.04

=

0.159

=

w x 11200

. 0.159 x 8 E qu1val ent we1'ght of meta 1= 0.04

Volume of� at NTP

(b) Oxideformation method E

=

= Wt . of metal x 8 Wt. ofoxygen

Weight ofoxygen

=

weight ofmetal oxide

11. 4.215 g ofa metallic carbonate was heated in a hard glass tube and the co2 evolved was found to measure 1336 ml at 27C ° and 700 mm p ressure.

What is the equivalent weight ofthe metal?

- weight ofmet.al

Solution Metallic carbonate -+ metallic oxide+ C02

(c) Chloridefonnation method: E

=

Wt ofmetal x 35 5 Wt. ofchloride ·

Weight ofchloride

Wt.ofco2

= weight ofmetal

=

chloride - weight ofmetaJ (d) Double decomposition methods Eq. wt. ofsalt taken Eq. wt. ofsalt ppt.

_

-

�""

Wl ofsalt taken Wt. ofsalt ppt.

By law ofequivalent :

E+C03" E + 0,,

=i:�g

Hence, E (eq. wt. ofmetal)= 12.58

w2

Illustrations 9. Find the equivalent mass ofR:iP04 in the reaction: Ca(OH)2 + �P04 -Ca HP04 + 2Hp

Solution As in this reaction only two hydrogen

atoms are replaced

so

its equivalent

mass

will be

given by the following expression:

Chemical Formula A chemical folUlula represents the combination of

atoms ofall the elements which makes up a compound

It represents the reJative ratio ofatoms ofits coostitu­

eni elements. In case ofa compound. it represents one

molecule, one mole, one gram molecular weight of

the

compound

Example, S1 represents one molecule

ofphosphorous and one mole ofphosphorous .

Equivalent mass ofH,P04

Example, CuS04.5H10 represents one molecule one , mole and one gram molecular weight of hydrated

Molecular mass of�P04

copper sulphate.

2

10. On heating

224 00

Wt. ofmetallic oxide= 4.125- 2.2= 2.015 g

WI

98 = = 49 2

44 x 1120

=2.2g

(e) Metal displacement methods El

3 1.8

•

0. 199 g ofa metallic oxide in a current

ofhydrogen 0.045 g ofwater is formed. Find the

equivalent weight ofthe metal.

So 02 present in 0.045g H,O

""

=

I 8 g Hp

�

04 5 16 x · 1

=

Empirical Formula lt gives the simplest ratio ofthe number ofatoms of

Solution Weight ofMetallic oxide= 0.199 g As 16 g oxygen is present in

Emp irical and Molecular Fonnulas

004 . g

different elements present in one molecule ofa com­

pound It does not represent the actual number of

atoms of different elements present in one molecule

ofthe compound.

Calculation of the Empiric.al Formula: It

involves the following steps, one by one:

Mole Concept and Stoichiometry • 1.7 (i) First determine the percentage composition by weight of each element present in the compound. wt. ofC02 12 x1OO C% = 44x wt. of organic comp. wt of8i0 2 x100 H%=- x 18 18 wt. of org. comp.

Duma's method

Volume ofN2 atS.T.P N %= 28 x x 1OO 22400 wt. of Org. comp OR Vol ofN2 atS.T.P. N% = 8x Wt ofOrg comp. KJeldahl's Method

N% =

1.4 xNx V Wt. of org. compound

HereN ammonia

=

normality ofacid the used to neutralize

V = volume ofacid the used to neutralize ammonia (ii) Now the percentage ofeach element is divided by its atomic weight to get the relative number ofatoms of each element (iii) These relative numbers obtained are divided by smaHest number to get the simplest ratio numbers. (iv) If the simplest ratio number is not a whole number it should be multiplied by a suitable integer to get a whole number. (v) The ratio ofthese simple whole numbers gives· the empirical formula ofthe compound. Molecular Formula It shows the actual number of atoms ofdifferent ele­ ments present in one molecule of the compound. Molecular weight Empirical formula weight

•

n

•

Molecular formula = empirical formulax n.

=

•

Molecular weight -0f a substance can be deter­ mined by methods Like Vapour density method, Elevation in boiling point method, Depression in freezing point methods etc.

•

The molecular weight of volatile compounds is determined by Victor Mayer's method.

•

Molecular Weight = 2

•

Molecular weight=empirical formula wt x'n'.

•

For some compounds the molecular formula and empirical formula may be same also.

x

Vapour density

Illustrations 12. A certain organic compound was found to contain

33.6 % C, 5.6% H and 49.6 % chlorine. Find the empirical formula ofthe organic compound.

Solution

Element

%

Atomic ratio

33.8 33.8/12

c

=

Ratio

2.8 2.8/0.7 = 4

5.6 5.6/ 1 = 5.6

H

Simplest

5.6/0.7 = 8

Cl

49.6 49.6/35.5 =1.4 l.4/0.7 =2

0

0.7/0.7=1 Thus C4H1C�O is the empirical formula of the compound.

13. A well-known antibiotic penicillin, contains C

57.49 %, H 5.39 %, N 8.39 5, S 9.58 %; the remainder being oxygen. Given that penicillin contains one atom ofsuJphur per molecule, cal­ culate the molecular fonnula of the antibiotic.

Solution % c= 57.49

% H=5.39 %N=8.39 %S=9.58 % 0 = 100-(57.49 + 5.39 + 8.39 + 9.58) = 19.15 Empirical formula

1.8 • Mole Conceet and Stoichiometry Element

%

Relative no.

Simplest

of atoms

Ratio 16

57.49/12 - 4.79

c

57.49

H

5.39

5.39/1 = 5.39

8.39

8.39/14 = 0.599

s

9.58

9.58/32= 0.299

0

19.15

19.15116 = 1.196

18 2

Empirical fonnula is C16 H I

4 iso,

As the given antibiotic contains only one atom of sul­ phur per molecule. The calculated empirical fonnula C16H11N1SO,, contains only one atom ofsulphur. Hence the same is the molecular fonnula of the antibiotic.

% S=9.8 % Na= 7.0 %0= 14.7

Element

o/e

Relative no.

Simplest

of atoms

Ratio

c

51.4

51.4/12 = 4.28

14

H

4.3

4.3/1 - 4.3

14

N

12.8

12.8/14 ... 0.914

3

s

9.8

9.8/32= 0.306

Na

7.0

1n3 = 0.304

0

14.7

14.7/16= 0.918

3

14. A compound has molar mass of 147 g. It contains 49 % C and 2. 72 % H. On Cari us estimation. 2.561 mg of the compound gave 5 mg of AgCI. Find the empirical and molecular fonnula.

Solution o/o

Element

Simplest

Atomic ratio

Rado c

49

3

H

2.72 2.72/1 - 2.72

Cl

(35.51143.5) )( 5 2.581

49112 94.1

"'47.9

.(

47.9/35.5

2

Expression of Concentration of Solution The solution whose concentration is exactly known is referred as standard solution and such solutions are prepared in volumetric or standard flask. Concentration

It is the amount of solute present in one litre of solu­ tion. It is denoted by C or S.

100 _

C or S �

1.34

c.11lc1 is the empirical formuJa. n ... 147173.5 - 2

C=N> of aqueous NaOH containing

40.0 g of NaOH per litre. What is the basic­

ity

of the acid?

50000 ml

"' 50000

-

Nonnality ofacid

Normality of NaOH • Now, Mew. of acid

40 40

-

/:x1

82

x

I 000 • l 1000

Meq. of NaOH

•

39 n x 100 82

So volume of water to be added

JOO

•

I

x

95

So n .. 2;

• 49900 ml a

20

S .. N

As

Solution

5

x

N2

18. The formula weight of an acid is 82.0 1 0 0 cm>

16. Calculate che volume of water to be added ro a

•

0

•

So

Illustrations

Vl

25

V2 (NaOH)

49.91

Hence here acid is dibasic.

gu wiU be of HzS04 solu­ tion bring down the acid normajjcy to 0.2 N?

19. What volume at NfP of

17. 10 ml of 0.02 M KMn04 is required to oxidize 20

ml of oxalic acid of cemlin mcngth. 25 ml of the

required co

be passed into

ammonia

30 ml

same oxalic acid is required to neutralize 20 mJ of

NaOH of unknown strength. Find the amount of

NaOH in a litre of che solution. (Molecular weight of NaOH .. Solution

40)

In aciclic medium

l M KMn04 • 5 N KMn04 0.02 M KMn04 • 0.1 N KMn04 According to normality equation, N, V1 (KMn04)

•

N,V, (Oxalic acid)

Meq. of HzS04 after passing NH • 30 x 0.20· 6 > Meq. of H,S04 lost• 30 - 6• 24 Meq. of NH, passed

�

N Z"'

x

20

0.1 .. . 10 0 05 N ; 0

Meq. of Hz S04 lose

1000. 24

WNli)• 0.408 8

Volume of NH = 0.5376 litre =

0.1 x 10"' NZ

x

•

\

0 ac STP• 22.4 x .408 17

537.6 ml

20. What volume

of a solution of hydrochloric acid for

contairung 73 g of acid litre would suffice

Mole Concept and Stoichiometry the exact neutralization of sodium hydroxide obtained by allowing 0.46 g of metallic sodium to act upon water?

