Chapterwise Topicwise

Solved Papers
2021-1979

IITJEE
JEE Main & Advanced

Chemistry
Ranjeet Shahi

Arihant Prakashan (Series), Meerut


Arihant Prakashan (Series), Meerut
All Rights Reserved

© Author

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CONTENTS
1-22

19. Extraction of Metals

282-293

2. Atomic Structure

23-40

20. Qualitative Analysis

294-306

3. Periodic Classification and
Periodic Properties

21. Organic Chemistry Basics

307-331


41-47

22. Hydrocarbons

332-349

4. Chemical Bonding

48-66

23. Alkyl Halides

350-363

5. States of Matter

67-83

24. Alcohols and Ethers

364-377

25. Aldehydes and Ketones

378-396

26. Carboxylic Acids and
their Derivatives


397-412

1. Some Basic Concepts of Chemistry

6. Chemical and Ionic Equilibrium

84-108

7. Thermodynamics and
Thermochemistry

109-129

8. Solid State

130-139

9. Solutions and Colligative Properties

140-155

27. Aliphatic Compounds
Containing Nitrogen

413-422

10. Electrochemistry

156-177


28. Benzene and Alkyl Benzene

423-440

11. Chemical Kinetics

178-195

12. Nuclear Chemistry

196-199

29. Aromatic Compounds
Containing Nitrogen

441-457

13. Surface Chemistry

200-206

30. Aryl Halides and Phenols

458-470

14. s-Block Elements

207-217

31. Aromatic Aldehydes, Ketones

and Acids

471-484

15. p-Block Elements-I

218-227

16. p-Block Elements-II

228-248

32. Biomolecules and Chemistry
in Everyday Life

485-501

33. Environmental Chemistry

502-504

17. Transition and
Inner-Transition Elements

249-258

18. Coordination Compounds

259-281


JEE Advanced Solved Paper 2021

1-16


SYLLABUS
JEE MAIN

Section A : PHYSICAL CHEMISTRY
UNIT I Some Basic Concepts in Chemistry
Matter and its nature, Dalton's atomic theory; Concept of atom,
molecule, element and compound; Physical quantities and their
measurements in Chemistry, precision and accuracy, significant
figures, S.I. Units, dimensional analysis; Laws of chemical
combination; Atomic and molecular masses, mole concept, molar
mass, percentage composition, empirical and molecular formulae;
Chemical equations and stoichiometry.
UNIT II States of Matter
Classification of matter into solid, liquid and gaseous states.
Gaseous State Measurable properties of gases; Gas laws - Boyle's law,
Charle's law, Graham's law of diffusion, Avogadro's law, Dalton's law of
partial pressure; Concept of Absolute scale of temperature; Ideal gas
equation, Kinetic theory of gases (only postulates); Concept of average,
root mean square and most probable velocities; Real gases, deviation
from Ideal behaviour, compressibility factor, van der Waals' equation,
liquefaction of gases, critical constants.
Liquid State Properties of liquids - vapour pressure, viscosity and
surface tension and effect of temperature on them (qualitative
treatment only).
Solid State Classification of solids: molecular, ionic, covalent and

metallic solids, amorphous and crystalline solids (elementary idea);
Bragg's Law and its applications, Unit cell and lattices, packing in solids
(fcc, bcc and hcp lattices), voids, calculations involving unit cell
parameters, imperfection in solids; electrical, magnetic and dielectric
properties.
UNIT III Atomic Structure
Discovery of sub-atomic particles (electron, proton and neutron);
Thomson and Rutherford atomic models and their limitations; Nature of
electromagnetic radiation, photoelectric effect; spectrum of hydrogen
atom, Bohr model of hydrogen atom - its postulates, derivation of the
relations for energy of the electron and radii of the different orbits,
limitations of Bohr's model; dual nature of matter, de-Broglie's
relationship, Heisenberg uncertainty principle.
Elementary ideas of quantum mechanics, quantum mechanical model
of atom, its important features,
ψ and ψ2, concept of atomic orbitals as one electron wave functions;
Variation of ψ and ψ2 with r for 1s and 2s orbitals; various quantum
numbers (principal, angular momentum and magnetic quantum
numbers) and their significance; shapes of s, p and d - orbitals, electron
spin and spin quantum number; rules for filling electrons in orbitals –
aufbau principle, Pauli's exclusion principle and Hund's rule,
electronic configuration of elements, extra stability of half-filled and
completely filled orbitals.
UNIT IV Chemical Bonding and Molecular Structure
Kossel Lewis approach to chemical bond formation, concept of ionic
and covalent bonds.
Ionic Bonding Formation of ionic bonds, factors affecting the formation
of ionic bonds; calculation of lattice enthalpy.
Covalent Bonding Concept of electronegativity, Fajan's rule, dipole
moment; Valence Shell Electron Pair Repulsion (VSEPR) theory and

