Mole Concept

PAGE # 18

MOLE CONCEPT

ATOMS

All the matter is made up of atoms. An atom is thesmallest particle of an element that can take part in achemical reaction. Atoms of most of the elementsare very reactive and do not exist in the free state (assingle atom).They exist in combination with the atomsof the same element or another element.Atoms are very, very small in size. The size of an atomis indicated by its radius which is called "atomicradius" (radius of an atom). Atomic radius ismeasured in "nanometres"(nm).1 metre = 109 nanometre or 1nm = 10-9 m.Atoms are so small that we cannot see them underthe most powerful optical microscope.

Note :

Hydrogen atom is the smallest atom of all , having anatomic radius 0.037nm.

(a) Symbols of Elements :

A symbol is a short hand notation of an element whichcan be represented by a sketch or letter etc.Dalton was the first to use symbols to representelements in a short way but Dalton's symbols forelement were difficult to draw and inconvenient touse, so Dalton's symbols are only of historicalimportance. They are not used at all.

It was J.J. Berzelius who proposed the modernsystem of representing en element.The symbol of an element is the "first letter" or the"first letter and another letter" of the English name orthe Latin name of the element.

e.g. The symbol of Hydrogen is H.The symbol of Oxygen is O.There are some elements whose names begin withthe same letter. For example, the names of elementsCarbon, Calcium, Chlorine and Copper all begin withthe letter C. In such cases, one of the elements isgiven a "one letter "symbol but all other elements aregiven a "first letter and another letter" symbol of theEnglish or Latin name of the element. This is to benoted that "another letter" may or may not be the"second letter" of the name. Thus,The symbol of Carbon is C.The symbol of Calcium is Ca.The symbol of Chlorine is Cl.

The symbol of Copper is Cu (from its Latin nameCuprum)

It should be noted that in a "two letter" symbol, thefirst letter is the "capital letter" but the second letter isthe small letter

English Name of the Element

Symbol

Hydrogen HHelium HeLithium LiBoron B

Carbon CNitrogen NOxygen OFluorine F

Neon NeMagnesium MgAluminium Al

Silicon SiPhosphorous P

Sulphur SChlorine ClArgon Ar

Calcium Ca

Symbol Derived from English Names

Symbols Derived from Latin Names

English Name of the Element

SymbolLatin Name of the Element

Sodium Na Natrium

Potassium K Kalium

(b) Significance of The Symbol of an

Element :

(i) Symbol represents name of the element.

(ii) Symbol represents one atom of the element.

(iii) Symbol also represents one mole of the element.That is, symbol also represent 6.023 × 1023 atoms ofthe element.

(iv) Symbol represent a definite mass of the element

i.e. atomic mass of the element.

Example :(i) Symbol H represents hydrogen element.

(ii) Symbol H also represents one atom of hydrogenelement.

(iii) Symbol H also represents one mole of hydrogenatom.(iv) Symbol H also represents one gram hydrogen

atom.

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PAGE # 19

IONS

An ion is a positively or negatively charged atom orgroup of atoms.Every atom contains equal number of electrons(negatively charged) and protons (positively charged).Both charges balance each other, hence atom iselectrically neutral.

(a) Cation :

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

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

(b) Anion :

If an atom has more number of electrons than that ofneutral atom, then it gets negatively charged and anegatively charged ion is known as anion.e.g. Chloride ion (Cl¯), oxide ion (O2-) etc.

An anion bears that much units of negative charge asthere are the number of electrons gained by theneutral atom to form that anion.e.g. A nitrogen atom gains 3 electrons to form nitrideion, so nitride ion bears 3 units of negative chargeand it is represented as N3-.

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

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

(ii) Polyatomic ions : Those ions which are formedfrom group of atoms joined together are calledpolyatomic ions or compound ions.e.g. Ammonium ion (NH

4+) , hydroxide ion (OH�) etc.

which are formed by the joining of two types of atoms,nitrogen and hydrogen in the first case and oxygen andhydrogen in the second.

(d) Valency of ions :The valency of an ion is same as the charge presenton the ion.If an ion has 1 unit of positive charge, its valency is 1and it is known as a monovalent cation. If an ion has2 units of negative charge, its valency is 2 and it isknown as a divalent anion.

LIST OF COMMON ELECTROVALENT POSITIVE RADICALS

LIST OF COMMON ELECTROVALENT NEGATIVE RADICALS

Monovalent Electronegative Bivalent Electronegative

Trivalent Electronegative

Tetravalent Electronegative

1. Fluoride F� 1. Sulphate SO 42- 1. Nitride N3- 1. Carbide C4-

2. Chloride Cl� 2. Sulphite SO 32- 2. Phosphide P3-

3. Bromide Br� 3. Sulphide S2-3. Phosphite PO3

3-

4. Iodide I 4. Thiosulphate S2O32- 4. Phosphate PO4

3-

5. Hydride H� 5. Zincate ZnO22-

6. Hydroxide OH� 6. Oxide O2-

7. Nitrite NO2� 7. Peroxide O2

2-

8.Nitrate NO3� 8. Dichromate Cr2O7

2-

9. Bicarbonate or Hydrogen carbonate HCO3� 9. Carbonate CO3

2-

10. Bisulphite or Hydrogen sulphite HSO3� 10. Silicate SiO 3

2-

11. Bisulphide or Hydrogen sulphide HS�

12. Bisulphate or Hydrogen sulphate HSO4�

13. Acetate CH COO3�

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

PAGE # 20

LAWS OF CHEMICAL COMBINATION

The laws of chemical combination are theexperimental laws which led to the idea ofatoms being the smallest unit of matter. The laws ofchemical combination played a significant role in the

development of Dalton�s atomic theory of matter.

There are two important laws of chemical combination.These are:

(i) Law of conservation of mass(ii) Law of constant proportions

(a) Law of Conservation of Mass or Matter :

This law was given by Lavoisier in 1774 . According tothe law of conservation of mass, matter can neitherbe created nor be destroyed in a chemical reaction.

OrThe law of conservation of mass means that in achemical reaction, the total mass of products is equalto the total mass of the reactants. There is no changein mass during a chemical reaction.Suppose we carry out a chemical reaction between Aand B and if the products formed are C and D then,A + B C + D

Suppose 'a' g of A and 'b' g of B react to produce 'c' g ofC and 'd' g of D. Then, according to the law ofconservation of mass, we have,

a + b = c + dExample :When Calcium Carbonate (CaCO

3) is heated, a

chemical reaction takes place to form Calcium Oxide(CaO) and Carbon dioxide (CO

2). It has been found

by experiments that if 100 grams of calcium carbonateis decomposed completely, then 56 grams of CalciumOxide and 44 grams of Carbon dioxide are formed.

Since the total mass of products (100g ) is equal tothe total mass of the reactants (100g), there is nochange in the mass during this chemical reaction.The mass remains same or conserved.

(b) Law of Constant Proportions / Law of

Definite Proportions :

Proust, in 1779, analysed the chemical composition(types of elements present and percentage ofelements present ) of a large number of compoundsand came to the conclusion that the proportion ofeach element in a compound is constant (or fixed).According to the law of constant proportions: Achemical compound always consists of the sameelements combined together in the same proportionby mass.

Note :The chemical composition of a pure substance isnot dependent on the source from which it is obtained.

Example :

Water is a compound of hydrogen and oxygen. It canbe obtained from various sources (like river, sea, welletc.) or even synthesized in the laboratory. Fromwhatever source we may get it, 9 parts by weight ofwater is always found to contain 1 part by weight ofhydrogen and 8 parts by weight of oxygen. Thus, inwater, this proportion of hydrogen and oxygen alwaysremains constant.

Note :The converse of Law of definite proportions that whensame elements combine in the same proportion, thesame compound will be formed, is not always true.

(c) Law of Multiple Proportions :

According to it, when one element combines with theother element to form two or more different compounds,the mass of one element, which combines with aconstant mass of the other, bears a simple ratio toone another.

Simple ratio means the ratio between small naturalnumbers, such as 1 : 1, 1 : 2, 1 : 3e.g.Carbon and oxygen when combine, can form twooxides that are CO (carbon monoxide), CO

2 (carbon

dioxide).In CO,12 g carbon combined with 16 g of oxygen.In CO

2,12 g carbon combined with 32 g of oxygen.

Thus, we can see the mass of oxygen which combinewith a constant mass of carbon (12 g) bear simpleratio of 16 : 32 or 1 : 2

Note :The law of multiple proportion was given by Dalton in1808.

Sample Problem :

1. Carbon is found to form two oxides, which contain42.8% and 27.27% of carbon respectively. Show thatthese figures illustrate the law of multiple proportions.

Sol. % of carbon in first oxide = 42.8% of oxygen in first oxide = 100 - 42.8 = 57.2% of carbon in second oxide = 27.27 % of oxygen in second oxide = 100 - 27.27 = 72.73

For the first oxide -Mass of oxygen in grams that combines with 42.8 gof carbon = 57.2 Mass of oxygen that combines with 1 g of carbon =

1.3442.857.2

g

For the second oxide -Mass of oxygen in grams that combines with 27.27 gof carbon = 72.73 Mass of oxygen that combines with 1 g of carbon =

2.6827.2772.73

g

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

PAGE # 21

(d) Law of Reciprocal Proportions :

According to this law the ratio of the weights of twoelement A and B which combine separately with afixed weight of the third element C is either the sameor some simple multiple of the ratio of the weights inwhich A and B combine directly with each other.e.g.

The elements C and O combine separately with thethird element H to form CH

4 and H

2O and they combine

directly with each other to form CO2.

H 24

CH4 H O2

C O16

12

12

CO2 32

In CH4, 12 parts by weight of carbon combine with 4

parts by weight of hydrogen. In H2O, 2 parts by weight

of hydrogen combine with 16 parts by weight ofoxygen. Thus the weight of C and O which combinewith fixed weight of hydrogen (say 4 parts by weight)are 12 and 32 i.e. they are in the ratio 12 : 32 or 3 : 8.Now in CO

2, 12 parts by weight of carbon combine

directly with 32 parts by weight of oxygen i.e. theycombine directly in the ratio 12 : 32 or 3 : 8 which isthe same as the first ratio.

Note :

The law of reciprocal proportion was put forward byRitcher in 1794.

Sample Problem :2. Ammonia contains 82.35% of nitrogen and 17.65%

of hydrogen. Water contains 88.90% of oxygen and11.10% of hydrogen. Nitrogen trioxide contains63.15% of oxygen and 36.85% of nitrogen. Show thatthese data illustrate the law of reciprocal proportions.