Thus, meq. ofNa= Mcq. ofNaOH formed= Meq. ofHCI used

ChemlcaJ Equations and Numerical

•

A balanced cbemicaJ reaction represents a stoi­ chiometric equation.

•

In a stoichiometric equation, the coefficient of reactants and products represents their stoichio­ metric amounts.

•

The reactant which is completely used dur­ ing an irreversible reaction is called limiting reagent while the reactant left is called excess reagent.

0.46 x 1000 .. ..lL x v 23

1.11

Calculations

Solution Na+ �o- NaOH + Yl � NaOH+ HCI - NaCl+ �o

•

36.5

(Mcq. ofHCI= N x V)

For example, if 20 g of caJcium is burnt in

= 10 mJ

32 g of01 then Ca is Hmiting reagent while 02 is

excess reagent.

Limiting Reagent

•

When two reagents or reactants react to give various product then the number of moles of product fonn according to the reagent or reactant which is com­ pletely consumed during the reaction. The reagent or reactant which is completely consumed during the reaction is caJled limiting reagent that is, the reagent which gives the least number of moles ofthe product is called limiting reagent.

Stoichiometric

calculations

help

in

finding

whether the production of a particular substance is economically cheap or not. •

These stoichiometric calculations ing four types:

are

of follow­

{a) Calculations based on weight: weight rela­ tionship

(b) Calculations based on weight: volume rela­ tionships. {c) CaJculations based on volume: volume relationships

Illustrations 21. 2 g � reacts with 10 g 02 to form water. How much water is formed? Solution Hydrogen reacts with oxygen as follows: 2� + 02- 2H.p Number of moles ofH2 = Number of moles of 02

•

� = I moles ig 0.3125 moles •

As 2 moles of � reacts with I mole of 02 to form 2 moles of H,P so, for I mole of Hi only 0.3125 moles of 01 is present. Thus 02 is the limit­ ing reagent here. As 1 mole of02 gives 2 moles of�O. So 0.3125 mole of02 gives =

f x 0.3125

0.625 moles of �O

Weight ofHzO formed= Moles x MoL weight

= 0.625 x 18 = l l.25 g H20

{d) Calculations based on weight: volume energy relationships •

lf the amount of the reactant in a particular reac­ tion is known, then the amount of the other sub­ stance needed in the reaction or the amount ofthe product formed in the reaction can be found out.

•

For stoichiometric calculations, the following steps must be considered:

(a) A baJanced chemical equation using chemi­ cal formulas of reactants and products must be written.

(b) Here the coefficients of balanced chemical equation provide the mole ratio of the reac­ tants and products.

(c) This mole ratio is convertible into weight weight ratio {w/W), weight-volume {wN) ratio or volwne-volume {vN) ratio These arc called percentage by weight, peroentage by volume and percentage by strength respectively. .

1.12

• Mole Concep � and Stoichiometry

Gravimetric Analysis It is an analycical technique

.

based on the measure­

ments of mass of solid subsrances or the volume

of the gaseous species. It is also divided i n t o three categories:

(ii) .Mass-Volume relacion

On solving, we gee X Percentage of NaCl

•

0.518

•

57. 36

Solution 2Fe

\.Wghtr-V+� Relationship

with the mass of another reactant or produce.

problems, one should proceed

according co the following instructions given below:

As 2

of the reacranrs and product. We muse also write the

x

r-ep1

159.7 8

55.85 g Fe is present in 159.7 g Fe20,

So 2.78 kg Fe is present in 159.7 g )( 2.78 kg 2 )( 55.85 g

senr che chemical change.

• Now write che number of moles below rhe formula

e

2 )( 55.85 8

• First, write down che balanced equacion co repre­

= 3.97 kg Fep,

relauve weights of the reacranrs and products(which

As 5 kg ore contains

below the respective formula.

So 100 kg ore contains•

are cakubced from the respective molecular formula)

finally, apply the unitary method co caJculace the unknown fuccor{n or s).

Illustrations

co give a solution of fel· ion. This solution titrated

co the

end point with 35.6 ml of 0.1 N KMn04, Pc2· co fie)·. What is the mass per­

SoJucion Reacrion of MnO,-and Fe2• in addic medium is us follows: Mn0 - + SH· 4

H2SO� gave 1.0784 g of a minure of Na1S04 and

KJS01. Find the percentage composition of the compound.

For(0.9031 -

%

centage of iron in the mixrure?

72 g

For X g NaCl

79.4

•

ocher elements) was dis.solved in hydrochloric acid

- Pe104 + 4H1

23. 0.903 I g of NaCl and KCI on reaction with

.

79.4 kg FezO,

which oxidized

So by 18 g

SoJuaon

; LOO

3.97

25. 0.256 g sample of iron aJJoy (a mixture of iron with

As by 72 g steam the weight of Pe oxidized= 168 g

•

3.97 kg Fe101

Thus % of Fe20 J in ore

its oxide by the action of 18 g of steam on it.

.. L 68 x L 8 72

•

•

22. Find weight of iron which will be convened into

3 )( �6 4 )( 18

%

Fe, what per cent of ore is Fe101?

Problems lrM>Mng Mass-Mass or

Solution 3Fe + 4Hp

%

tion of FePr If 5.0 kg of ore concaios 2.78 kg of

This relationship relates the mass ofa reactant or produce

•

174 (0.9031 - X> .. 1.0784 g 149

+

24. The mineral haematite is Fe20 . Haematite ore , contain unwanted material called gangue in addi­

(iii) Volume -Volume relation

For solving such

142 X 117

Percenrage of KCl - 42.64

(i) Mass-Mass (weight-weight) relation

(i)

charn,

•

X) g KCl

:�: (0.903 L

5Fe2' -

Mn2'

+

4H20

or 5000 ml N KMnO�

•

5

><

X)

5Fe3'

55.8 g of Fe

35.6 mL ofO.l N KMn04 5 )( 55.8 )( 35.6 )( 0.1 5000

-

+

So Moo.- - 5Fe2•

142 x 1f7 •

+

•

0.2 g Fe

As 0.256 g

alloy contains

•

0.2 g Fe

Mole Concept and Stoichiometry . 0.2 x I 00 S o 100 g iron . aIIoy contains = 0.256

One should not forget that if other conditions

(II) Problems Involving

Mass-Volume

or Weight-Volume Relationship This relationship relates the mass of a reactant or product with the volume of another gaseous reac­ tant or product involved in a chemical reaction. For example, Weight ofa solid substance can be compared with the volume ofgases with the help of the fact that 1 mole or 1 gram molecule ofa gas occupies 22.4 litres or 22400 c.c. at N.T.P. So in brief, the weight-volume relationship can be represented as follows:

By mole

2

By amu

40

73

113

2

By gm.wt.

40g

73 g

113 g

2g

40g

or vol.

73 g

113 g

22.4

Illustrations 26. A mixture ofaluminium and zinc weighing 1.67 grams was completely dissolved in acid and the evolved 1.69 il tres ofhydrogen gas measured at 273 K and one atmosphere pressure. What was the mass of aluminium in the original mixture?

Solution Let the mass ofaluminium in the sample The volume of8i at N.T.P. given by Al _ "" 3 x 22.4 x A L 2 )( 27 The volume of� at NTP given by Zn

=

(1.67-A) 22.4 65.4

L

lit at

(l.67-A) 22.4 3 x 22.4 x A + = I .69 65.4 54

NTP.

142.2 x A= 176.26

For solving such problems one should proceed according to the following instructions given below: •

First, write down the relevant balanced chemical equation (s).

•

Now write the weights of various solid reactants and products.

•

Oases are nonnally expressed in tenns of vol­ umes. In case the volume ofthe gas is measured at the room temperature and pressure (that is, any condition other th� NTP), convert it into N.T.P. by applying gas laws.

The volume of a gas at any temperature and pres­ sure can be converted into its weight an vice versa by using this relation, PV= (g/M) x RT Here g is weight ofgas, M is molecular weight of gas and R is gas constant.

•

not mentioned. the chemical reaction that 1s assumed to occur at N.T.P. that is. at O"C (273 K) temperature and 760 mm (1 Bbn< 0.040

g NaOH

meq .NaOH •

0.225g NaOH

The per cenc of NaOH in the sample is as follows: 0.225g NaOH 0.311g sample

><

100

•

72.3 % NaOH.

