MOLE CONCEPT # 1

Mole ConceptIntroduction :

There are a large number of objects around us which we can see and feel.Anything that occupies space and has mass is called matter.

Ancient Indian and Greek Philosopher�s beleived that the wide variety of object around us are made from

combination of five basic elements : Earth, Fire, Water, Air and Sky.The Indian Philosopher kanad (600 BC) was of the view that matter was composed of very small, indivisibleparticle called �parmanus�.Ancient Greek Philosophers also believed that all matter was composed of tiny building blocks which were hardand indivisible.The Greek philosopher Democritus named these building blocks as atoms, meaning indivisible.All these people have their philosophical view about matter, they were never put to experimental tests, nor everexplain any scientific truth.It was John Dalton who firstly developed a theory on the structure of matter, later on which is known asDalton�s atomic theory.

DALTON�S ATOMIC THEORY :

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

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

Atoms cannot be created or destroyed by any chemical process.

Atoms of different elements are different in nature.

Classification of matter

on the basis of physical behaviour on the basis of chemical behaviour

Solids Liquids Gases Pure substances Mixtures

Element Compound

Basic Definitions :

Relative atomic mass :One of the most important concept come out from Dalton�s atomic theory was that of relative atomic mass or

relative atomic weight. This is done by expressing mass of one atom with respect to a fixed standard. Daltonused hydrogen as the standard (H = 1). Later on oxygen (O = 16) replaced hydrogen as the reference. Thereforerelative atomic mass is given asOn hydrogen scale :

Relative atomic mass (R.A.M) = atomhydrogenoneofmasselementanofatomoneofMass

On oxygen scale :

Relative atomic mass (R.A.M) = atomoxygenoneofmass

161

elementanofatomoneofMass

MOLE CONCEPT # 2

The present standard unit which was adopted internationally in 1961, is based on the mass of one

carbon-12 atom.

Relative atomic mass (R.A.M) = atom12Coneofmass

121

elementanofatomoneofMass

Atomic mass unit (or amu) :

The atomic mass unit (amu) is equal to th

121

mass of one atom of carbon-12 isotope.

1 amu = 121

× mass of one C-12 atom

~ mass of one nucleon in C-12 atom.= 1.66 × 10�24 gm or 1.66 × 10�27 kg

one amu is also called one Dalton (Da).

Today, amu has been replaced by �u� which is known as unified mass

Atomic & molecular mass :It is the mass of 1 atom of a substance it is expressed in amu.

Atomic mass = R.A.M × 1 amu

Relative molecular mass = atom12Coneofmass

121

cetansubstheofmoleculeoneofmass

Molecular mass = Relative molecular mass × 1 amu

Note :Relative atomic mass is nothing but the number of nucleons present in the atom.

Example-1 Find the relative atomic mass of �O� atom and its atomic mass.

Solution The number of nucleons present in �O� atom is 16.

relative atomic mass of �O� atom = 16.

Atomic mass = R.A.M × 1 amu = 16 × 1 amu = 16 amu

Mole : The Mass / Number RelationshipMole is a chemical counting S unit and defined as follows :A mole is the amount of a substance that contains as many entities (atoms, molecules or other particles)as there are atoms in exactly 0.012 kg (or 12 gm) of the carbon-12 isotope.From mass spectrometer we found that there are 6.023 × 1023 atoms present in 12 gm of C-12 isotope.The number of entities in 1 mol is so important that it is given a separate name and symbol known as Avogadroconstant denoted by NA.i.e. on the whole we can say that 1 mole is the collection of 6.02 × 1023 entities. Here entities may representatoms, ions, molecules or even pens, chair, paper etc also include in this but as this number (NA) is very largetherefore it is used only for very small things.

MOLE CONCEPT # 3

HOW BIG IS A MOLE ?

602,200,000,000,000,000,000,000

Amount of water inworld's oceans (litres)

Age of earth (seconds)Population of earth

Distance from earthto sun (centimeters)

Avogadro's number

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

Gram Atomic Mass :The atomic mass of an element expressed in gram is called gram atomic mass of the element.

orIt is also defined as mass of 6.02 × 1023 atoms.

orIt is also defined as the mass of one mole atoms.

For example for oxygen atom :

Atomic mass of �O� atom = mass of one �O� atom = 16 amu

gram atomic mass = mass of 6.02 × 1023 �O� atoms

= 16 amu × 6.02 × 1023

= 16 × 1.66 × 10�24 g × 6.02 ×1023 = 16 g( 1.66 × 10�24 × 6.02 × 1023 ~ 1 )

Example-2 How many atoms of oxygen are their in 16 g oxygen.Solution Let x atoms of oxygen are present

So, 16 × 1.66 × 10�24 × x = 16 g

x = 2410x66.1

1 = NA

Gram molecular mass :The molecular mass of a substance expressed in gram is called the gram-molecular mass of the substance.

orIt is also defined as mass of 6.02 × 1023 molecules

orIt is also defined as the mass of one mole molecules.

For example for �O2� molecule :

Molecular mass of �O2� molecule = mass of one �O2� molecule

= 2 × mass of one �O� atom

= 2 × 16 amu

= 32 amugram molecular mass = mass of 6.02 × 1023 �O2� molecules = 32 amu × 6.02 × 1023

= 32 × 1.66 × 10�24 gm × 6.02 × 1023 = 32 gm

MOLE CONCEPT # 4

Example-3 The molecular mass of H2SO4 is 98 amu. Calculate the number of moles of each element in 294 g ofH2SO4.

Solution Gram molecular mass of H2SO4 = 98 gm

moles of H2SO4 = 98294

= 3 moles

H2SO4 H S OOne molecule 2 atom one atom 4 atom1 × NA 2 × NA atoms 1 × NA atoms 4 × NA atoms one mole 2 mole one mole 4 mole 3 mole 6 mole 3 mole 12 mole

Gay-Lussac�s Law of Combining Volume :

According to him elements combine in a simple ratio of atoms, gases combine in a simple ratio of their volumesprovided all measurements should be done at the same temperature and pressure

H2 (g) + Cl2 (g) 2HCl

1 vol 1 vol 2 vol

Avogadro�s hypothesis :

Equal volume of all gases have equal number of molecules (not atoms) at same temperature and pressurecondition.

S.T.P. (Standard Temperature and Pressure)

At S.T.P. condition : temperature = 0°C or 273 K

pressure = 1 atm = 760 mm of Hgand volume of one mole of gas at STP is found to be experimentally equal to 22.4 litres which is known as molarvolume.

Note : Measuring the volume is equivalent to counting the number of molecules of the gas.

Example-4 Calculate the volume in litres of 20 g hydrogen gas at STP.

Solution No. of moles of hydrogen gas = massMolecularMass

= gm2gm20

= 10 mol

volume of hydrogen gas at STP = 10 × 22.4 lt.

Y-map : Interconversion of mole - volume, mass and number of particles :

Mole

× 22.4 lt

22.4 lt

Volume at STP

NA

× NA

Number

× mol. wt.

× At. wt.

At. wt. mol. wt.

Mass

MOLE CONCEPT # 5

The laws of chemical combination :Atoine Lavoisier, John Dalton and other scientists formulate certain law concerning the composition of matterand chemical reactions. These laws are known as the law of chemical combination.

(i) The law of conservation of mass :In a chemical change total mass remains conserved.i.e. mass before reaction is always equal to mass after reaction.

Example-5 H2 (g) + 21

O2 (g) H2O (l)

Solution H2 (g) + 21

O2 (g) H2O (l)

Before reaction initially 1 mole 21

mole 0

After the reaction 0 0 1 mole

mass before reaction = mass of 1 mole H2 (g) + 21

mole O2 (g)

= 2 + 16 = 18 gmmass after reaction = mass of 1 mole water = 18 gm

(ii) Law of constant or Definite proportion :All chemical compounds are found to have constant composition irrespective of their method of preparation orsources.Example : In water (H2O), Hydrogen and Oxygen combine in 2 : 1 molar ratio, this ratio remains constantwhether it is tap water, river water or sea water or produced by any chemical reaction.

Example-6 1.80 g of a certain metal burnt in oxygen gave 3.0 g of its oxide. 1.50 g of the same metal heated insteam gave 2.50 g of its oxide. Show that these results illustrate the law of constant proportion.

Solution In the first sample of the oxide,Wt. of metal = 1.80 g,Wt. of oxygen = (3.0 � 1.80) g = 1.2 g

5.1g2.1g80.1

oxygenof.wtmetalof.wt

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

5.1g1

g50.1oxygenof.wtmetalof.wt

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

(iii) The law of multiple proportion :When one element combines with the other element to form two or more different compounds, the mass of oneelement, which combines with a constant mass of the other, bear a simple ratio to one another.

Note : Simple ratio here means the ratio between small natural numbers, such as 1 : 1, 1 : 2, 1 : 3, later onthis simple ratio becomes the valency and then oxidation state of the element.See oxidation number of carbon also have same ratio 1 : 2 in both the oxide.

MOLE CONCEPT # 6

Example-7 Carbon is found to form two oxides, which contain 42.9% and 27.3% of carbon respectively. Show thatthese figures illustrate the law of multiple proportions.

Solution Step-1To calculate the percentage composition of carbon and oxygen in each of the two oxides.

First oxide Second oxideCarbon 42.9 % 27.3 % (Given)Oxygen 57.1% 72.7 %(by difference)Step-2To calculate the masses of carbon which combine with a fixed mass i.e., one part by mass of oxygenin each of the two oxides.n the first oxide, 57.1 parts by mass of oxygen combine with carbon = 42.9 parts.

1 part by mass of oxygen will combine with carbon = 1.579.42

= 0.751.

n the second oxide. 72.7 parts by mass of oxygen combine with carbon = 27.3 parts.

1 part by mass of oxygen will combine with carbon = 7.723.27

= 0.376

Step-3.To compare the masses of carbon which combine with the same mass of oxygen in both the oxides.The ratio of the masses of carbon that combine with the same mass of oxygen (1 part) is .0.751 : 0.376 or 2 : 1Since this is simple whole number ratio, so the above data illustrate the law of multiple proportions.

Percentage Composition :Here we are going to find out the percentage of each element in the compound by knowing the molecular formulaof compound.We know that according to law of definite proportions any sample of a pure compound always possess constantratio with their combining elements.

Example-8 Every molecule of ammonia always has formula NH3 irrespective of method of preparation or sources.i.e. 1 mole of ammonia always contains 1 mol of N and 3 mole of H. In other words 17 gm of NH3 alwayscontains 14 gm of N and 3 gm of H. Now find out % of each element in the compound.

Solution Mass % of N in NH3 = 100NHofmol1ofMass

NHmol1inNofMass

3

3 = 17

gm14 × 100 = 82.35 %

Mass % of H in NH3 = 100NHofemol1ofMass

NHmol1isHofMass

3

3 = 173

× 100 = 17.65 %

Empirical and molecular formula :We have just seen that knowing the molecular formula of the compound we can calculate percentage compositionof the elements. Conversely if we know the percentage composition of the elements initially, we can calculatethe relative number of atoms of each element in the molecules of the compound. This gives us the empiricalformula of the compound. Further if the molecular mass is known then the molecular formula can easily bedetermined.

MOLE CONCEPT # 7

The empirical formula of a compound is a chemical formula showing the relative number of atoms in the simplestratio. An empirical formula represents the simplest whole number ratio of various atoms present in a compound.The molecular formula gives the actual number of atoms of each element in a molecule. The molecular formulashows the exact number of different types of atoms present in a molecule of a compound.The molecular formula is an integral multiple of the empirical formula.i.e. molecular formula = empirical formula × n

where n = massformulaempiricalmassformulamolecular

Example-9 Acetylene and benzene both have the empirical formula CH. The molecular masses of acetylene andbenzene are 26 and 78 respectively. Deduce their molecular formulae.

Solution Empirical Formula is CHStep-1 The empirical formula of the compound is CH Empirical formula mass = (1 × 12) + 1 = 13.

Molecular mass = 26Step-2 To calculate the value of �n�

n = massformulaEmpiricalmassMolecular

= 1326

= 2

Step-3 To calculate the molecular formula of the compound.Molecular formula = n × (Empirical formula of the compound)

= 2 × CH = C2 H2

Thus the molecular formula is C2 H2

Similarly for benzeneTo calculate the value of �n�

n = massformulaEmpiricalmassMolecular

= 1378

= 6

thus the molecular formula is 6 × CH = C6H6

Example-10 An organic substance containing carbon, hydrogen and oxygen gave the following percentage composition.C = 40.684% ; H = 5.085% and O = 54.228%The molecular weight of the compound is 118 gm. Calculate the molecular formula of the compound.

Solution Step-1To calculate the empirical formula of the compound.

Element Symbol

Carbon C

Hydrogen H

Oxygen O

Percentage of element

40.687

5.085

54.228

At. massof element

12

1

16

Relative no.Percentage

At. massof atoms =

40.687 12

= 3.390

5.085

1= 5.085

54.228

16= 3.389

Simplestatomic ratio

3.3903.389

=1

5.0853.389

=1.5

3.3893.389

=1

Simplest wholeno. atomic ratio

2

3

2

Empirical Formula is C2 H3 O2

Step-2 To calculate the empirical formula mass.The empirical formula of the compound is C2 H3 O2 .

Empirical formula mass = (2 × 12) + (3 × 1) + (2 × 16) = 59.

MOLE CONCEPT # 8

Step-3 To calculate the value of �n�

n = massformulaEmpiricalmassMolecular

= 59

118 = 2

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

Thus the molecular formula is C4 H6 O4.

Chemical Reaction :It is the process in which two or more than two substances interact with each other where old bonds are brokenand new bonds are formed.

Chemical Equation :All chemical reaction are represented by chemical equations by using chemical formula of reactants and products.Qualitatively a chemical equation simply describes what the reactants and products are. However, a balanced

chemical equation gives us a lot of quantitative information. Mainly the molar ratio in which reactants combine

and the molar ratio in which products are formed.

Attributes of a balanced chemical equation:(a) It contains an equal number of atoms of each element on both sides of equation.(POAC)

(b) It should follow law of charge conservation on either side.(c) Physical states of all the reagents should be included in brackets.

(d) All reagents should be written in their standard molecular forms (not as atoms )

(e) The coefficients give the relative molar ratios of each reagent.

Example-11 Write a balance chemical equation for following reaction :When potassium chlorate (KClO3) is heated it gives potassium chloride (KCl) and oxygen (O2).

Solution KClO3 (s) KCl (s) + O2 (g) (unbalanced chemical equation )

2KClO3 (s) 2 KCl (s) + 3 O2 (g) (balanced chemical equation)

Remember a balanced chemical equation is one which contains an equal number of atoms of each

element on both sides of equation.

Interpretation of balanced chemical equations :Once we get a balanced chemical equation then we can interpret a chemical equation by following

ways

Mass - mass analysis Mass - volume analysis

Mole - mole analysis Vol - Vol analysis (separately discussed as eudiometry or gas analysis)

Now you can understand the above analysis by following example

Mass-mass analysis :

Consider the reaction

2KClO3 2KCl + 3O2 According to stoichiometry of the reaction

mass-mass ratio: 2 × 122.5 : 2 × 74.5 : 3 × 32

MOLE CONCEPT # 9

or KClofMass

KClOofMass 3 =

5.7425.1222

2

3

OofMass

KClOofMass =

3235.1222

Example-12 367.5 gram KClO3 (M = 122.5) when heated. How many gram KCl and oxygen is produced.Solution Balance chemical equation for heating of KClO3 is

2KClO3 2KCl + 3O2

mass-mass ratio : 2 × 122.5 gm : 2 × 74.5 gm : 3 × 32 gm

KClofmass

KClOofmass 3 =

5.7425.1222

W5.367

= 5.745.122

W = 3 × 74.5 = 223.5 gm

2

3

OofMass

KClOofMass =

3235.1222

W5.367

= 323

5.1222

W = 144 gm

Mass - volume analysis :

Now again consider decomposition of KClO3

2KClO3 2KCl + 3O2

mass volume ratio : 2 × 122.5 gm : 2 × 74.5 gm : 3 × 22.4 lt. at STP

we can use two relation for volume of oxygen

STPatOofvolume

KClOofMass

2

3 = lt4.223

5.1222

...(i)

and STPatOofvolumeKClofMass

2 = lt4.223

5.742

...(ii)

Example-13 367.5 gm KClO3 (M = 122.5) when heated, how many litre of oxygen gas is produced at STP.Solution You can use here equation (1)

STPatOofvolume

KClOofmass

2

3 = lt4.223

5.1222

V5.367

= lt4.2235.1222

V = 3 × 3 × 11.2 V = 100.8 lt

Mole-mole analysis :

This analysis is very much important for quantitative analysis point of view. Students are advised toclearly understand this analysis.Now consider again the decomposition of KClO3 .

2KClO3 2KCl + 3O2

In very first step of mole-mole analysis you should read the balanced chemical equation like2 moles KClO3 on decomposition gives you 2 moles KCl and 3 moles O2. and from the stoichiometryof reaction we can write

MOLE CONCEPT # 10

2

KClOofMoles 3 = 2

KClofMoles =

3

OofMoles 2

Now for any general balance chemical equation like

a A + b B c C + d D

you can write.

areactedAofMoles

=b

reactedBofmoles =

cproducedCofmoles

= d

producedDofmoles

Note : In fact mass-mass and mass-vol analysis are also interpreted in terms of mole-mole analysis you can

use following chart also.

Mass At. wt. / Mol. Wt.Mole

Mole-mole relationship of equation

Mole

× 22.4 lt

Volume at STP

× mol. wt./A

t. wt.

Mass

Limiting reagent :The reactant which is consumed first and limits the amount of product formed in the reaction, and is therefore,called limiting reagent.Limiting reagent is present in least stoichiometric amount and therefore, controls amount of product.The remaining or left out reactant is called the excess reagent.When you are dealing with balance chemical equation then if number of moles of reactants are not in the ratioof stoichiometric coefficient of balanced chemical equation, then there should be one reactant which is limitingreactant.

Example-14 Three mole of Na2 CO3 is reacted with 6 moles of HCl solution. Find the volume of CO2 gas producedat STP. The reaction is

Na2 CO3 + 2HCl 2 NaCl + CO2 + H2O

Solution From the reaction : Na2 CO3 + 2HCl 2 NaCl + CO2 + H2O

given moles 3 mol 6 molgiven mole ratio 1 : 2Stoichiometric coefficient ratio 1 : 2

See here given moles of reactant are in stoichiometric coefficient ratio therefore none reactant left over.