Sol

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

1g of H combine with N = 17.65

82.35 g = 4.67 g

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

1 g H combine with O = 11.10

88.90g = 8.01 g

Ratio of the weights of N and O which combinewith fixed weight (=1g) of H= 4.67 : 8.01 = 1 : 1.72

In N2O

3, ratio of weights of N and O which combine

with each other = 36.85 : 63.15 = 1 : 1.71

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

(e) Gay Lussac�s Law of Gaseous Volumes :

Gay Lussac found that there exists a definiterelationship among the volumes of the gaseousreactants and their products. In 1808, he put forwarda generalization known as the Gay Lussac�s Law of

combining volumes. This may be stated as follows :

When gases react together, they always do so in

volumes which bear a simple ratio to one another

and to the volumes of the product, if these are also

gases, provided all measurements of volumes are

done under similar conditions of temperature and

pressure.

e.g.Combination between hydrogen and chlorine to form

hydrogen chloride gas. One volume of hydrogen and

one volume of chlorine always combine to form two

volumes of hydrochloric acid gas.

H2 (g) + Cl

2 (g) 2HCl (g)

1vol. 1 vol. 2 vol.

The ratio between the volume of the reactants and

the product in this reaction is simple, i.e., 1 : 1 : 2.

Hence it illustrates the Law of combining volumes.

(f) Avogadro�s Hypothesis :

This states that equal volumes of all gases under

similar conditions of temperature and pressure

contain equal number of molecules.

This hypothesis has been found to explain elegantly

all the gaseous reactions and is now widely

recognized as a law or a principle known as Avogadro�s

Law or Avogadro�s principle.

The reaction between hydrogen and chlorine can be

explained on the basis of Avogadro�s Law as follows :

Hydrogen + Chlorine Hydrogen chloride gas 1 vol. (By experiment)1 vol. 2 vol.

n molecules. n molecules. 2n molecules.(By Avogadro's Law)

21 molecules. molecules. 1 molecules. (By dividing throughout by 2n)

1 Atom 1 Atom 1 Molecule (Applying Avogadro's hypothesis)

21

It implies that one molecule of hydrogen chloride gas

is made up of 1 atom of hydrogen and 1 atom of

chlorine.

(i) Applications of Avogadro�s hypothesis :

(A) In the calculation of atomicity of elementarygases.

e.g.

2 volumes of hydrogen combine with 1 volume of

oxygen to form two volumes of water vapours.Hydrogen + Oxygen Water vapours2 vol. 1 vol. 2 vol.

PAGE # 22

Applying Avogadro�s hypothesis

Hydrogen + Oxygen Water vapours2 n molecules n molecules 2 n molecules

or 1 molecule 2

1 molecule 1 molecule

Thus1 molecule of water contains 2

1 molecule of

oxygen. But 1 molecule of water contains 1 atom of

oxygen. Hence. 2

1 molecule of oxygen = 1 atom of

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

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

Vapour density (V.D.) = hydrogenofDensitygas ofDensity

=

epressur and temp. same the at hydrogen of volume same the of Mass

gas the of volume certain a of Mass

If n molecules are present in the given volume of a gasand hydrogen under similar conditions of temperatureand pressure.

V.D. = hydrogen of molecules n of Massgas the of molecules n of Mass

= hydrogen of molecule 1 of Massgas the of molecule 1 of Mass

= hydrogen of mass Moleculargas the of mass Molecular

= 2

mass Molecular

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

ATOMIC MASS UNIT

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

RELATIVE ATOMIC MASS

The atomic mass of an element is a relative quantityand it is the mass of one atom of the element relativeto one -twelfth (1/12) of the mass of one carbon-12atom. Thus, Relative atomic mass

= atom12Coneofmass

12

1

elementtheofatomoneofMass

[1/12 the mass of one C-12 atom = 1 amu, 1 amu =1.66 × 10�24 g = 1.66 × 10�27 kg.]

Note :One amu is also called one dalton (Da).

GRAM-ATOMIC MASS

The atomic mass of an element expressed in grams

is called the Gram Atomic Mass of the element.

The number of gram -atoms

= elementtheofmassAtomicGram

gramsinelementtheofMass

e.g.

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

and (ii) 64g of sulphur.

(i) The atomic mass of oxygen = 16.

Gram-Atomic Mass of oxygen (O) = 16 g.

No. of Gram-Atoms = 1616

= 1

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

= sulphurofMass AtomicGram64

= 32

64= 2

AtomicNumber Element Symbol

Atomicmass

1 Hydrogen H 12 Helium He 43 Lithium Li 74 Beryllium Be 95 Boron B 116 Carbon C 127 Nitrogen N 148 Oxygen O 169 Fluorine F 1910 Neon Ne 2011 Sodium Na 2312 Magnesium Mg 2413 Aluminium Al 2714 Silicon Si 2815 Phosphorus P 3116 Sulphur S 3217 Chlorine Cl 35.518 Argon Ar 4019 Potassium K 3920 Calcium Ca 40

RELATIVE MOLECULAR MASS

The relative molecular mass of a substance is themass of a molecule of the substance as comparedto one-twelfth of the mass of one carbon -12 atomi.e.,Relative molecular mass

= atom12Coneofmass

12

1

substancetheofmoleculeoneofMass

The molecular mass of a molecule, thus, representsthe number of times it is heavier than 1/12 of themass of an atom of carbon-12 isotope.

PAGE # 23

GRAM MOLECULAR MASS

The molecular mass of a substance expressed in

grams is called the Gram Molecular Mass of the

substance . The number of gram molecules

= ancetsubstheofmassmolecularGram

gramsintancesubstheofMass

e.g.

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

Gram Molecular Mass of hydrogen (H2) = 2 g .

(ii) Molecular mass of methane (CH4) = 16u

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

e.g. the number of gram molecules present in 64 g of

methane (CH4).

= 4CHofmassmolecularGram

64 =

16

64 = 4.

(a) Calculation of Molecular Mass :

The molecular mass of a substance is the sum of

the atomic masses of its constituent atoms present

in a molecule.

Ex.1 Calculate the molecular mass of water.

(Atomic masses : H = 1u, O = 16u).

Sol. The molecular formula of water is H2O.

Molecular mass of water = ( 2 × atomic mass of H)

+ (1 × atomic mass of O)

= 2 × 1 + 1 × 16 = 18

i.e., molecular mass of water = 18 amu.

Ex.2 Find out the molecular mass of sulphuric acid.

(Atomic mass : H = 1u, O = 16u, S = 32u).

Sol. The molecular formula of sulphuric acid is H2SO

4.

Molecular mass of H2SO

4

= (2 × atomic mass of H) + ( 1 × atomic mass of S)

+ ( 4 × atomic mass of O)

= (2 × 1) + (1× 32) + (4×16) = 2 + 32 + 64 = 98

i.e., Molecular mass of H2SO

4= 98 amu.

FORMULA MASS

The term �formula mass� is used for ionic compounds

and others where discrete molecules do not exist,

e.g., sodium chloride, which is best represented as

(Na+Cl�)n, but for reasons of simplicity as NaCl or

Na+Cl�. Here, formula mass means the sum of the

masses of all the species in the formula.

Thus, the formula mass of sodium chloride = (atomic

mass of sodium) + (atomic mass of chlorine)

= 23 + 35.5

= 58.5 amu

EQUIVALENT MASS

(a) Definition :

Equivalent mass of an element is the mass of theelement which combine with or displaces 1.008 partsby mass of hydrogen or 8 parts by mass of oxygen or35.5 parts by mass of chlorine.

(b) Formulae of Equivalent Masses of differentsubstances :(i) Equivalent mass of an element =

element the ofValency element the of wt.Atomic

(ii) Eq. mass an acid = acid the ofBasicity acid the of wt. Mol.

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

(iii) Eq. Mass of a base = base the ofAcidity base the of wt. Mol.

Acidity is the number of replaceable OH� ions fromone molecule of the base

(iv) Eq. mass of a salt

= metal ofvalency atoms metal of Numbersalt the of wt. Mol.

(v) Eq. mass of an ion = ion the on Charge

ion the of wt.Formula

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

=

substance the ofmolecule

oneby gained or lost electrons of No.

wt At. or wt.Mol

Equivalent weight of some compounds are given inthe table :

S.No. CompoundEquivalent

weight

1 HCl 36.5

2 H2SO4 49

3 HNO3 63

4 45

5 .2H2O63

6 NaOH 40

7 KOH 56

8 CaCO3 50

9 NaCl 58.5

10 Na2CO3 53

COOH

COOH

PAGE # 24

In Latin, mole means heap or collection or pile. Amole of atoms is a collection of atoms whose totalmass is the number of grams equal to the atomicmass in magnitude. Since an equal number of molesof different elements contain an equal number ofatoms, it becomes convenient to express theamounts of the elements in terms of moles. A molerepresents a definite number of particles, viz, atoms,molecules, ions or electrons. This definite number iscalled the Avogadro Number (now called the Avogadroconstant) which is equal to 6.023 × 1023.

A mole is defined as the amount of a substance thatcontains as many atoms, molecules, ions, electronsor other elementary particles as there are atoms inexactly 12 g of carbon -12 (12C).

(a) Moles of Atoms :

(i) 1 mole atoms of any element occupy a mass whichis equal to the Gram Atomic Mass of that element.

e.g. 1 Mole of oxygen atoms weigh equal to GramAtomic Mass of oxygen, i.e. 16 grams.

(ii) The symbol of an element represents 6.023 x 1023

atoms (1 mole of atoms) of that element.

e.g : Symbol N represents 1 mole of nitrogen atomsand 2N represents 2 moles of nitrogen atoms.

(b) Moles of Molecules :

(i) 1 mole molecules of any substance occupy a masswhich is equal to the Gram Molecular Mass of thatsubstance.

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

(ii) The symbol of a compound represents 6.023 x1023 molecules (1 mole of molecules) of thatcompound.

e.g. : Symbol H2O represents 1 mole of watermolecules and 2 H2O represents 2 moles of watermolecules.

Note :

The symbol H2O does not represent 1 mole of H2molecules and 1 mole of O atoms. Instead, itrepresents 2 moles of hydrogen atoms and 1 moleof oxygen atoms.

Note :The SI unit of the amount of a substance is Mole.

(c) Mole in Terms of Volume :

Volume occupied by 1 Gram Molecular Mass or 1mole of a gas under standard conditions oftemperature and pressure (273 K and 1atm.pressure) is called Gram Molecular Volume. Its valueis 22.4 litres for each gas.Volume of 1 mole = 22.4 litre (at STP)

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

SOME IMPORTANT RELATIONS AND FORMULAE

(i) 1 mole of atoms = Gram Atomic mass = mass of6.023 × 1023 atoms

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

= elementofMassAtomicGram

gramsinelementofMass

(iv) Number of moles of molecules

= substanceofMassMolecularGram

gramsinsubstanceofMass

(v) Number of moles of molecules

= AN

N

numberAvogadro

elementofmolecules ofNo.

Ex.3 To calculate the number of moles in 16 grams ofSulphur (Atomic mass of Sulphur = 32 u).

Sol. 1 mole of atoms = Gram Atomic Mass.So, 1 mole of Sulphur atoms = Gram Atomic Mass ofSulphur = 32 grams.Now, 32 grams of Sulphur = 1 mole of SulphurSo, 16 grams of Sulphur= (1/32) x 16 = 0.5 molesThus, 16 grams of Sulphur constitute 0.5 mole ofSulphur.