shapes of simple molecules.

Quantum mechanical approach to covalent bonding Valence bond
theory - Its important features, concept of hybridization involving s, p
and d orbitals; Resonance.
Molecular Orbital Theory Its important features, LCAOs, types of
molecular orbitals (bonding, antibonding), sigma and pi-bonds,
molecular orbital electronic configurations of homonuclear diatomic
molecules, concept of bond order, bond length and bond energy.
Elementary idea of metallic bonding. Hydrogen bonding and its
applications.
UNIT V Chemical Thermodynamics
Fundamentals of thermodynamics System and surroundings, extensive
and intensive properties, state functions, types of processes.
First law of thermodynamics Concept of work, heat internal energy and
enthalpy, heat capacity, molar heat capacity, Hess's law of constant heat
summation; Enthalpies of bond dissociation, combustion, formation,
atomization, sublimation, phase transition, hydration, ionization and
solution.
Second law of thermodynamics Spontaneity of processes; ΔS of the
universe and ΔG of the system as criteria for spontaneity, ΔGo
(Standard Gibb's energy change) and equilibrium constant.
UNIT VI Solutions
Different methods for expressing concentration of solution - molality,
molarity, mole fraction, percentage (by volume and mass both), vapour
pressure of solutions and Raoult's Law - Ideal and non-ideal solutions,
vapour pressure - composition plots for ideal and non-ideal solutions.
Colligative properties of dilute solutions - relative lowering of vapour
pressure, depression of freezing point, elevation of boiling point and
osmotic pressure; Determination of molecular mass using colligative

properties; Abnormal value of molar mass, van't Hoff factor and its
significance.
UNIT VII Equilibrium
Meaning of equilibrium, concept of dynamic equilibrium.
Equilibria involving physical processes Solid -liquid, liquid - gas and
solid - gas equilibria, Henry's law, general characteristics of equilibrium
involving physical processes.
Equilibria involving chemical processes Law of chemical equilibrium,
equilibrium constants (K and K) and their significance, significance of
ΔG and ΔGo in chemical equilibria, factors affecting equilibrium
concentration, pressure, temperature, effect of catalyst; Le -Chatelier's
principle.
Ionic equilibrium Weak and strong electrolytes, ionization of
electrolytes, various concepts of acids and bases (Arrhenius, Bronsted Lowry and Lewis) and their ionization, acid-base equilibria (including
multistage ionization) and ionization constants, ionization of water, pH
scale, common ion effect, hydrolysis of salts and pH of their solutions,
solubility of sparingly soluble salts and solubility products, buffer
solutions.
UNIT VIII Redox Reactions and Electrochemistry
Electronic concepts of oxidation and reduction, redox reactions,
oxidation number, rules for assigning oxidation number, balancing of
redox reactions.


Eectrolytic and metallic conduction, conductance in electrolytic
solutions, specific and molar conductivities and their variation with
concentration: Kohlrausch's law and its applications.

temperature on rate of reactions - Arrhenius theory, activation energy
and its calculation, collision theory of bimolecular gaseous reactions

(no derivation).

Electrochemical cells - Electrolytic and Galvanic cells, different types of
electrodes, electrode potentials including standard electrode
potential, half - cell and cell reactions, emf of a Galvanic cell and its
measurement; Nernst equation and its applications; Relationship
between cell potential and Gibbs' energy change; Dry cell and lead
accumulator; Fuel cells; Corrosion and its prevention.