29. A volume of 22.5 ml of 2.50 N NaOH was

required co neutralize 10.5 ml of a solution of H2S04 of unknown sucngrh . The dcnsicy of the

H2S04 solurion

was

l .16 g per ml. Calcuh.re the

per cenc by weighc of H2S04in rbe solution . Solution In a riaacion using NaOHand H?O.. Meq . ofNaOH

•

meq. of H2S04

Cr,O/•

•

56.3meq NaOH

><

14 H·

+

follows:

6 Fe2• - 2Cr'· 6.00 moles

6.00 g-cq Wt

6.00g-eq Wt

According

ro

+

+ 7H20

6Fel'

chis equarion, 1.00 g -eq

wt

2 of Fc •

ion will originate from 1.00g-eq wr of FeS04, we get meq. ofCr,O/- .. meq .ofFe1' .. meq. of FeS04

mcq. ofCr10/ is: 0.100

meq Cr 0 2-

11.1

m 1SO UtJOn

x33.6mlsolution

.. 3.36meq . Cr,o,iThe normalicy of the FcS04 solution is:

3.36meq . FeS04 . •0.156N FeS04 J 21.6m so Iunon As FeS04 solution is 0.156N with respect to Fe1• ion.

and the number of mg Fe1· per ml is as follows:

=

8.71

2 , meq F e _ . >< 55.85 mg Fe ' ml soluaon meq. Fe2'

•

mg Fe2·

ml solution

31. What should be the weight of NaNO; to make

50 ml of an aqueous solution so that it contains

The 56.3 meq . of H2S04: 56.3 meq . H S04 2

+

as

1.00 mole

0.156

meq NaOH . 1 >< 22.5ml soluc1on . m 1so ut1 on

The equation is

Solution

The meq .of NaOH: 2.50

22.6 % H1S04 by weight

0.0491

g

tt,so.

,

meq . H SO • •

!1 ·12.2g

The per cent H2S04 is

Solution

Molrcu!ar mass ofNaNO, ., 23 + 14+ 3 x16

27 . 6g H1S04

The weight ofrhe 10.5ml of H1S04: 10.5 ml>< l.16

70 mg Na' ml-1?

•

85 g mol-1

As 23 mg Na is present in 85 mg ofNaNO, So 70mg Na is present in

85 x 70 23 •

258.7 mg NaNO,

As I ml solution contains 258.7 mg NaN03 So 50 ml solution contains

I/6th. Calculate the molar ratio of two acids

Solution

I ml

Reactions

= 13935 mg

H,SO,• H,O + CO t

HCOOH

13.935 g

a mole

(Ill) Problems Based on Volume-Volume Relationship This relationship relates the volume ofgaseous reactant

orproduct with the volume ofanother gaseous reactant

or product involved in a chemical reaction. For solving

such problems, ooe should proceed according to the

following instructions given below: •

solution the volume of the gas decreases by i n the original mixture.

258.7 mg x 50 ml

=

Mole Concept and Stoichiometry • 1.15

First write down the relevant balanced chemical

equation.

• Now write down the volume ofreactants and prod­

8iCP. b mole

a mole

H,S O•• H,O + co

+ col

b mole

b mole

Total number of moles of CO and C02 =

a + b + b = a + 2b

Total number ofmoles ofC02

=

b

So Mole fraction ofC02 absorbed in KOH =

-

b a + 2b

_ _ b =

a + 2b

l

6 ·

ucts below the fonnula to each reactant and prod­

a = 4b

gaseous substance occupies 22.4 litres at N.T.P.

Thus ratio ofHCOOH and HzCp4 is 4 : I .

uct using the fact that one gram molecule of every

• If volume of the gas is measured under particular

or room temperature, convert it to N.T.P. with the help of ideal gas equation.

• Now use Avogadro's hypothesis "gases under simi­

lar condjtions of temperature and pressure contain

the same number of molecules". Thus under simitar

conditions of temperature and pressure, the volume of reacting gases are proportional to the number of moles of the gases in the balanoed equation. Example, Nl (g) + I mole 22.4 lit

I volume Ix volume

38.z (g)

-

3 moles

3 x 22.4 lit

3 volumes

3 x volume

2NH3 (g) 2 moles 2 x 22.4 lit

2 volumes

2 x volumes

From the above equation, it is clear that 22.4

litres of N2 will react with 3x22.4 litres of H, to fonn 2x22.4 litres ofNH3.

Illustrations 32. A mixture of HCOOH and (COOH)2 is heated with concentrated HzSO•. The gas produced

is collected and on its treatment with KOH

33. A gas mixture of 3 litre of produce (C3HJ and butane (C4H1J on complete combustion at 25°C produced I 0 litre C01• Find out the composi­

tion of gas mixture

Solution C3H, + 502 - 3C02 + 4Hp (I) C4H 10 + 1312 02

-

4C02 + 5�0 (1)

Supposexlitre ofC3H1 and y litre ofC4H10arc pres­ ent in the mixture. So

x+y=3

------(1 )

As volume of COl fonned = JO= C02 formed by CJHI + col formed by C,H10 -----(2) I 0 = 3x + 4b Solving Eqs. I . and 2. y = llitre x

=

2Litre

n-Factor or Valence Factor It is very important for both redox reactions and non-redox reactions by wruch we can obtain follow­

ing informations:

1.16

•

Mole Concept and Stoichiometry

(i) It calculates the molar ratio of the species taking part in reactions that is, reactants. The reciprocal ofn-factor's ratio of the reactants represents the molar ratio of the reactants. Example, IfA (having n-factor =a) reacts with B (having n-factor = b) then its n-factor's ratio is a : b, so molar ratio of A to B is b : a.

(n = 3) that is,

tribasic acid

KOH

+ OH-

(n = I) that is, monoacidic base

(n = 2) that is, Al(OH}1

bA

(n = 3) that is,

+

aB -+

Product

(n-factor = b)

Molecula r weight . . (ii) Equivalent weight = n-factor or Atomic weight n-factor

(ii)

Before calculating the n-factor of any of the reactant in a given cbemjcal reaction we must have a clear idea about the type of reaction. The reaction may be any of these types:

n=2

Crp/-

(ii) Redox reaction

n•6

Acid-Base or Neutralization Reactions:

As we know that according to the Arrhenius concept, "An acid provides W ion(s) while a base provides OH ion(s) in neutraliza­ tion these W and OH- ion/ions combines together".

The number ofH· ion(s) and OH ion(s) represent n­ factor for acid and base respectively, that is, basicity and acidity respectively. Example, HX

-+

H'

+ c1-

(n = I } that is, monobasic acid H2so.

-

2tt·

(n .. 2) that is,

+

so,2-

dibasic acid

30H-

triacidic base

-+

2C02

n-factor = I (+4) x 2 - (+3) x 2 1 "' 2

(i) Acid base or neutralization reaction

decomposition

+

Example (1) Wben onJy one atom undergoes oxida­ tion or reduction Cp/-

or double

AP•

Redox Reactions: These reactions involve oxidation and reduction simultaneously. Herc exchange of electrons occurs. To find n-factor for oxidizing or reducing agent we must find out the change in oxidation state ofthese species.

Calculation of n-Factor

(iii) Precipitation reaction

diacidic base

-+

It can be represented as follows:

(n-factor = a)

(I)

K·

-+

-+

2Cr1..

n-factor = I (+3)

x

2 - (+6)

x

21=6

Example (2) For the salt which react io such a way that one atom undergoes change in oxidation state but appear in two product having same oxidation state. •6. l

CrPt

-+

•)

cr3· + cr3·

n-factor = I (+6)

x

2 - (+3)

x

2 I=6

Example (3) For the salts which react io such a way that one atom undergoes change in oxidation state but appear in two product having different oxidation state .

3MnO,-

-+

2Mn,.

+

Mn..

... 1 2 x (+2)- 2 x (+7) I + 1 1 x 4 - 14 I + 1 6 - 7 1 = 1 1

(+6)- I (+7) I= I

Hence n-factor = 1 113.