6.023 × 10

(N ) Atoms

23

A

6.023 × 10

(N ) molecules

23

A

1 Mole

1 gram atomof element

1 gram molecule of substance

1 gram formula mass of substance

In terms of particles

In terms of mass

22.4 litre

In term ofvolume

PROBLEMS BASED ON THE MOLE CONCEPT

Ex.4 Calculate the number of moles in 5.75 g of sodium. (Atomic mass of sodium = 23 u)

Sol. Number of moles

= elementofMassAtomicGram

gramsinelementtheofMass =

23

5.75= 0.25 mole

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

5.75 g of sodium = 23

5.75mole of sodium = 0.25 mole

PAGE # 25

Ex.5 What is the mass in grams of a single atom ofchlorine ? (Atomic mass of chlorine = 35.5u)

Sol. Mass of 6.022 × 1023 atoms of Cl = Gram AtomicMass of Cl = 35.5 g.

Mass of 1 atom of Cl =23106.022

g35.5

= 5.9 × 10�23 g.

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

Sol. Mass of mercury (Hg) in grams = Density(g cm�3)× Volume (cm3)= 13.6 g cm�3 × 1000 cm3 = 13600 g.

Number of moles of mercury

= mercuryofMassAtomicGramgramsinmercuryofMass

=200

13600 = 68

Ex.7 The mass of a single atom of an element M is3.15× 10�23 g . What is its atomic mass ? Whatcould the element be ?

Sol. Gram Atomic Mass = mass of 6.022 × 1023 atoms= mass of 1 atom × 6.022 × 1023

= (3.15 × 10�23g) × 6.022 × 1023

= 3.15 × 6.022 g = 18.97 g.

Atomic Mass of the element = 18.97uThus, the element is most likely to be fluorine.

Ex.8 An atom of neon has a mass of 3.35 × 10�23 g.How many atoms of neon are there in 20 g of thegas ?

Sol. Number of atoms

= atom1ofMass

massTotal = 23�103.35

02

= 5.97 × 1023

Ex.9 How many grams of sodium will have the samenumber of atoms as atoms present in 6 g ofmagnesium ?(Atomic masses : Na = 23u ; Mg =24u)

Sol. Number of gram -atom of Mg

= MassAtomicGramgramsinMgofMass

= 246

= 4

1

Gram Atoms of sodium should be = 4

1

1 Gram Atom of sodium = 23 g

4

1 gram atoms of sodium = 23 ×

4

1 = 5.75 g

Ex.10 How many moles of Cr are there in 85g of Cr2S

3 ?

(Atomic masses : Cr = 52 u , S =32 u)Sol. Molecular mass of Cr

2S

3 = 2 × 52 + 3 × 32 = 104

+ 96 = 200 u.200g of Cr

2S

3 contains = 104 g of Cr.

85 g of Cr2S

3 contains =

200

85104 g of Cr = 44.2g

Thus, number of moles of Cr = 52

44.2 = 0.85 .

Ex.11 What mass in grams is represented by

(a) 0.40 mol of CO2,

(b) 3.00 mol of NH3,

(c) 5.14 mol of H5IO

6

(Atomic masses : C=12 u, O=16 u, N=14 u,

H=1 u and I = 127 u)

Sol. Weight in grams = number of moles × molecular

mass.

Hence,

(a) mass of CO2 = 0.40 × 44 = 17.6 g

(b) mass of NH3 = 3.00 × 17 = 51.0 g

(c) mass of H5IO

6 = 5.14 × 228 = 1171.92g

Ex.12 Calculate the volume in litres of 20 g of hydrogen

gas at STP.

Sol. Number of moles of hydrogen

= hydrogenofMassMolecularGram

grams in hydrogenofMass =

220

= 10

Volume of hydrogen = number of moles × Gram

Molecular Volume.

= 10 ×22.4 = 224 litres.

Ex.13 The molecular mass of H2SO

4 is 98 amu.

Calculate the number of moles of each elementin 294 g of H

2SO

4.

Sol. Number of moles of H2SO

4 =

98

294= 3 .

The formula H2SO

4 indicates that 1 molecule of

H2SO

4 contains 2 atoms of H, 1 atom of S and 4

atoms of O. Thus, 1 mole of H2SO

4 will contain 2

moles of H,1 mole of S and 4 moles of O atomsTherefore, in 3 moles of H

2SO

4 :

Number of moles of H = 2 × 3 = 6

Number of moles of S = 1 × 3 = 3

Number of moles of O = 4 × 3 = 12

Ex.14 Find the mass of oxygen contained in 1 kg ofpotassium nitrate (KNO

3).

Sol. Since 1 molecule of KNO3 contains 3 atoms of

oxygen, 1 mol of KNO3 contains 3 moles of

oxygen atoms. Moles of oxygen atoms = 3 × moles of KNO

3

= 3 × 101

1000 = 29.7

(Gram Molecular Mass of KNO3 = 101 g)

Mass of oxygen = Number of moles × Atomic

mass= 29.7 × 16 = 475.2 g

Ex.15 You are asked by your teacher to buy 10 moles ofdistilled water from a shop where small bottleseach containing 20 g of such water are available.How many bottles will you buy ?

Sol. Gram Molecular Mass of water (H2O) = 18 g

10 mol of distilled water = 18 × 10 = 180 g.

Because 20 g distilled water is contained in 1bottle,

180 g of distilled water is contained in =20

180

bottles = 9 bottles.

Number of bottles to be bought = 9

PAGE # 26

Ex.16 6.022 × 1023 molecules of oxygen (O2) is equal tohow many moles ?

Sol. No. of moles =

AN

N

moleculesofno.sAvogadro'

oxygenofmoleculesofNo. =

23

23

106.023

106.023

= 1

PERCENTAGE COMPOSITION

The percentage composition of elements in acompound is calculated from the molecular formulaof the compound.The molecular mass of the compound is calculatedfrom the atomic masses of the various elementspresent in the compound. The percentage by massof each element is then computed with the help of thefollowing relations.Percentage mass of the element in the compound

= massMolecular

elementtheofmassTotal× 100

Ex.17 What is the percentage of calcium in calciumcarbonate (CaCO

3) ?

Sol. Molecular mass of CaCO3 = 40 + 12 + 3 × 16

= 100 amu.Mass of calcium in 1 mol of CaCO

3 = 40g.

Percentage of calcium = 100

10040 = 40 %

Ex.18 What is the percentage of sulphur in sulphuricacid (H

2SO

4) ?

Sol. Molecular mass of H2SO

4 = 1 × 2 + 32 + 16 × 4 = 98

amu.

Percentage of sulphur = 98

10032 = 32.65 %

Ex.19 What are the percentage compositions ofhydrogen and oxygen in water (H

2O) ?

(Atomic masses : H = 1 u, O = 16 u)

Sol. Molecular mass of water, H2O = 2 + 16 = 18 amu.

H2O has two atoms of hydrogen.

So, total mass of hydrogen in H2O = 2 amu.

Percentage of H = 18

1002 = 11.11 %

Similarly,

percentage of oxygen = 18

10016 = 88.88 %

The following steps are involved in determining theempirical formula of a compound :(i) The percentage composition of each element isdivided by its atomic mass. It gives atomic ratio of theelements present in the compound.

(ii) The atomic ratio of each element is divided by theminimum value of atomic ratio as to get the simplestratio of the atoms of elements present in thecompound.

(iii) If the simplest ratio is fractional, then values ofsimplest ratio of each element is multiplied bysmallest integer to get the simplest whole numberfor each of the element.

(iv) To get the empirical formula, symbols of variouselements present are written side by side with theirrespective whole number ratio as a subscript to thelower right hand corner of the symbol.

(v) The molecular formula of a substance may bedetermined from the empirical formula if the molecularmass of the substance is known. The molecularformula is always a simple multiple of empiricalformula and the value of simple multiple (n) isobtained by dividing molecular mass with empiricalformula mass.

n = MassFormulaEmpiricalMassMolecular

Ex-20 A compound of carbon, hydrogen and nitrogencontains these elements in the ratio of 9:1:3.5 respectively.Calculate the empirical formula. If its molecular mass is108, what is the molecular formula ?

Sol.

ElementMassRatio

AtomicMass

Relative Numberof Atoms

Simplest Ratio

Carbon 9 12 0.75×4 = 3

Hydrogen 1 1 1 × 4 = 4

Nitrogen 3.5 14 0.25 ×4 = 1

0.75129

111

0.25143.5

Empirical ratio = C3H

4N

Empirical formula mass = (3 × 12) + (4× 1) + 14 = 54

n = MassFormulaEmpiricalMassMolecular

= 254

108

Thus, molecular formula of the compound

= (Empirical formula)2

= (C3H

4N)

2

= C6H

8N

2

Ex.21 A compound on analysis, was found to have thefollowing composition :(i) Sodium = 14.31%, (ii) Sulphur = 9.97%, (iii) Oxygen= 69.50%, (iv) Hydrogen = 6.22%. Calculate themolecular formula of the compound assuming thatwhole hydrogen in the compound is present as waterof crystallisation. Molecular mass of the compoundis 322.

Sol. Element PercentageAtomic mass

Relative Number of atoms

Simplest ratio

Sodium 14.31 23 0.622

Sulphur 9.97 32 0.311

Hydrogen 6.22 1 6.22

Oxygen 69.50 16 4.34

20.311

0.622

10.311

0.311

200.311

6.22

140.311

4.34

23

31.14

1

22.6

16

50.69

32

97.9

PAGE # 27

The empirical formula = Na2SH

20O

14

Empirical formula mass= (2 × 23) + 32 + (20 × 1) + (14 × 16)

= 322Molecular mass = 322Molecular formula = Na

2SH

20O

14

Whole of the hydrogen is present in the form of waterof crystallisation. Thus, 10 water molecules arepresent in the molecule.So, molecular formula = Na

2SO

4. 10H

2O

CONCENTRATION OF SOLUTIONS

(a) Strength in g/L :

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

Strength in g/L = litreinsolutionofVolume

graminsoluteofWeight

(b) Molarity :

Molarity of a solution is defined as the number ofmoles of the solute dissolved per litre (or dm3) ofsolution. It is denoted by �M�. Mathematically,

M = litreinsolutiontheofVolume

soluteofmolesofNumber

litreinsolutionofVolume

soluteofMassMoleculargram/GraminsoluteofMass

M can be calculated from the strength as given below :

M = solute of mass Molecular

litre per grams in Strength

If �w� gram of the solute is present in V cm3 of a givensolution , then

M = massMolecular

w ×

V

1000

e.g. a solution of sulphuric acid having 4.9 grams of itdissolved in 500 cm3 of solution will have its molarity,

M = massMolecular

w×

V

1000

M = 98

4.9 ×

500

1000 = 0.1

(c) Formality :

In case of ionic compounds like NaCl, Na2CO

3 etc.,

formality is used in place of molarity. The formality ofa solution is defined as the number of gram formulamasses of the solute dissolved per litre of thesolution. It is represented by the symbol �F�. The term

formula mass is used in place of molecular massbecause ionic compounds exist as ions and not asmolecules. Formula mass is the sum of the atomicmasses of the atoms in the formula of the compound.

litreinsolutionofVolume

soluteofMasslagram/FormuinsoluteofMass

(d) Normality :

Normality of a solution is defined as the number ofgram equivalents of the solute dissolved per litre (dm3)of given solution. It is denoted by �N�.Mathematically,

N = litre in solution the of Volume

solute of sequivalent gram of Number

N = litre in solution the of Volume

solute of weight equivalent/ graminsolute of Weight

N can be calculated from the strength as given below :

N = solute of mass Equivalent

litre per grams in Strength=

E

S

If �w� gram of the solute is present in V cm3 of a givensolution.