UNIT X Surface Chemistry
Adsorption - Physisorption and chemisorption and their
characteristics, factors affecting adsorption of gases on solidsFreundlich and Langmuir adsorption isotherms, adsorption from
solutions.
Catalysis Homogeneous and heterogeneous, activity and selectivity of
solid catalysts, enzyme catalysis and its mechanism.
Colloidal state distinction among true solutions, colloids and
suspensions, classification of colloids - lyophilic, lyophobic; multi
molecular, macromole-cular and associated colloids (micelles),
preparation and properties of colloids Tyndall effect, Brownian
movement, electrophoresis, dialysis, coagulation and
flocculation; Emulsions and their characteristics.

UNIT IX Chemical Kinetics
Rate of a chemical reaction, factors affecting the rate of reactions
concentration, temperature, pressure and catalyst; elementary and
complex reactions, order and molecularity of reactions, rate law, rate
constant and its units, differential and integral forms of zero and first
order reactions, their characteristics and half - lives, effect of

Section B : INORGANIC CHEMISTRY

UNIT XI Classification of Elements and
Periodicity in Properties
Periodic Law and Present Form of the Periodic Table, s, p, d and f Block
Elements, Periodic Trends in Properties of Elementsatomic and Ionic Radii,
Ionization Enthalpy, Electron Gain Enthalpy, Valence, Oxidation States and
Chemical Reactivity.
UNIT XII General Principles and Processes of Isolation of Metals
Modes of occurrence of elements in nature, minerals, ores; steps involved
in the extraction of metals - concentration, reduction (chemical and
electrolytic methods) and refining with special reference to the extraction
of Al, Cu, Zn and Fe; Thermodynamic and electrochemical principles
involved in the extraction of metals.
UNIT XIII Hydrogen
Position of hydrogen in periodic table, isotopes, preparation, properties
and uses of hydrogen; physical and chemical properties of water and
heavy water; Structure, preparation, reactions and uses of hydrogen
peroxide; Classification of hydrides ionic, covalent and interstitial;
Hydrogen as a fuel.
UNIT XIV s - Block Elements (Alkali and Alkaline Earth Metals)
Group 1 and 2 Elements
General introduction, electronic configuration and general trends in
physical and chemical properties of elements, anomalous properties of the
first element of each group, diagonal relationships.
Preparation and properties of some important compounds - sodium
carbonate, sodium chloride, sodium hydroxide and sodium hydrogen
carbonate; Industrial uses of lime, limestone, Plaster of Paris and cement;
Biological significance of Na, K, Mg and Ca.
UNIT XV p - Block Elements
Group 13 to Group 18 Elements
General Introduction Electronic configuration and general trends in

physical and chemical properties of elements across the periods and
down the groups; unique behaviour of the first element in each
group.Group wise study of the p – block elements
Group 13 Preparation, properties and uses of boron and aluminium;
structure, properties and uses of borax, boric acid, diborane, boron
trifluoride, aluminium chloride and alums.
Group 14 Tendency for catenation; Structure, properties and uses of
allotropes and oxides of carbon, silicon tetrachloride, silicates, zeolites and
silicones.
Group 15 Properties and uses of nitrogen and phosphorus; Allotrophic
forms of phosphorus; Preparation, properties, structure and uses of
ammonia nitric acid, phosphine and phosphorus halides,(PCl3, PCl5);
Structures of oxides and oxoacids of nitrogen and phosphorus.