Example (4) For the salts which react in such a way that one atom undergoes change in oxidation state in

Mole Concept and Stoichiometry • 1.17 two products one with changed oxidation state from that of reactant and another having same oxidation state as that of reactant. K,Crp7 + 14HC1 - 2KCI + 2CrCl3

+ JC� + mp

Here, out of 14 moles of c1- (in HCl) only 6 moles of c1- are changing the oxidation state from i the product Cl2 while the oxidation state -I to 0 n of remaining 8 c1- ions remains the same in HCl and CrC�. Hence the total number of moles of electrons lost by 14 moles of HCl is 6 here. So each mole of HCI takes up 6/14 that is, 3n moles of electrons and hence n-factor of HCI is 3/7. Example (5) For the salt which react in such a way that two or more atoms undergoes change in oxida­ tion states. Feep. - Fe3• + 2C02

= 1 1 x (+2) - I x (+3)l + l2 x (+3) - 2 x (+4)1 = 1 +2=3 Hence n-factor of FeC20. is 3. Example (6) Salts or compounds which undergo iotramolecuJar redox reaction that is, in which an atom undergoes oxidation while another atom undergoes reduction. Here n-factor can be find out by knowing the balanced chemical reaction and considering only one process. ( I)

(OJ

2KCl01 - 2KCI + 302 In this reaction, oxidation atom is getting oxidized while chlorine atom is getting reduced so n-factor of KC101 considering oxidation =

1 3 x (-2)- 3x(O) I = 6

a-factor of KCI03 considering reduction

=1 1 x 5 - I

x

(-1)

2Hp1 - 2Hp + 02

Here out of these two 8z02 molecules shown by the reaction one mole of fi,02 is oxidized and one mole is e r duced.. For oxidation reaction (-1 ) ' 2

l•O

ff,01 - 02

o-factor = 1 2

x 0 - (-1) x 2 1 = 2

For reduction reaction 8z02 -

2 �o

n-factor = I (-2)

Here, the oxidation state ofFe in this form +2 to +3 while that ofC change from +3 to +4. Here n-factor can be find out by calculating the total change in oxidation state per mole ofthe salt. Here I mole of FeC204 have one ofFe1� and one mole of c20.2· so the total change in oxidation state is given as

., (-2)

undergoes oxidation as well as reduction. Here, n-fac­ tor can be found out by knowing the balanced chemi­ cal reaction and considering only one process.

I= 6

Example (7) Salts or compounds which undergo dis..

proportion reaction that is, in which a single species

x 2 - (-1 ) x 2 1 = 2

Hence o-factor of Hp2 in both oxidation and Ieduction reaction is same that is, 2. lo the disproponion reactions in which moles of compounds undergoing oxidation and reduction are not same like. 6� +

120H

-t

IOX + 2XOJ + 6Hp

Here out of6 moles of� , 5 moles of"1 undergo reduction and accept I0 moles of electrons (oxidiz­ ing agent) while I mole of� underg�s oxidation by loosing I 0 moles of electrons (reducing agents) �

-+

5� +

� +

2X5• + 10 eI 0 e- -+ I 0 x5X2

-

( n =lO) (n = 2) Reducing Oxidizing agent

1ox-

+ 2xs..

(n = I )

(n = 5)

agent

Hence a-factor of"1 acting as oxidizing agent is 2 and that � acting as reducing agent has a-factor JO. (Ill) Precipitation or Double decomposition Reaction: It is the reaction in which there

is no change in oxidation state for any atom. Here n-factor of the salt used in the reaction can be found out by multiplying the oxidation state of the cation or anion by the total number of atoms per molecule of the salt.

1.18

•

Mole Concept and Stoichiometry

BaC11 + J moles ofMnO�· for the oxidation of A,... to A03- in an acidic medium.

What is the value of n?

Solution As J .61 x 10-3 M KMn04 =

For KzSO� n-factor = Oxidation state of Kxnumber of K­ atoms in one molecule ofKzS04

2 68 x 10-1 M solution of A.,.. .

10-1 M x M 26 So M/5 KMnO4 = . 8 x 5 1.61 x .I 0-3

= 0.33 M solution ofAn+

=(+ J ) x 2 = 2

Laws of Equivalence According to law ofequivalence, for each and every reactant and product, Equivalents of each reactant reacted lents of each product fom1ed.

=

Equiva­

Example.

Suppose the reaction is taking place as foJJows: P+Q-R+S

According to law of equivalence, Equivalents of P reacted = Equivalents of Q reacted = Equivalents of R produced = Equivalents ofS produced Equivalents of any substance Weight of substance (in g) Equivalent weight = Normality (N) x Volume (V) (lo litre) Normality (N) = a-Factor x Molarity (M) Normality and molarity are temperature dependent. As on changing the temperature, the volume ofsolu­ tion changes, so normality and molarity change. POAC (Principle of Atom Conservation) Method

For any chemical reaction, mole atom ofany element remain conserved during the chemical reaction.

0.33 M =

M 5-n

-

I

:>

.

n 3 - = 0.33 = n=2

35. One g of impure N8iCO) is djssolved in water

and the solution is made upto 250 ml. To 50 ml of this made up solution, 50 ml of 0. I N HCl is added and the mixture after shaking well re.quired l 0 ml of 0.16 N sodium hydroxide solution for complete neutralization. Calculate

the per cent purity of the sample ofN�C01•

Solution Strength of the N�C01 solution =4 g L- 1

Suppose the nonnality of Na2C01 solution

=

Nx

As after mixing N8iC03 and HCl solution, NaOH solution is added so according to normality

equation

50 x Nx + 0.16 Nx = 0.068 N

x

I 0 = 50 x 0.1

Strength (g L 1 ) = Normality x Equivalent mass

==

0.068 x 53

=

3.6 g L-1

So purity ofN8iC01 =

�IOO

3 ·6

= 90 %. 36. 0.257 g of a nitrogeneous compound was

digested with sulphuric acid and then distilled with excess of strong alkali. The gas evolved was absorbed in 50 ml N/1 0 l\S04• At the end of the experiment, the acid required 20.2 ml of

Molle Concept and Stoichiometry • 1.19 N/10 NaOH for neutralization. Determine the percentage of nitrogen in the substance.

Solution 23.2 ml ofN/10 NaOH = 23.2 ml ofN/10

�so. Volume ofN/10 H1S04 neutralized by NH,

the solution of unknown concentration is taken in a

flask and the essential reagents are ended in it. Now

the solution ofknown concentration is added from the

burette in this solution till complete reaction occurs

between them and the end point or equivalence point of the two reacting species are equal.

For example,

Ifwe take two solution X and Y then at the end point

= 50 - 23.2 = 26.8 ml

Nx Vx=Nv V,

26.8 ml ofN/10 H2S04 s 26.8 ml ofN/10 � =

Here Nx

26.8 mJ ofN/10 nitrogen

x

26.8

I 000

Vv = Volume of solution Y

g nitrogen

g of the organic substance.

14

26.8

1000

x

Double Titratlon This titration is used for specific compounds using

So% of nitrogen

Tii x

various indicators. When the solution having NaOH 100 0 .257

'"" 1 4·6 %

and NBiC03 is titrated using phenolphthalein indica­ tor, at phenolphthalein end point following reactions take place: NaOH + HCl - NaCl + �O

Volumetric Analysis (Titration)

NBiCOJ -r HCl - NaHCOJ + NaCl

Volumetric analysis is an analytical method used

Here Equivalents of NaOH + Vi equivalents of

to find the concentration of a substance in a solu­ tion by adding exactly same number equivalents of some another substance present in a solution of known concentration (standard solution). Volumetric analysis is also called titrimetric

N�C03 = Equivalents of HCI

to find the concentration of unknown solution is as

titrant and the substance whose concen­

tration is to be calculated is known as titrate.

Types of Titration Titration or volumetric analysis is of following types:

1.

Simple titration

2. Double titration

3. Back titration

4. Iodimetric and iodometric titration

Simple Titration

The purpose of this titration is to find the concentra­ tion of an unknown solution by using the known con­

centration of another solution. Here a known volume

. . . . .... (i)

When methyl orange is used. N�C03 is con­

vened into NaCl + C02 + Hp

analysis. Here the substance whose solution is used known

X

Nv - Normality of solution Y

.

The mass ofnitrogen was originally present in 0.257

=

Nonnality of solution X

VX = Volume of solution

26.8 ml ofN/10 nitrogen wilJ contain 14 1000

=

=

Equivalents ofNaOH + Equivalents of N�CO J . . . . . . . . . (ii)

Equivalents of HCl

These titrations are carried out by using phenol­

phthalein and methyl orange in continuation as well as separately also.

We apply law of equivalents to find the percent­

age composition of the mi xture with the help of equation (i) and (ii) if the HCl consumption in two different steps is given. • Phenolphthalein indicates end point when Na2C03

is converted into NaHCOr

Back Titration

Suppose we have an impure solid substance 'O' weighing 'w' gm and we have to find the percent­

age purity of 'O' in the given sample. We have also

1.20 • Mole Concept and Stoichiometry given two solutions 'M', 'N' here the concentration of 'N' is known 'N, · and chat of 'M' is unknown. For a back titration co be carried ouc; che following condi­ tions muse be satisfied. (i) Compounds 'M', 'N' and ·o· must be such that 'M' and 'N' can react with each other.