Now use Mole-mole analysis to calculate volume of CO2 produced at STP

1

CONaofMoles 32 = 1

oducedPrCOofMole 2

Moles of CO2 produced = 3volume of CO2 produced at STP = 3 × 22.4 L = 67.2 L

Example-15 6 moles of Na2 CO3 is reacted with 4 moles of HCl solution. Find the volume of CO2 gas produced atSTP. The reaction is

Na2 CO3 + 2HCl 2 NaCl + CO2 + H2O

MOLE CONCEPT # 11

Solution From the reaction : Na2 CO3 + 2HCl 2 NaCl + CO2 + H2Ogiven mole of reactant 6 : 4given molar ratio 3 : 2Stoichiometric coefficient ratio 1 : 2

See here given number of moles of reactants are not in stoichiometric coefficient ratio. Therefore there

should be one reactant which consumed first and becomes limiting reagent.But the question is how to find which reactant is limiting, it is not very difficult you can easily find it.According to the following method.

How to find limiting reagent :

Step : Divide the given moles of reactant by the respective stoichiometric coefficient of that reactant.Step : See for which reactant this division come out to be minimum. The reactant having minimum

value is limiting reagent for you.Step : Now once you find limiting reagent then your focus should be on limiting reagent

From Step & Na2 CO3 HCl

16

= 624

= 2 (division is minimum)

HCl is limiting reagentFrom Step

From 2

HClofMole =

1producedCOofMoles 2

mole of CO2 produced = 2 moles volume of CO2 produced at S.T.P. = 2 × 22.4 = 44.8 lt.

Principle of Atom Conservation (POAC) :POAC is conservation of mass. Atoms are conserved, moles of atoms shall also be conserved in a chemical

reaction (but not in nuclear reactions.)

This principle is fruitful for the students when they don�t get the idea of balanced chemical equation in the

problem.

The strategy here will be around a particular atom. We focus on a atom and conserve it in that reaction. This

principle can be understand by the following example.

Consider the decomposition of KClO3 (s) KCl (s) + O2 (g) (unbalanced chemical reaction)

Apply the principle of atom conservation (POAC) for K atoms.

Moles of K atoms in reactant = moles of K atoms in products

or moles of K atoms in KClO3 = moles of K atoms in KCl.

Now, since 1 molecule of KClO3 contains 1 atom of K

or 1 mole of KClO3 contains 1 mole of K, similarly,1 mole of KCl contains 1 mole of K.

Thus, moles of K atoms in KClO3 = 1 × moles of KClO3

and moles of K atoms in KCl = 1 × moles of KCl.

moles of KClO3 = moles of KCl

or3

3

KClOofwt.mol.ginKClOofwt.

= KClofwt.mol.ginKClofwt.

The above equation gives the mass-mass relationship between KClO3 and KCl which is important in

stoichiometric calculations.

MOLE CONCEPT # 12

Again, applying the principle of atom conservation for O atoms,

moles of O in KClO3 = 3 × moles of KClO3

moles of O in O2 = 2 × moles of O2

3 × moles of KClO3 = 2 × moles of O2

or 3 × 3

3

KClOof.wt.molKClOof.wt

= 2 × .)lt4.22(.volmolardardtans

NTPatOof.vol 2

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

Example-16 27.6 g K2CO3 was treated by a series of reagents so as to convert all of its carbon to K2 Zn3 [Fe(CN)6]2.Calculate the weight of the product.[mol. wt. of K2CO3 = 138 and mol. wt. of K2Zn3 [Fe(CN)6]2 = 698]

Solution Here we have not knowledge about series of chemical reactionsbut we know about initial reactant and final product accordingly

K2CO3 StepsSeveral

K2Zn3 [Fe(CN)6]2

Since C atoms are conserved, applying POAC for C atoms,moles of C in K2CO3 = moles of C in K2Zn3 [Fe(CN)6]2

1 × moles of K2CO3 = 12 × moles of K2Zn3 [Fe(CN)6]2

( 1 mole of K2CO3 contains 1 moles of C)

32

32

COKof.wt.mol

COKof.wt = 12 × productof.wt.mol

producttheof.wt

wt. of K2Zn3 [Fe(CN)6]2 = 138

6.27 ×

12698

= 11.6 g

Miscellaneous :

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

Mathematically, average atomic mass of X (Ax) = 100

xa.....xaxa nn2211

Where :a1, a2, a3 ........... atomic mass of isotopes.

and x1, x2, x3 ........... mole % of isotopes.

Example-17 Naturally occuring chlorine is 75% Cl35 which has an atomic mass of 35 amu and 25% Cl37 whichhas a mass of 37 amu. Calculate the average atomic mass of chlorine -

(A) 35.5 amu (B) 36.5 amu (C) 71 amu (D) 72 amu

Solution (A) Average atomic mass = 100

massatomicitsxisotopeIof%massatomsitsxisotopeof%

= 100

37x2535x75 = 35.5 amu

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

MOLE CONCEPT # 13

MEAN MOLAR MASS OR MOLECULAR MASS:

The average molar mass of the different substance present in the container = n21

nn2211

n....nnMn......MnMn

.

Where :

M1, M2, M3 ........... are molar masses.n1, n2, n3 ........... moles of substances.

Example-18 The molar composition of polluted air is as follows :Gas At. wt. mole percentage compositionOxygen 16 16%Nitrogen 14 80%Carbon dioxide - 03%Sulphurdioxide - 01%What is the average molecular weight of the given polluted air ? (Given, atomic weights of C and S are12 and 32 respectively.

Solution Mavg =

nj

1jj

nj

1jjj

n

Mn

Here

nj

1jjn = 100

Mavg = 100

1x643x4428x8032x16 =

100641322240512

= 1002948

= 29.48 Ans.

Oxidation & Reduction

Let us do a comparative study of oxidation and reduction :

Oxidation Reduction1. Addition of Oxygen 1. Removal of Oxygene.g. 2Mg + O2 2MgO e.g. CuO + C Cu + CO

2. Removal of Hydrogen 2. Addition of Hydrogene.g. H2S + Cl2 2HCl + S e.g. S + H2 H2S

3. Increase in positive charge 3. Decrease in positive charge e.g. Fe2+ Fe3+ + e� e.g. Fe3+ + e� Fe2+

4. Increase in oxidation number 4. Decrease in oxidation number (+2) (+4) (+7) (+2)e.g. SnCl2 SnCl4 e.g. MnO4

� Mn2+

5. Removal of electron 5. Addition of electrone.g. Sn2+ Sn4+ + 2e� e.g. Fe3+ + e� Fe2+

MOLE CONCEPT # 14

Oxidation Number It is an imaginary or apparent charge developed over atom of an element when it goes from its elemental

free state to combined state in molecules.

It is calculated on basis of an arbitrary set of rules.

It is a relative charge in a particular bonded state.

In order to keep track of electron-shifts in chemical reactions involving formation of compounds, a morepractical method of using oxidation number has been developed.

In this method, it is always assumed that there is a complete transfer of electron from a less electronegativeatom to a more electronegative atom.

Rules governing oxidation numberThe following rules are helpful in calculating oxidation number of the elements in their different compounds.It is to be remembered that the basis of these rule is the electronegativity of the element .

Fluorine atom :

Fluorine is most electronegative atom (known). It always has oxidation number equal to �1 in all its

compounds

Oxygen atom :

In general and as well as in its oxides , oxygen atom has oxidation number equal to �2.

In case of (i) peroxide (e.g. H2O2, , Na2O2 ) is �1,

(ii) super oxide (e.g. KO2) is �1/2

(iii) ozonide (e.g. KO3) is �1/3

(iv) in OF2 is + 2 & in O2F2 is +1

Hydrogen atom :

In general, H atom has oxidation number equal to +1. But in metallic hydrides ( e.g. NaH, KH), it is �1.

Halogen atom :

In general, all halogen atoms (Cl, Br , I) have oxidation number equal to �1.

But if halogen atom is attached with a more electronegative atom than halogen atom, then it will showpositive oxidation numbers.

e.g. 3

5ClOK

, 3

5OHI

,

4

7OHCI

,

5

3KBrO

Metals :

(a) Alkali metal (Li , Na, K, Rb, .......) always have oxidation number +1(b) Alkaline earth metal (Be , Mg , Ca .......) always have oxidation number +2.(c) Aluminium always has +3 oxidation number

Note : Metal may have negative or zero oxidation number Oxidation number of an element in free state or in allotropic forms is always zero

e.g. 0

3

0

4

0

8

0

2 O,P,S,O

Sum of the oxidation numbers of atoms of all elements in a molecule is zero.

Sum of the oxidation numbers of atoms of all elements in an ion is equal to the charge on the ion .

If the group number of an element in modern periodic table is n, then its oxidation number may vary from(n � 10) to (n � 18) (but it is mainly applicable for p-block elements )

e.g. N- atom belongs to 15th group in the periodic table, therefore as per rule, its oxidation number mayvary from

�3 to +5 ( 2

3

3NO,HN

, 32

3ON

, 2

4ON

, 52

5ON

)

The maximum possible oxidation number of any element in a compound is never more than thenumber of electrons in valence shell.(but it is mainly applicable for p-block elements )

MOLE CONCEPT # 15

Calculation of average oxidation number :

Example-19 Calculate oxidation number of underlined element :(a) Na2 S2O3 (b) Na2 S 4O6

Solution. (a) Let oxidation number of S-atom is x. Now work accordingly with the rules given before .(+1) × 2 + (x) × 2 + (�2) ×3 =0

x = + 2

(b) Let oxidation number of S-atom is x (+1) × 2 + (x) × 4 + (�2) × 6 = 0

x = + 2.5

It is important to note here that Na2S2O3 have two S-atoms and there are four S-atom in Na2S4O6.

However none of the sulphur atoms in both the compounds have + 2 or + 2.5 oxidation number, it is theaverage of oxidation number, which reside on each sulphur atom. Therefore, we should work to calculatethe individual oxidation number of each sulphur atom in these compounds.

Calculation of individual oxidation numberIt is important to note that to calculate individual oxidation number of the element in its compound one shouldknow the structure of the compound and use the following guidelines.Formula :Oxidation Number = Number of electrons in the valence shell � Number of electrons taken up after bonding

Guidelines : It is based on electronegativity of elements.

1. If there is a bond between similar type of atom and each atom has same type of hybridisation, thenbonded pair electrons are equally shared by each element.

Example : Calculate oxidation number of each Cl-atom in Cl2 molecule

Structure :

: Number of electrons in the valence shell = 7.Number of electrons taken up after bonding = 7.

oxidation number = 7 � 7 = 0.

: similarly, oxidation number = 7 � 7 = 0

2. If there is a bond between different type of atoms :e.g. A � B (if B is more electronegative than A)

Then after bonding, bonded pair of electrons are counted with B - atom .

Example : Calculate oxidation number of each atom in HCl molecule

Structure :

Note : Electron of H-atom is now counted with Cl-atom, because Cl-atom is more electronegative than H-atom

H : Number of electrons in the valence shell = 1 Number of electrons taken up after bonding = 0 Oxidation number of H = 1 � 0 = + 1

MOLE CONCEPT # 16

Cl : Number of electrons in the valence shell = 7 Number of electrons taken up after bonding = 8 Oxidation number of Cl = 7� 8 = � 1

Example-20 Calculate individual oxidation number of each S-atom in Na2S2O3 (sodium thiosulphate) with the help ofits structure .

Solution. Structure :

Note : (central S-atom) is sp3 hybridised (25% s-character) and (terminal S-atom) is sp2

hydbridised (33% s-character). Therefore, terminal sulphur atom is more electronegativethan central sulphur atom. Now, the shared pair of electrons are counted with terminal S-atom.

, S-atom : Number of electrons in the valence shell = 6Number of electrons left after bonding = 0Oxidation number of central S-atom = 6 � 0 = + 6

, S-atom : Number of electrons in the valence shell = 6Number of electrons left after bonding = 8Oxidation number of terminal S-atom = 6 � 8 = � 2

Now, you can also calculate Average Oxidation number of S = 2

)2(6 = + 2 (as we have calculated before)

Miscellaneous Examples : In order to determine the exact or individual oxidation number we need to take help from the structures of the

molecules. Some special cases are discussed as follows:

The structure of CrO5 is

O O

O O

Cr||O

From the structure, it is evident that in CrO5 there are two peroxide linkages and one double bond. The contributionof each peroxide linkage is �2. Let the oxidation number of Cr is x.

x + (�2)2 + (�2) = 0 or x = 6

Oxidation number of Cr = + 6 Ans

The structure of H2SO5 is H O O S

O

OOH

From the structure, it is evident that in H2SO5, there is one peroxide linkage, two sulphur-oxygen double bondsand one OH group. Let the oxidation number of S = x. (+ 1) + (� 2) + x + (�2) 2+ (�2) + 1 = 0

or x + 2 � 8 = 0 or x � 6 = 0 or x = 6 Oxidation number of S in H2SO5 is + 6 Ans.

MOLE CONCEPT # 17

Paradox of fractional oxidation numberFractional oxidation number is the average of oxidation state of all atoms of element under examination and thestructural parameters reveal that the atoms of element for whom fractional oxidation state is realised aactually present in different oxidation states. Structure of the species C3O2, Br3O8 and S4O6

2� reveal the followingbonding situations :

The element marked with asterisk (*) in each species is exhibiting different oxidation number from restof the atoms of the same element in each of the species. This reveals that in C3O2, two carbon atomsare present in +2 oxidation state each whereas the third one is present in zero oxidation state and theaverage is + 4/3. However, the realistic picture is +2 for two terminal carbons and zero for the middlecarbon.

OC*CCO202

Structure of C3O2

(Carbon suboxide) Likewise in Br3O8, each of the two terminal bromine atoms are present in +6 oxidation state and the middle

bromine is present in +4 oxidation state. Once again the average, that is different from reality, is + 16/3.

In the same fashion, in the species S4O62�, average oxidation number of S is + 2.5, whereas the reality

being +5,0,0 and +5 oxidation number respectively for respective sulphur atoms.

In general, the conclusion is that the idea of fractional oxidation state should be taken with care andthe reality is revealed by the structures only.

Oxidising and reducing agentOxidising agent or Oxidant :

Oxidising agents are those compounds which can oxidise others and reduce itself during the chemicalreaction. Those reagents in which for an element, oxidation number decreases or which undergoesgain of electrons in a redox reaction are termed as oxidants.

e.g. KMnO4 , K2Cr2O7 , HNO3, conc.H2SO4 etc are powerful oxidising agents .

Reducing agent or Reductant :Reducing agents are those compounds which can reduce other and oxidise itself during the chemicalreaction. Those reagents in which for an element, oxidation number increases or which undergoes lossof electrons in a redox reaction are termed as reductants.e.g. K , Na2S2O3 etc are the powerful reducing agents.

Note : There are some compounds also which can work both as oxidising agent and reducing agente.g. H2O2, NO2

�

MOLE CONCEPT # 18

HOW TO IDENTIFY WHETHER A PARTICULAR SUBSTANCE IS AN OXIDISING OR A REDUCING AGENT

Redox reactionA reaction in which oxidation and reduction simultaneously take place is called a redox reactionIn all redox reactions, the total increase in oxidation number must be equal to the total decrease in oxidationnumber.

e.g. 10 4

2SOFe

+ 5

4KMnO2

+ 8H2SO4 342

3SOFe5

+ 4

2SOMn2

+ K2SO4 + 8H2O

Disproportionation Reaction :A redox reaction in which same element present in a particular compound in a definite oxidation state isoxidized as well as reduced simultaneously is a disproportionation reaction.Disproportionation reactions are a special type of redox reactions. One of the reactants in a disproportionationreaction always contains an element that can exist in at least three oxidation states. The element in theform of reacting substance is in the intermediate oxidation state and both higher and lower oxidation states ofthat element are formed in the reaction. For example :

)aq(OH21

22

)(OH2

2

2

+ )g(O2

0

)s(S0

8 + 12OH¯(aq) )aq(S4

22

+ )aq(OS2

2232

+ 6H2O ()

)g(Cl0

2 + 2OH¯(aq) )aq(ClO

1 + )aq(Cl

1 + H2O ()

Consider the following reactions :

(a) 2KClO3 2KCl + 3O2

KClO3 plays a role of oxidant and reductant both. Here, Cl present in KClO3 is reduced and O presentin KClO3 is oxidized. Since same element is not oxidized and reduced, so it is not a disproportionationreaction, although it looks like one.

(b) NH4NO2 N2 + 2H2ONitrogen in this compound has -3 and +3 oxidation number, which is not a definite value. So it is not adisproportionation reaction. It is an example of comproportionation reaction, which is a class of redoxreaction in which an element from two different oxidation state gets converted into a single oxidationstate.

(c)5

3KClO4

7

4KClO3

+ 1�

KCl

It is a case of disproportionation reaction and Cl atom is disproportionating.

List of some important disproportionation reactions

1. H2O2 H2O + O2

2. X2 + OH�(dil.) X¯ + XO¯ (X = Cl, Br, I)

3. X2 + OH�(conc.) X¯ + XO3¯

MOLE CONCEPT # 19

F2 does not undergo disproportionation as it is the most electronegative element.

F2 + NaOH(dil.) F� + OF2

F2 + NaOH(conc.) F� + O2

4. (CN)2 + OH� CN� + OCN�

5. P4 + OH� PH3 + H2PO2¯

6. S8 + OH� S2� + S2O32�

7. MnO42� MnO4¯ + MnO2

8. NH2OH N2O + NH3

NH2OH N2 + NH3

9. Oxyacids of Phosphorus ( +1, +3 oxidation number)

H3PO2 PH3 + H3PO3

H3PO3 PH3 + H3PO4

10. Oxyacids of Chlorine( Halogens)( +1, +3, +5 Oxidation number)

ClO� Cl� + ClO2�

ClO2� Cl� + ClO3

�

ClO3� Cl� + ClO4

�

11. HNO2 NO + HNO3

Reverse of disproportionation is called Comproportionation. In some of the disproportionation reactions,by changing the medium (from acidic to basic or reverse), the reaction goes in backward direction andcan be taken as an example of Comproportionation reaction.

¯ + O3¯ + H+ 2 + H2O

Balancing of redox reactionsAll balanced equations must satisfy two criteria.1. Atom balance (mass balance ) :

There should be the same number of atoms of each kind on reactant and product side.

2. Charge balance :The sum of actual charges on both sides of the equation must be equal.There are two methods for balancing the redox equations :1. Oxidation - number change method2. Ion electron method or half cell method

Since First method is not very much fruitful for the balancing of redox reactions, students are advised touse second method (Ion electron method ) to balance the redox reactions

Ion electron method : By this method redox equations are balanced in two different medium.(a) Acidic medium (b) Basic medium

Balancing in acidic mediumStudents are adviced to follow the following steps to balance the redox reactions by Ion electron methodin acidic medium

Example-21 Balance the following redox reaction :

FeSO4 + KMnO4 + H2SO4 Fe2(SO4)3 + MnSO4 + H2O + K2SO4

Solution. Step� Assign the oxidation number to each element present in the reaction.