Group 16 Preparation, properties, structures and uses of dioxygen
and ozone; Allotropic forms of sulphur; Preparation, properties,
structures and uses of sulphur dioxide, sulphuric acid (including
its industrial preparation); Structures of oxoacids of sulphur.
Group 17 Preparation, properties and uses of chlorine and
hydrochloric acid; Trends in the acidic nature of hydrogen halides;
Structures of Interhalogen compounds and oxides and oxoacids of
halogens.
Group 18 Occurrence and uses of noble gases; Structures of
fluorides and oxides of xenon.
UNIT XVI d–and f–Block Elements
Transition Elements General introduction, electronic
configuration, occurrence and characteristics, general trends in
properties of the first row transition elements - physical properties,
ionization enthalpy, oxidation states, atomic radii, colour, catalytic
behaviour, magnetic properties, complex formation, interstitial

compounds, alloy formation; Preparation, properties and uses of
K2 Cr2 O7 and KMnO4.
Inner Transition Elements
Lanthanoids - Electronic configuration, oxidation states, chemical
reactivity and lanthanoid contraction. Actinoids - Electronic
configuration and oxidation states.
UNIT XVII Coordination Compounds
Introduction to coordination compounds, Werner's theory; ligands,
coordination number, denticity, chelation; IUPAC nomenclature of
mononuclear coordination compounds, isomerism; Bonding
Valence bond approach and basic ideas of Crystal field theory,
colour and magnetic properties; importance of coordination
compounds (in qualitative analysis, extraction of metals and in
biological systems).
UNIT XVIII Environmental Chemistry
Environmental pollution Atmospheric, water and soil.
Atmospheric pollution - Tropospheric and stratospheric.
Tropospheric pollutants Gaseous pollutants Oxides of carbon,
nitrogen and sulphur, hydrocarbons; their sources, harmful effects
and prevention; Green house effect and Global warming; Acid rain;
Particulate pollutants Smoke, dust, smog, fumes, mist; their
sources, harmful effects and prevention.
Stratospheric pollution Formation and breakdown of ozone,
depletion of ozone layer - its mechanism and effects.
Water pollution Major pollutants such as, pathogens, organic
wastes and chemical pollutants their harmful effects and
prevention.
Soil pollution Major pollutants such as: Pesticides (insecticides,
herbicides and fungicides), their harmful effects and prevention.
Strategies to control environmental pollution.



Section C : ORGANIC CHEMISTRY
UNIT XIX Purification & Characterisation of Organic Compounds
Purification Crystallisation, sublimation, distillation, differential
extraction and chromatography principles and their applications.
Qualitative analysis Detection of nitrogen, sulphur, phosphorus and
halogens.
Quantitative analysis (basic principles only) Estimation of carbon,
hydrogen, nitrogen, halogens, sulphur, phosphorus.
Calculations of empirical formulae and molecular formulae;
Numerical problems in organic quantitative analysis.
UNIT XX Some Basic Principles of Organic Chemistry
Tetravalency of carbon; Shapes of simple molecules hybridization
(s and p); Classification of organic compounds based on functional
groups: —C=C—,—C=C— and those containing halogens, oxygen,
nitrogen and sulphur, Homologous series; Isomerism - structural and
stereoisomerism.
Nomenclature (Trivial and IUPAC)
Covalent bond fission Homolytic and heterolytic free radicals,
carbocations and carbanions; stability of carbocations and free radicals,
electrophiles and nucleophiles.
Electronic displacement in a covalent bond Inductive effect,
electromeric effect, resonance and hyperconjugation.
Common types of organic reactions Substitution, addition,
elimination and rearrangement.
UNIT XXI Hydrocarbons
Classification, isomerism, IUPAC nomenclature, general methods of
preparation, properties and reactions.
Alkanes Conformations: Sawhorse and Newman projections (of

ethane); Mechanism of halogenation of alkanes.
Alkenes Geometrical isomerism; Mechanism of electrophilic addition:
addition of hydrogen, halogens, water, hydrogen halides
(Markownikoff's and peroxide effect); Ozonolysis, oxidation, and
polymerization.
Alkenes acidic character; addition of hydrogen, halogens, water and
hydrogen halides; polymerization.
Aromatic hydrocarbons Nomenclature, benzene structure and
aromaticity; Mechanism of electrophilic substitution: halogenation,
nitration, Friedel – Craft's alkylation and acylation, directive influence of
functional group in mono-substituted benzene.