N = solute the of mass Equivalentw

× V

1000

e.g. A solution of sulphuric acid having 0.49 gram ofit dissolved in 250 cm3 of solution will have itsnormality,

N = solute the of mass Equivalentw

× V

1000

N = 49

0.49×

250

1000 = 0.04

(Eq. mass of sulphuric acid = 49).

Solution Seminormal

Decinormal

Centinormal

Normality101

1001

21

Some Important Formulae :

(i) Milli equivalent of substance = N × V

where , N normality of solutionV Volume of solution in mL

(ii) If weight of substance is given,

milli equivalent (NV) = E1000w

Where, W Weight of substance in gramE Equivalent weight of substance

(iii) S = N × E

S Strength in g/LN Normality of solutionE Equivalent weight

(iv) Calculation of normality of mixture :

Ex.22 100 ml of 10

NHCl is mixed with 50 ml of

5

NH

2SO

4 .

Find out the normality of the mixture.Sol. Milli equivalent of HCl + milli equivalent H

2SO

4

= milli equivalent of mixtureN

1 V

1 + N

2 V

2 = N

3 V

3 { where, V

3 =V

1 + V

2 )

50

51

100101 N

3 × 150

N3 =

150

20 =

15

2= 0.133

PAGE # 28

Ex.23 100 ml of 10

NHCl is mixed with 25 ml of

5

NNaOH.

Find out the normality of the mixture.

Sol. Milli equivalent of HCl � milli equivalent of NaOH

= milli equivalent of mixtureN

1 V

1 � N

2 V

2 = N

3 V

3 { where, V

3 =V

1 + V

2 )

25

51

�100101

= N3 × 125

N3 =

251

Note :

1 milli equivalent of an acid neutralizes 1 milliequivalent of a base.

(e) Molality :

Molality of a solution is defined as the number ofmoles of the solute dissolved in 1000 grams of thesolvent. It is denoted by �m�.Mathematically,

m = gram in solvent the of Weightsolute the of moles of Number

× 1000 �m� can

be calculated from the strength as given below :

m = solute of mass Molecularsolvent of gram 1000 per Strength

If �w� gram of the solute is dissolved in �W� gram of the

solvent then

m = solute the of mass Mol.

w×

W1000

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

m = 250

1000106

1.325 = 0.05

Note :

Relationship Between Normality and Molarity of aSolution :Normality of an acid = Molarity × Basicity

Normality of base = Molarity × Acidity

Ex.24 Calculate the molarity and normality of a solutioncontaining 0.5 g of NaOH dissolved in 500 cm3

of solvent.Sol. Weight of NaOH dissolved = 0.5 g

Volume of the solution = 500 cm3

(i) Calculation of molarity :Molecular weight of NaOH = 23 + 16 + 1 = 40

Molarity =litre in solution of Volume

solute of weightmolecularsolute/ of Weight

= 500/1000

0.5/40 = 0.025

(ii) Calculation of normality :Normality

=litre in solution of Volume

solute of weightequivalentsolute/ of Weight

=500/1000

0.5/40 = 0.025

Ex.25 Find the molarity and molality of a 15% solution

of H2SO

4 (density of H

2SO

4 solution = 1.02 g/cm3)

(Atomic mass : H = 1u, O = 16u , S = 32 u)

Sol. 15% solution of H2SO

4 means 15g of H

2SO

4 are

present in 100g of the solution i.e.

Wt. of H2SO

4 dissolved = 15 g

Weight of the solution = 100 g

Density of the solution = 1.02 g/cm3 (Given)

Calculation of molality :

Weight of solution = 100 g

Weight of H2SO

4 = 15 g

Wt. of water (solvent) = 100 � 15 = 85 g

Molecular weight of H2SO

4 = 98

15 g H2SO

4 =

98

15= 0.153 moles

Thus ,85 g of the solvent contain 0.153 moles .

1000 g of the solvent contain= 85

0.153× 1000 = 1.8 mole

Hence ,the molality of H2SO

4 solution = 1.8 m

Calculation of molarity :

15 g of H2SO

4 = 0.153 moles

Vol. of solution = solution ofDensity

solution of Wt.

= 1.02

100 = 98.04 cm3

This 98.04 cm3 of solution contain H2SO

4 = 0.153

moles

1000 cm3 of solution contain H2SO

4

= 98.04

0.153 × 1000 = 1.56 moles

Hence the molarity of H2SO

4 solution = 1.56 M

(f) Mole Fraction :

The ratio between the moles of solute or solvent to

the total moles of solution is called mole fraction.

mole fraction of solute = Nn

n

solutionofMoles

soluteofMoles

= W/Mw/m

w/m

Mole fraction of solvent = Nn

N

solutionofMoles

solventofMoles

=

W/Mw/m

W/M

where,

n number of moles of solute

N number of moles of solvent

m molecular weight of solute

M molecular weight of solvent

w weight of solute

W weight of solvent

PAGE # 29

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

mole fraction of solute = solutionofMoles

soluteofMoles =

W/Mw/m

w/m

=

18/9060/10

60/10

= 0.032

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

STOICHIOMETRY

(a) Quantitative Relations in Chemical

Reactions :

Stoichiometry is the calculation of the quantities ofreactants and products involved in a chemicalreaction.

It is based on the chemical equation and on therelationship between mass and moles.N

2(g) + 3H

2(g) 2NH

3(g)

A chemical equation can be interpreted as follows -

1 molecule N2 + 3 molecules H

2 2 molecules

NH3(Molecular interpretation)

1 mol N2 + 3 mol H

2 2 mol NH

3

(Molar interpretation)

28 g N2 + 6 g H

2 34 g NH

3

(Mass interpretation)

1 volume N2 + 3 volume H

2 2 volume NH

3

(Volume interpretation)

Thus, calculations based on chemical equations aredivided into four types -

(i) Calculations based on mole-mole relationship.

(ii) Calculations based on mass-mass relationship.

(iii) Calculations based on mass-volume relationship.

(iv) Calculations based on volume -volumerelationship.

(i) Calculations based on mole-mole relationship :In such calculations, number of moles of reactantsare given and those of products are required.Conversely, if number of moles of products are given,then number of moles of reactants are required.

Ex.27 Oxygen is prepared by catalytic decompositionof potassium chlorate (KClO

3). Decomposition

of potassium chlorate gives potassium chloride(KCl) and oxygen (O

2). How many moles and how

many grams of KClO3 are required to produce

2.4 mole O2.

Sol. Decomposition of KClO3 takes place as,

2KClO3(s) 2KCl(s) + 3O

2(g)

2 mole KClO3 3 mole O

2

3 mole O2 formed by 2 mole KClO

3

2.4 mole O2 will be formed by

4.2

32

mole KClO3 = 1.6 mole KClO

3

Mass of KClO3 = Number of moles × molar mass

= 1.6 × 122.5 = 196 g

(ii) Calculations based on mass-mass relationship:In making necessary calculation, following steps are

followed -

(a) Write down the balanced chemical equation.

(b) Write down theoretical amount of reactants and

products involved in the reaction.

(c) The unknown amount of substance is calculated

using unitary method.

Ex.28 Calculate the mass of CaO that can be prepared

by heating 200 kg of limestone CaCO3 which is

95% pure.

Sol. Amount of pure CaCO3 = 200

10095

= 190 kg

= 190000 g

CaCO3(s) CaO(s) + CO

2(g)

1 mole CaCO3 1 mole CaO

100 g CaCO3 56 g CaO

100 g CaCO3 give 56 g CaO

190000 g CaCO3 will give=

10056

× 190000 g CaO

= 106400 g = 106.4 kg

Ex.29 Chlorine is prepared in the laboratory by treatingmanganese dioxide (MnO

2) with aqueous

hydrochloric acid according to the reaction -

MnO2 + 4HCl MnCl

2 + Cl

2 + 2H

2O

How many grams of HCl will react with 5 g MnO2 ?

Sol. 1 mole MnO2 reacts with 4 mole HCl

or 87 g MnO2 react with 146 g HCl

5 g MnO2 will react with =

87146

× 5 g HCl = 8.39 g HCl

Ex.30 How many grams of oxygen are required to burncompletely 570 g of octane ?

Sol. Balanced equation

2C H + 25O 8 18 2 16CO + 18H O2 2

2 mole2 × 114

25 mole25 × 32

First method : For burning 2 × 114 g of the octane,

oxygen required = 25 × 32 g

For burning 1 g of octane, oxygen required =1142

3225

g

Thus, for burning 570 g of octane, oxygen required

= 1142

3225

× 570 g = 2000 g

PAGE # 30

Mole Method : Number of moles of octane in 570grams

114570

= 5.0

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

For burning 5 moles of octane, oxygen required

= 0.23225

× 5.0 g = 2000 g

Proportion Method : Let x g of oxygen be required forburning 570 g of octane. It is known that 2 × 114 g of

the octane requires 25 × 32 g of oxygen; then, the

proportion.

etanocg1142oxygeng3225

= etanocg570

x

x = 1142

5703225

= 2000 g

Ex.31 How many kilograms of pure H2SO

4 could be

obtained from 1 kg of iron pyrites (FeS2) according to

the following reactions ?4FeS

2 + 11O

2 2Fe

2O

3 + 8SO

2

2SO2 + O

2 2SO

3

SO3 + H

2O H

2SO

4

Sol. Final balanced equation,4FeS + 15O + 8H O2 2 2

2Fe O + 8H SO2 3 2 4

8 mole8 × 98 g

4 mole4 × 120 g

4 × 120 g of FeS2 yield H

2SO

4 = 8 × 98 g

1000 g of FeS2 will yield H

2SO

4 =

1204988

× 1000

= 1633.3 g

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

PV = mw

RT

Ex-32. What volume of NH3 can be obtained from 26.75 g

of NH4Cl at 27ºC and 1 atmosphere pressure.

Sol. The balanced equation is -

NH Cl(s) 4 NH (g) + HCl(g)3

1 mol1 mol53.5 g

53.5 g NH4Cl give 1 mole NH

3

26.75 g NH4Cl will give

5.531

× 26.75 mole NH3

= 0.5 molePV = nRT1 ×V = 0.5 × 0.0821 × 300

V = 12.315 litre

Ex-33 What quantity of copper (II) oxide will react with2.80 litre of hydrogen at STP ?