4

262OSFe

+ 4

2� 7 1MnOK

+ 4

26

2

1OSH

34

26

2

3)OS(Fe

+ 4

262OSMn

+ 2

2

1OH

MOLE CONCEPT # 20

Step :Now convert the reaction in Ionic form by eliminating the elements or species, which are not undergoingeither oxidation or reduction.

Fe2+ +

4

7OMn Fe3+ + Mn2+

Step :Now identify the oxidation / reduction occuring in the reaction

Step V : Spilt the Ionic reaction in two half, one for oxidation and other for reduction.

Fe2+ oxidation

Fe3+

2ductionRe

4 MnMnO

Step V :Balance the atom other than oxygen and hydrogen atom in both half reactions

Fe2+ Fe3+ MnO4 � Mn2+

Fe & Mn atoms are balanced on both side.

Step V :Now balance O & H atom by H2O & H+ respectively by the following way : For one excess oxygenatom, add one H2O on the other side and two H+ on the same side.Fe2+ Fe3+ (no oxygen atom ) .................(i)8H+ + MnO4

� Mn2+ + 4H2O ................(ii)

Step V :Equation (i) & (ii) are balanced atomwise. Now balance both equations chargewise. To balance thecharge, add electrons to the electrically positive side.

Fe2+ oxidation Fe3+ + e� ............(1)

5e� + 8H+ + MnO4�

ductionRe Mn2+ + 4H2O ............(2)

Step V :The number of electrons gained and lost in each half -reaction are equalised by multiplying both the halfreactions with a suitable factor and finally the half reactions are added to give the overall balancedreaction.Here, we multiply equation (1) by 5 and (2) by 1 and add them :

Fe2+ Fe3+ + e� ..........(1) × 5

OH4MnFe5MnOH8Fe5

1)2.........(OH4MnMnOH8e5

223

42

22

4

(Here, at his stage, you will get balanced redox reaction in Ionic form)

Step X :Now convert the Ionic reaction into molecular form by adding the elements or species, which areremoved in step (2).Now, by some manipulation, you will get :

5 FeSO4 + KMnO4 + 4H2SO4 25

Fe2 (SO4)3 + MnSO4 + 4H2O + 21

K2SO4 or

10FeSO4 + 2KMnO4 + 8H2SO4 5Fe2(SO4)3 + 2MnSO4 + 8H2O + K2SO4.

MOLE CONCEPT # 21

Balancing in basic medium :In this case, except step VI, all the steps are same. We can understand it by the following example:

Example-22 Balance the following redox reaction in basic medium : ClO� + CrO2

� + OH� Cl� + CrO42� + H2O

Solution. By using upto step V, we will get :

24

6Oxidation

2

3�ductionRe

1OCrOCrClOCl

Now, students are advised to follow step VI to balance �O� and �H� atom.

2H+ + ClO� Cl� + H2O | 2H2O+ CrO2� CrO4

2� + 4H+

Now, since we are balancing in basic medium, therefore add as many as OH� on both side of equationas there are H+ ions in the equation.

2OH� + 2H+ + ClO� Cl� + H2O +2OH� 4OH� + 2H2O + CrO2� CrO4

2� + 4H+ + 4OH�

Finally you will get Finally you will get

H2O + ClO� Cl� + 2OH� ...........(i) 4OH� + CrO2� CrO4

2� + 2H2O ........... (ii)

Now see equation (i) and (ii) in which O and H atoms are balanced by OH� and H2ONow from step VIII

2e� + H2O + ClO� Cl� + 2OH� ............. (i) ×3

4OH� + CrO2� CrO4

2� + 2H2O + 3e� ............. (ii) ×2

���������������������������������������������������

Adding : 3ClO� + 2CrO2� + 2OH� 3Cl� + 2CrO4

2� + H2O

Solutions :A mixture of two or more substances can be a solution. We can also say that �a solution is a homogeneous

mixture of two or more substances,�� �Homogeneous� means �uniform throughout�. Thus a homogeneous mixture,

i.e., a solution, will have uniform composition throughout.

Properties of a solution : A solution is clear and transparent. For example, a solution of sodium chloride in water is clear and

tranparent. The solute in a solution does not settle down even after the solution is kept undisturbed for some time. In a solution, the solute particle cannot be distinguished from the solvent particles or molecules even

under a microscope. In a true solution, the particles of the solute disappear into the space between thesolvent molecules.

The components of a solution cannot be separated by filtration.

Concentration terms :The following concentration terms are used to expressed the concentration of a solution. These are

Molarity (M) Molality (m) Mole fraction (x) % calculation Normality (N) ppm

Remember that all of these concentration terms are related to one another. By knowing one concentration

term you can also find the other concentration terms. Let us discuss all of them one by one.

MOLE CONCEPT # 22

Molarity (M) :The number of moles of a solute dissolved in 1 L (1000 ml) of the solution is known as the molarity

of the solution.

i.e., Molarity of solution = litre in solution of volume

soluteofmoles of number

Let a solution is prepared by dissolving w gm of solute of mol.wt. M in V ml water.

Number of moles of solute dissolved = Mw

V ml water have Mw

mole of solute

1000 ml water have mlVM

1000w

Molarity (M) =

mlV)solute of wt .Mol(1000w

Some other relations may also useful.

Number of millimoles = 1000)solute of .wt .Mol(

soluteofmass = (Molarity of solution × Vml)

Molarity of solution may also given as :

ml in solution of volume Totalsolute of millimole of Number

Molarity is a unit that depends upon temperature. It varies inversely with temperature .

Mathematically : Molarity decreases as temperature increases.

Molarity etemperatur1

volume

1

If a particular solution having volume V1 and molarity = M1 is diluted upto volume V2 mL thanM1V1 = M2V2

M2 : Resultant molarity

If a solution having volume V1 and molarity M1 is mixed with another solution of same solute havingvolume V2 mL & molarity M2

then M1V1 + M2V2 = MR (V1 + V2)

MR = Resultant molarity = 21

2211

VVVMVM

Example-23 149 gm of potassium chloride (KCl) is dissolved in 10 Lt of an aqueous solution. Determine the molarityof the solution (K = 39, Cl = 35.5)

Solution Molecular mass of KCl = 39 + 35.5 = 74.5 gm

Moles of KCl = gm 5.74gm 149

= 2

Molarity of the solution = 102

= 0.2 M

Molality (m) :The number of moles of solute dissolved in1000 gm (1 kg) of a solvent is known as the molality of thesolution.

i.e., molality = 1000gram in solvent of masssolute of moles of number

MOLE CONCEPT # 23

Let Y gm of a solute is dissolved in X gm of a solvent. The molecular mass of the solute is M0. Then Y/M0 moleof the solute are dissolved in X gm of the solvent. Hence

Molality = 1000XM

Y

0

Molality is independent of temperature changes.

Example-24 225 gm of an aqueous solution contains 5 gm of urea. What is the concentration of the solution in termsof molality. (Mol. wt. of urea = 60)

Solution Mass of urea = 5 gm Molecular mass of urea = 60

Number of moles of urea = 605

= 0.083

Mass of solvent = (255 � 5) = 250 gm

Molality of the solution = gram in solvent of Masssolute of moles of Number

× 1000 = 250083.0

× 1000= 0.332.

Mole fraction (x) :The ratio of number of moles of the solute or solvent present in the solution and the total number ofmoles present in the solution is known as the mole fraction of substances concerned.

Let number of moles of solute in solution = nNumber of moles of solvent in solution = N

Mole fraction of solute (x1) = Nn

n

Mole fraction of solvent (x2) = Nn

N

also x1 + x2 = 1

Mole fraction is a pure number. It will remain independent of temperature changes.

% calculation :The concentration of a solution may also expressed in terms of percentage in the following way.

% weight by weight (w/w) : It is given as mass of solute present in per 100 gm of solution.

i.e. % w/w = 100gm in solution of massgm in solute of mass

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

i.e., % w/v = 100ml in solution of olumev

gm in solute of mass

% volume by volume (v/v) : It is given as volume of solute present in per 100 ml solution.

i.e., % v/v = mlinsolution of volumeml in solute of volume

× 100

Example-25 0.5 g of a substance is dissolved in 25 g of a solvent. Calculate the percentage amount of the substancein the solution.

Solution Mass of substance = 0.5 gMass of solvent = 25 g

percentage of the substance (w/w) = 100255.0

5.0

= 1.96

MOLE CONCEPT # 24

Example-26 20 cm3 of an alcohol is dissolved in80 cm3 of water. Calculate the percentage of alcohol in solution.Solution Volume of alcohol = 20 cm3

Volume of water = 80 cm3

Percentage of alcohol = 1008020

20

= 20.

Hydrogen peroxide (H2O2)

H2O2 can behave both like oxidising and reducing agent in both the mediums (acidic and basic).

Oxidising agent : (H2O2 H2O)

(a) Acidic medium : 2e� + 2H+ + H2O2 2H2O v.f. = 2

(b) Basic medium : 2e� + H2O2 2OH�

v.f = 2 Reducing agent : (H2O2

O2)(a) Acidic medium : H2O2 O2 + 2H+ + 2e�

v.f = 2(b) Basic medium : 2OH� + H2O2 O2 + 2H2O + 2e�

v.f = 2

Volume strength of H2O2 : Strength of H2O2 is represented as 10V , 20 V , 30 V etc.

20V H2O2 means one litre of this sample of H2O2 on decomposition gives 20L of O2 gas at STP.Decomposition of H2O2 is given as :

H2O2 H2O + 21

O2

1 mole21

× 22.4 L O2 at STP

= 34g = 11.2 L O2 at STP

Molarity of H2O2 (M) = 11.2

OHofstrengthVolume 22

Strength (in g/L) : Denoted by SStrength = Molarity × Mol. wt = Molarity × 34

Hardness of water (Hard water does not give lather with soap)Temporary hardness - due to bicarbonates of Ca & MgPermanent hardness - due to chlorides & sulphates of Ca & Mg. There are some method by which we can softenthe water sample.

(a) By boiling : 2HCO3� H2O + CO2 + CO3

2� or

By Slaked lime : Ca(HCO3)2 + Ca(OH)2 CaCO3 + 2H2O

Ca2+ + CO32� CaCO3

(b) By Washing Soda : CaCl2 + Na2CO3 CaCO3 + 2NaCl

(c) By ion exchange resins : Na2R + Ca2+ CaR + 2Na+

(d) By adding chelating agents like (PO3�)3 etc.

MOLE CONCEPT # 25

Parts Per Million (ppm)When the solute is present in very less amount, then this concentration term is used. It is defined as the numberof parts of the solute present in every 1 million parts of the solution. ppm can both be in terms of mass or interms of moles. If nothing has been specified, we take ppm to be in terms of mass. Hence, a 100 ppm solutionmeans that 100 g of solute is present in every 1000000 g of solution.

ppmA = 610massTotal

Aofmass = mass fraction × 106

Measurement of Hardness :Hardness is measured in terms of ppm (parts per million) of CaCO3 or equivalent to it.

Hardness in ppm = 610solutionofmassTotal

3CaCOofmass

Example-27 0.00012% MgSO4 and 0.000111% CaCl2 is present in water. What is the measured hardness ofwater and millimoles of washing soda required to purify water 1000 L water ?

Solution. Basis of calculation = 100 g hard water

MgSO4 = 0.00012g = 12000012.0

mole

CaCl2 = 0.000111g = 111

000111.0mole

equivalent moles of CaCO3 =

111000111.0

12000012.0

mole

mass of CaCO3 =

111000111.0

12000012.0

× 100 = 2 × 10�4 g

Hardness (in terms of ppm of CaCO3) = 64

10100102

= 2 ppm

CaCl2 + Na2CO3 CaCO3 + 2NaCl

NaSO4 + Na2CO3 MgCO3 + Na2SO4

Required Na2CO3 for 100g of water =

111000111.0

12000012.0

mole

= 2 × 10�6 mole

Required Na2CO3 for 1000 litre water = 100

210

100102 6

6

mole ( d = 1g/mL)

= 1000

20 mole = 20 m mole

Strength of Oleum :

Oleum is SO3 dissolved in 100% H2SO4. Sometimes, oleum is reported as more than 100% by weight, say y%(where y > 100). This means that (y 100) grams of water, when added to 100 g of given oleum sample, willcombine with all the free SO3 in the oleum to give 100% sulphuric acid.Hence, weight % of free SO3 in oleum = 80(y 100)/18

MOLE CONCEPT # 26

Example-28 What volume of water is required (in mL) to prepare 1 L of 1 M solution of H2SO4 (density = 1.5g/mL) byusing 109% oleum and water only (Take density of pure water = 1 g/mL).

Solution. 1 mole H2SO4 in 1L solution = 98 g H2SO4 in 1500 g solution = 98 g H2SO4 in 1402 g water.Also, in 109% oleum, 9 g H2O is required to form 109 g pure H2SO4 & so, to prepare 98 g H2SO4, waterneeded is 9/109 × 98 = 8.09 g.

Total water needed = 1402 + 8.09 = 1410.09 g = 1410.09 mL

MISCELLANEOUS SOLVED PROBLEMS (MSPS)

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

Sol. (i) 23, 23 amu (ii) 19, 19 amu (iii) 1, 1.008 amu , (iv) 40, 40 amu, (v) 108, 108 amu.

2. A sample of (C2H6) ethane has the same mass as 107 molecules of methane. How many C2H6 molecules doesthe sample contain ?

Sol. Moles of CH4 = A

7

N10

Mass of CH4 = A

7

N10

× 16 = mass of C2H6

So Moles of C2H6 = 30N1610

A

7

So No. of molecules of C2H6 = 30N1610

A

7

× NA = 5.34 × 106.

3. From 160 g of SO2 (g) sample, 1.2046 x 1024 molecules of SO2 are removed then find out the volume of left overSO2 (g) at STP.

Sol. Given moles = 64

160 = 2.5.

Removed moles = 23

24

10023.6

102046.1

= 2.

so left moles = 0.5.volume left at STP = 0.5 × 22.4 = 11.2 lit.

4. 14 g of Nitrogen gas and 22 g of CO2 gas are mixed together. Find the volume of gaseous mixture at STP.

Sol. Moles of N2 = 2814

= 0.5.

moles of CO2 = 4422

= 0.5.

so total moles = 0.5 + 0.5 = 1.so vol. at STP = 1 × 22.4 = 22.4 lit.

5. Show that in the reaction N2 (g) + 3H2(g) 2NH3 (g), mass is conserved.

Sol. N2 (g) + 3H2(g) 2NH3 (g)moles before reaction 1 3 0moles after reaction 0 0 2Mass before reaction = mass of 1 mole N2(g) + mass of 3 mole H2(g)

= 14 x 2 + 3 x 2 = 34 g

MOLE CONCEPT # 27

mass after reaction = mass of 2 mole NH3= 2 x 17 = 34 g.

6. When x gram of a certain metal brunt in 1.5 g oxygen to give 3.0 g of its oxide. 1.20 g of the same metal heatedin a steam gave 2.40 g of its oxide. shows the these result illustrate the law of constant or definite proportion

Sol. Wt. of metal = 3.0 � 1.5 = 1.5 g

so wt. of metal : wt of oxygen = 1.5 : 1.5 = 1 : 1similarly in second case ,wt. of oxygen = 2.4 � 1.2 = 1.2 g

so wt. of metal : wt of oxygen = 1.2 : 1.2 = 1 : 1so these results illustrate the law of constant proportion.

7. Find out % of O & H in H2O compound.

Sol. % of O = 1816

× 100 = 88.89%

% of H = 182

× 100 = 11.11%

8. Acetylene & butene have empirical formula CH & CH2 respectively. The molecular mass of acetylene and buteneare 26 & 56 respectively deduce their molecular formula.

Ans. C2H2 & C4H8

Sol. n = massformulaEmpiricalmassMolecular

For Acetylene :

n = 1326

= 2

Molecular formula = C2H2

For Butene :

n = 1456

= 4

Molecular formula = C4H8 .

9. An oxide of nitrogen gave the following percentage composition :N = 25.94

and O = 74.06Calculate the empirical formula of the compound.

Ans. N2O5

Sol. Element % / Atomic mass Simple ratio Simple intiger ratio

N 1 2

O 2.5 5

85.114

94.25

4.6316

74.06

So empirical formula is N2O5.

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

Sol. R.D. = ONof.wt.MCOof.wt.M

2

2 =

4444

= 1.

11. Find the vapour density of N2O5

Sol. V.D. = 2

ONof.wt.Mol 52 = 54.

MOLE CONCEPT # 28

12. Write a balance chemical equation for following reaction :When ammonia (NH3) decompose into nitrogen (N2) gas & hydrogen (H2) gas.

Sol. NH3 21

N2 + 23

H2 or 2NH3 N2 + 3H2 .

13. When 170 g NH3 (M =17) decomposes how many grams of N2 & H2 is produced.

Sol. NH3 21

N2 + 23

H2

1

NHofmoles 3 = 2/1

Nofmoles 2 = 2/3

Hofmoles 2 .

So moles of N2 = 21

× 17

170 = 5. So wt. of N2 = 5 × 28 = 140 g.

Similarly moles of H2 = 23

× 17

170 = 15.

So wt. of H2 = 15 × 2 = 30 g.

14. 340 g NH3 (M = 17) when decompose how many litres of nitrogen gas is produced at STP.

Sol. NH3 21

N2 + 23

H2

moles of NH3 = 17340

= 20.

So moles of N2 = 21

× 20 = 10.

vol. of N2 at STP = 10 × 22.4 = 224 lit.

15. 4 mole of MgCO3 is reacted with 6 moles of HCl solution. Find the volume of CO2 gas produced at STP, thereaction is

MgCO3 + 2HCl MgCl2 + CO2 + H2O.Sol. Here HCl is limiting reagent. So moles of CO2 formed = 3.

So vol. at STP = 3 × 22.4 = 67.2 lit.

16. 117 gm NaCl is dissolved in 500 ml aqueous solution. Find the molarity of the solution.

Sol. Molarity = 1000/500

5.58/117 = 4M.

17. 0.32 mole of LiAlH4 in ether solution was placed in a flask and 74 g (1 moles) of t-butyl alcohol was added. Theproduct is LiAlHC12H27O3 . Find the weight of the product if lithium atoms are conserved.

[Li = 7, Al = 27, H = 1, C = 12, O = 16]Sol. Applying POAC on Li

1 × moles of LiAlH4 = 1× moles of LiAlH C12H27O3

254 × 0.32 = 1 × wt. of LiAlH C12H27O3.wt. of LiAlH C12H27O3 = 81.28 gm.

18. Calculate individual oxidation number of each S-atom in Na2S4O6 (sodium tetrathionate) with the help of itsstructure .

Sol.