UNIT XXIV Organic Compounds Containing Nitrogen
General methods of preparation, properties, reactions and uses.
Amines Nomenclature, classification, structure basic character and
identification of primary, secondary and tertiary amines and their basic
character.
Diazonium Salts Importance in synthetic organic chemistry.
UNIT XXV Polymers
General introduction and classification of polymers, general methods
of polymerization-addition and condensation, copolymerization;
Natural and synthetic rubber and vulcanization; some important
polymers with emphasis on their monomers and uses - polythene,
nylon, polyester and bakelite.
UNIT XXVI Biomolecules
General introduction and importance of biomolecules.
Carbohydrates Classification aldoses and ketoses; monosaccharides
(glucose and fructose), constituent monosaccharides of
oligosacchorides (sucrose, lactose, maltose) and polysaccharides
(starch, cellulose, glycogen).

Proteins Elementary Idea of α-amino acids, peptide bond, .
polypeptides; proteins: primary, secondary, tertiary and quaternary
structure (qualitative idea only), denaturation of proteins, enzymes.
Vitamins Classification and functions.
Nucleic Acids Chemical constitution of DNA and RNA. Biological
functions of Nucleic acids.
UNIT XXVII Chemistry in Everyday Life
Chemicals in medicines Analgesics, tranquilizers, antiseptics,
disinfectants, antimicrobials, antifertility drugs, antibiotics, antacids,
antihistamins - their meaning and common examples.
Chemicals in food Preservatives, artificial sweetening agents - common
examples.
Cleansing agents Soaps and detergents, cleansing action.
Unit XXVIII Principles Related to
Practical Chemistry
—

Detection of extra elements (N, S, halogens) in organic
compounds; Detection of the following functional groups:
hydroxyl (alcoholic and phenolic), carbonyl (aldehyde and ketone),
carboxyl and amino groups in organic compounds.

UNIT XXII Organic Compounds Containing Halogens
General methods of preparation, properties and reactions; Nature of
C—X bond; Mechanisms of substitution reactions.

—

Inorganic compounds Mohr's salt, potash alum.


Uses/environmental effects of chloroform, iodoform, freons and DDT.

—

UNIT XXIII Organic Compounds Containing Oxygen
General methods of preparation, properties, reactions and uses.
Alcohols, Phenols and Ethers

Organic compounds Acetanilide,
p-nitroacetan ilide, aniline yellow, iodoform.

—

Chemistry involved in the titrimetric excercises - Acids bases and
the use of indicators, oxali acid vs KMnO4, Mohr's salt vs KMnO4.

Alcohols Identification of primary, secondary and tertiary alcohols;
mechanism of dehydration.

Chemistry involved in the preparation of the following

—

Chemical principles involved in the qualitative salt analysis

—

Cations — Pb2+ , Cu2+, Al3+, Fe3+, Zn2+, Ni2+, Ca2+, Ba2+ , Mg2+ NH4+.
Anions – CO32-, S2-, SO42-, NO2, NO3, Cl -, Br-, I- (Insoluble salts
excluded).


Phenols Acidic nature, electrophilic substitution reactions:
halogenation, nitration and sulphonation, Reimer - Tiemann reaction.
Ethers: Structure
Aldehyde and Ketones Nature of carbonyl group;
Nucleophilic addition to >C=O group, relative reactivities of aldehydes
and ketones; Important reactions such as - Nucleophilic addition
reactions (addition of HCN, NH3 and its derivatives), Grignard reagent;
oxidation; reduction (Wolff Kishner and Clemmensen); acidity of α hydrogen, aldol condensation, Cannizzaro reaction, Haloform reaction;
Chemical tests to distinguish between aldehydes and Ketones.
Carboxylic Acids Acidic strength & factors affecting it.

—

Chemical principles involved in the following experiments
1. Enthalpy of solution of CuSO4
2. Enthalpy of neutralization of strong acid and strong base.
3. Preparation of lyophilic and lyophobic sols.
4. Kinetic study of reaction of iodide ion with hydrogen peroxide
at room temperature.