(ii) 'M' and pure ·o· can also react wich each other however che impurities present in ·o· muse not react with 'M'. (iii) The product of 'M' and 'O' should not reacr wich 'N'. Now we will take a certain volume of 'M' in 11 flask but remember the equivalents of 'M' taken muse be greater or equal co equivalencs of pure 'O' in the sample and perform a simple titration using 'N'. Here we assume that che volume of 'N' use is ·v ,· litre. Equivalents of 'N' reacted with 'M' Hence initial equiv11lents of'M'

•

•

N1 V1

N1 V1

Now we will take same volume of'M' in another flask however now ·o· is added in che flask. Here pure pare of ·o· reacts wirh 'M' and excess of 'M' is

back cicraced with 'N'. Suppose che volume of 'N' consumed is V litre here.

,

Equiv;tJenrs of 'N' reacted with excess of 'M' = N, V2 Hence equivalent of'M' in excess = N, V1 chat is, equivalent of 'M' reacred pure ·o·

•

(N I v I - N I vl)

Equivalents of pure 'O' . (N1 v1 - N1

V,)

Suppose che n-factor of ·o· in its reaction with 'M' is 'a' then che moles of pure ·o· (N1

V1 - N, V2)

(N 1 V 1

-

a

N V) 1 2 x

Molar mass of'O'

Percentage purity of 'O' (N1

V, - N , V�) x

When in redox titracions iodine is used a.s an oxidiz­ ing agent these titration arc called iodine titration. These are of two types:

lodimeric Titration

In such cicracions iodine solution is used as an oxi­ dant and iodine is directly citrated against a reduc­ ing agenr. This type of cicrations arc used for the determination of strength of reducing agents like sulphides, arsenides, thiosulphates etc., by titrating chem against a standard soluuon of iodine. Trus cype of titration ipvolves free iodine. Here iodine solution is created with known sodium thio­ sulphace solution having a normality, N and volume V litre.

12

+

(n = 2)

2Na2s,0 1 -+ 2Nal (n = I)

Equivalents of 12

•

+

Na2S406

Equivalents of Na1Sp1 used

•NxV Moles of I..

=

Nx2 V -

Mass of free 11 in the solution

•

(

N; V x 254) g

lodometric Titration It is an indirecr method o( estimacion of iodine. In chis titration an oxidizing agenc is used co liber­ ate from iodine solution and the librated iodine is creaced with a scandard solution of a reducing agent added from a burecce. Here a neucrnl or an acidic solution of an oxidizing agent is used and che amount of liberated 11 is equal co che quantity of chis oxidizing agent. These rirrarions are used co derermine rhe con­ cencracion of K Cr207, KMnO�, CuS04, Ferric 1 ions, H202 etc.

a Hence mass of pure ·o· ..

lodlmetrlc and lodometrlc Tltratlons

Molar mass of·o· w

x 1 00

Oxidizing agcnr (X)

+

Kl -+ 11

j

+

reduced state of oxidant

2Na1S10i

Equivalencs of (X) of

Na2s20, used

•

•

Equi.valents of 12

•

Nx v

Equi\•aJents

Mole Concept and Stoichiometry • 1.21

spt

•

NxV

Let the n-factor of (X) in its reaction with KI

•

+

H20 (not balanced)

Solution Sp,Z- - 2SO/"

be x, then

Mass of (X) consumed

Mno.-

Mn02 (s) + SOt + OH-

Equivalents of 11 liberated from KI = N x V Equivalents of (X)

+

N

;

V

Change in oxidation number of sulphur per molecule of

x MX

spt =

(Here Mx is the molar mass of (X)).

2 )(

(6 - 2) = 8

Change in oxidation number of Mn per molecule of Mn04= 7-4 = 3

REMEMBER ...

I:

No. of moles in 0.158 g of Na2S20 1

Kl - Kl,,

=

(complex soluble in water and provide K• and 11· )

0.15 8 = 1 )( 1 0-1 158

No. of equi valents = 8 x

i o-'

Normalfry ofO.l M KMnO, solution = 0 . 1 )(

Illustrations 37. How many mL of a 0.05 M K.MnO, solution are required to ox.ide 2.0 g of FeS04 in a dilute solu­ tion (acidic)?

Solution

IOFeSO, + 2KMnO, + 88iSO, -

Klso, +

2MnSO,+ 5Fe2(SO,)l + 8H

I0 >< U 1.8

2 g ofFeS04 will

require KMn04

•

2 x 158 g

�;;���.�

Suppose V ml of KMnO, solution (0.05 M required .

is

"'

2 x 1 58 x 2 10 )( 15 1 .8

used for the following titra­

tion. What volume of the solution in ml will be rcacc

�

0 3

)( 1 0-1 )( 101

39. {a) Give a half-equation for the oxidation of the ethanedioate ion, cp.i- (aq), to carbon diox­ ide co2 (g).

a half equation for the reduction of the {VII) ion, Mn04- (aq), co man­ ganese (II) ions, Mn2' (aq), in acidic condi­ tions.

(c) Calculate the volume of an

acidified solu­ (VU),

tion of 0.02 M porassium manganate

which would be needed to '>xjdizc

t00 cml of a saruruted solution of magnesium

V = 52.7 ml

required to

v ..

KMn04,

1 58 )( 0.05 )( v 1000

38. 0.1 M K.Mn04 is

)( 0.3 - 8 )( 1 0-1

manganace

1 58 )( 0.05 )( v 1000

Thus,

�

1( 0

(b) Give

Amount of KMn04 in this solution =

0.3

= 26.7 ml

p •

•

Suppose V ml of KMn04 is required then

So

2 x I 58

10 x 15 l . 8 g of FeSO, require KMn04

3

with 0.158 g of Na,Sp/

etbanedioare, MgCP�·

Solubiliry of magnesium ethancdioace at 20"C is 9.3 x 1 O-J mot dm-J.

Solution (a) Cp,/" ('lq) -

2C02 (g) + 2e-

(b) Mno.- (aq) + SH' (aq) + 5e Mn2• (aq) + 4H

p (I)

1.22

•

Mole Concept and Stoichiometry

2 (c) 5Cp4 - (aq) + 2MnO.- (aq) + 16W (aq)

- l OC02 (g) + 8� (I) + 2MnJ• (aq)

I 00 cmJ of saturated MgCp4 solution at 200C con­

tains 9.3 x 10"4 mot.

So 2.5 mole As34 reacts with I mole Mno.­ fn this equation, .

Rat10

25

5

. 1 = n _3

According 10 the equation,

Oxidation state of arsenic in the product = +5.

As 5 mol orcp/- react with 2 mol MnO;.

Volume Strength of H202 Solution

So 9.3 x lo-4 mo! ofCp/- react with 2 x 9.3 x lo-4

5

The concentration of Hp2 is usually represented

moI

'X' volume, it means that I volume ofHzD2 solution

As 1 000 cm1 of 0.02 M Mn04- contains 0.02 mol Mno.-.

S0

A

2�0

+

02

2 x 34 g

22.4 L at STP

As 22400 ml of 02 gas is liberated by 68 g of �02

5

solution

So X ml of02 gas will be liberated by

1 8.6 cm3

40. A solution of arsenic (Ill) oxide contruning 0.248 g required 50 cm3 of acidified potassium manganate (VII) solution (0.02 mol dm-1) for complete oxidation. What is the oxidation

state of arsenic in the product? [A, (0) = 116, A, (As)= 75)

Solution As 1.,. (aq) - As•+ (aq) + (n - 3)e-

Mn2• (aq) + 4Hp

( 1)

0.248

Number of moles Asp1 = 1 0 + 48 5

It means that I 7X/5600 g of Hp1 will be present in

I ml of solution.

1000 ml of solution contains H102 17X x IOOO 5600

c:

17X 5.6

1J� = N �4 (n-factor of H10� x

X = 5.6 x N that is,

Volume

mality

strength of H101

=

=

2)

5.6 x Nor­

Illustrations

= 0.00125 moles

41. The label on a Hi02 bvttJe reads as I 0 Vol . Find

0.0025 moles

50 cm' of potassium pennanganate (VII) (0.02 mol

dm-l) contains

the concentration of the H202%.

Solution 2�02

50 x 0.02 moles 1000

-

2Hp + 02

I 0 vol. means 1 vol H202 = I 0 ml 02

= 0.00 I moles potassium permanganate (VII) As 0.0025 moles As3' reacts with 0.001

17 x g of H202 5600

=

Strength (g L-1) = Normality x Equivalent wt

5As3...(aq) + (n - 3) Mn04- (aq) + 8(n - 3)H"(aq) - (n - 3) Mn?. (aq) + 4(n - 3)Hp (1) + 5Asn+ (aq)

=

68 x

= 22400

=

Mno.- (aq) + &H• (aq) + Se -

MnQ4-

decomposition.