MOLE CONCEPT # 29

19. Find the average and individual oxidation number of Fe & Pb in Fe3O4 & Pb3O4, which are mixed oxides.Sol. (i) Fe3O4 is mixture of FeO & Fe2O3 in 1 : 1 ratio

so, individual oxidation number of Fe = +2 & +3

& average oxidation number = 3

)3(2)2(1 = 8/3

(ii) Pb3O4 is a mixture of PbO & PbO2 in 2 : 1 molar ratioso, individual oxidation number of Pb are +2 & +4

& average oxidation number of Pb = 3/83

)4(1)2(2

20. Balance the following equations :

(a) H2O2 + MnO4� Mn+2 + O2 (acidic medium)

(b) Zn + HNO3 (dil) Zn(NO3)2 + H2O + NH4NO3

(c) Cr3 + KOH + Cl2 K2CrO4 + KO4 + KCl + H2O.

(d) P2H4 PH3 + P4

(e) Ca3 (PO4)2 + SiO2 + C CaSiO3 + P4 + CO

Ans. (a) 6H+ + 5H2O2 + 2MnO4� 2Mn+2 + 5O2 + 8H2O

(b) 4Zn + 10HNO3 (dil) 4Zn(NO3)2 + 3H2O + NH4NO3

(c) 2Cr3 + 64KOH + 27Cl2 2K2CrO4 + 6KO4 + 54KCl + 32H2O.

(d) 6P2H4 8PH3 + P4

(e) 2Ca3 (PO4)2 + 6SiO2 + 10C 6CaSiO3 + P4 + 10CO

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

Ans. (a) 0.4 M (b) 1.33 M (c) 1 M (d) 0.83 M.

Sol. (a) Final molarity = 300200

01200

= 0.4 M.

(b) Final molarity = 33.11500

5.36100025.18

11500

M

(c) Final molarity of H+ = 100200

25.01001200

= 1 M.

(d) Final molarity = 100200

5.01001200

= 0.83 M.

22. 518 gm of an aqueous solution contains 18 gm of glucose (mol.wt. = 180). What is the molality of the solution.

Sol. wt. of solvent = 518 � 18 = 500 gm. so molarity = 1000/500180/18

= 0.2.

23. 0.25 of a substance is dissolved in 6.25 g of a solvent. Calculate the percentage amount of the substance in thesolution.

Sol. wt. of solution = 0.25 + 6.25 = 6.50.

so % (w/w) = 50.625.0

× 100 = 3.8%.

MOLE CONCEPT # 30

24. A fresh H2O2 solution is labelled 11.2 V. This solution has the same concentration as a solution which is :(A) 3.4% (w / w) (B) 3.4% (v / v) (C) 3.4% (w / v) (D) None of these

Sol. Molarity of H2O2 = 2.112.11

2.11strength.vol

= 1

Now, %(w/v) = mLinsolutionof.wtginsoluteof.wt

× 100

= Molarity × Mol. wt. of solute × 101

= 1 × 34 × 101

= 3.4% Ans. (C)

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

Ans. (a) 0.4 M (b) 1.33 M (c) 1 M (d) 0.83 M.

Sol. (a) Final molarity = 300200

01200

= 0.4 M.

(b) Final molarity = 33.11500

5.36100025.18

11500

M

(c) Final molarity of H+ = 100200

25.01001200

= 1 M.

(d) Final molarity = 100200

5.01001200

= 0.83 M.

26. 518 gm of an aqueous solution contains 18 gm of glucose (mol.wt. = 180). What is the molality of the solution.

Sol. wt. of solvent = 518 � 18 = 500 gm. so molarity = 1000/500180/18

= 0.2.

27. 0.25 of a substance is dissolved in 6.25 g of a solvent. Calculate the percentage amount of the substance in thesolution.

Sol. wt. of solution = 0.25 + 6.25 = 6.50.

so % (w/w) = 50.625.0

× 100 = 3.8%.

28. A fresh H2O2 solution is labelled 11.2 V. This solution has the same concentration as a solution which is :(A) 3.4% (w / w) (B) 3.4% (v / v) (C) 3.4% (w / v) (D) None of these

Sol. Molarity of H2O2 = 2.112.11

2.11strength.vol

= 1

Now, %(w/v) = mLinsolutionof.wtginsoluteof.wt

× 100

= Molarity × Mol. wt. of solute × 101

= 1 × 34 × 101

= 3.4% Ans. (C)

MOLE CONCEPT # 31

Type (I) : Very Short Answer Type Questions : [01 Mark]

1 Define relative atomic mass ?

2. Calculate the molecular mass of the following :(i) H

2O (ii) CO

2(iii) CH

4

3. What will be the mass of one C-12 atom in g ?

4. According to common practice, if you have Avogadro number of Hydrogen than its mass is equal to

5. Match the following prefixes with their multiples: Prefixes Multiples(i) micro 106

(ii) deca 109

(iii) mega 10�6

(iv) giga 10�15

(v) femto 10

6. How are 0.50 mol Na2CO

3 and 0.50 M Na

2CO

3 different?

7. If ten volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapourwould be produced?

Type (II) : Short Answer Type Questions : [02 Marks]

8. Calculate the mass per cent of different elements present in sodium sulphate (Na2SO

4).

9. Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.

10. How much copper can be obtained from 100 g of copper sulphate (CuSO4) ?

11. Write a note on Avogadro number and mole ?

12. What do you mean by significant figures ?

13. How many significant figures are present in the following?(i) 0.0025 (ii) 208 (iii) 5005 (iv) 126,000 (v) 500.0 (vi) 2.0034

14. Round up the following upto three significant figures:(i) 34.216 (ii) 10.4107 (iii) 0.04597 (iv) 2808

Type (III) : Long Answer Type Questions: [03 Marks]

15. Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution.

Molar mass of sodium acetate is 82.0245 g mol�1.

16. Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL�1 and themass percent of nitric acid in it being 69%.

17. What is the concentration of sugar (C12

H22

O11

) in mol L�1 if its 20 g are dissolved in enough water to make afinal volume up to 2L?

18. Express the following in the scientific notation:(i) 0.0048 (ii) 234,000 (iii) 8008 (iv) 500.0 (v) 6.0012

19. Which one of the following will have largest number of atoms?(i) 1 g Au (s) (ii) 1 g Na (s) (iii) 1 g Li (s) (iv) 1 g of Cl

2(g)

MOLE CONCEPT # 32

20. How many significant figures should be present in the answer of the following calculations?

(i) 5785.0

112.015.29802856.0 (ii) 5 × 5.364 (iii) 0.0125 + 0.7864 + 0.0215

21. Calculate the number of atoms in each of the following (i) 52 moles of Ar (ii) 52 u of He (iii) 52 g of He.

22. Determine the simplest formula of a compound containing 50% of element X (atomic mass = 10) and 50% of

the element Y (atomic mass = 20) by weight.

23. From 64 g oxygen, 11.2 lt. Oxygen gas at S.T.P and 6.02 × 1023 oxygen atoms are removed from the oxygencontainer. Find the mass of the oxygen gas left.

24. Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040.

25. Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:Isotope Isotopic molar mass Abundance36Ar 35.96755 g mol�1 0.337%38Ar 37.96272 g mol�1 0.063%40Ar 39.9624 g mol�1 99.600%

26. Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid

according to the reaction4 HCl (aq)+MnO

2(s)2H

2O (l)+MnCl

2(aq)+ Cl

2 (g)

How many grams of HCl react with 5.0 g of manganese dioxide?

Type (IV) : Very Long Answer Type Questions: [05 Marks]

27. Define the laws of chemical combination and explain them .

28. Calculate the amount of carbon dioxide that could be produced when(i) 1 mole of carbon is burnt in air.(ii) 1 mole of carbon is burnt in 16 g of dioxygen.(iii) 2 moles of carbon are burnt in 16 g of dioxygen.

29. If the density of methanol is 0.793 kg L�1, what is its volume needed for making 2.5 L of its 0.25 M solution?

30. Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shownbelow:1Pa = 1N m�2

If mass of air at sea level is 1034 g cm�2, calculate the pressure in pascal.

31. A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be

carcinogenic in nature. The level of contamination was 15 ppm (by mass).(i) Express this in percent by mass.(ii) Determine the molality of chloroform in the water sample.

32. Define any of the five concentration terms and their definition with formulae ?

33. In a reactionA + B

2 AB

2

Identify the limiting reagent, if any, in the following reaction mixtures.(i) 300 atoms of A + 200 molecules of B

2(ii) 2 mol A + 3 mol B

2

(iii) 100 atoms of A + 100 molecules of B2

(iv) 5 mol A + 2.5 mol B2

(v) 2.5 mol A + 5 mol B2

34. Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemicalequation:N

2 (g) + H

2 (g) 2NH

3 (g)

(i) Calculate the mass of ammonia produced if 2.00 × 103 g dinitrogen reacts with 1.00 × 103 g of dihydrogen.(ii) Will any of the two reactants remain unreacted(iii) If yes, which one and what would be its mass?

MOLE CONCEPT # 33

35. Convert the following into basic units:(i) 28.7 pm (ii) 15.15 µs (iii) 25365 mg

36. A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 gcarbon dioxide , 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this weldinggas is found to weigh 11.6 g. Calculate (i) empirical formula, (ii) molar mass of the gas, and(iii) molecular formula.

37. Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO

2 according to the reaction,

CaCO3 (s) + 2 HCl (aq) CaCl

2 (aq) + CO

2(g) + H

2O(l). What mass of CaCO

3 is required to react completely

with 25mL of 0.75 M HCl?

38. The following data are obtained when dinitrogen and dioxygen react together to form different compounds:Mass of dinitrogen Mass of dioxygen

(i) 14 g 16 g(ii) 14 g 32 g(iii) 28 g 32 g(iv) 28 g 80 g

(a) Which law of chemical combination is obeyed by the above experimental data?Give its statement.

(b) Fill in the blanks in the following conversions:(i) 1 km = ...................... mm = ...................... pm(ii) 1 mg = ...................... kg = ...................... ng(iii) 1 mL = ...................... L = ...................... dm3

PART - I : SUBJECTIVE QUESTIONS

Section (A) : Basic definitions, Atomic mass unit, NA & Mole, Moles of atoms and moleculesA-1. Find the total number of H,

S and 'O' atoms in the following :

(a) 196 gm H2SO

4(b) 196 amu H

2SO

4

(c) 5 mole H2S

2O

8(d) 3 molecules H

2S

2O

6 .

A-2. If, from 10 moles NH3 and 5 moles of H

2SO

4 , all the H-atoms are removed in order to form H

2 gas, then find

the number of H2 molecules formed.

A-3. If from 3 moles MgSO4.7H

2O, all the 'O' atoms are taken out and converted into ozone find the number of O

3

molecules formed.

Section (B) : Laws of chemical combination, Molar volume of ideal gases at STP, Averagemolar mass

B-1. Two compounds each containing only tin and oxygen had the following composition :Mass % of tin Mass % of oxygen

Compound A 78.77 21.23Compound B 88.12 11.88Show these data illustrate the law of multiple proportions.

B-2. The measured density at N.T.P. of a gaseous sample of a compound was found to be 1.20 g/liter. What is theweight of 1 mole of the gaseous sample.

B-3. Three oxides of nitrogen N2O, NO2 and N2O3 are mixed in a molar ratio of 3 : 2 : 1. Find the average molarmass of gaseous mixture.

MOLE CONCEPT # 34

Section (C) : Empirical Formula, % Composition of a given compound by mass, % By mole,Minimum molecular mass determination.

C-1. A compound contains 25% hydrogen and 75% carbon by mass. Determine the empirical formula of thecompound.

C-2. In a gaseous mixture 2mol of CO2, 1 mol of H

2 and 2 mol of He are present than determine mole percentage

of CO2.

C-3. A compound has haemoglobin like structure. It has one Fe. It contain 4.6% of Fe. Determine its approximatemolecular mass.

Section (D) : Stoichiometry, Equation based calculations (Elementary level single equationor 2)

D-1. Calculate the residue obtained on strongly heating 2.76 g Ag2CO

3.

Ag2CO

3

2Ag + CO2 +

21

O2

D-2. Calculate the weight of iron which will be converted into its oxide by the action of 18g of steam.

Unbalanced reaction : Fe + H2O

Fe3O

4 + H

2.

D-3. A sample of KClO3 on decomposition yielded 448 mL of oxygen gas at NTP.

Calculate(i) Weight of oxygen product,(ii) Weight of KClO

3 originally taken, and

(iii) Weight of KCl produced. (K = 39 , Cl = 35.5 and O = 16)

Section (E) : Limiting reagent, % Excess, % Yield / EfficiencyE-1. 50 g of CaCO

3 is allowed to react with 73.5 g of H

3PO

4. Calculate :

(i) Amount of Ca3(PO

4)

2 formed (in moles)

(ii) Amount of unreacted reagent (in moles)

E-2. Carbon disulphide, CS2, can be made from by-product SO

2. The overall reaction is

5C + 2SO2 CS

2 + 4CO

How much CS2 can be produced from 440 kg of waste SO

2 with 60 kg of coke if the SO

2 conversion is 80%?

E-3. The percent yield for the following reaction carried out in carbon tetrachloride (CCI4) solution is 80%

Br2 + CI

2 2BrCI

(a) What amount of BrCI would be formed from the reaction of 0.025 mol Br2 and 0.025 mol CI

2?

(b) What amoumt of Br2 is left unchanged?

Section (F) : Principle of atom conservation (POAC), Reactions in sequence & parallel,Mixture analysis, % Purity

F-1. What amount of CaO will be produced by 1 g of Calcium ?

F-2. The following process has been used to obtain iodine from oil-field brines in California.

Nal + AgNO3 Agl + NaNO

3; 2AgI + Fe Fel

2 + 2Ag

2FeI2 + 3Cl

2 2FeCl

3 + 2I

2

How many grams of AgNO3 are required in the first step for every 254 kg

2 produced in the third step.

MOLE CONCEPT # 35

F-3. KClO3 decomposes by two parallel reaction

(i) 2KClO3

2KCl + 3O2

(ii) 4KClO3

3KClO4 + KCl

If 3 moles of O2 and 1 mol of KClO4 is produced along with other products then determine initial moles ofKClO3.

F-4. A 2 gm sample containing Na2CO

3 and NaHCO

3 losses 0.248 g when heated to 3000 C , the temperature at

which NaHCO3 decomposes to Na

2CO

3 , CO

2 and H

2O.

What is the percentage of Na

2CO

3 in the given

mixture ? (Na = 23 , C = 12 , O = 16 and H = 1)

Section (G) : Basics of oxidation number

G-1. Calculate the oxidation number of underlined elements in the following compounds :(a) K[Co(C2O4)2(NH3)2] (b) K4P2O7 (c) CrO2Cl2(d) Na2[Fe(CN)5(NO+)] (e) Mn3O4 (f) Ca(ClO

2)

2

(g) [Fe(NO+) (H2O)5]SO4 (h) ZnO22� (i) Fe0.93O

G-2. Identify the oxidant and the reductant in the following reactions :(a) KMnO

4 + KCl + H

2SO

4 MnSO

4 + K

2SO

4 + H

2O + Cl

2

(b) FeCl2 + H

2O

2 + HCl FeCl

3 + H

2O

(c) Cu + HNO3 (dil) Cu (NO

3)

2 + H

2O + NO

(d) Na2HAsO

3 + KBrO

3 + HCl NaCl + KBr + H

3AsO

4

(e) I2 + Na

2S

2O

3 Na

2S

4O

6 + NaI

Section (H) : Balancing redox reactionsH-1. Write balanced net ionic equations for the following reactions in acidic solution :

(a) S4O62�(aq) + Al(s) H2S(aq) + Al3+(aq)

(b) S2O32�(aq) + Cr2O7

2�(aq) S4O62�(aq) + Cr3+(aq)

(c) ClO3� (aq) + As2S3(s) Cl¯(aq) + H2AsO4¯ (aq) + HSO4¯(aq)

(d) O3¯(aq) + Re(s) ReO4¯(aq) + ¯(aq)

(e) HSO4¯(aq) + As4(s) + Pb3O4(s) PbSO4(s) + H2AsO4¯ (aq)

(f) HNO2(aq) NO3¯ + NO(g)

H-2. Write balanced net ionic equations for the following reactions in basic solution :(a) C4H4O6

2�(aq) + ClO3¯(aq) CO32�(aq) + Cl¯(aq)

(b) Al(s) + BiONO3(s) Bi(s) + NH3(aq) + [Al(OH)4]¯ (aq)

(c) H2O2(aq) + Cl2O7(aq) ClO2¯(aq) + O2(g)

(d) Tl2O3(s) + NH2OH(aq) TlOH(s) + N2(g)

(e) [Cu(NH3)4]2+ (aq) + S2O4

2�(aq) SO32�(aq) + Cu(s) + NH3(aq)

(f) Mn(OH)2(s) + MnO4¯(aq) MnO2(s)

Section (I) : Units of concentration measurement, Interconversion of concentration unitsI-1. Find the mass of KOH needed to prepare 100 ml 1 M KOH solution. [At. mass K = 39]

I-2. How much NaNO3 must be weighed out to make 50 ml of an aqueous solution containing 70 mg of Na+

per mL ?

MOLE CONCEPT # 36

I-3. (i) If you are given a 2M NaOH solution having density 1 gm/ml, then find the molality of solution.(ii) Find the molarity of 5m (molal) NaOH solution having density 1.5 gm/ml.(iii) Find the mole fraction of solute in problem (i)

(iv) Find the mole fraction of solute in problem (ii)

(v) Find the % (w/w) of NaOH in solution in problem (i)

(vi) Find the % (w/w) of NaOH in solution in problem (ii)

(vii)Find the % (w/v) of NaOH in solution in problem (ii)(viii) A 300 gm, 30% (w/w) NaOH solution is mixed with 500 gm 40% (w/w) NaOH solution. Find the mass percentage

(w/w) of final solution.(ix) What is % (w/v) NaOH in problem (viii) if density of final solution is 2 gm/ml ?(x) What is the molality of final solution obtained in problem (viii) ?

Section (J) : Dilution & Mixing of two liquidsJ-1. Find the Cl� concentration in solution which is obtained by mixing one mole each of BaCl

2, NaCl and HCl in 500

ml water.

J-2. What volume of water should be added to 50 ml of HNO3 having density 1.5 g ml�1 and 63.0% by weight to

have one molar solution.

J-3. What maximum volume of 3 M solution of KOH can be prepared from 1 L each of 1 M KOH and 6 M KOH

solutions by using water ?

PART - II : OBJECTIVE QUESTIONS

* Marked Questions may have more than one correct option.