Sol. CuO + H 2 Cu + H O2

1 mol79.5 g

1 mol22.4 litre at NTP

22.4 litre of hydrogen at STP reduce CuO = 79.5 g2.80 litre of hydrogen at STP will reduce CuO

= 4.225.79

× 2.80 g = 9.93 g

Ex-34 Calculate the volume of carbon dioxide at STPevolved by strong heating of 20 g calcium carbonate.

Sol. The balanced equation is -

CaCO 3 CaO + CO 2

1 mol = 22.4 litre at STP

1 mol 100 g

100 g of CaCO3 evolve carbon dioxide = 22.4 litre

20 g CaCO3 will evolve carbon dioxide

= 100

4.22 × 20 = 4.48 litre

Ex.35 Calculate the volume of hydrogen liberated at 27ºC

and 760 mm pressure by heating 1.2 g of magnesiumwith excess of hydrochloric acid.

Sol. The balanced equation is

Mg + 2HCl MgCl + H2 2

1 mol 24 g

24 g of Mg liberate hydrogen = 1 mole

1.2 g Mg will liberate hydrogen = 0.05 mole

PV = nRT1 × V = 0.05 × 0.0821 × 300

V = 1.2315 litre

(iv) Calculations based on volume volumerelationship :These calculations are based on two laws :(i) Avogadro�s law (ii) Gay-Lussac�s Law

e.g.N (g) + 3H (g)2 2 2NH (g) (Avogadro's law)3

2 mol 2 × 22.4 L

1 mol1 × 22.4 L

3 mol3 × 22.4 L

(under similar conditions of temperature andpressure, equal moles of gases occupy equalvolumes)N (g) + 3H (g)2 2 2NH (g)3

1 vol 3 vol 2 vol(Gay- Lussac's Law)

(under similar conditions of temperature andpressure, ratio of coefficients by mole is equal to ratioof coefficient by volume).

Ex-36 One litre mixture of CO and CO2 is taken. This is

passed through a tube containing red hot charcoal.The volume now becomes 1.6 litre. The volume aremeasured under the same conditions. Find thecomposition of mixture by volume.

Sol. Let there be x mL CO in the mixture , hence, there willbe (1000 � x) mL CO

2. The reaction of CO

2 with red

hot charcoal may be given as -

CO (g) + C(s)2 2CO(g) 2 vol.2(1000 � x)

1 vol.(1000 �x)

Total volume of the gas becomes = x + 2(1000 � x)

x + 2000 � 2x = 1600

x = 400 mL

volume of CO = 400 mL and volume of CO2 = 600 mL

PAGE # 31

Ex-37 What volume of air containing 21% oxygen by volume

is required to completely burn 1kg of carbon containing

100% combustible substance ?

Sol. Combustion of carbon may be given as,

C(s) + O (g) 2 CO (g)2

1 mol12 g

1 mol32 g

12 g carbon requires 1 mole O2 for complete

combustion

1000 g carbon will require 1000121 mole O

2 for

combustion, i.e. , 83.33 mole O2

Volume of O2 at STP = 83.33 × 22.4 litre

= 1866.66 litre

21 litre O2 is present in 100 litre air

1866.66 litre O2 will be present in

21100

× 1866.66 litre air

= 8888.88 litre or 8.89 × 103 litre

Ex-38 An impure sample of calcium carbonate contains80% pure calcium carbonate 25 g of the impuresample reacted with excess of hydrochloric acid.Calculate the volume of carbon dioxide at STPobtained from this sample.

Sol. 100 g of impure calcium carbonate contains = 80 gpure calcium carbonate25 g of impure calcium carbonate sample will contain

= 10080

× 25 = 20 g pure calcium carbonate

The desired equation is -

CaCO + 2HCl 3 CaCl + CO + H O2 2 2

1 mol100 g

22.4 litre at STP

100 g pure CaCO3 liberate = 22.4 litre CO

2.

20 g pure CaCO3 liberate = 20

1004.22

= 4.48 litre CO2

VOLUMETRIC CALCULATIONS

The quantitative analysis in chemistry is primarily

carried out by two methods, viz, volumetric analysis

and gravimetric analysis.In the first method the mass

of a chemical species is measured by measurement

of volume, whereas in the second method it is deter-

mined by taking the weight.

The strength of a solution in volumetric analysis is

generally expressed in terms of normality, i.e., num-

ber of equivalents per litre but since the volume in the

volumetric analysis is generally taken in millilitres

(mL), the normality is expressed by milliequivalents

per millilitre.

USEFUL FORMULAE FOR

VOLUMETRIC CALCULATIONS

(i) milliequivalents = normality × volume in millilitres.

(ii) At the end point of titration, the two titrants, say 1

and 2, have the same number of milliequivalents,

i.e., N1V

1 = N

2V

2, volume being in mL.

(iii) No. of equivalents = 1000

.e.m.

(iv) No. of equivalents for a gas =

)STPat.eq1of.vol(volumeequivalentSTPatVolume

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

weight.

(vi) (a) Normality = molarity × factor relating mol. wt.

and eq. wt.

(b) No. of equivalents = no. of moles × factor relat

ing mol. wt. and eq. wt.

Ex.39 Calculate the number of milli equivalent of H2SO

4

present in 10 mL of N/2 H2SO

4 solution.

Sol. Number of m.e. = normality × volume in mL =21

× 10 = 5.

Ex.40 Calculate the number of m.e. and equivalents of

NaOH present in 1 litre of N/10 NaOH solution.

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

= 101

× 1000 = 100

Number of equivalents = 1000

.e.mof.no =

1000100

= 0.10

Ex.41 Calculate number of m.e. of the acids present in

(i) 100 mL of 0.5 M oxalic acid solution.

(ii) 50 mL of 0.1 M sulphuric acid solution.

Sol. Normality = molarity × basicity of acid

(i) Normality of oxalic acid = 0.5 × 2 = 1 N

m.e. of oxalic acid = normality × vol. in mL = 1 × 100

= 100.

(ii) Normality of sulphuric acid = 0.1 × 2 = 0.2 N

m.e. of sulphuric acid = 0.2 × 50 = 10

Ex.42 A 100 mL solution of KOH contains 10

milliequivalents of KOH. Calculate its strength in nor-

mality and grams/litre.

Sol. Normality = mLinvolume.e.mof.no

= 1.010010

strength of the solution = N/10

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

= 56101 = 5.6 gram/litre.

56

156

acidity.wtmolecular

KOHof.wt.eq

PAGE # 32

Ex.43 What is strength in gram/litre of a solution of H2SO

4,

12 cc of which neutralises 15 cc of 10N

NaOH

solution ?

Sol. m.e. of NaOH solution = 101

× 15 = 1.5

m.e. of 12 cc of H2SO

4 = 1.5

normality of H2SO

4 =

125.1

Strength in grams/litre = normality × eq. wt.

= 12

5.1 × 49 grams/litre

= 6.125 grams/litre.

49

298

basicitywt.molecular

SOHofwt.eq. 42

Ex.44 What weight of KMnO4 will be required to prepare

250 mL of its 10N

solution if eq. wt. of KMnO4 is 31.6 ?

Sol. Equivalent weight of KMnO4 = 31.6

Normality of solution (N) = 101

Volume of solution (V) = 250 ml

1000NEV

W ; W = 1000

2506.31101 79.0

406.31 g

Ex.45 100 mL of 0.6 N H2SO

4 and 200 mL of 0.3 N HCl

were mixed together. What will be the normality of theresulting solution ?

Sol. m.e. of H2SO

4 solution = 0.6 × 100 = 60

m.e. of HCl solution = 0.3 × 200 = 60

m.e. of 300 mL (100 + 200) of acidic mixture= 60 + 60 = 120.

Normality of the resulting solution = .voltotal.e.m

= 300120

= 52

N.

Ex.46 A sample of Na2CO

3. H

2O weighing 0.62 g is added

to 100 mL of 0.1 N H2SO

4. Will the resulting solution

be acidic, basic or neutral ?

Sol. Equivalents of Na2CO

3. H

2O =

6262.0

= 0.01

62

2124

OH.CONaof.wt.eq 232

m.e. of Na2CO

3. H

2O = 0.01 × 1000 = 10

m.e. of H2SO

4 = 0.1 × 100 = 10

Since the m.e. of Na2CO

3. H

2O is equal to that of H

2SO

4,

the resulting solution will be neutral.

(a) Introduction :

Volumetric analysis is a method of quantitativeanalysis. It involves the measurement of the volumeof a known solution required to bring about thecompletion of the reaction with a measured volumeof the unknown solution whose concentration orstrength is to be determined. By knowing the volumeof the known solution, the concentration of the solutionunder investigation can be calculated. Volumetricanalysis is also termed as titrimetric analysis.

(b) Important terms used in volumetric

analysis :

(i) Titration : The process of addition of the knownsolution from the burette to the measured volume ofsolution of the substance to be estimated until thereaction between the two is just complete, is termedas titration. Thus, a titration involves two solutions:

(a) Unknown solution and (b) Known solution or stan-dard solution.

(ii) Titrant : The reagent or substance whose solu-tion is employed to estimate the concentration of un-known solution is termed titrant. There are two typesof reagents or titrants:

(A) Primary titrants : These reagents can beaccurately weighed and their solutions are not to bestandardised before use. Oxalic acid, potassiumdichromate, silver nitrate, copper sulphate, ferrousammonium sulphate, sodium thiosulphates etc., arethe examples of primary titrants.

(B) Secondary titrants : These reagents cannotaccurately weighed and their solutions are to bestandardised before use. Sodium hydroxide,potassium hydroxide, hydrochloric acid, sulphuricacid, iodine, potassium permanganate etc. are theexamples of secondary titrants.

(iii) Standard solution : The solution of exactly knownconcentration of the titrant is called the standardsolution.

(iv) Titrate : The solution consisting the substance tobe estimated is termed unknown solution. Thesubstance is termed titrate.

(v) Equivalence point : The point at which the reagent(titrant) and the substance (titrate) under investigationare chemically equivalent is termed equivalence pointor end point.

(vi) Indicator : It is the auxiliary substance used forphysical (visual) detection of the completion of titrationor detection of end point is termed as indicator.Indicators show change in colour or turbidity at thestage of completion of titration.

(c) Concentraion representation of solution

(A) Strength of solution : Grams of solute dissolvedper litre of solution is called strength of solution'

(B) Parts Per Million (ppm) : Grams of solutedissolved per 106 grams of solvent is calledconcentration of solution in the unit of Parts Per Million(ppm). This unit is used to represent hardness ofwater and concentration of very dilute solutions.

(C) Percentage by mass : Grams of solute dissolvedper 100 grams of solution is called percentage bymass.

(D) Percentage by volume : Millilitres of solute per100 mL of solution is called percentage by volume.For example, if 25 mL ethyl alcohol is diluted withwater to make 100 mL solution then the solution thusobtained is 25% ethyl alcohol by volume.

(E) Mass by volume percentage :Grams of solutepresent per 100 mL of solution is called percentagemass by volume.For example, let 25 g glucose is dissolved in water tomake 100 mL solution then the solution is 25% massby volume glucose.