JEE ADVANCED
PHYSICAL CHEMISTRY
General Topics Concept of atoms and molecules,
Dalton's atomic theory, Mole concept, Chemical formulae,
Balanced chemical equations, Calculations (based on
mole concept) involving common oxidation-reduction,
neutralisation, and displacement reactions, Concentration
in terms of mole fraction, molarity, molality and normality.

Gaseous and Liquid States Absolute scale of
temperature, ideal gas equation, Deviation from ideality,
van der Waals' equation, Kinetic theory of gases, average,
root mean square and most probable velocities and their
relation with temperature, Law of partial pressures,
Vapour pressure, Diffusion of gases.
Atomic Structure and Chemical Bonding Bohr model,
spectrum of hydrogen atom, quantum numbers,
Wave-particle duality, de-Broglie hypothesis, Uncertainty
principle, Qualitative quantum mechanical picture of
hydrogen atom, shapes of s, p and d orbitals, Electronic
configurations of elements (up to atomic number 36),
Aufbau principle, Pauli's exclusion principle and Hund's
rule, Orbital overlap and covalent bond; Hybridisation
involving s, p and d orbitals only, Orbital energy diagrams
for homonuclear diatomic species, Hydrogen bond,
Polarity in molecules, dipole moment (qualitative aspects
only), VSEPR model and shapes of molecules (linear,
angular, triangular, square planar, pyramidal, square
pyramidal, trigonal bipyramidal, tetrahedral and
octahedral).
Energetics First law of thermodynamics, Internal energy,
work and heat, pressure-volume work, Enthalpy, Hess's
law, Heat of reaction, fusion and vaporization, Second law
of thermodynamics, Entropy, Free energy, Criterion of
spontaneity.
Chemical Equilibrium Law of mass action, Equilibrium
constant, Le-Chatelier's principle (effect of concentration,
temperature and pressure), Significance of DG and DGo in
chemical equilibrium, Solubility product, common ion

effect, pH and buffer solutions, Acids and bases (Bronsted
and Lewis concepts), Hydrolysis of salts.
Electrochemistry Electrochemical cells and cell reactions,
Standard electrode potentials, Nernst equation and its
relation to DG, Electrochemical series, emf of galvanic
cells, Faraday's laws of electrolysis, Electrolytic
conductance, specific, equivalent and molar conductivity,
Kohlrausch's law, Concentration cells.

Chemical Kinetics Rates of chemical reactions, Order of
reactions, Rate constant, First order reactions,
Temperature dependence of rate constant (Arrhenius
equation).
Solid State Classification of solids, crystalline state, seven
crystal systems (cell parameters a, b, c), close packed
structure of solids (cubic), packing in fcc, bcc and hcp
lattices, Nearest neighbours, ionic radii, simple ionic
compounds, point defects.
Solutions Raoult's law, Molecular weight determination
from lowering of vapour pressure, elevation of boiling
point and depression of freezing point.
Surface Chemistry Elementary concepts of adsorption
(excluding adsorption isotherms), Colloids, types,
methods of preparation and general properties,
Elementary ideas of emulsions, surfactants and micelles
(only definitions and examples).
Nuclear Chemistry Radioactivity, isotopes and isobars,
Properties of rays, Kinetics of radioactive decay (decay
series excluded), carbon dating, Stability of nuclei with
respect to proton-neutron ratio, Brief discussion on fission

and fusion reactions.

INORGANIC CHEMISTRY
Isolation/Preparation and Properties of the following
Non-metals Boron, silicon, nitrogen, phosphorus, oxygen,
sulphur and halogens, Properties of allotropes of carbon
(only diamond and graphite), phosphorus and sulphur.
Preparation and Properties of the following
Compounds Oxides, peroxides, hydroxides, carbonates,
bicarbonates, chlorides and sulphates of sodium,
potassium, magnesium and calcium, Boron, diborane,
boric acid and borax, Aluminium, alumina, aluminium
chloride and alums, Carbon, oxides and oxyacid (carbonic
acid), Silicon, silicones, silicates and silicon carbide,
Nitrogen, oxides, oxyacids and ammonia, Phosphorus,
oxides, oxyacids (phosphorus acid, phosphoric acid) and
phosphine, Oxygen, ozone and hydrogen peroxide,
Sulphur, hydrogen sulphide, oxides, sulphurous acid,
sulphuric acid and sodium thiosulphate, Halogens,
hydrohalic acids, oxides and oxyacids of chlorine,
bleaching powder, Xenon fluorides.
Transition Elements (3d series) Definition, general
characteristics, oxidation states and their stabilities,
colour (excluding the details of electronic transitions) and