Section (A) : Basic definitions, Atomic mass unit, NA & Mole, Moles of atoms and moleculesA-1. The modern atomic weight scale is based on :

(A) C12 (B) O16 (C) H1 (D) N14

A-2. Which of the following expressions is correct (n = no. of moles of the gas, NA = Avogadro constant, m = mass

of 1 molecule of the gas, N = no. of molecules of the gas)?(A) n = m N

A(B) m = N

A(C) N = nN

A(D) m = mn/N

A

A-3. The charge on 1 gram ions of Al3+ is : (NA = Avogadro number, e = charge on one electron)

(A) 271

NAe coulomb (B)

31

× NAe

coulomb (C)

91

× NAe

coulomb (D) 3 × N

Ae

coulomb

A-4. Which of the following contains the greatest number of atoms ?(A) 1.0 g of butane (C

4H

10) (B) 1.0 g of nitrogen (N

2)

(C) 1.0 g of silver (Ag) (D) 1.0 g of water (H2O)

A-5. A gaseous mixture contains CO2(g) and N

2O(g) in 2 : 5 ratio by mass. The ratio of the number of molecules of

CO2(g) and N

2O(g) is :

(A) 5 :2 (B) 2 : 5 (C) 1 : 2 (D) 5 : 4

A-6. The weight of a molecule of the compound C60

H22

is :(A) 1.09 × 10�21 g (B) 1.24 × 10�21 g (C) 5.025 × 10�23 g (D) 16.023 × 10�23 g

Section (B) : Laws of chemical combination, Molar volume of ideal gases at STP, Averagemolar mass

B-1. Four 1-1 litre flasks are separately filled with the gases H2, He, O2 and O3 at the same temperature andpressure. The ratio of total number of atoms of these gases present in different flask would be :(A) 1 : 1 : 1 : 1 (B) 1 : 2 : 2 : 3 (C) 2 : 1 : 2 : 3 (D) 3 : 2 : 2 : 1

MOLE CONCEPT # 37

B-2. Under the same conditions, two gases have the same number of molecules. They must(A) be noble gases (B) have equal volumes(C) have a volume of 22.4 dm3 each (D) have an equal number of atoms

B-3 3g of a hydrocarbon on combustion in excess of oxygen produces 8.8 g of CO2 and 5.4 g of H2O. The dataillustrates the law of :(A) conservation of mass (B) multiple proportions(C) constant proportions (D) none of these

B-4. 16 g of an ideal gas SOx occupies 5.6 L. at STP. The value of x is(A) x = 3 (B) x = 2 (C) x = 4 (D) none

B-5 Boron has two stable isotopes, 10B (relative abundance = 19%) and 11B (relative abundance = 81%). Theatomic mass (in amu) that should appear for boron in the periodic table is :(A) 10.8 (B) 10.2 (C) 11.2 (D) 10.6

Section (C) : Empirical Formula, % Composition of a given compound by mass, % By mole,Minimum molecular mass determination.

C-1. The empirical formula of a compound of molecular mass 120 is CH2O. The molecular formula of the compound

is :

(A) C2H

4O

2(B) C

4H

8O

4(C) C

3H

6O

3(D) all of these

C-2. Calculate the molecular formula of compound which contains 20% Ca and 80% Br (by wt.) if molecular weightof compound is 200. (Atomic wt. Ca = 40, Br = 80)(A) Ca

1/2Br (B) CaBr

2(C) CaBr (D) Ca

2Br

C-3.*_ Which is/are correct statements about 1.7 gm of NH3

(A) It contain 0.3 mol H � atom (B) it contain 2.408 1023 atoms(C) Mass % of hydrogen is 17.65% (D) It contains 0.3 mol N-atom

C-4 A compound possess 8% sulphur by mass. The least molecular mass is :(A) 200 (B) 400 (C) 155 (D) 355

C-5. Cortisone is a molecular substance containing 21 atoms of carbon per molecule. The mass percentage ofcarbon in cortisone is 69.98%. Its molar mass is :(A) 176.5 (B) 252.2 (C) 287.6 (D) 360.1

Section (D) : Stoichiometry, Equation based calculations (Elementary level single equationor 2)

D-1 12 g of alkaline earth metal gives 14.8 g of its nitride. Atomic weight of metal is -(A) 12 (B) 20 (C) 40 (D) 14.8

D-2 For the reaction 2P + Q R, 8 mol of P and excess of Q will produce :(A) 8 mol of R (B) 5 mol of R (C) 4 mol of R (D) 13 mol of R

D-3 If 1.5 moles of oxygen combine with Al to form Al2O

3, the weight of Al used in the reaction is :

(A) 27 g (B) 40.5 g (C) 54g (D) 81 g

D-4 How many liters of CO2 at STP will be formed when 0.01 mol of H

2SO

4 reacts with excess of Na

2CO

3 .

Na2CO

3 + H

2SO

4 Na

2SO

4 + CO

2 + H

2O

(A) 22.4 L (B) 2.24 L (C) 0.224 L (D) 1.12 L

D-5. When 100g of ethylene polymerises entirely to polyethene, the weight of polyethene formed as per the equationn(C

2H

4) (CH

2CH

2)

n is :

(A) (n/2)g (B) 100g (C) (100/n)g (D) 100ng

MOLE CONCEPT # 38

D-6. How many moles of potassium chlorate need to be heated to produce 11.2 litre oxygen at N.T.P.

(A) 21

mol (B) 31

mol (C) 41

mol (D) 32

mol

Section (E) : Limiting reagent, % Excess, % Yield / EfficiencyE-1. Calculate the amount of Ni needed in the Mond's process given below

Ni + 4CO Ni(CO)4

If CO used in this process is obtained through a process, in which 6 g of carbon is mixed with 44 g CO2 .

(A) 14.675 g (B) 29 g (C) 58 g (D) 28 g

E-2. For the reaction 2P + Q R, 8 mol of P and 5 mol of Q will produce(A) 8 mol of R (B) 5 mol of R (C) 4 mol of R (D) 13 mol of R

E-3. How many mole of Zn(FeS2) can be made from 2 mole zinc, 3 mole iron and 5 mole sulphur.

(A) 2 mole (B) 3 mole (C) 4 mole (D) 5 mole

E-4. Equal weight of 'X' (At. wt. = 36) and 'Y' (At. wt. = 24) are reacted to form the compound X2Y3. Then :(A) X is the limiting reagent(B) Y is the limiting reagent(C) No reactant is left over and mass of X2Y3 formed is double the mass of �X� taken

(D) none of these

E-5. 0.5 mole of H2SO

4 is mixed with 0.2 mole of Ca (OH)

2. The maximum number of moles of CaSO

4 formed is

(A) 0.2 (B) 0.5 (C) 0.4 (D) 1.5

E-6. The mass of 70% H2SO

4 required for neutralisation of 1 mol of NaOH.

(A) 49 gm (B) 98 gm (C) 70 gm (D) 34.3 gm

E-7. In a certain operation 358 g of TiCl4 is reacted with 96 g of Mg. Calculate % yield of Ti if 32 g of Ti is actually

obtained [At. wt. Ti = 48, Mg = 24](A) 35.38 % (B) 66.6 % (C) 100 % (D) 60 %

Section (F) : Principle of atom conservation (POAC), Reactions in sequence & parallel,Mixture analysis, % Purity

F-1. 25.4 g of iodine and 14.2g of chlorine are made to react completely to yield a mixture of Cl and Cl3. Calculate

the number of moles of Cl and Cl3 formed.

(A) 0.1 mole, 0.1 mole (B) 0.1 mole, 0.2 mole(C) 0.5 mole, 0.5 mole (D) 0.2 mole, 0.2 mole

F-2. What weights of P4O

6 and P

4O

10 will be produced by the combustion of 31g of P

4 in 32g of oxygen leaving no

P4 and O

2.

(A) 2.75g, 219.5g (B) 27.5g, 35.5g (C) 55g, 71g (D) 17.5g, 190.5g

F-3. 0.05 mole of LiAlH4 in ether solution was placed in a flask containing 74g (1 mole) of t-butyl alcohol. The

product LiAlHC12

H27

O3 weighed 12.7 g. If Li atoms are conserved, the percentage yield is : (Li = 7, Al = 27, H

= 1, C = 12, O = 16).(A) 25% (B) 75% (C) 100% (D) 15%

F-4 What weight of CaCO3 must be decomposed to produce the sufficient quantity of carbon dioxide to convert21.2 kg of Na2CO3 completely in to NaHCO3. [Atomic mass Na = 23, Ca = 40]

CaCO3 CaO + CO2

Na2 CO3 + CO2 + H2O 2NaHCO3

(A) 100 Kg (B) 20 Kg (C) 120 Kg (D) 30 Kg

F-5 NX is produced by the following step of reactionsM + X2 M X2 ; 3MX2 + X2 M3X8 ; M3 X8 + N2CO3 NX + CO2 + M3O4

How much M (metal) is consumed to produce 206 gm of NX. (Take at wt of M = 56, N=23, X = 80)

(A) 42 gm (B) 56 gm (C) 3

14 gm (D)

47

gm

MOLE CONCEPT # 39

F-6.* A sample of a mixture of CaCl2 and NaCl weighing 4.44 gm was treated to precipitate all the Ca as CaCO

3,

which was then heated and quantitatively converted to 1.12g of CaO. (At . wt. Ca = 40, Na = 23, Cl = 35.5)(A) Mixture contains 50% NaCl (B) Mixture contains 60% CaCl

2

(C) Mass of CaCl2 is 2.22 g (D) Mass of CaCl

2 1.11 g

Section (G) : Basics of oxidation numberG-1. The oxidation number of Oxygen in Na2O2 is :

(A) + 1 (B) + 2 (C) � 2 (D) � 1

G-2. In FeCr2O4, the oxidation numbers of Fe and Cr are :(A) + 2 and + 3 (B) 0 and + 2 (C) + 2 and + 6 (D) + 3 and + 6

G-3. The oxidation number of Phosphorus in Mg2P2O7 is :(A) + 3 (B) + 2 (C) + 5 (D) � 3

G-4. The oxidation states of Sulphur in the anions SO32� , S2O4

2� and S2O62� follow the order :

(A) S2O62� < S2O4

2 < SO32� (B) S2O4

2� < SO32� < S2O6

2�

(C) SO32� < S2O4

2� < S2O62� (D) S2O4

2 < S2O62� < SO3

2�

G-5. Match List-I (Compounds) with List-II (Oxidation states of Nitrogen) and select answer using the codes givenbelow the lists :

List-I List-II(a) NaN

3(1) +5

(b) N2H

2(2) +2

(c) NO (3) �1/3

(d) N2O

5(4) �1

(Code) :(a) (b) (c) (d) (a) (b) (c) (d)

(A) 3 4 2 1 (B) 4 3 2 1(C) 3 4 1 2 (D) 4 3 1 2

G-6. 1 mole of N2H4 loses ten moles of electrons to form a new compound Y. Assuming that all the nitrogen appearsin the new compound, what is the oxidation state of nitrogen in Y? (There is no change in the oxidation state ofhydrogen).(A) � 1 (B) � 3 (C) + 3 (D) + 5

G-7. The average oxidation state of Fe in Fe3O

4 is :

(A) 2 and 3 (B) 8/3 (C) 2 (D) 3

G-8.* Which of the following are examples of disproportionation reaction :(A) HgO Hg + O2 (B) KClO3 KCl + O2

(C) KClO3 KClO4 + KCl (D) Cl2 + OH� ClO� + Cl� + H2O

Section (H) : Balancing redox reactionsH-1. In the reaction xHI + yHNO3 NO + 2 + H2O, upon balancing with whole number coefficients :

(A) x = 3, y = 2 (B) x= 2, y = 3 (C) x = 6, y = 2 (D) x = 6, y = 1

H-2. For the redox reaction MnO4� + C2O4

2� + H+ Mn2+ + CO2 + H2O,

the correct whole number stoichiometric coefficients of MnO4�, C2O4

2� and H+ are respectively:(A) 2, 5, 16 (B) 16, 5, 2 (C) 5, 16, 2 (D) 2, 16, 5

H-3. For the redox reaction xP4 + yHNO3 H3PO4 + NO2 + H2O, upon balancing with whole number coefficients:

(A) x = 1, y = 5 (B) x = 2, y = 10 (C) x = 1, y = 20 (D) x = 1, y = 15

MOLE CONCEPT # 40

H-4. In the reaction X � + XO3

� + H+ X2 + H2O, the molar ratio in which X � and XO3

� react is :(A) 1 : 5 (B) 5 : 1 (C) 2 : 3 (D) 3 : 2

H-5. CN� is oxidised by NO3� in presence of acid :

a CN� + b NO3� + c H+ (a + b) NO + a CO2 +

2c

H2O

What are the whole number values of a, b, c in that order :(A) 3, 7, 7 (B) 3, 10, 7 (C) 3, 10, 10 (D) 3, 7, 10

H-6.* In the following reaction : Cr(OH)3 + OH� + IO3� CrO4

2� + H2O + I�

(A) IO3� is oxidising agent (B) Cr(OH)3 is oxidised

(C) 6e� are being taken per iodine atom (D) None of these

Section (I) : Units of concentration measurement, Interconversion of concentration units-1. 500 mL of a glucose solution contains 6.02 × 1022 molecules. The concentration of the solution is

(A) 0.1 M (B) 1.0 M (C) 0.2 M (D) 2.0 M

-2. Equal moles of H2O and NaCl are present in a solution. Hence, molality of NaCl solution is :

(A) 0.55 (B) 55.5 (C) 1.00 (D) 0.18

-3. Decreasing order of mass of pure NaOH in each of the aqueous solution.(I) 50 g of 40% (W/W) NaOH(II) 50 ml of 50% (W/V) NaOH (d

sol = 1.2 g/ml).

(III) 50 g of 15 M NaOH (dsol

= 1 g/ml).(A) I, II, III (B) III, II, I (C) II, III, I (D) III = II = I.

-4. Mole fraction of A in H2O is 0.2. The molality of A in H

2O is :

(A) 13.9 (B) 15.5 (C) 14.5 (D) 16.8

-5. What is the molarity of H2SO4 solution that has a density of 1.84 g/cc and contains 98% by mass of H2SO4?(Given atomic mass of S = 32)(A) 4.18 M (B) 8.14 M (C) 18.4 M (D) 18 M

-6. The molarity of the solution containing 2.8%( mass / volume) solution of KOH is : (Given atomic mass ofK = 39 ) is :(A) 0.1 M (B) 0.5 M (C) 0.2 M (D) 1 M

-7. A solution of FeCl3 is

30M

its molarity for Cl� ion will be :

(A) 90M

(B) 30M

(C) 10M

(D) 5M

Section (J) : Dilution & Mixing of two liquidsJ-1. If 500 ml of 1 M solution of glucose is mixed with 500 ml of 1 M solution of glucose final molarity of solution will

be :(A) 1 M (B) 0.5 M (C) 2 M (D) 1.5 M

J-2. The volume of water that must be added to a mixture of 250 ml of 0.6 M HCl and 750 ml of0.2 M HCl to obtain 0.25 M solution of HCl is :(A) 750 ml (B) 100 ml (C) 200 m (D) 300 m

J-3. What volume of a 0.8 M solution contains 100 milli moles of the solute?(A) 100 mL (B) 125 mL (C) 500 mL (D) 62.5 mL

J-4. The molarity of Cl¯ in an aqueous solution which was (w/V) 2% NaCl, 4% CaCI2 and 6% NH

4Cl will be

(A) 0.342 (B) 0.721 (C) 1.12 (D) 2.18

MOLE CONCEPT # 41

J-5. 2M of 100 ml Na2 SO

4 is mixed with 3M of 100 ml NaCl solution and 1M of 200 ml CaCl

2 solution. Then the

ratio of the concentration of cation and anion.(A) 1/2 (B) 2 (C) 1.5 (D) 1

PART - III : ASSERTION / REASONING

Each question has 5 choices (A), (B), (C), (D) and (E) out of which ONLY ONE is correct.(A) STATEMENT-1 is true, STATEMENT-2 is true and STATEMENT-2 is correct explanation forSTATEMENT-1.(B) STATEMENT-1 is true, STATEMENT-2 is true and STATEMENT-2 is not correct explanation forSTATEMENT-1.(C) STATEMENT-1 is true, STATEMENT-2 is false.(D) STATEMENT-1 is false, STATEMENT-2 is true.(E) Both STATEMENTS are false.

1. Statement-1 : The weight percentage of a compound A in a solution is given by

% of A = 100solutionofmassTotal

AMass

Statement-2 : The mole fraction of a component A is given by,

Mole fraction of A = componentsallofmolesof.noTotalAofmolesof.No

2. Statement-1 : The ratio of the mass of 100 billion atoms of magnesium to the mass of 100 billion atoms of

lead can be expressed as 20724

.

Statement-2 : Atomic weights are relative masses.

3. Statement-1 : The average mass of one Mg atom is 24.305 amu, which is not the actual mass of one Mgatom.Statement-2 : Three isotopes, 24Mg, 25Mg and 26Mg, of Mg are found in nature.

4. Statement-1 : A molecule of butane, C4 H

10 has a mass of 58.12 amu.

Statement-2 : One mole of butane contains 6.022 × 1023 molecules and has a mass of 58.12 g.

5. Statement-1 : Both 12 g. of carbon and 27 g. of aluminium will have 6.02 × 1023 atoms.Statement-2 : Gram atomic mass of an element contains Avogadro�s number of atoms.

6. Statement-1 : The molality and molarity of very dilute aqueous solutions differ very little.Statement-2 : The density of water is about 1.0 g cm�3 at room temperature.

7. Statement-1 : For calculating the molality or the mole fraction of solute, if the molarity is known, it is necessaryto know the density of the solution.Statement-2 : Molality, molarity and the mole fraction of solute can be calculated from the weight percentageand the density of the solution

8. Statement-1 : Laboratory reagents are usually made up to a specific molarity rather than a given molality.Statement-2: The volume of a liquid is more easily measured than its mass.

9. Statement-1 : Molality and mole fraction concentration units do not change with temperature.Statement-2 : These units are not defined in terms of any volume.

10. Statement-1 : A one molal solution prepared at 20°C will retain the same molality at 100°C, provided there is

no loss of solute or solvent on heating.Statement-2 : Molality is independent of temperature.

MOLE CONCEPT # 42

PART - I : SUBJECTIVE QUESTIONS

1. The nonstoichiometry compound �titanium monoxide� has a continuous range of composition from Ti0.75

O toTiO

0.69. What are the maximum and minimum values of the percent by weight of oxygen in this compound?

[At. wt of Ti = 48]

2. Assume that the nucleus of the F atom is a sphere of radius 5 × 10�13 cm. Calculate the density of matter in Fnucleus.

3. A plant virus is found to consist of uniform cylindrical particles of 150 Å in diameter and 5000 Å long. The

specific volume of the virus is 0.75 cm3/g. If the virus is considered to be a single partical, find its molecularweight.