PAGE # 33

(d) Classification of reactions involved in

volumetric analysis

(A) Neutralisation reactions

The reaction in which acids and bases react to formsalt called neutralisation.

e.g., HCI + NaOH NaCI + H2O

H+(acid)

+ OH�

(base) H

2O (feebly ionised)

The titration based on neutralisation is calledacidimetry or alkalimetry.

(B) Oxidation-reduction reactions

The reactions involving simultaneous loss and gainof electrons among the reacting species are calledoxidation reduction or redox reactions, e.g., let usconsider oxidation of ferrous sulphate (Fe2+ ion) bypotassium permanganate (MnO

4� ion) in acidic

medium.

MnO4� + 8H+ + 5e� Mn2+ + 4H

2O

(Gain of electrons or reduction)5 [Fe2+ Fe3+ + e�](Loss of electrons or oxidation)

MnO4� + 5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H

2O

________________________________________________________________

In the above reaction, MnO4� acts as oxidising agent

and Fe2+ acts as reducing agent.

The titrations involving redox reactions are called redoxtitrations. These titrations are also called accordingto the reagent used in the titration, e.g., iodometric,cerimetric, permanganometric and dichromometrictitrations

(C) Precipitation reactions :A chemical reaction in which cations and anionscombine to form a compound of very low solubility (inthe form of residue or precipitate), is calledprecipitation.BaCl

2 + Na

2SO

4 BaSO

4 + 2NaCl

(white precipitate)

The titrations involving precipitation reactions arecalled precipitation titrations.

(D) Complex formation reactions :

These are ion combination reactions in which asoluble sl ightly dissociated complex ion orcompound is formed.Complex compounds retain their identity in thesolution and have the properties of the constituentions and molecules.e.g. CuSO

4 + 4NH

3 [Cu(NH

3)

4]SO

4

(complex compound)AgNO

3 + 2KCN K[Ag(CN)

2] + KNO

3

(complex compound)

2CuSO4 + K

4[Fe(CN)

6] Cu

2[Fe(CN)

6] + 2K

2SO

4

(complex compound)The titrations involving complex formation reactionsare called complexometric titrations.

The determination of concentration of bases bytitration with a standard acid is called acidimetry andthe determination of concentration of acid by titrationwith a standard base is called alkalimetry.The substances which give different colours withacids and base are called acid base indicators. Theseindicators are used in the visual detection of theequivalence point in acid-base titrations.The acid-base indicators are also called pHindicators because their colour change according tothe pH of the solution.

In the selection of indicator for a titration, followingtwo informations are taken into consideration :

(i) pH range of indicator(ii) pH change near the equivalence point in thetitration.

The indicator whose pH range is included in the pHchange of the solution near the equivalence point, istaken as suitable indicator for the titration.

(i) Strong acid-strong base titration : In the titrationof HCl with NaOH, the equivalence point lies in thepH change of 4�10. Thus, methyl orange, methyl red

and phenolphthalein will be suitable indicators.

(ii) Weak acid-strong base titration : In the titrationof CH

3COOH with NaOH the equivalence point lies

between 7.5 and 10. Hence, phenolphthalein (8.3�10) will be the suitable indicator.

(iii) Weak base-strong acid titration : In the titrationof NH

4OH (weak base) against HCl (strong acid) the

pH at equivalence point is about 6.5 and 4. Thus,methyl orange (3.1�4.4) or methyl red (4.2�6.3) will

be suitable indicators.

(iv) Weak acid-weak base titration : In the titration ofa weak acid (CH

3COOH) with weak base (NH

4OH)

the pH at the equivalence point is about 7, i.e., liesbetween 6.5 and 7.5 but no sharp change in pH isobserved in these titrations. Thus, no simple indica-tor can be employed for the detection of the equiva-lence point.

(v) Titration of a salt of a weak acid and a strongbase with strong acid:

H2CO

3 + 2NaOH Na

2CO

3 + 2H

2O

Weak acid Strong base

Na2CO

3 when titrated with HCl, the following two

stages are involved :Na

2CO

3 + HCl NaHCO

3 + NaCl (First stage)

pH = 8.3, near equivalence pointNaHCO

3 + HCl NaCl + H

2CO

3 (Second stage)

pH = 4, near equivalence point

For first stage, phenolphthalein and for second stage,methyl orange will be the suitable indicator.

PAGE # 34

TITRATION OF MIXTURE OF NaOH, Na2CO3 ANDNaHCO3 BY STRONG ACID LIKE HCl

In this titration the following indicators are mainly used :

(i) Phenolphthalein (weak organic acid) : It showscolour change in the pH range (8 � 10)

(ii) Methyl orange (weak organic base) : It shows

colour change in the pH range (3.1 � 4.4). Due to

lower pH range, it indicates complete neutralisation

of whole of the base.

Let for complete neutralisation of Na2CO

3, NaHCO

3 and NaOH, x,y and z mL of standard HCl are required. The

titration of the mixture may be carried by two methods as summarised below :

Mixture Phenlphthalein Methyl orange Phenolphthalein Methyl orange from from from after first end beginning beginning point

1. NaOH z + (x/2) (x + z) z + x/2 x/2 (for remaining 50% + Na CO2 3

beginning

Na CO )

2. NaOH z + 0 (z + y) z + 0 y (for remaining 100% + NaHCO NaHCO

3. Na CO (x/2) + 0 (x + y) (x/2) + 0 x/2 + y (for remaining 50% + NaHCO of Na CO and 100% NaHCO are indicated)

2 3

3 3

2 3

3 2 3

3

Volume of HClused with

Volume of HCl used

An indicator is a substance which is used to deter-mine the end point in a titration. In acid-base titra-tions organic substances (weak acids or weak bases)are generally used as indicators. They change theircolour within a certain pH range. The colour changeand the pH range of some common indicators aretabulated below:________________________________________Indicator pH range Colour

change________________________________________

Methyl orange 3.2 � 4.5 Orange to red

Methyl red 4.4 � 6.5 Red to yellow

Litmus 5.5 � 7.5 Red to blue

Phenol red 6.8 � 8.4 Yellow to red

Phenolphthalein 8.3 � 10.5 Colourless to pink________________________________________

Theory of acid-base indicators : Two theories havebeen proposed to explain the change of colour ofacid-base indicators with change in pH.

1. Ostwald's theory:According to this theory

(a) The colour change is due to ionisation of the acid-base indicator. The unionised form has differentcolour than the ionised form.

(b) The ionisation of the indicator is largely affected inacids and bases as it is either a weak acid or a weakbase. In case, the indicator is a weak acid, itsionisation is very much low in acids due to commonions while it is fairly ionised in alkalies. Similarly if theindicator is a weak base, its ionisation is large inacids and low in alkalies due to common ions.

Considering two important indicators phenolphtha-lein (a weak acid) and methyl orange (a weak base),Ostwald's theory can be illustrated as follows:

PAGE # 35

Phenolphthalein: It can be represented as HPh. Itionises in solution to a small extent as:

HPh H+ + Ph�

Colourless Pink

Applying law of mass action,

K = ]HPh[

]Ph][H[

The undissociated molecules of phenolphthalein arecolourless while ph� ions are pink in colour. In pres-ence of an acid, the ionisation of HPh is practicallynegligible as the equilibrium shifts to left hand sidedue to high concentration of H+ ions. Thus, the solu-tion would remain colourless. On addition of alkali,hydrogen ions are removed by OH� ions in the form ofwater molecules and the equilibrium shifts to righthand side. Thus, the concentration of ph� ions in-creases in solution and they impart pink colour to thesolution.

Let us derive Hendetson's equation for an indicator

HIn + H2O H

3+O + In�

'Acid form' 'Base form'

Conjugate acid-base pair

KIn =

]HIn[]OH][In[ 3

KIn = Ionization constant of indicator

[H3

+O] = KIn ]In[

]HIn[

pH = �log10

[H3

+O] = �log10

[KIn] � log

10 ]In[

]HIn[

pH = pKIn + log

10

]HIn[]In[

(Henderson's equation for

indicator)At equivalence point ;[In�] = [HIn] and pH = pK

In

Methyl orange : It is a very weak base and can berepresented as MeOH. It is ionised in solution to giveMe+ and OH� ions.

MeOH Me+ +OH�

Orange Red

Applying law of mass action,

K = ]MeOH[

]OH][Me[

In presence of an acid, OH� ions are removed in theform of water molecules and the above equilibriumshifts to right hand side. Thus, sufficient Me+ ions areproduced which impart red colour to the solution. Onaddition of alkali, the concentration of OH� ions in-creases in the solution and the equilibrium shifts toleft hand side, i.e., the ionisation of MeOH is practi-cally negligible. Thus, the solution acquires the colourof unionised methyl orange molecules, i.e. orange.

This theory also explains the reason why phenol-phthalein is not a suitable indicator for titrating a weakbase against strong acid. The OH� ions furnished bya weak base are not sufficient to shift the equilibriumtowards right hand side considerably, i.e., pH is notreached to 8.3. Thus, the solution does not attainpink colour. Similarly, it can be explained why methylorange is not a suitable indicator for the titration ofweak acid with strong base.

SOLUBILITY

The solubility of a solute in a solution is alwaysexpressed with respect to the saturated solution.

(a) Definition :

The maximum amount of the solute which can bedissolved in 100g (0.1kg) of the solvent to form asaturated solution at a given temperature.Suppose w gram of a solute is dissolved in W gramof a solvent to make a saturated solution at a fixedtemperature and pressure. The solubility of the solutewill be given by -

W

w× 100 =

solventtheofMass

solutetheofMass× 100

For example, the solubility of potassium chloride inwater at 20ºC and 1 atm. is 34.7 g per 100g of water.

This means that under normal conditions 100 g ofwater at 20ºC and 1 atm. cannot dissolve more than

34.7g of KCl.

(b) Effect of Temperature and Pressure on

Solubility of a Solids :

The solubility of a substance in liquids generallyincreases with rise in temperature but hardly changeswith the change in pressure. The effect of temperaturedepends upon the heat energy changes whichaccompany the process.

Note :If heat energy is needed or absorbed in the process,it is of endothermic nature. If heat energy is evolvedor released in the process, it is of exothermic nature.

(i) Effect of temperature on endothermic dissolutionprocess : Most of the salts like sodium chloride,potassium chloride, sodium nitrate, ammoniumchloride etc. dissolve in water with the absorption ofheat. In all these salts the solubility increases withrise in temperature. This means that sodium chloridebecomes more soluble in water upon heating.

(ii) Effect of temperature on exothermic dissolutionprocess : Few salts like lithium carbonate, sodiumcarbonate monohydrate, cerium sulphate etc.dissolve in water with the evolution of heat. Thismeans that the process is of exothermic nature. Inthese salts the solubility in water decreases with risein temperature.

PAGE # 36

Note :

1. While expressing the solubility, the solution mustbe saturated but for expressing concentration (masspercent or volume percent), the solution need not tobe saturated in nature.

2. While expressing solubility, mass of solvent isconsidered but for expressing concentration themass or volume of the solution may be taken intoconsideration.