2Hp2

. 2 x 9.3 x l o-4 mot contains

Number of moles As3•

gives 'X' volumes of 01 gas at STP on complete Consider tne decomposition of �02 as

lOOO x 2 x 9.3 x J Q-4 J cm 0.02 x 5

of 0.02 M MnQ4•

=

in

terms of volume, If a sample of H202 is labelled as

mole

I g Hl02

=

400 68

22

=

329 ml

I litre of I 0 vol. means = 10000 ml of 02

Weight ofHz01 to give 10000 ml 1

=

329

)( 10000

-

30.4 g

�

=

dissolved in sulphuric acid. When water is added to

oleum, S01 reacts with �O to fonn H,SO� as a result

mass of the solution increases.

Wt. ofS03 x 1 00

IOO

The hardness of water is due to the presence of bicarbonates, chlorides and sulphates of calcium

and magnesium. Bicarbonates causes temporary hardness while chlorides and sulphates causes

permanent hardness. The extent of hardness is

called degree of hardness. It is defined as the number of parts by weight of calcium carbonate

SOJ + Hp - Hlso,

The total mass of H2SO, obtained by diluting

I 00 g of sample of oleum with desired amount

of water, is equal to the percentage labelling of oleum.

present in one million parts by weight of water

that is, in ppm (milligram per litre) of CaC01. Hardness o f water =

gm of calcium carbonate 1 tv. v · gm of water

Total mass of H,S04

present in oleum after dilution. =

0.66 x 80 = 52.8

Hardness of Water

Oleum or fuming sulphuric acid contains S01 gas

=

=-

= 52.8 %

3.04 %

Percentage Labelling of Oleum

% labelling of oleum

Wt. of S03

% ofS01 =

Cone. = 30.4 g/lit Cone. % = 30.4 x 1

Mole Concept and Stoichiometry • 1.23

mass of H2SO, initially present + mass of

H,SO, produced on dilution.

Suppose the mass of oleum sample be I00 g,

which on dilution becomes 109 g. This implies that 9 g of H20 was added.

sol + Hp - HlSO,

Moles ofH,O added "' 9/ 1 8 • Moles ofSOJ pres­

ent in oleum sample.

9 Mass ofS03 in oleum = 1 x 80 = 40 g 8

Thus, oleum sample contained 40 % sol and (>() %

1-f,SO,.

42. Find the percentage of free S01 in an Oleum sample which is labelled I 12 % H,SO,.

Solution Oleum = H2SO, + so, = H,S,07 If initial weight of labelled H,S207 = I 00 gm Weight ofH1SO,, after dilution = 1 1 2

Moles of Hp = moles of S01 =

43. 50 litres of water containing Ca(HCOJ)2 when

convened into soft water required 22.2 g Ca(OH)» Calculate the amount of Ca(HC0 } 12

per litre of bard water.

Solution Reaction Ca(HC03)2 + Ca(OH)2 - 2CaCO, + 2Hp l62 g

74g

As 74 g Ca(OH}2 reacts with 162 g Ca(HC01)2 So 22.2 g Ca(OH)1 will react with 1 62 x 2 2.2 --= = = ..-...74

Illustrations

Wt. ofH20 - 1 2 gm

Illustrations

:�

= 48.6 g Ca(HC01)1 As 50 L waler contains = 48.6 g Ca(HC01)2 So I L water contains

6

...

4ff0

= 0.972 g L I 44. One litre of a sample of hard water contains

0.9 mg ofCaCl2 and 0.9 mg ofMgClr Find the

= 0.66

totaJ hardness in terms of parts of eaco, per 10-parts of water by mass.

1.24 • Mole Concept and Stoichiometry

Solutlon

Mol. mass ofCaCl2

= 111

•

Eq. wt. of R-COOH

= Eq.

Mol. mass ofMgCl2 "' 95 111

g ofCaCl2

=

0.9 mg ofCaC� =

100 g ofCaC03

=

�??

x

0.9

0.9

g ofMgC�

= 0.94

s =

ofR-COOAg - 107

In Redox Titration

V (oxidizing agent)

mg ofCaC01

><

M (oxidizing agent)-

V (reducing agent) >< M (reducing agent)

0.81 mg ofCaC03

95 g ofMgC�

wt.

100 g ofCaC03

><

�x 0.9 ofCaC03

1 9

No. of moles of oxjdizing agent No. of moles of reducing agent

In Acid Base Titratlon

mg ofCaC03

Thus, one litre of hard water contains (0.81 + 1.75 mg of CaC03

0.94)

-

One litre of water l O' g

=

l 06 mg

Degree of hardness = l.75 ppm

V (acid) >< M (acid)

=

No. of moles of acid V (base) x (M base) x No. ofmoles of base z, M 1 v1 = Z2 M2 v2

Here Z1 = acidity or basicity ofsubstance - l

and Zz = acidity or basicity ofsubstance - 2

• I

mole of H20 #; 22400 mJ or cc. of H,O (since it is liquid)

l mol ofl\O •

1 8 cc. ofH20 (as density of 8i0 = !gm/cc)

=

Mass of one mole of e

mass

0.55 mg.

Some substances like CuS0,.5Hp, Na2co,. I OH20 have a tendency to lose water in air. These are called effiorescent and this ten­ dency is called effiorescence.

•

Some solid substances like NaOH, KOH, which have a tendency to absorb moisture greatly from air and to get wet are called 'Deliquescent' and this tendency is called Deliquesceoce.

•

Hygroscopic substances like quickJime (CaO) anhydrous Pp5 etc., absorb moisture from air.

•

To find equivalent weight of an acid: It is possible by Silver Salt Formation Method. Eq. wt. of R-COOAg 1 08

=

If all the three acids nre mooobasic or bases monoacidic,

are

ofone e- >< NA

= 9.1 x 10-31 x 6.02 x 1023 =

•

""

For a mixture ofsolution ofnon reacting substances.

Wt. of R-COOAg wt. ofAg

lfwe mix HCI and H2SO. then, 'M'ofH,SO, will be multiplied by 2 because H2SO, is dibasic acid. Law of equivalence ts used only In following con­ ditions

If either a substance or solution reacts completely with a known volume of a standard solution, then Number of equivalents of substance = Number of equivalents of standard solution. WWI

Eq. wt.5..,

=

NV I 000

·

Number ofeq. of metaJ = Number ofeq. of solution W Eq. wt.

=

NV soIut1on. .

1000

Mole Concept and Stoichiometry • 1.25 Solved Problems from the Ills 1. A mixture of �C20, and NaHC,O, weighing 2.02 g was dissolved in water and the solu­ tion made up 10 one htre.

ml of Ibis solu­

10

tion required 3.0 ml ofO. I N NaOH solution for complete neutralization. In another experiment

10

ml of same solurion in hot dilute �SO, medium required 4 mJ of 0.1 N KMnO, for

complete neutralization. Calculate the amount of�C,O, and NaHC20, in the mixture.

fOT 1 9901

Solution Suppose mass of H2C20, present in the

mixture is

'X'

g in I litre and mass of NaHC20,

present in the mixture is 'Y' g in I litre.

Equivalent mass of H1Cp, =

9� = 45

Na,cp, + �o

\

Mo mass

10 ml solution

112

= 151.2 ..... (i)

For redox reaction

�= 45. 1 2 Equivalent mass ofNaHC,O, = � = 56 9

(Change in oxidation number of carbon per mol­ -

2C4�)

4 x 0.1 1000

=

On solving equarions (i) and (ii), we get

X = 0.9 g Y = l.l2g

(5.0

g) consisting of lead

.

(1IT 1 9901

Solution Le1 the amoun1 of NaN03 in the mixture be

' X' g.

= (5.0 - X) g. Heat 2NaN03 --- 2NaN02 + 02

y x 10 - 3 x 0.1 11 2 x 1000 - 1000

I x 1 1 2 X + 45 y ... 3 x O. �ri5

Ct

Y x I0 56 x 1 000

56 x + 45 y = 100.8 .. . . . . (ii)

(2

Equivalent mass ofH2C,O, =

+

The amount of Pb(N03)2 in the mixture

3 x 0.1 1000

ecule = 2;

X x I0 45 x 1000

amount of lead nitrate and sodium nitrate in

Equivalent of�C,04 in I 0 mJ solution + Equiv­

+

.

the mixture

= 112

x x 10 45 )( 1000

4 x 0.1 1000

stant. If the loss in mass is

Mot mass . 2

Equivalent mass ofNaHC,O, =

=

=

600"C until the mass of the residue was con­ 28.0 %. Find the

+ 2H,O

=

alents ofNaHC,04 in

10 mJ solution + Equiva­

nitrate and sodium nitrate was heated below

�c,o, + 2NaOH - Nalc,o,

-

Equivalents of �C,O, in

lents ofNaHC,O, in 1 0 ml solution

2. A solid mixture

For acid-base reaction

NaHC,O, + NaOH

Now,

x 85)g

32 g

Heat 2Pb(N03)2 --� 2Pb6 + 4N02 + 02

(2 x 331) g = 662 g

(4 .)( 46) 32 g = 216 g

170 g ofNaN03 evolve oxygen = 32 g X g ofNaN03 evolve oxygen =

?lo x X g

662 g of Pb(N0»2 evolve gas - 216 g (50 - X) g of Pb(NOJ2 evolve gas =

�� x (5.0 - X) g

1.26

•·

Mole Concept and Stoichiometry

Total loss= Loss given

32 170

2 in the problem = 1

216 + 662 , + 24HJO Compound (B)

tions ofCz86 and C2H4 in the mixture.