4. A sample of chalk contains clay as impurity. The clay impurity loses 11% of its weight as moisture on prolongheating. 5 gram sample of chalk on heating shows a loss in weight (due to evolution of CO

2 and water) by 1.1

g. Calculate % of chalk (CaCO3) in the sample. [Hint : Chalk (CaCO

3) release CO

2 on heating]

5. Consider the following reaction involved in the preparation of teflon polymer .

XeF6 + + HF + XeF

4.

Determine the moles of XeF6 required for preparation of 100 g Teflon.

6. (i) K4Fe(CN)

6 + 3H

2SO

4 2K

2SO

4 + FeSO

4 + 6HCN

(ii) 6HCN + 12H2O 6HCOOH + 6NH

3

(iii) (a) 6NH3 + 3H

2SO

4 3(NH

4)

2 SO

4(b) 6HCOOH 42SOH 6CO + 6H

2O

Above steps of reactions occur in a container starting with one mole of K4[Fe(CN)

6], 5 mole of H

2SO

4 and

enough water. Find out the limiting reagent in step (i) and calculate maximum moles of CO gas and (NH4)

2 SO

4

that can be produced.

7. A2 + 2B2 A2B4 ; 23

A2 + 2B2 A3B4

Two substance A2 & B2 react in the above manner. When A2 is limited it gives A2B4 in excess gives A3B4. A2B4can be converted to A3B4 when reacted with A2. Using this information calculate the composition of the finalmixture when the mentioned amount of A2 & B2 are taken

(a) 4 moles A2 & 4 moles B2 (b) 21

moles AA2 & 2 moles B2 (c) 1.25 moles A2 & 2 moles B2

8. When 1 mol of A reacts with 21

mol of B2 (A +

21

B2 AB), 100 Kcal heat is liberated and when 1 mol of

A reacted with 2 mol of B2 (A + 2B

2 AB

4) , 200 Kcal heat is liberated. When 1 mol of A is completely

reacted with excess of B2 to form AB as well as AB

4, 140 Kcal heat is liberated Calculate the mol of B

2 used.

[Write your answer as (No. of moles of B2 used 1000)]

9. In a gravimetric determination of P, an aqueous solution of dihydrogen phosphate ion H2PO

4� is treated with a

mixture of ammonium and magnesium ions to precipitate magnesium ammonium phosphate,Mg(NH

4)PO

4 . 6H

2O. This is heated and decomposed to magnesium pyrophosphate , Mg

2P

2O

7 , which is

weighed. A solution of H2PO

4� yielded 1.054 g of Mg

2P

2O

7. What weight of NaH

2PO

4 was present originally?

(Na = 23 , H = 1 , P = 31 , O = 16 , Mg = 24)

10. 1 g of dry green algae absorbs 4.7 x 10�3

mole of CO

2 per hour by photosynthesis. If the fixed carbon atoms

were all stored after photosynthesis as starch (C6H

10 O

5)

n how long would it take for the algae to double their

own weight assuming photosynthesis takes place at a constant rate ?

MOLE CONCEPT # 43

11. Cl2 + KOH

%60 KCl + KClO + H

2O

KClO %50

KCl + KClO3

KClO3

%80 KClO

4 + KCl

112 L Cl2 gas at STP is passed in 10 L KOH solution, containing 1 mole of potassium hydroxide per liter.

Calculate the total moles of KCl produced, rounding it off to nearest whole number. (Yield of chemical reactionsare written above the arrow () of respective reaction)

12. The 'roasting' of 100.0 g of a copper ore yielded 71.8 g pure copper. If the ore is composed of Cu2S and CuS

with 4.5 % inert impurity, calculate the percent of Cu2S in the ore.

The reactions areCu

2S + O

2 2Cu + SO

2and CuS + O

2 Cu + SO

2 [Atomic mass Cu = 63.5, S = 32]

13. NaBr, used to produce AgBr for use in photography can itself be prepared as follows :

Fe + Br2 FeBr

2

FeBr2 + Br

2 Fe

3 Br

8(not balanced)

Fe3Br

8 + Na

2CO

3 NaBr + CO

2 + Fe

3O

4(not balanced)

How much Fe, in kg, is consumed to produce 2.50 x 103 kg NaBr ?

14. How many grams of 90% pure Na2SO4 can be produced from 250 g of 95% pure NaCl.

15. Find the sum of average oxidation number of S in H2SO

5 (peroxy monosulphuric acid) and Na

2S

2O

3 (sodium

thiosulphate).

16. The reaction Cl2(g) + S

2O

32� SO

42� + Cl� is to be carried out in basic medium. Starting with 0.15

mole of Cl2, 0.01 mole S

2O

32� and 0.3 mole of OH�, how many moles of OH� will be left in solution after

the reaction is complete. Assume no other reaction occurs.

17. HCl solution (50% by mass and d = 1.25 g/ml) is used to prepare one litre 0.5 M HCl. What is the volume ofsample used for this purpose.

18. What volume (in ml) of 0.2 M H2SO4 solution should be mixed with the 40 ml of 0.1 M NaOH solution such that

the resulting solution has the concentration of H2SO4 as 556

M.

19. To 100 ml of 5 M NaOH solution (density 1.2 g/ml) were added 200 ml of another NaOH solution which has adensity of 1.5 g/ml and contains 20 mass percent of NaOH. What will be the volume of the gas (at STP) inlitres liberated when aluminium reacts with this (final) solution.The reaction is Al + NaOH + H

2O NaAlO

2 + H

2 (At.wt. Na = 23, 0 = 16, H = 1)

20. A drop (0.05 mL) of 12 M HCl is spread over a thin sheet of aluminium foil (thickness 0.10 mm and density ofAl = 2.70 g/mL). Assuming whole of the HCl is used to dissolve Al, what will be the maximum area of holeproduced in foil ?

PART - II : OBJECTIVE QUESTIONS

Single choice type1. Which of the following will contain same number of atoms as 20g of calcium?

(A) 24g magnesium (B) 12g carbon (C) 8g oxygen gas (D) 16 g oxygen atom

2. A sample of ammonium phosphate (NH4)

3PO

4 contains 3.18 mol of H atoms. The number of mol of

O atoms in the sample is :(A) 0.265 (B) 0.795 (C) 1.06 (D) 3.18

3. The molar mass of normal water is .......... as compared to heavy water.(A) 10% less (B) 10% high (C) 2% less (D) zero% less

4. The ratio of the weight of one litre of a gas to the weight of 1.0 L oxygen gas both measured at S.T.P. is 2.22.The molecular weight of the gas would be :(A) 14.002 (B) 35.52 (C) 71.04 (D) 55.56

MOLE CONCEPT # 44

5. One mole of a mixture of N2, NO

2 and N

2O

4 has a mean molar mass of 55.4. On heating to a temperature at

which all the N2O

4 may be presumed to have dissociated : N

2O

4 2NO

2, the mean molar mass tends to

the lower value of 39.6. What is the mole ratio of N2 : NO

2 : N

2O

4 in the original mixture ?

(A) 0.5 : 0.1 : 0.4 (B) 0.6 : 0.1 : 0.3 (C) 0.5 : 0.2 : 0.3 (D) 0.6 : 0.2 : 0.2

6. 112.0 mL of NO2 at STP was liquefied, the density of the liquid being 1.15 g mL�1. Calculate the volume and the

number of molecules in the liquid NO2

(At. wt. of N = 14)(A) 0.10 mL and 3.01 × 1022 (B) 0.20 mL and 3.01 × 1021

(C) 0.20 mL and 6.02 × 1023 (D) 0.40 mL and 6.02 × 1021

7. A certain organic substance used as a solvent in many reactions contains carbon, hydrogen, oxygen andsulphur. Weight % of hydrogen in the compound is 7.7. The weight ratio C : O : S = 3 : 2 : 4. What is the leastpossible molar mass of the compound ?(A) 86 (B) 63 (C) 94 (D) 78

8. The oxides of a certain (hypothetical) element contain 27.28%, 42.86% and 52.94% oxygen. What is the ratioof the valancies of the element in the 3 oxides ?(A) 2 : 3 : 4 (B) 1 : 3 : 4 (C) 1 : 2 : 4 (D) 1 : 2 : 3

9. Formation of polyethene from calcium carbide takes place as follows :CaC

2+H

2O Ca(OH)

2 + C

2H

2 ; C

2H

2 + H

2 C

2H

4

n(C2H

4) (CH

2CH

2)

n.

The amount of polyethylene possibly obtainable from 64.0 kg CaC2 can be

(A) 28Kg (B) 14kg (C) 21kg (D) 42 kg

10. The hourly energy requirement of an astronaut can be satisfied by the energy released when 34 g of sucrose(C

12H

22O

11) are burnt in his body. How many g of oxygen would be needed to be carried in space capsule to

meet his requirement for one day :(A) 916.2 gm (B) 91.62 gm (C) 8.162 gm (D) 9.162 gm.

11. If a piece of iron gains 10% of its weight due to partial rusting into Fe2O

3 the percentage of total iron that has

rusted is :(A) 23 (B) 13 (C) 23.3 (D) 25.67

12. If 10 g of Ag reacts with 1 g of sulphur , the amount of Ag2S formed will be [Atomic weight of Ag = 108,

S = 32] ?(A) 7.75 g (B) 0.775 g (C) 11 g (D) 10 g

13. 100 g impure CaCO3 on heating gives 5.6 lt. CO2 gas at STP. Find the percentage of calcium in the lime stonesample. [At. wt. : Ca = 40 ; C = 12 ; O = 16](A) 10 (B) 20 (C) 1 (D) 30

14. When a 12 g mixture of carbon and sulphur is burnt in air, then a mixture of CO2 and SO

2 is produced, in which

the number of moles of SO2 is half that of CO

2. The mass of the carbon in the mixture is :

(A) 4.08 g (B) 5.14 g (C) 8.74 g (D) 1.54 g

15. When x grams of carbon are heated with y grams of oxygen in a closed vessel, no solid residue is left behind.Which of the following statements is correct ?(A) y/x must lie between 1.33 and 2.67 (B) y/x must be greater than or equal 2.67.(C) y/x must be less than or equal 1.33 (D) y/x must be greater than or equal 1.33.

16. XeF6 fluorinates

2 to F

7 and liberates Xenon(g). 210 mmol of XeF

6 can yield a maximum of_____ mmol of IF

7

(A) 420 (B) 180 (C) 210 (D) 245

17. 1 mol of iron (Fe) reacts completely with 0.65 mol O2 to give a mixture of only FeO and Fe

2O

3. Mole ratio of

ferrous oxide to ferric oxide is :

(A) 3 : 2 (B) 4 : 3 (C) 20 : 13 (D) none of these

18. Composition of a sample is Fe0.93

O1.00

. If Fe is present in +2 & +3 oxidation state in this sample then % of Fepresent in +3 oxidation state(A) 85% (B) 30% (C) 15% (D) 60%

MOLE CONCEPT # 45

19. When ZnS is boiled with strong nitric acid, the products are zinc nitrate, sulphuric acid and nitrogen dioxide.What are the changes in the oxidation numbers of Zn, S and N :(A) + 2, + 4, � 1 (B) + 2, + 6, � 2 (C) 0, + 4, � 2 (D) 0, + 8, � 1

20. When arsenic sulphide is boiled with NaOH, sodium arsenite and sodium thioarsenite are formed according toreaction :

x As2S3+ y NaOH x Na3AsO3 + x Na3AsS3 + 2y

H2O. What are the values of x and y?

(A) 1, 6 (B) 2, 8 (C) 2, 6 (D) 1, 4

21. The following equations are balanced atomwise and chargewise.(i) Cr2O7

2� + 8H+ + 3H2O2 2Cr3+ + 7H2O + 3O2

(ii) Cr2O72� + 8H+ + 5H2O2 2Cr3+ + 9H2O + 4O2

(iii) Cr2O72� + 8H+ + 7H2O2 2Cr3+ + 11H2O + 5O2

The precise equation/equations representing the oxidation of H2O2 is/are :

(A) (i) only (B) (ii) only (C) (iii) only (D) all the three

22. xNO3� + yI� + zH+

2NO + 3I2 + 4H2O x , y z respectively in the above equation are :(A) 2, 6, 8 (B) 1, 6 , 4 (C) 0, 6 , 8 (D) 2 , 3 , 4

. Molarity of H2SO

4 is 18 M. Its density is 1.8 g/cm3, hence molality is

(A) 18 (B) 100 (C) 36 (D) 500

. A solution of glucose received from some research laboratory has been marked mole fraction x and molality(m) at 10ºC. When you will calculate its molality and mole fraction in your laboratory at 24ºC you will find

(A) mole fraction (x) and molality (m) (B) mole fraction (2x) and molality (2m)(C) mole fraction (x/2) and molality (m/2) (D) mole fraction (x) and (m ± dm) molality

. Mole fraction of ethyl alcohol in aqueous ethyl alcohol (C2H

5OH) solution is 0.25. Hence percentage of ethyl

alcohol by weight is :(A) 54% (B) 25% (C) 75% (D) 46%

. A mineral water sample was analysed and found to contain 1 × 10�3 % ammonia (w/w). The mole of dissolvedammonia gas in one litre water bottle is (d

water 1 gm/ml)

(A) 5.8 ×10�4 mol (B) 1 × 10�2 mol (C) 0.58 ×10�2 mol (D) same as w/w

27. What is the quantity of water that should be added to 16 g. methanol to make the mole fraction of methanol as0.25 :(A) 27 g. (B) 12 g. (C) 18 g. (D) 36 g.

28. Calculate the mass percent (w/w) of sulphuric acid in a solution prepared by dissolving 4 g of sulphur trioxidein a 100 ml sulphuric acid solution containing 80 mass percent (w/w) of H2SO4 and having a density of 1.96 g/ml. (molecular weight of H2SO4 = 98). Take reaction SO

3 + H

2O H

2SO

4

(A) 80.8% (B) 84% (C) 41.65% (D) None of these

29. On mixing 15.0 ml of ethyl alcohol of density 0.792 g ml 1 with 15 ml of pure water at 40 C, the resulting solutionis found to have a density of 0.924 g ml 1. The percentage contraction in volume is :(A) 8 % (B) 2 % (C) 3 % (D) 4 %

More than one choice type

30. In which of the following pairs do 1 g of each have an equal number of molecules?(A) N

2O and CO (B) N

2 and C

3O

2(C) N

2 and CO (D) N

2O and CO

2

31. Silver metal in ore is dissolved by potassium cyanide solution in the presence of air by the reaction4 Ag + 8 KCN + O

2 + 2H

2O 4 K[Ag (CN)

2] + 4 KOH

(A) The amount of KCN required to dissolve 100 g of pure Ag is 120 g.(B) The amount of oxygen used in this process is 0.742 g (for 100 gm pure Ag)(C) The amount of oxygen used in this process is 7.40 g (for 100 gm pure Ag)(D) The volume of oxygen used at STP is 5.20 litres.

MOLE CONCEPT # 46

32. If 27 g of Carbon is mixed with 88 g of Oxygen and is allowed to burn to produce CO2 , then :

(A) Oxygen is the limiting reagent. (B) Volume of CO2 gas produced at NTP is 50.4 L.

(C) C and O combine in mass ratio 3 : 8. (D) Volume of unreacted O2 at STP is 11.2 L.

33. A + B A3B

2 (unbalanced)

A3B

2 + C A

3B

2C

2 (unbalanced)

Above two reactions are carried out by taking 3 moles each of A and B and one mole of C. Then which optionis/are correct ?

(A) 1 mole of A3B

2C

2 is formed (B) 21 mole of AA

3B

2C

2 is formed

(C) 1/2 mole of A3B

2 is formed (D) 1/2 mole of A

3B

2 is left finally

34. Consider the redox reaction 2S2O32� + 2 S4O6

2� + 2 � :(A) S2O3

2� gets reduced to S4O62� (B) S2O3

2� gets oxidised to S4O62�

(C) 2 gets reduced to � (D) 2 gets oxidised to �

35.* If 100 ml of 1M H2SO

4 solution is mixed with 100 ml of 9.8%(w/w) H

2SO

4 solution (d = 1 g/ml) then :

(A) concentration of solution remains same (B) volume of solution become 200 ml(C) mass of H

2SO

4 in the solution is 98 gm (D) mass of H

2SO

4 in the solution is 19.6 gm

36. Equal volume of 0.1M NaCl and 0.1M FeCl2 are mixed with no change in volume due to mixing. Which of thefollowing will be true for the final solution. (No precipitation occurs). Assume complete dissociation of salts andneglect any hydrolysis.(A) [Na+] = 0.05 M (B) [Fe2+] = 0.05M (C) [Cl] = 0.3M (D) [Cl] = 0.15M

PART - III : MATCH THE COLUMN & COMPREHENSION

1. Column - I Column - II(A) A gaseous organic compound containing (p) One mole of compound contains 4N

A atoms of

C = 52.17%, H = 13.04% & O = 34.78% Hydrogen. (by weight) having molar mass 46 g/mol.

(B) 0.3 g of an organic compound containing (q) The empirical formula of the compound is same as C, H and O on combustion yields 0.44 g its molecule formula. of CO

2 and 0.18 g of H

2O, with two O

atoms per molecule.

(C) A hydrocarbon containing C = 42.857% (r) Combustion products of one mole of compound and H = 57.143% (by mole) containing 3C contains larger number of moles of CO

2 than that of

atoms per molecule. H2O.

(D) A hydrocarbon containing 10.5 g carbon (s) CO2 gas produced by the combustion of 0.25 mole

per gram of hydrogen having vapour of compound occupies a volume of 11.2 L at NTP. density 46.

2. Column I Column II

(A) Zn(s) + 2HCl(aq) ZnCl2(s) + H2(g) (p) 50% of excess reagent left above reaction is carried out by taking 2 moles each of Zn and HCl

(B) AgNO3(aq) + HCl(aq) AgCl(s) + HNO3(g) (q) 22.4 L of gas at STP is liberated above reaction is carried out by taking 170 g AgNO3 and 18.25 g HCl (Ag = 108)

(C) CaCO3(s) CaO(s) + CO2(g) (r) 1 moles of solid (product) obtained. 100 g CaCO3 is decomposed

(D) 2KClO3(s) 2KCl(s) + 3O2(g) (s) HCl is the limiting reagent 2/3 moles of KClO3 decomposed

MOLE CONCEPT # 47

3. Column-I Column-II(A) Molarity (p) Dependent on temperature

(B) Molality (q) BBAA

AA

MnMnnM

x 100

(C) Mole fraction (r) Independent of temperature

(D) Mass % (s) BB

A

MXX

× 1000

Where MA , M

B are molar masses, n

A , n

B are no of moles & X

A , X

B are mole fractions of solute and solvent

respectively.