(c) Effect of Temperature on the Solubility

of a Gas

(i) The solubility of a gas in a liquid decreases withthe rise in temperature.

(ii) The solubility of gases in liquids increases onincreasing the pressure and decreases on decreas-ing the pressure.

SAMPLE PROBLEMS

Ex.47 12 grams of potassium sulphate dissolves in75 grams of water at 60ºC. What is the solubility

of potassium sulphate in water at that temperature ?

Sol. Solubility = 100solventofmasssoluteofmass

= 75

12×100 = 16 g

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

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

solubility of the solute.

Sol. Solubility = solventofMass

soluteofMass× 100

Mass of solute = 4 gMass of solvent = 40 g

Solubility = 404

× 100 = 10 g

Ex.49 (a) What mass of potassium chloride would beneeded to form a saturated solution in 50 g ofwater at 298 K ? Given that solubility of the salt is46g per 100g at this temperature.

(b) What will happen if this solution is cooled ?

Sol. (a) Mass of potassium chloride in 100 g of waterin saturated solution = 46 gMass of potassium chloride in 50 g of water insaturated solution.

= 10046

× (50g) = 23 g

(b) When the solution is cooled, the solubility ofsalt in water will decrease. This means, that uponcooling, it will start separating from the solutionin crystalline form.

EXERCISE

1. The solubility of K2SO

4 in water is 16 g at 50ºC. The

minimum amount of water required to dissolve 4 g

K2SO

4 is -

(A) 10 g (B) 25 g

(C) 50 g (D) 75 g

2. Molarity of H2SO

4 (density 1.8g/mL) is 18M. The

molality of this solution is -

(A)36 (B) 200

(C) 500 (D) 18

3. 8g of sulphur are burnt to form SO2, which is oxidised

by Cl2 water. The solution is treated with BaCl

2

solution. The amount of BaSO4 precipitated is -

(A) 1.0 mole (B) 0.5 mole

(C) 0.75 mole (D) 0.25 mole

4. In a compound AxB

y -

(A) Mole of A = Mole of B = mole of AxB

y

(B) Eq. of A = Eq. of B = Eq. of AxB

y

(C) X × mole of A = y × mole of B = (x + y) × mole of AxB

y

(D) X × mole of A = y × mole of B

5. The percentage of sodium in a breakfast cereal

labelled as 110 mg of sodium per 100 g of cereal is -

(A) 11% (B) 1.10%

(C) 0.110% (D) 110%

6. Two elements A (at. wt. 75) and B (at. wt. 16) combine

to yield a compound. The % by weight of A in the

compound was found to be 75.08. The empirical

formula of the compound is -

(A) A2B (B) A

2B

3

(C) AB (D) AB2

7. No. of oxalic acid molecules in 100 mL of 0.02 N

oxalic acid are -

(A) 6.023 × 1020 (B) 6.023 × 1021

(C) 6.023 × 1022 (D) 6.023 × 1023

8. Which of the following sample contains the maximum

number of atoms -

(A) 1 mg of C4H

10(B) 1 mg of N

2

(C) 1 mg of Na (D) 1 mL of water

9. The total number of protons, electrons and neutrons

in 12 g of C126 is -

(A) 1.084 × 1025 (B) 6.022 × 1023

(C) 6.022 × 1022 (D) 18

10. 4.4 g of CO2 and 2.24 litre of H

2 at STP are mixed in a

container. The total number of molecules present in

the container will be -

(A) 6.022 × 1023 (B) 1.2044 × 1023

(C) 2 mole (D) 6.023 × 1024

PAGE # 37

11. Which is not a molecular formula ?

(A) C6H

12O

6(B) Ca(NO

3)

2

(C) C2H

4O

2(D) N

2O

12. The hydrated salt, Na2SO

4. nH

2O undergoes 55.9%

loss in weight on heating and becomes anhydrous.

The value of n will be -

(A) 5 (B) 3

(C) 7 (D) 10

13. Which of the following mode of expressing

concentration is independent of temperature -

(A) Molarity (B) Molality

(C) Formality (D) Normality

14. The haemoglobin of the red blood corpuscles of most

of the mammals contains approximately 0.33% of

iron by weight. The molecular weight of haemoglobin

is 67,200. The number of iron atoms in each molecule

of haemoglobin is (Atomic weight of iron = 56) -

(A) 2 (B) 3

(C) 4 (D) 5

15. An oxide of metal have 20% oxygen, the eq. wt. of

metal oxide is -

(A) 32 (B) 40

(C) 48 (D) 52

16. How much water is to be added to dilute 10 mL of 10

N HCl to make it decinormal ?

(A) 990 mL (B) 1010 mL

(C) 100 mL (D) 1000 mL

17. The pair of compounds which cannot exist in solution is -

(A) NaHCO3 and NaOH

(B) Na2SO

3 and NaHCO

3

(C) Na2CO

3 and NaOH

(D) NaHCO3 and NaCl

18. If 250 mL of a solution contains 24.5 g H2SO

4 the

molarity and normality respectively are -

(A) 1 M, 2 N (B) 1M,0.5 N

(C) 0.5 M, 1N (D) 2M, 1N

19. The mole fraction of NaCl, in a solution containing 1

mole of NaCl in 1000 g of water is -

(A) 0.0177 (B) 0.001

(C) 0.5 (D) 0.244

20. 3.0 molal NaOH solution has a density of 1.110 g/

mL. The molarity of the solution is -

(A) 2.9732 (B) 3.05

(C) 3.64 (D) 3.0504

21. How many atoms are contained in a mole of Ca(OH)2 -

(A) 30 × 6.02 × 1023 atoms/mol

(B) 5 × 6.02 × 1023 atoms/mol

(C) 6 × 6.02 × 1023 atoms/mol

(D) None of these

22. Insulin contains 3.4% sulphur. The minimum

molecular weight of insulin is -

(A) 941.176 u (B) 944 u

(C) 945.27 u (D) None of these

23. Number of moles present in 1 m3 of a gas at NTP are -

(A) 44.6 (B) 40.6

(C) 42.6 (D) 48.6

24. Weight of oxygen in Fe2O

3 and FeO is in the simple

ratio of -

(A) 3 : 2 (B) 1 : 2

(C) 2 : 1 (D) 3 : 1

25. 2.76 g of silver carbonate on being strongly heated

yield a residue weighing -

(A) 2.16g (B) 2.48 g

(C) 2.32 g (D) 2.64 g

26. How many gram of KCl would have to be dissolved in

60 g of H2O to give 20% by weight of solution -

(A) 15 g (B) 1.5 g

(C) 11.5 g (D) 31.5 g

27. When the same amount of zinc is treated separately

with excess of H2SO

4 and excess of NaOH, the ratio

of volumes of H2 evolved is -

(A) 1 : 1 (B) 1 : 2

(C) 2 : 1 (D) 9 : 4

28. Amount of oxygen required for combustion of 1 kg of a

mixture of butane and isobutane is -

(A) 1.8 kg (B) 2.7 kg

(C) 4.5 kg (D) 3.58 kg

29. Rakesh needs 1.71 g of sugar (C12

H22

O11

) to sweeten

his tea. What would be the number of carbon atoms

present in his tea ?

(A) 3.6 × 1022 (B) 7.2 × 1021

(C) 0.05 × 1023 (D) 6.6 × 1022

30. The total number of AlF3 molecule in a sample of AlF

3

containing 3.01 × 1023 ions of F� is -

(A) 9.0 × 1024 (B) 3.0 × 1024

(C) 7.5 × 1023 (D)1023

31. The volume occupied by one molecule of water

(density 1 g/cm3) is -

(A) 18 cm3 (B) 22400 cm3

(C) 6.023 × 10�23 (D) 3.0 × 10�23 cm3

PAGE # 38

32. 224 mL of a triatomic gas weigh 1 g at 273 K and 1

atm. The mass of one atom of this gas is -

(A) 8.30 × 10�23 g (B) 2.08 × 10�23 g

(C) 5.53 × 10�23 g (D) 6.24 × 10�23 g

33. The percentage of P2O

5 in diammonium hydrogen

phosphate is -

(A) 77.58 (B) 46.96

(C) 53.78 (D) 23.48

34. The mole fraction of water in 20% (wt. /wt.) aqueous

solution of H2O

2 is -

(A) 68

77(B)

77

68

(C) 80

20(D)

20

80

35. Which of the following has the maximum mass ?

(A) 25 g of Hg

(B) 2 moles of H2O

(C) 2 moles of CO2

(D) 4 g atom of oxygen

36. Total mass of neutrons in 7mg of 14C is -

(A) 3 × 1020 kg (B) 4 × 10�6 kg

(C) 5 × 10�7 kg (D) 4 × 10�7 kg

37. Vapour density of a metal chloride is 66. Its oxide

contains 53% metal. The atomic weight of metal is -

(A) 21 (B) 54

(C) 26.74 (D) 2.086

38. The number of atoms in 4.25 g NH3 is approximately -

(A) 1 × 1023 (B) 1.5 × 1023

(C) 2 × 1023 (D) 6 × 1023

39. The modern atomic weight scale is based on -

(A) C12 (B) O16

(C) H1 (D) C13

40. Amount of oxygen in 32.2g of Na2SO

4. 10H

2O is -

(A) 20.8 g (B) 22.4 g

(C) 2.24 g (D) 2.08 g

41. Which of the followings does not change on dilution ?

(A) Molarity of solution

(B) Molality of solution

(C) Millimoles and milli equivalent of solute

(D) Mole fraction of solute

42. Equal masses of O2, H

2 and CH

4 are taken in a

container. The respective mole ratio of these gases

in container is -

(A) 1 : 16 : 2 (B) 16 : 1 : 2

(C) 1 : 2 : 16 (D) 16 : 2 : 1

43. The number of molecules present in 11.2 litre CO2 at

STP is -

(A) 6.023 × 1032 (B) 6.023 × 1023

(C) 3.011 × 1023 (D) None of these

44. 250 ml of 0.1 N solution of AgNO3 are added to 250

ml of a 0.1 N solution of NaCl. The concentration of

nitrate ion in the resulting solution will be -

(A) 0.1N (B) 1.2 N

(C) 0.01 N (D) 0.05 N

45. Amount of BaSO4 formed on mixing the aqueous

solution of 2.08 g BaCl2 and excess of dilute H

2SO

4 is -

(A) 2.33 g (B) 2.08 g

(C) 1.04 g (D) 1.165 g

46. 2g of NaOH and 4.9 g of H2SO

4 were mixed and

volume is made 1 litre. The normality of the resulting

solution will be -

(A) 1N (B) 0.05 N

(C) 0.5 N (D) 0.1N

47. 1g of a metal carbonate neutralises completely 200

mL of 0.1N HCl. The equivalent weight of metal

carbonate is -

(A) 25 (B) 50

(C) 100 (D) 75

48. 100 mL of 0.5 N NaOH were added to 20 ml of 1N

HCl and 10 mL of 3 N H2SO

4. The solution is -

(A) acidic (B) basic

(C) neutral (D) none of these

49. 1M solution of H2SO

4 is diluted from 1 litre to 5 litres ,

the normality of the resulting solution will be -

(A) 0.2 N (B) 0.1 N

(C) 0.4 N (D) 0.5 N

50. The volume of 7g of N2 at S.T.P. is -

(A) 11.2 L (B) 22.4 L

(C) 5.6 L (D) 6.5 L

51. One mole of calcium phosphide on reaction with

excess of water gives -

(A) three moles of phosphine

(B) one mole phosphoric acid

(C) two moles of phosphine

(D) one mole of P2O

5

52. Mg (OH)2 in the form of milk of magnesia is used to

neutralize excess stomach acid. How many moles of

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

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

(A) 0.0171 (B) 0.0343

(C) 0.686 (D) 1.25

PAGE # 39

53. Calcium carbonate decomposes on heating

according to the following equation -


2(g)

How many moles of CO2 will be obtained by

decomposition of 50 g CaCO3 ?