K2SO,.Al2(S04)1.24Hp

(IIT 1995]

Compound (C) As 0.321 g ofsulphur react with 1.422 g of(A)

So 32.1 g of sulphur react with

142.2 g wt. is related to two moles ofMOx or KOx. 2 (39 + 16X] • 142.2 142.2

Volume of echene

So compound (A) is K02 {Potassium superoxide) which is a binary compound.

Compound (C)

•

K1SO,.Al2(S0•),.24Hp

the form ofFe(III)? [UT 1994]

-

perccoiagc <

�=�

�

= 6.023

x I 011

Sites occupied by nitrogen molecule (200/o) =

2 1

�x 6.023 x I017 = 1.2046 x I017

Number of nitrogen molecule = T= 298 K, PP= 0.001 atm

(iii) 2KMnO,

V = 2.46 cm3

+ 3H2so, + sHp2 -

R

S mote

mole

K,SO, + 2MnSO, + 8H20 + 501 mM ofH_p2 = mM ofKMnO, =

M

x 20 .. 2

� x t =2

x

t

=

0.082 1

=

How many grams of CaO are required to nen­ traliz.e 852 g of Pp10? l flT 200SI

�i x NA

2.46 x 1 0-1 L

L-atm/K/mol

Number of nitrogen molecule =

=

O.OOt x 2.46 x 10-1 x 6 023 x 0.0821 x298 ·

6.02 x

tOlJ

1016

Thus number of sites occupied by each nitrogen molecule Sites occupied by N2 molecules Number ofN2 molecules -

_

20.

18 moles ofCaO.

cm1• Density of surface sites s i 6.023 x I 0''/cm2 and surface area is I 000 cm2, find out the number of surface sites occupied per molecule ofN,.

(i) Mn02 L + N8zC20, + 2H2SO,

Solution The reaction is as follows:

=

1.2046 x I 011 =2 6.02 x I 016

1.34 • Mole Concept and Stoichiometry

MULTIPLE-CHOICE QUESTIONS

Straight Objective Type Questions

8.

a.

{Single Choice only) I.

If 0.50 mole of BaCI, is mixed "'ith 0.20 mol of NaIPO', the ma.'limum number of moles of Ba1(P04)2 thnt cnn be fonned is

c. 0.30 2.

d. 0.40

One mole of N1H4 loses ten moles of electrons to fonn a new compound Y. Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in Y? (There is no change in the oxidation state of hydrogen).

The equivalent weight of MnS04 is half iis molecu-

lar weight when it is convened to a.

b. Mno.1-

MnO

e. Mn01 4.

b. MolaJity

One litre hard water contnins 12.00 mg Mg1•• Milliequivalents of washing soda required to remove its hardness is a.

c.

6.

b. 12.16

I x 10·1

II.

8.67

b. 6.87 d. 4.26

c. 5.67 7.

I c.c. Np at NTP contains

�i� b. 2�J� �·]J 1.

c.

x I 0·1 atoms

,I(

"

I0

d. all of above

The mass of I x 10" molecules ofCuS04.5H20 is a. 41 .59 g

b. 415.9 g

d. none of these

The weight ofone molecule ofa compound CllllH•u is a. 1.2 x I 0-.111 gm

d. 6.023

12.

c.

13.

x I 021 gm

The percentnge weight of Zn in white vitriol [ZnS04.7Hp] is approx:.nately equal to (Zn = 65, S = 32, 0 = 16 and H • I ) ••

33.65 •10

23.65%

a. 1100 ml

14.

d. 3240 ml

d. 8.5 gm

Number of atoms in 4.25 gm ofN� is approximately a. c.

6 x 1011 1.5

x I 021

b. 15 x 1021

d. 2.5

x

102'

The number of molecules in 4.25 g of ammonia is 1.

c.

17.

b. 1120 mJ

b. 7.6gm

c. 6.5 gm

16.

d. 22.65%

The amount of Zinc (atomic weight .. 65) neces­ sary to produce 224 ml of H1 by the reaction with an acid will be a. 0.65 gm

IS.

b. 32.56%

The volume ofcarbon dioxide gas evolved at STP by heating 7.3 gm ofMg(HC01)1 will be

c. 2230 ml

molecules

JO·• electrons

d. 0.5 g molecule

b. 1.4 x I0-11 gm

d. 12.16 x 10-1

At STP the density of CCl4 vapour in g/1 will be nearest to

b. 5 g molecuJe

c. 5.025 x 1011 gm

d. Nonnnlity

c. Molarity s.

11.

d. MnO,

Formality

I g molecule

c. 4.159 g

In which mode of expression, the concentration of a solution remains independent of temperature? a.

The number of grarn-mol�c of oxygen which con­ tain 6.02 x Ioz• CO molecules is c.

10.

2

d. 8.4

a. I 0 g molecule

d. +4

c. ·13 3.

9.

b. -2

a. +2

b. 4.8

3.0

c. 8.0

b. 0.20

a. 0.10

The volume strengthof1.5 N Hp solution is:

1.5 x

1(}!1

3.5x 1011

b. 2.5

d. 15

x I Ql}

x 102l

The weight of a single atom ofoxygen is

•. 5.057 x

1021 g

c. .2.656 x 10-11 g

b. 1.556 x 1 011 g

d. 4.538 x I o-u g

l8.

Mole Concept and Stoichiometry • 1.35 ln the reaction

2:6.

4NH) (g) + 502 (g) _. 4NO (g) + 6Hp(t),

when I mole of ammonia and I mole of 01

made 10 react 10 completion

a. 3.325

17.

b. 1.0 mole of NO will be produ.ced

c.

.alenl weight ofthe metal is

are

a. 1.0 llJOle of H,O is produced all the oxygen will be consumed

b. 20 litres ammonia, 20 litres nitrogen, 20 litres

89

b. 98

59

d. 29

a.

c.

released at STP on heari ng 9.85 g of BaC03 (atomic Ba "' 1 37) will be

•. 2.24 l

c.

b. 4.96 l

is treated with excess dilu� HCI?

is

What is the volume

b. 15.68

a.

89.6

c. 98.4

b. 189.6

JO.

2.584 x 1019

b. 6.023 x I�

ll.

a. 1.3 x I � g

c.

3.72 x 101.1 g

d. l.4 x lQ-l' g

one Fe. h contain 4.6 % of Fe. The approximate molecular mass is 1 100 g moJ-1

b. 1000 g mol·1

d. 1200 g mol-1

The maximum number of molecules is present in a.

15 L of H1 gas al STP

b. 5 L ofN1 gas at STP c.

0.5 g of H1 gas

d. 10 gof01 gas 32.

d. 6.023 x I01t

b. 5.01 x 10-11 g

b. 67.2 L

d. 22.4 L

A compound has haemoglobin like structure. It has

c.

To neutralize completely 20 ml of 0.1 M aqueous solution of phosphorus acid, the volume of 0.1 M aqueous KOH solution required is a.

The weight of one molecule of a compound Ct.oHiu is

44.8 L

a. 1400 g mol·1

d. 169.5

water (volume 0.0018 ml) at room temperature? c.

25.

c.

How many water molecules are th.ere in one drop of a. 4.86 x 101'

What volume ofhydrogtm gas, at 273 K and I atm pn:ssurc will be consumed in obtaining 21.6 g of

a. 89.6 L

at STP

required for complete combustion of 32 g of CH. (mot. wt. of CH" = 16)

b. 0.14

d. 0.35

(atomic mass = 10.8) from the reduction of boron lrichloride by hydrogen?

d. 1.568 x Io�

(in litres) of oxygen

0.28

elemental boron

is 0.5 % by weight (at. wt = 78.4) then minimum

c.. 2.136 x 10'

with 0.1 molar solution of hydrochloric add gave a

c.

Percentage of Se in peroxidase anhydrase enzyme

a. 1.568 x I 0.1

25 ml of a solution of barium hydroxide on titration

a. O.o?

d. 4.26

0.448

I 0 litreS ammonia, 25 litres nitrogen, 15 litres

titre va.lue of35 ml. The molarity of barium hydrox­ ide solution was

29.

b. 2. 1 2

molecular weight of peroxidase anbydrase enzyme

24.

28.