4. Column-I Column-II(A) 100 ml of 0.2 M AlCl3 solution + 400 ml (p) Total concentration of cation(s) = 0.12 M of 0.1 M HCl solution(B) 50 ml of 0.4 M KCl + 50 ml H2O (q) [SO4

2�] = 0.06 M(C) 30 ml of 0.2 M K2SO4 + 70 ml H2O (r) [SO4

2�] = 2.5 M(D) 200 ml 24.5% (w/v) H2SO4 (s) [Cl¯] = 0.2 M

PART - IV : COMPREHENSION

Read the following comprehension carefully and answer the questions.

Comprehension # 1A chemist decided to determine the molecular formula of an unknown compound. He collects followinginformations :(I) Compounds contains 2 : 1 'H' to 'O' atoms(number of atoms).(II) Compounds has 40% C by mass(III) Approximate molecular mass of the compound is 178 g(IV) Compound contains C, H and O only.

1. What is the % by mass of oxygen in the compound(A) 53.33% (B) 88.88% (C) 33.33% (D) None of these

2. What is the empirical formula of the compound(A) CH

3O (B) CH

2O (C) C

2H

2O (D) CH

3O

2

3. Which of the following could be molecular formula of compound(A) C

6H

6O

6(B) C

6H

12O

6(C) C

6H

14O

12(D) C

6H

14O

6

Comprehension # 2According to the Avogadro�s law, equal number of moles of gases occupy the same volume at identical condition

of temperature and pressure. Even if we have a mixture of non-reacting gases then Avogadro�s law is still

obeyed by assuming mixture as a new gas.Now let us assume air to consist of 80% by volume of Nitrogen (N

2) and 20% by volume of oxygen (O

2). If air

is taken at STP then its 1 mol would occupy 22.4 L. 1 mol of air would contain 0.8 mol of N2 and 0.2 mol of O

2

hence the mole fractions of N2 and O

2 are given by 8.0X

2N , 2.0X2O

4. Volume occupied by air at NTP containing exactly 11.2 gm of Nitrogen :(A) 22.4 L (B) 8.96 L (C) 11.2 L (D) 2.24 L

5. If air is treated as a solution of O2 and N

2 then % W/W of oxygen is :

(A) 9

10(B)

9200

(C) 9

700(D)

9350

6. Density of air at NTP is :

(A) 1 g/L (B) 79

g/L (C) 72

g/L (D) can�t be determined

MOLE CONCEPT # 48

Comprehension # 3

The concentrations of solutions can be expressed in number of ways; viz : mass fraction of solute (or masspercent), Molar concentration (Molarity) and Molal concentration (molality). These terms are known asconcentration terms and also they are related with each other i.e. knowing one concentration term for thesolution, we can find other concentration terms also. The definition of different concentration terms are givenbelow :Molarity : It is number of moles of solute present in one litre of the solution.Molality : It is the number of moles of solute present in one kg of the solvent

Mole Fraction = solventofmolessoluteofmolessoluteofmoles

If molality of the solution is given as 'a' then mole fraction of the solute can be calculated by

Mole Fraction =

solventM1000

a

a

; = )1000Ma(

Ma

solvent

solvent

where a = molality and Msolvent

= Molar mass of solventWe can change : Mole fraction Molality Molarity

7. 60 gm of solution containing 40% by mass of NaCl are mixed with 100 gm of a solution containing 15% bymass NaCl. Determine the mass percent of sodium chloride in the final solution.(A) 24.4% (B) 78% (C) 48.8% (D) 19.68%

8. What is the molality of the above solution.

(A) 4.4 m (B) 5.5 m (C) 24.4 m (D) none

9. What is the molarity of solution if density of solution is 1.6 gm/ml

(A) 5.5 M (B) 6.67 M (C) 2.59 M (D) none

Comprehension # 4In chemistry, oxidation and reduction are taken as two mutually exclusive events. For example, if life is oxidationthen death is taken as reduction, taking off a flight is oxidation then standing would be reduction and so manyother. In brief it is used as redox in chemical science.There are so many conceptual facts regarding redox such as adding oxygen or oxygenation, removing hydrogenor dehydrogenation, removing electron or dielectronation are fixed for oxidation and their correspondingantonyms would be reduction processes. Simple way of judging whether a monatomic species has undergone oxidation or reduction is to note if the charge number of species has changed. It is possible to assign toan atom in polyatomic species an operative charge number called their oxidation number or state. (O. N. or O.S.). There is no standard symbol for this quantity so we say it is . An O. N. is assigned to an element in acompound by assuming that it is present as ion with a characteristic charge for instance oxygen is present asO(-II) and fluorine as F(�I) and some time it may be hypothetical also. For example

For ZnO

For NH3

In continuation to our study, species promoting oxidation are named as oxidant and those promoting reductionare termed as reductant. At the same time their equivalent weights is the ratio of their molecular weight andchange is O. N. () involving one molecule/formula unit of the reactant i.e., molecular weight divided bynumber of electrons lost or gained by one molecule/formula during their respective action.Based on the above discussion answer the following objective question having one best answer.

10. Which corresponds to oxidation action(A) = 0 (B) = 0 (C) > 0 (D) < 0

MOLE CONCEPT # 49

11. A compound contain P(II), Q(V) R(-II). The possible formula of the compound is(A) PQR

2(B) Q

2(PR

3)

2(C) P

3[QR

4]2

(D) P3(Q

4R)

2

12. A compound has number of carbon, number of hydrogen and number of oxygen their equation of findingoxidation number (x) of carbon will be

(A) 3 + 4x2 + = 0 (B) x + � 2 = 0 (C) x + x

� 3

2 = 0 (D) none of these

PART - I : IIT-JEE PROBLEMS (PREVIOUS YEARS)

* Marked Questions may have more than one correct option.

1. How many moles of electron weigh one kilogram : [JEE-2002, 3/150]

(A) 6.023 × 1023 (B) 108.91

× 1031 (C) 108.9023.6

× 1054 (D) 023.6108.9

1

× 108

2. Which has maximum number of atoms : [JEE-2003, 3/144](A) 24 g of C (12) (B) 56 g of Fe (56) (C) 27 g of Al (27) (D) 108 g Ag (108)

3. Amongst the following, the pair having both the metals in their highest oxidation state is :[JEE 2004, 3/84]

(A) [Fe(CN)6]3� and [Co(CN)6]

3� (B) CrO2Cl2 and MnO4�

(C) TiO2 and MnO2 (D) [MnCl4]2� and [NiF6]

2�

4. Paragraph for Question Nos. (i) to (iii)

Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules(approximately 6.023 × 1023) are present in a few grams of any chemical compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concepthas implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry.The following example illustrates a typical case, involving chemical / electrochemical reaction, which requiresa clear understanding of the mole concept.

A 4.0 molar aqueous solution of NaCl is prepared and 500 mL of this solution is electrolysed. This leads to theevolution of chlorine gas at one of the electrodes (atomic mass : Na = 23, Hg = 200 ; 1 Faraday = 96500coulombs).

**[At the anode : 2Cl� Cl2 + 2e �

At the cathode : Na+ + e� NaNa + Hg NaHg (sodium amalgam)]

** (These reactions were not present in IIT-JEE paper)

(i) The total number of moles of chlorine gas evolved is : [JEE-2007, 4/162](A) 0.5 (B) 1.0 (C) 2.0 (D) 3.0

(ii) If the cathode is a Hg electrode, the maximum weight (g) of amalgam formed from this solution is :

[JEE-2007, 4/162](A) 200 (B) 225 (C) 400 (D) 446

(iii) The total charge (coulombs) required for complete electrolysis is : [JEE-2007, 4/162](A) 24125 (B) 48250 (C) 96500 (D) 193000

5. Given that the abundances of isotopes 54Fe, 56Fe and 57Fe are 5%, 90% and 5%, respectively, the atomic massof Fe is : [JEE-2009, 3/160](A) 55.85 (B) 55.95 (C) 55.75 (D) 56.05

MOLE CONCEPT # 50

6. A student performs a titration with different burettes and finds titre values of 25.2 mL, 25.25 mL, and 25.0 mL.The number of significant figures in the average titre value is : [JEE 2010, 3/163]

7. Among the following, the number of elements showing only one non-zero oxidation state is : [JEE 2010, 3/163]

O, Cl, F, N, P, Sn, Tl, Na, T i

8. Reaction of Br2 with Na

2CO

3 in aqueous solution gives sodium bromide and sodium bromate with evolution of

CO2 gas. The number of sodium bromide molecules involved in the balanced chemical equation is

[JEE 2011, 4/180]

9. Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity ofthe solution is : [JEE 2011, 3/160](A) 1.78 M (B) 2.00 M (C) 2.05 M (D) 2.22 M

10. 29.2% (w/w) HCl stock solution has a density of 1.25 g mL�1. The molecular weight of HCl is 36.5 g mol�1. Thevolume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is :

[JEE 2012, 4/136]

PART - II : AIEEE PROBLEMS (PREVIOUS YEARS)

1. Number of atoms in 560g of Fe (atomic mass 56g mol�1) is : [AIEEE 2002](1) Twice that of 70g N (2) Half that of 20g H (3) Both (1) and (2) (4) None of these

2. In an organic compound of molar mass 108 g mol�1 C,H and N atoms are present in 9 : 1 : 3.5 by weight.Molecular formula can be : [AIEEE 2002](1) C

6H

8N

2(2) C

7H

10N (3) C

5H

6N

3(4) C

4H

18N

3

3. When KMnO4 acts as an oxidising agent and ultimately forms MnO42�, MnO2, Mn2O3 and Mn2+, then the

number of electrons transferred in each case is : [AIEEE 2002](1) 4, 3, 1, 5 (2) 1, 5, 3, 7 (3) 1, 3, 4, 5 (4) 3, 5, 7, 1

4. Which of the following is a redox reaction? [AIEEE 2002](1) NaCl + KNO3 NaNO3 + KCl

(2) CaC2O4 + 2 HCl CaCl2 + H2C2O4

(3) Mg (OH)2 + 2NH4Cl MgCl2 + 2NH4OH

(4) Zn + 2 AgCN 2 Ag + Zn (CN)2

5. Which of the following concentration factor is affected by change in temperature ? [AIEEE 2002](1) Molarity (2) Molality (3) Mole fraction(4) Weight fraction

6. What volume of hydrogen gas at 273 K and 1 atm pressure will be consumed in obtaining 21.6 gm of elementalboron (atomic mass = 10.8) from the reduction of boron trichloride by hydrogen- [AIEEE 2003](1) 44.8 lit. (2) 22.4 lit. (3) 89.6 lit. (4) 67.2 lit.

7. 6.02 × 1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is -[AIEEE 2004]

(1) 0.001 M (2) 0.01 M (3) 0.02 M (4) 0.1 M

8. If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, themass of one mole of a substance will [AIEEE 2005](1) decrease twice (2) increase two fold(3) remain unchanged (4) be a function of the molecular mass of the substance

9. The oxidation state of Cr in [Cr(NH3)

4Cl

2]+ is : [AIEEE 2005]

(1) + 3 (2) + 2 (3) + 1 (4) 0

10. The oxidation state of chromium in the final product formed by the reaction between KI and acidified potassiumdichromate solution is : [AIEEE 2005](1) + 4 (2) + 6 (3) + 2 (4) + 3

MOLE CONCEPT # 51

11. Two solution of a substance (non electrolyte) are mixed in the following manner. 480 ml of 1.5M first solution+ 250 ml of 1.2M second solution.What is the molarity of the final mixture ? [AIEEE 2005](1) 2.70M (2) 1.344M (3) 1.50M (4) 1.20M

12. Which of the following chemical reactions depicts the oxidizing behaviour of H2SO

4? [AIEEE 2006]

(1) 2HI + H2SO

4 I

2 + SO

2 + 2H

2O (2) Ca(OH)

2 + H

2SO

4 CaSO

4 + 2H

2O

(3) NaCl + H2SO

4 NaHSO

4 + HCl (4) 2PCl

5 + H

2SO

4 2POCl

3 + 2HCl + SO

2Cl

2

13. How many moles of magnesium phosphate, Mg3(PO

4)

2 will contain 0.25 mole of oxygen atoms ?

[AIEEE-2006](1) 0.02 (2) 3.125 × 10�2 (3) 1.25 × 10�2 (4) 2.5 × 10�2

14. Density of a 2.05M solution of acetic acid in water is 1.02 g/ml. The molality of the solution is :[AIEEE-2006]

(1) 1.14 mol kg�1 (2) 3.28 mol kg�1 (3) 2.28 mol kg�1 (4) 0.44 mol kg�1

15. In the reaction [AIEEE-2007, 3/120]2AI

(s) + 6HCl

(aq) 2Al3+

(aq) + 6Cl�

(aq) + 3H

2 (g)

(1) 6L HCl(aq)

is consumed for every 3L H2 produced.

(2) 33.6 L H2(g)

is produced regardless temperature and pressure for every moles that reacts.(3) 67.2 L H

2(g) at STP is produced for every mole of Al that reacts .

(4) 11.2 L H2(g)

at STP is produced for every mole of HCl(aq)

consumed.

16. The density (in g mL�1) of a 3.60 M sulphuric acid solution that is 29% (H2SO

4 molar mass = 98 g mol�1) by

mass will be : [AIEEE-2007, 3/120](1) 1.22 (2) 1.45 (3) 1.64 (4) 1.88

17. A 5.2 molal aqueous solution of methyl alcohol, CH3OH, is supplied. What is the mole fraction of methyl

alcohol in the solution? [AIEEE-2011, 3/120](1) 0.100 (2) 0.190 (3) 0.086 (4) 0.050

18. The molality of a urea solution in which 0.0100 g of urea, [(NH2)2CO] is added to 0.3000 dm3 of water at STPis: [Re. Paper AIEEE-2011, 3/120](1) 5.55 × 10�4 (2) 33.3 m (3) 3.33 × 10�2 m (4) 0.555 m

19. The density of a solution prepared by dissolving 120 g of urea (mol. mass = 60 u) in 1000 g of water is1.15 g/mL. The molarity of this solution is : [AIEEE-2012, 4/120](1) 0.50 M (2) 1.78 M (3) 1.02 M (4) 2.05 M

MOLE CONCEPT # 52

BOARD LEVEL EXERCISE : HINT & SOLUTIONS

1 Relative atomic mass : It is the mass of one atom with respect to a fixed standard. Currently, it is 1/12th ofmass of one carbon-12 atom.

Relative atomic mass (R.A.M) = atom12Coneofmass

121

elementanofatomoneofMass

2. (i) Molecular mass of H2O = 2 + 16 = 18 amu

(ii) Molecular mass of CO2 = 12 + 2 × 16 = 44 amu

(iii) Molecular mass of CH4 = 12 + 4 = 16 amu

3. 1 mole of 12C atoms = 6.022 × 1023 atoms = 12gThus, 6.022 × 1023 atoms of 12C have mass = 12 g

1 atom of 12C will have mass = 2310022.6

12

g = 1.9927 × 10�23 g.

4. Moles of H2 =

A

A

NN

= 1 mass of 1 mole H2 = 2 g.

5. micro = 10�6, deca = 10 , mega = 106,giga = 109, femto = 10�15

6. Molar mass of Na2CO

3 = 2 × 23 + 12 + 3 × 16 = 106 g mol�1

0.50 mol Na2CO

3 means 0.50 × 106 g = 53 g

0.50 M Na2CO

3 means 0.50 mol, i.e. 53 g Na

2CO

3 are present in 1 litre of the solution.

7. H2 and O

2 react according to the equation

2H2(g) + O

2 (g) 2H

2O(g)

Thus, 2 volumes of H2 react with 1 volume of O

2 to produce 2 volumes of water vapour. Hence , 10 volumes of

H2 will react completely with 5 volumes of O

2 to produce 10 volumes of water vapour.

8. Mass % of an element = compoundtheofmassMolarcompoundtheinelementthatofMass

Now, molar mass of Na2 SO

4 = 2 (23.0) + 32.0 + 4 × 16.0 = 142 g mol�1

Mass percent of sodium = 14246

× 100 = 32.39%

Mass percent of sulphur = 14232

× 100 = 22. 54%

Mass percent of oxygen = 14264

× 100 = 45.07%

9. Element Symbol % by mass

Atomic mass

Moles of the element

(Relative no. of moles

Simple molar ratio

Simplest whole

number molar ratio

Iron Fe 69.9 55.85

85.559.69

=1.25 25.125.1

=1 2

Oxygen O 30.1 16.00

00.161.30

= 1.88 25.188.1

= 1.5 3

Empirical formula = Fe2O

3

MOLE CONCEPT # 53

10. 1 mole of CuSO4 contains 1 mole (1g atom) of Cu

Molar, Cu that can be obtained from 159.5 g of CuSO4 = 63.5 g

Cu that can be obtained from 100 g of CuSO4 =

5.1595.63

× 100 g = 69.81 g

11. Refer notes.

12. The total number of digits in a number including the last digit whose value is uncertain is called the number ofsignificant figures.

13. (i) 2 (ii) 3 (iii) 4 (iv) 3 (v) 4 (vi) 5

14. (i) 34.2 (ii) 10.4 (iii) 0.0460 (v) 2810

15. 0.375 M aqueous solution mean that 1000 mL of the solution contain sodium acetate = 0.375 mole

500 mL of the solution should contain sodium acetate = 2375.0

mole

Molar mass of sodium acetate = 82.0245gmol�1

Mass of sodium acetate acquired = 2375.0

mole × 82.0245 g mol�1 = 15.380g

16. Mass percent of 69% means that 100 g of nitric acid solution contain 69 g of nitric acid by mass.Molar mass of nitric acid (HNO

3) = 1 + 14 + 48 = 63 g mol�1

Moles in 69 g HNO3 = 1moleg63

g69 =1.095 mole [1 Mark]

Volume of 100 g nitric acid solution = 1mLg41.1

g100 = 70.92 mL = 0.07092 L [1 Mark]

Conc. of HNO3 in moles per litre = L07092.0

mole095.1 = 15.44 M [1 Mark]

17. Molar mass of sugar (C12

H22

O11

) = 12 × 12 + 22 × 1 + 11 × 16 = 342 g mol�1

No. of moles in 20 g of sugar = 1moleg342

g20

= 0.0585 mole [2 Mark]

Molar concentration = LinsolofVolumesoluteofMoles

= L2

mol0585.0 = 0.0293 mol L�1 = 0.0293 M [1 Mark]

18. (i) 4.8 × 10�3 (ii) 2.34 × 105 (iii) 8.008 × 103 (iv) 5.000 × 102 (v) 6.0012 × 10º

19. (i) 1 g Au = 197

1 mol =

1971

× 6.02 × 1023 atoms ; (ii) 1 g Na = 231

mol = 231

× 6.02 × 1023 atoms

(iii) 1 g Li = 71

mol = 71

× 6.02 × 1023 atoms

(iv) 1g Cl2 =

711

mol = 711

× 6.02 × 1023 molecules =712

× 6.02 × 1023 atoms.