(A) 23

(B) 25

(C) 21

(D) 1

54. Sulphur trioxide is prepared by the following two

reactions -

S8(s) + 8O

2(g) 8SO

2(g)

2SO2(g) + O

2(g) 2SO

3(g)

How many grams of SO3 are produced from 1 mole

of S8 ?

(A) 1280 (B) 640

(C) 960 (D) 320

55. PH3(g) decomposes on heating to produce

phosphorous and hydrogen. The change in volume

when 100 mL of such gas decomposed is -

(A) + 50 mL (B) + 500 mL

(C) � 50 mL (D) � 500 mL

56. What amount of BaSO4 can be obtained on mixing

0.5 mole BaCl2 with 1 mole of H

2SO

4 ?

(A) 0.5 mol (B) 0.15 mol

(C) 0.1 mol (D) 0.2 mol

57. In the reaction , CrO5 + H

2SO

4 Cr

2(SO

4)3 + H

2O + O

2

one mole of CrO5 will liberate how many moles of O

2 ?

(A) 5/2 (B) 5/4

(C) 9/2 (D) None

58. Calcium carbonate decomposes on heating

according to the equation -


2(g)

At STP the volume of CO2 obtained by thermal

decomposition of 50 g of CaCO3 will be -

(A) 22.4 litre (B) 44 litre

(C) 11.2 litre (D) 1 litre

59. When FeCl3 is ignited in an atmosphere of pure

oxygen, the following reaction takes place-

4FeCl3(s) + 3O

2(g) 2Fe

2O

3(s) + 6Cl

2(g)

If 3 moles of FeCl3 are ignited in the presence of 2

moles of O2 gas, how much of which reagent is

present in excess and therefore, remains unreacted ?

(A) 0.33 mole FeCl3 remains unreacted

(B) 0.67 mole FeCl3 remains unreacted

(C) 0.25 mole O2 remains unreacted

(D) 0.50 mole O2 remains unreacted

60. The volume of CO2 (in litres) liberated at STP when

10 g of 90% pure limestone is heated completely, is-

(A) 22.4 L (B) 2.24 L

(C) 20.16 L (D) 2.016 L

61. A metal oxide has the formula Z2O

3. It can be reduced

by hydrogen to give free metal and water. 0.1596 g of

the metal requires 6 mg of hydrogen for complete

reduction. The atomic mass of the metal is -

(A) 27.9 (B) 159.6

(C) 79.8 (D) 55.8

Question number 62, 63, 64 and 65 are based on the

following information :

Q. Dissolved oxygen in water is determined by using a

redox reaction. Following equations describe the

procedure -

I 2Mn2+(aq) + 4OH�(aq) + O2 (g) 2MnO2(s) + 2H2O( )

II MnO2(s)+2I�(aq)+4H+(aq) Mn2+(aq)+I2(aq) + 2H2O( )

III �232OS2 (aq) + I2(aq) �2

64OS2 (aq) + 2I�(aq)

62. How many moles of �232OS are equivalent to each

mole of O2 ?

(A) 0.5 B) 1

(C) 2 (D) 4

63. What amount of I2 will be liberated from 8 g dissolved

oxygen ?

(A) 127 g (B) 254 g

(C) 504 g (D) 1008 g

64. 3 × 10�3 moles O2 is dissolved per litre of water, then

what will be molarity of I� produced in the given

reaction ?

(A) 3 × 10�3 M (B) 4 × 3 × 10�3 M

(C) 2 × 3 × 10�3 M (D) 3�10321

M

65. 8 mg dissolved oxygen will consume -

(A) 5 × 10�4 mol Mn+2

(B) 2.5 × 10�4 mol Mn2+

(C) 10 mol Mn2+

(D) 2 mol Mn2+

66. 2 g of a base whose eq. wt. is 40 reacts with 3 g of an

acid. The eq. wt. of the acid is-

(A) 40 (B) 60

(C) 10 (D) 80

67. Normality of 1% H2SO

4 solution is nearly -

(A) 2.5 (B) 0.1

(C) 0.2 (D) 1

68. What volume of 0.1 N HNO3 solution can be prepared

from 6.3 g of HNO3 ?

(A) 1 litre (B) 2 litres

(C) 0.5 litre (D) 4 litres

PAGE # 40

69. The volume of water to be added to 200 mL of

seminormal HCl solution to make it decinormal is -

(A) 200 mL (B) 400 mL

(C) 600 mL (D) 800 mL

70. 0.2 g of a sample of H2O

2 required 10 mL of 1N KMnO

4

in a titration in the presence of H2SO

4. Purity of H

2O

2 is-

(A) 25% (B) 85%

(C) 65% (D) 95%

71. Which of the following has the highest normality ?

(A) 1 M H2SO

4(B) 1 M H

3PO

3

(C) 1 M H3PO

4(D) 1 M HNO

3

72. The molarity of 98% H2SO

4(d = 1.8g/mL) by wt. is -

(A) 6 M (B) 18.74 M

(C) 10 M (D) 4 M

73. 0.7 g of Na2CO

3 . xH

2O is dissolved in 100 mL. 20 mL

of which required to neutralize 19.8 mL of 0.1 N HCl.

The value of x is -

(A) 4 (B) 3

(C) 2 (D) 1

74. 0.45 g of an acid of molecular weight 90 was

neutralised by 20 mL of 0.5 N caustic potash. The

basicity of the acid is -

(A) 1 (B) 2

(C) 3 (D) 4

75. 1 litre of 18 molar H2SO

4 has been diluted to 100

litres. The normality of the resulting solution is -

(A) 0.09 N (B) 0.18

(C) 1800 N (D) 0.36

76. 150 ml of 10N

HCl is required to react completely with

1.0 g of a sample of limestone. The percentage purity

of calcium carbonate is -

(A) 75% (B) 50%

(C) 80% (D) 90%

77. 50 ml of 10N

HCl is treated with 70 ml 10N

NaOH.

Resultant solution is neutralized by 100 ml of

sulphuric acid. The normality of H2SO

4 -

(A) N/50 (B) N/25

(C) N/30 (D) N/10

78. 200 mL of 10N

HCl were added to 1 g calcium car-

bonate, what would remain after the reaction ?

(A) CaCO3

(B) HCl

(C) Neither of the two (D) Part of both

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

MnSO4 is -

(A) M/5 (B) M/3

(C) M/6 (D) M/2

80. A 3 N solution of H2SO

4 in water is prepared from

Conc. H2SO

4 (36 N) by diluting [KVPY-PartII-2007]

(A) 20 ml of the conc. H2SO

4 to 240 ml

(B) 10 ml of the conc. H2SO

4 to 240 ml

(C) 1 ml of the conc. H2SO

4 to 36 ml

(D) 20 ml of the conc. H2SO

4 to 36 ml

81. The solubility curve of KNO3 as a function of tempera-

ture is given below [KVPY-Part-II-2007]

0 20 40 60 80 100

0

50

100

150

200

250

Temperature (°C)

Sol

ubili

ty (

g/10

0 m

l wat

er)

The amount of KNO3 that will crystallize when a satu-

rated solution of KNO3 in 100 ml of water is cooled

from 90°C to 30 °C, is

(A) 16 g (B) 100 g

(C) 56 g (D)160 g

82. The volume of 0.5 M aqueous NaOH solution required

to neutralize 10 ml of 2 M aqueous HCl solution is :

[KVPY-Part-I-2008](A) 20ml (B) 40ml

(C) 80ml (D) 120ml

83. 3.01×1023 molecules of elemental Sulphur will react

with 0.5 mole of oxygen gas completely to produce

[KVPY-Part-I-2008](A) 6.02 × 1023 molecules of SO

3

(B) 6.02 × 1023 molecules of SO2

(C) 3.01 × 1023 molecules of SO3

(D) 3.01 x 1023 molecules of SO2

84. The solubility of a gas in a solution is measured in

three cases as shown in the figure given below where

w is the weight of a solid slab placed on the top of the

cylinder lid. The solubility will follow the order :

[KVPY-Part-I-2008]

gas

solution

w

gas

solution

w w

gas

solution

w w w

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

(C) a = b = c (D) a >b < c

PAGE # 41

85. The density of a salt solution is1.13 g cm�3 and itcontains 18% of NaCI by weight. The volume of thesolution containing 36.0 g of the salt will be :

[KVPY-Part-II-2008](A) 200 cm3 (B) 217 cm3

(C) 177 cm3 (D) 157cm3

86. One mole of nitrogen gas on reaction with 3.01 x 1023

molecules of hydrogen gas produces - [KVPY-Part-I-2009]

(A) one mole of ammonia(B) 2.0 x 1023 molecules of ammonia(C) 2 moles of ammonia(D) 3.01 × 1023 molecules of ammonia

87. [KVPY-Part-I-2009]Solubility g/I

20 40 60 80 100

KNO3

KCl

Temperature (ºC)

50

100

150

200

250

Given the solubility curves of KNO3 and KCl, which of

the following statements is not true ?(A) At room temperature the solubility of KNO

3 and

KCI are not equal(B) The solubilities of both KNO

3 and KCI increase

with temperature(C) The solubility of KCI decreases with temperature(D) The solubility of KNO

3 increases much more com-

pared to that of KCl with increase in temperature

88. 10 ml of an aqueous solution containing 222 mg ofcalcium chloride (mol. wt. = 111) is diluted to 100 ml.The concentration of chloride ion in the resulting so-lution is - [KVPY-Part-II-2009](A) 0.02 mol/lit. (B) 0.01 mol/lit.(C) 0.04 mol/lit (D) 2.0 mol/lit.

89. Aluminium reduces manganese dioxide to manga-nese at high temperature. The amount of aluminiumrequired to reduce one gram mole of manganesedioxide is - [KVPY-Part-II-2009](A) 1/2 gram mole (B) 1 gram mole(C) 3/4 gram mole (D) 4/3 gram mole

90. One mole of oxalic acid is equivalent to[IJSO-2009]

(A) 0.5 mole of NaOH (B) 1 mole of NaOH(C) 1.5 mole of NaOH (D) 2 mole of NaOH

Mole Concept

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