What is the volume (in litres) ofC01 libera1ed at STP.

c.

hydrogen

hydrogen

when2.l2 gram ofsorlium carbonate (mol. wt. "' 106) a. 1 1 .2

hydrogen

d. 20 litres ammonia. I 0 litreS nitrogen, 30 litres

d. 0.84 1

1.12 1

20 litres ammonia, 25 litres nitrogen, 1 5 litres

hydrogen

Assuming fully decomposed, the volume of col mass,

23.

lo Haber process. 30 litres of dihydrogen and 30

litres of dinitrogen were taken for reaction which

NaOH + HfO, -+ NaH,PO, + Hp is c.

22.

d. 33.25

the aforesaid condition in the end?

a.

21.

b. 13.25

The equivalent weight ofphosphoric acid {l\PO,)

in the reaction:

20.

c. 23.52

yielded only 50% of the expected producl. What will be the composition of gaseous mixture under

d. all the ammonia will be consumed

19.

3 g of an oxide of a metal is converted to chloride

completely and it yielded 5 g ofchloride. The equiv­

c.

33.

I 0 ml

b. 40 ml

60 ml

d. 80 ml

6.02 x IO"° molecules ofurea are present in 100 mlof itS solution. The concentration ofurea solution is

1.36

• Mole Concept and Stoichiometry

a.

c.

34.

O.Q2 M

b. 0.001 M

0.01 M

d. 0.1 M

a. One

('. Three

g of CaCO) is completely decompos«l to X and CaO. X is passed into an aqueous solution con­

taining one mole of sodium carbonate. What is the

e. Five 42.

number of moles of sodium bicarbonate formed? ...

84)

a.

0.010

c. 0.4 JS.

d. 1 0

44.

acid. The copper nitrate on strong heating gave 5 g

of its oxide. The equivalent weight of copper is

37.

c. 32

b. 1 6

weight of methane in the mixture?

83 %

46.

b. 73.0 %

number, is

39.

47. d. 0.05 A

a. 270 kg

40.

90 kg

c. e.

41.

II g

33 g

b. 31.6 d. 79

500 ml of NH contains 6.0 x JO?! n_iolecules at J STP. How many molecules are present m 100 ml of a.

kg of nlu-

b. 540 kg

48.

b. 22 g

d. 44 g

5.5 g

An alkaloid contains 17.28 % of nitrogen and its

molecular mass is 162. The number of nitrogen atoms present in one molecule of the alkaloid is

6x

102'

1 . 2 x 10''

b. J.5 x l()ll

d. none of these

1n the reaction, 4NH, + 502 -+ 4NO + 6fl,O, when

one mole of ammonia and one mole of oxygen are made to react to completion, then

d. 180 kg

a.

1.0 mole ofH10 is produced

b. all the oxygen is consumed

c. 1.5 mole of NO is formed

would be the weight ofC01 liberated after the coma.

d. 7

.. = 158) oxidizes oxalic acid in KMn0 (mol. wt 4 acidic medium 10 C01 and water as follows.

c.

100 g CaC01 is treated with I litre ofN HCI. What

pletion ofthe reaction?

b. 3

6

C01 at STP?

minium metal from bauxite by the Hall process is (al

c.

One mole of acidified �Crp, on reaction with excess Kl will liberate . . . . . . moles (s) ofl2

c. 39.5

The mass of carbon anode consumed (giving Of!IY mass ofAl = 27)

b. 28 g d . 20 g

a. 158

b. 0.5 A/18

carbon dioxide) n i the production of 270

56 g

What is the equivalent weight of KMn0."7

I ml ofwa1er has 20 drops and 'A' is Avogadro's

c. 0.05 A/18

kg-' kg-1

5Cp;'" +2Mn0; + 16H' �1oco1+2Mn1• +8Hp

d. 80.0 %

Number ofwater molecules in the drop ofwater, if

a. 0.5A

d. 0.44 mol

c. 4 2 g

c.

e. 90.1 %

38.

b. 3.28 mol

gave I I .2 dm' of CO, gas at STP. The mass of

a. 2

A gas mixture contains 50 % helium and 50 %

a.

45.

d. 34

methane by volume. What is the per cent by

c. 18.01 %

1 . 14 mol kg-'

2.28 mot kg-'

The decomposition of a certain mass of eaco,

a.

4 g of co·pper was dissolved in concentrated nitric

12

d. I O m l ofH1

KOH required to completely neutrali2e the gas is

d. CH

a.

b. 5 ml ofH2

is 1.02 g/ml. The molality of the solution s i c.

An organic compound containing C and H has

92.3% of carbon. Its empirical formula is

10ml of02

Density ofa 2.05 M solution ofacetic acid in water a.

b. CH1

36.

what is left at the end of the reaction? c.

43.

b. 0.2

l f 30 ml of Hz and 20 ml of 01 react tO form water, a. 5 ml of02

(mol. wt. ofCaCO, = l 00, NazC01 = J 06, NaHC01

b. Two d. Four

d. all the ammonia is consumed

49.

4 moles

each of S01 and 02 gases are allowed to

react to form SO, in a closed vessel. At equilibrium,

25 % of01 is used up. The total number of moles of all the gases at equilibrium is a.

2.0

c. 7.0

b. 6.0

d. 8.0

50.

Mole Concept and Stoichiometry • 1.37 1.520 g ofhydrox.ide ofa metal on ignition gave 0.995

g ofoxide. The equivalent weight ofthe metal is ••

c.

St.

0.995

b. 190

1.90

d. 9

g ofNaCl n i I00 ml ofwater is c.

52.

0.0472 moles

(mol. wt. 244) gave 0.195 gram of platinum on

a.

4

b. 3

2

d. I

0.5 N

b. I 0.0 N

59.

200x : 90%

c.

60.

6J.

The weighted average atomic

55.

mass

is closest to

62.

b. 201 amu

b. 10.625

d. . 9.876

volume of' 1 0 volume' Hp1 required will be 10.3

b. 22.7

c. 23.0

d. 44.8

An unknown element fonns an oxide. What will be

the equivalent weight of the element if tbe oxygen

loses 55.9 % of its weight. The formula of the

2

4.0 N H,S04

medium will be

c.

NllzS04.7Hp

c.

e.

N�S04.2Hp

a.

64.

�

7.5 moles 0.6 moles

b. 0.2 moles

d. 0.4 moles

How many grams ofcarbon dioxide can form when

a mixture of 4.95 g ethylene (C2H4) and 3.25 g of oxygen is ignited, assuming complete combustion

1 0 litres of02 gas is reacted with 30 litres of CO at

STP. The volume of each gas present at the end of

to form carbon dioxide and water?

a. 02 = 1 0 litres, CO:: 30 litres

c. 1.98 g

a.

the reaction are

b. 02 = 1 0 liues, co2 = 20 litres CO= 10 litres, C02 = 20 litres

d. CO • 20 litres, C02 "' I 0 litres

e.

02"' 10 litres, CO= 1 0 litres

.

d. 2 N H2so.

Number of moles of Mno.- required to oxidize

N3:.S04.10Hp

d. N�SO,.SH,O

b. I N H so

one mole of ferrous oxalate completely in acidic

a.

b. Na1S04.6H20

d. 54

as a

c. o.5 N H1so. 63.

b. 32

2 N HCI solution will have same molar cone. a.

d. 199 amu

The crystalline salt Na2SO,.xH20 on heating

c.

presence

In transforming 0.01 mole of PbS to PbS04 the

c.

of the naturally

crystalline salt is

56.

5.550 9.000

a. 14

mx : 2.0%

202 amu

n i

content is 20% by weight?

1"X : 8.0%

c.

Sodium nitrates on reduction with Zn

of NaOH solution produces NH1• Mass of sodium

a.

tion;

a. 200amu

a. 3.65 g ofHCl

d. 0.365 g o f HCl

d. 5.0N

An element, X has the following isotopic composi­

occurring element X

5 g of CaCO, comple�ely reaclS with

a.

acid is diluted to 10 litres with water. What is the

c.

d. 46

nitrate absorbing I mole ofelectron will be

normality of the resulting solution? 1.0 N

b. 60

c. 36.5 g of HCI

One litre solution containing 490 g of sulphuric

1.

30

b. 7.35 g ofHCl

d. 0.00427 moles

ignition. The number of nitrogen atoms per mol­

54.

58.

b. 427 moles

ecule of base is

53.

a. 20

0.532 gram of chloroplatinate of an organic base

c.

0.3 g of an acid is neutralized by 40 cm' of 0.125 N

NaOH. Equivalent mass ofthe acid is c.

The moJarity ofthe solution obtained by dissolving 2.5 1. 0.427 moles

57.

65.

5.96 g

b. 1.49 g d. 2.98 g

Hardness of water sample is 300 ppm CaCO,.

Hence its molarity is u.

c.

0.30 M

0.030 M

b. 0.003 M d. 0.0013 M