Thus 1 g of Li has the largest number of atoms.

MOLE CONCEPT # 54

20. (i) The least precise term has 3 significant figures (i.e. , in 0.112). Hence the answer should have 3 significantfigures.(ii) Leaving the exact number (5), the second terms has 4 significant figures. Hence the answer should have4 significant figures.(iii) 0.0125 + 0.7864 + 0.0215 = 0.8204. It has 4 significant figures.

21. (i) 1 mole of Ar = 6.022 × 1023 atoms 52 mole of Ar = 52 × 6.022 × 1023 atoms = 3.131 × 1025 atoms(ii) 1 atom of He = 4u of He

4 u of He = 1 atom of He

52u of He = 41

× 52 atoms = 13 atoms.

(iii) 1 mole of He = 4g = 6.022 × 1023 atoms

52 g of He = 4

10022.6 23

× 52 atoms = 7.8286 × 1024 atoms.

22. Element % % / Atomic mass Simple ratio Simple integer ratio X 50 50 / 10 = 5 2 2Y 50 50 / 20 = 2.5 1 1

Hence simplest formula is X2Y

23. Removed mass = 4.222.11

×32 + 23

23

1002.6

1002.6

× 16 = 32 g

mass left = 64 � 32 = 32 g.

24. OHHC 52X = )OH(n)OHHC(n

)OHHC(n

252

52

= 0.040 (Given)

The aim is to find number of moles of ethanol in 1 L of the solution which is nearly = 1L of water (since solutionis dilute).

No. of moles in 1L of water = 1molg18

g1000 = 55.55 moles [1 Mark]

Substituting n (H2O) = 55.55 in eqn (i) , we get

55.55)OHHC(n)OHHC(n

52

52

= 0.040

or 0.96 n (C2H

5OH) = 55.55 × 0.040 or n (C

2H

5OH) = 2.31 mole

Hence, molarity of the solution = 2.31 M [2 Mark]

25. Molar mass of Ar = 35.96755 × 0.00337 + 37.96272 × 0.00063 + 39.96924 × 0.99600

= 39.9544 g mol�1

26. 1 mole of MnO2, i.e., 55 + 32 = 87 g MnO

2 react with 4 moles of HCl , i.e., 4 × 36.5 g = 146 g of HCl.

5 g of MnO2 will react with HCl =

87146

× 5.0 g = 8.40 g

27. (1) The Law of Conservation of Mass : In a chemical change total mass remains conserved. i.e., massbefore reaction is always equal to mass after reaction.

Example : H2 (g) +

21

O2 (g) H

2O (l)

Before reaction 1 mole 21

mole

After the reaction 0 0 1 mole

mass before reaction = mass of 1 mole H2 (g) +

21

mole O2 (g)

= 2 + 16 = 18 gmmass after reaction = mass of 1 mole water = 18 gm

MOLE CONCEPT # 55

(2) Law of Constant or Definite Proportion: All chemical compounds are found to have constant compositionirrespective of their method of preparation or sources.Example : In water (H

2O), Hydrogen and Oxygen combine in 2 : 1 molar ratio, this ratio remains constant

whether it is tap water, river water or sea water or produced by any chemical reaction.(3) The Law Of Multiple Proportion : When one element combines with the other element to form two ormore different compounds, the mass of one element, which combines with a constant mass of the other, beara simple ratio to one another.Example : Carbon and oxygen when combine, can form two oxides viz CO (carbonmonoxide), CO

2

(Carbondioxides)In CO, 12 gm carbon combined with 16 gm of oxygen.In CO

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

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

28. The balanced equation for the combustion of carbon in dioxygen/air isC(s) + O

2 (g) CO

2 (g)

1 mole 1 mole 1 mole(32 g) (44 g)(i) In air,combustion is complete. Therefore, CO

2 produced from the combustion of 1 mole of carbon = 44 g.

(ii) As only 16 g of dioxygen is available, it can combine only with 0.5 mole of carbon i.e. dioxygen is the limitingreactant.Hence, CO

2 produced = 22g.

(iii) Here again, dioxygen is the limiting reactant. 16 g of dioxygen can combine only with 0.5 mole carbon. CO2

produced again is equal to 22.

29. Molar mass of methanol (CH3OH) = 32 g mol�1 =0.032 kg mol�1

Molarity of the given solution = 1

1

molkg032.0

Lkg793.0

= 24.78 mol L�1 [2 Mark]

Applying M1 × V

1= M

2V

2

(Given solution) (solution to be prepared)24.78 × V

1 = 0.25 × 2.5 L or V

1 = 0.02522 L = 25.22 mL [3 Mark]

30. Pressure is the force (i.e. weight) acting per unit areaBut weight = mg

Pressure = Weight per unit area = 2

2

cm

ms8.9g1034

= 2

2

cm

ms8.9g1034 × g1000

kg1 × m1m1

cm100cm100

× 2mskg

N1 × 2Nm1

Pa1 = 1.01332 × 105 Pa

31. (i) 15 ppm means 15 parts in million (106) parts

% by mass = 610

15 × 100 = 15 × 10�4 = 1.5 × 10�3 %

(ii) molar mass of chloroform (CHCl3) = 12 + 1 + 3 × 35.5 = 118.5 g mol�1

100 g of the sample contain chloroform = 1.5 × 10�3 g 1000 g (1kg) of the sample will contain chloroform = 1.5 × 10�2 g

= 5.118

105.1 2

= 1.266 × 10�4 mole

Molality = 1.266 × 10�4 m.

32. Refer notes.

33. (i) According to the given reaction, 1 atom of A reacts with 1 molecule of B2.

200 molecules of B2 will react with 200 atoms of A and 100 atoms of A will be left unreacted. Hence, B is the

limiting reagent while A is the excess reagent.(ii) According to the given reaction, 1 mole of A reacts with 1 mole of B

2

2 mole of A will react with 2 mole of B2. Hence , A is the limiting reactant.

(iii) No limiting reagent.

MOLE CONCEPT # 56

(iv) 2.5 mole of B2 will react with 2.5 mole of A. Hence , B

2 is the limiting reagent.

(v) 2.5 mole of A will react with 2.5 mole of B2. Hence , A is the limiting reagent.

34. (i) 1 mole of N2, i.e. 28 g react with 3 mole of H

2, i.e. 6g of H

2

2000 g of N2 will react with H

2 =

286

× 2000 g = 428.6g.

Thus , N2 is the limiting reagent while H

2 is the excess reagent.

2 mol of N2, i.e. 28 g of N

2 product NH

3 = 2mole = 34 g

2000 g of N2 will produce NH

3 =

2834

× 2000 g = 2428.57 g.

(ii) H2 will remain unreacted.

(iii) Mass left unreacted = 1000 g � 428.6 g = 571.4. g

35. (i) 28.7 pm = 28.7 pm = pm1

m10 12

= 2.87 × 10�111 m

(ii) 15.15 µs = 15.15 × 10�6 s = 1.515 × 10�5 s

(ii) 25365 mg = 25365 mg × mg1000g1

× g1000kg1

= 2.5365 × 10�2 kg

36. Amount of carbon in 3.38 g CO2 =

4412

× 3.38 g = 0.9218 g

Amount of hydrogen in 0.690g H2O =

182

× 0.690g = 0.0767 g

As compound contains only C and H, therefore, total mass of the compound= 0.9218 + 0.0767 g = 0.9985 g

% of C in the compound = 9985.09218.0

× 100 = 92.32

% of H in the compound = 9985.00767.0

× 100 = 7.68

Calculation of Empirical Formula

Element % by mass Atomic mass

Moles of the element

Simplest molar ratio

Simplest whole number molar ratio

C 92.32 12

1232.92

=7.69 1 1

H 7.68 1

168.7

=7.68 1 1

Empirical formula = CH

10.0 L of the gas at STP weighs = 11.6g

22.4 L of the gas at STP will weigh = 0.106.11

× 22.4 = 25.984 g 26g

Molar mass = 26 g mol�1

Empirical formula mass of CH = 12 + 1 = 13

n = mass.F.EmassMolecular

= 1326

= 2.

Since molecular formula = n factor × empirical formula

So, molecular formula = C2H

2.

37. Step 1. To calculate mass of HCl in 25mL of 0.75 M HCl1000 mL of 0.75 M HCl contain HCl= 0.75 mol = 0.75 × 36.5 g = 24.375 g

25 mL of 0.75 HCl will contain HCl = 1000

375.24 × 25 g = 0.6844 g [2 Mark]

MOLE CONCEPT # 57

Step 2. To calculate mass of CaCO3 reacting completely with 0.9125 g of HCl

CaCO3(s) + 2HCl (aq) CaCl

2(aq) + CO

2(g) + H

2O(l)

2 mol of HCl , i.e., 2 × 36.5g = 73 g HCl react completely with CaCO3= 1 mol = 100 g

0.6844 g HCl will react completely with CaCO3 =

73100

× 0.6844 g = 0.938 g.CaCO3

[3 Mark]

38. (a) Fixing the mass of dinitrogen as 28 g, masses of dioxygen combined will be 32, 64, 32 and 80 g in the givenfour oxides. These are in the ratio 1 : 2 : 1 : 5 which is a simple whole number ratio. Hence the given data obeythe law of multiple proportions.

(b) (i) 1 km = 1 km × km1m1000

× m1cm100

× cm1mm10

= 106mm

1 km = 1 km × km1km1000

× m10

pm112 = 1015 pm

(ii) 1 mg = 1mg × mg1000g1

× g1000kg1

= 10�6 kg

1 mg = 1 mg × mg1000g1

× g10

ng19 = 106 ng

(iii) 1 mL = 1 mL × mL1000L1

= 10�3 L

1 mL = 1 cm3 = 1cm3 × cm10cm10cm10dm1dm1dm1

= 10�3 dm3

EXERCISE - 1

PART - I

A-1. (a) H = 4NA , S = 2N

A , O = 8N

A atoms (b) H = 4 atoms, S = 2 atoms, O = 8 atoms.

(c) H = 10NA , S = 10N

A , O = 40 N

A atoms (d) H = 6 atoms, S = 6 atoms, O = 18 atoms.

A-2. 20 NA

A-3. 11 NA

B-1. In compound A21.23 parts of oxygen combine with 78.77 parts of tin

1 part of oxygen combine with 23.2177.78

= 3.7 parts of tin.

In compound B11.88 parts of oxygen combine with 88.12 parts of tin

1 part of oxygen combine with 88.1112.88

= 7.4 parts of tin.

Therefore, ratio of mass of tin with fixed mass of oxygen in compound A and B is, 3.4 : 7.4 i.e. 1 : 2.

B-2. 26.88 g B-3. 50 g/mol C-1. CH4

C-2. % CO2 =

2122

× 100 = 40%. C-3. 1200 g mol�1 D-1. 2.16 g

D-2. 42 g D-3. (i) 0.64 g, (ii) 1.64 g, (iii) 0.993 g.

E-1. (i) 1/6 mole (ii) 5/12 mole E-2. 76 kg of CS2

E-3. (a) Br2 + CI

2 2BrCI

1 mol 1mol 2mol0.025 0.025 0.050 mol

MOLE CONCEPT # 58

(b) Theoretical yield of BrCI = 0.050 mol

Actual yield is = 100

8005.0 = 0.04 mole

Br2 left unreacted = 0.025 � 0.02 = 0.005 mol.

F-1. m = 1.4 g F-2. 340 kg F-3.3

10moles. F-4. 66.4 %.

G-1. (a) +3 (b) +5 (c) +6 (d) +2 (e) +8/3 or (+2 and +3)(f) +3 (g) +1 (h) +2 (i) 200/93 = 2.15

G-2. (a) 4

)7(OMnK

+ )1(�

ClK + H2SO

4 4

)2(SOMn

+ K2SO

4 + H

2O +

)0(

2Cl .

4

)7(OMnK

(oxidant ) 4

)2(SOMn

(reduction half).

)1(�ClK (reductant)

)0(

2Cl (oxidant half).

(b) 2

)2(ClFe

+ )1(�

22OH + HCl 3

)3(ClFe

+ )2(�

2 OH (oxidation half)

2

)2(ClFe

(reductant) 3

)3(ClFe

(oxidation half).

)1(�

22OH (oxidant) H2O2� (reduction half).

(c))0(

Cu + )5(

3NOH

(dil) 2

Cu (NO3)

2 + H

2O +

2NO .

)0(Cu (reductant)

2Cu (NO

3)

2 (oxidation half).

3

5NOH

(oxidant) 2

NO

(reduction half).

(d) Na2 3

3OAsH

+ 3

5OBrK

+ HCl NaCl + 1�

BrK + 4

5

3 OAsH

Na2 3

3OAsH

(reductant) 4

5

3 OAsH

(oxidation half).

3

5OBrK

(oxidant) 1�

BrK .

(e)0

2I + Na2 3

2

2OS

Na2 6

5.2

4OS

+ 1�

NaI .

0

2I (oxidant) 1�

NaI (reductant half).

Na2 3

2

2OS

(reductant) Na2 6

5.2

4OS

(oxidant half).

H-1. (a) S4O62�(aq) + 6 Al(s) + 20 H+ 4H2S(aq) + 6Al3+(aq) + 6H2O

(b) 6S2O32�(aq) + Cr2O7

2�(aq) + 14 H+ 3S4O62�(aq) + 2Cr3+(aq) + 7H2O

(c) 14ClO3� (aq) + 3As2S3(s) + 18H2O 14Cl¯(aq) + 6H2AsO4¯ (aq) + 9HSO4¯(aq) + 15H+

(d) 7O3¯(aq) + 6Re(s) + 3H2O 6ReO4¯(aq) + 7¯(aq) + 6H+

(e) 26H+ + 30HSO4¯(aq) + As4(s) + 10 Pb3O4(s) 30 PbSO4(s) + 4H2AsO4¯ (aq) + 24H2O

(f) 3HNO2(aq) NO3¯ + 2NO(g) + H2O + H+

H-2. (a) 3C4H4O62�(aq) + 5ClO3¯(aq) + 18OH¯ 12 CO3

2�(aq) + 5 Cl¯(aq) + 15H2O

(b) 11Al(s) + 3BiONO3(s) + 21H2O + 11OH¯ 3Bi(s) + 3NH3(aq) + 11Al(OH)4¯(aq)

(c) 4H2O2(aq) + Cl2O7(aq) + 2OH¯ 2ClO2¯(aq) + 4O2(g) + 5H2O

(d) Tl2O3(s) + 4NH2OH(aq) 2TlOH(s) + 2N2(g) + 5H2O

(e) [Cu(NH3)4]2+ (aq) + S2O4

2�(aq) + 4OH� 2SO32�(aq) + Cu(s) + 4NH3(aq) + 2H2O

(f) 3Mn(OH)2(s) + 2MnO4¯(aq) 5MnO2(s) + 2H2O + 2OH¯

I-1. 5.6 gm I-2. 12.934 g

MOLE CONCEPT # 59

I-3. (i) 2.17 m , (ii) 6.25 M, (iii) 0.0376, (iv) 0.0826, (v) 8% (vi) 16.67%.

(vii) 25%, (viii) 36.25%, (ix) 72.5%, (x) 14.2 m.

J-1. 8 M J-2. 700 ml. J-3. 2.33 L

PART - II

A-1. (A) A-2. (C) A-3. (D) A-4. (A) A-5. (B) A-6. (B) B-1. (C)

B-2. (B) B-3 (A) B-4. (B) B-5. (A) C-1. (B) C-2. (B) C-3.*_ (ABC)

C-4 (B) C-5. (D) D-1 (C) D-2 (C) D-3 (C) D-4 (C) D-5. (B)

D-6. (B) E-1. (A) E-2. (C) E-3. (A) E-4. (C) E-5. (A) E-6. (C)

E-7. (A) F-1. (A) F-2. (B) F-3. (C) F-4 (B) F-5 (A) F-6.* (AC)

G-1. (D) G-2. (A) G-3. (C) G-4. (B) G-5. (A) G-6. (C) G-7. (B)

G-8.* (CD) H-1. (C) H-2. (A) H-3. (C) H-4. (B) H-5. (D) H-6.* (ABC)

-1. (C) -2. (B) -3. (B) -4. (A) -5. (C) -6. (B) -7. (C)

J-1. (A) J-2. (C) J-3. (B) J-4. (D) J-5. (D)

PART - III

1. (B) 2. (A) 3. (A) 4. (A) 5. (A) 6. (A) 7. (B)

8. (A) 9. (A) 10. (A)

EXERCISE - 2

PART - I

1. 30.77% & 18.7% 2. 6.02 × 1013 g/ml 3. 7.09 ×107 4. 33.33 %

5. 4 6. LR = K4Fe(CN)

6, 6 moles of CO, 2 moles of (NH

4)

2SO

4

7. (a) A3B4 = 2 & A2 = 1 ; (b) A2B4 = 0.5 & B2 = 1 ; (c) A2B4 = 0.5 & A3B4 = 0.5

8. 1100 9. 1.14 g 10. Approx 8 hours

11. 4 12. 62% 13. 509.7 kg

14. 320 g Na2SO4. 15. 8 16. 0.2 moles OH

17. 29.2 ml 18. 70 19. 67 20. 0.2 cm2

PART - II

1. (C) 2. (C) 3. (A) 4. (C) 5. (A) 6. (B) 7. (D)

8. (D) 9. (A) 10. (A) 11. (C) 12. (A) 13. (A) 14. (B)

15. (D) 16. (B) 17. (B) 18. (C) 19. (D) 20. (A) 21. (A)

22. (A) . (D) . (A) . (D) . (A) 27. (A) 28. (A)

29. (C) 30. (CD) 31. (ACD) 32. (BCD) 33. (BD) 34. (BC) 35. (ABD)

36. (ABD)

MOLE CONCEPT # 60

PART - III

1. (A) (q, s) ; (B) (p, s) ; (C) (p, q, r) ; (D) (q, r)

2. (A) (p, q, r, s) ; (B) (p, s) ; (C) (q, r) ; (D) - (q)

3. (A) p ; (B) r,s ; (C) r ; (D) r,q

4. (A) p, s ; (B) s ; (C) p, q ; (D) r

PART - IV

1. (A) 2. (B) 3. (B) 4. (C) 5. (B) 6. (B) 7. (A)

8. (B) 9. (B) 10. (C) 11. (C) 12. (B)

EXERCISE - 3

PART - I

1. (D) 2. (A) 3. (B) 4. (i) (B) (ii) (D) (iii) (D)

5. (B) 6. 3 7. 2 8. 5 9. (C) 10. 8 mL.

PART - II

1. (3) 2. (1) 3. (3) 4. (4) 5. (1) 6. (4) 7. (2)

8. (3) 9. (1) 10. (4) 11. (2) 12. (1) 13. (2) 14. (3)

15. (4) 16. (1) 17. (3) 18. (1) 19. (4)