calculation of spin-only magnetic moment; Coordination
compounds: nomenclature of mononuclear coordination
compounds, cis-trans and ionisation isomerisms,
hybridization and geometries of mononuclear

coordination compounds (linear, tetrahedral, square
planar and octahedral).
Preparation and Properties of the following
Compounds Oxides and chlorides of tin and lead, Oxides,
chlorides and sulphates of Fe2+, Cu2+ and Zn2+, Potassium
permanganate, potassium dichromate, silver oxide, silver
nitrate, silver thiosulphate.
Ores and Minerals Commonly occurring ores and
minerals of iron, copper, tin, lead, magnesium, aluminium,
zinc and silver.
Extractive Metallurgy Chemical principles and reactions
only (industrial details excluded), Carbon reduction
method (iron and tin), Self reduction method (copper and
lead), Electrolytic reduction method (magnesium and
aluminium), Cyanide process (silver and gold).
Principles of Qualitative Analysis Groups I to V (only
Ag+, Hg2+, Cu2+, Pb2+, Bi3+, Fe3+, Cr3+, Al3+, Ca2+, Ba2+, Zn2+, Mn2+
and Mg2+), Nitrate, halides (excluding fluoride), sulphate
and sulphide.

ORGANIC CHEMISTRY
Concepts Hybridisation of carbon, Sigma and pi-bonds,
Shapes of simple organic molecules, Structural and
geometrical isomerism, Optical isomerism of compounds
containing up to two asymmetric centres, (R,S and E,Z
nomenclature excluded), IUPAC nomenclature of simple
organic compounds (only hydrocarbons, mono-functional
and bi-functional compounds), Conformations of ethane
and butane (Newman projections), Resonance and
hyperconjugation, Keto-enol tautomerism, Determination

of empirical and molecular formulae of simple
compounds (only combustion method), Hydrogen bonds,
definition and their effects on physical properties of
alcohols and carboxylic acids, Inductive and resonance
effects on acidity and basicity of organic acids and bases,
Polarity and inductive effects in alkyl halides, Reactive
intermediates produced during homolytic and heterolytic
bond cleavage, Formation, structure and stability of
carbocations, carbanions and free radicals.
Preparation, Properties and Reactions of Alkanes
Homologous series, physical properties of alkanes
(melting points, boiling points and density), Combustion
and halogenation of alkanes, Preparation of alkanes by
Wurtz reaction and decarboxylation reactions.
Preparation, Properties and Reactions of Alkenes and
Alkynes Physical properties of alkenes and alkynes

(boiling points, density and dipole moments), Acidity of
alkynes, Acid catalysed hydration of alkenes and alkynes
(excluding the stereochemistry of addition and
elimination), Reactions of alkenes with KMnO4 and ozone,
Reduction of alkenes and alkynes, Preparation of alkenes
and alkynes by elimination reactions, Electrophilic
addition reactions of alkenes with X2, HX, HOX and H2O
(X=halogen), Addition reactions of alkynes, Metal
acetylides.
Reactions of Benzene Structure and aromaticity,
Electrophilic substitution reactions, halogenation,
nitration, sulphonation, Friedel-Crafts alkylation and
acylation Effect of o-, m- and p-directing groups in

monosubstituted benzenes.
Phenols Acidity, electrophilic substitution reactions
(halogenation, nitration and sulphonation), ReimerTiemann reaction, Kolbe reaction.
Characteristic Reactions of the following (including
those mentioned above) Alkyl halides, rearrangement
reactions of alkyl carbocation, Grignard reactions,
nucleophilic substitution reactions, Alcohols,
esterification, dehydration and oxidation, reaction with
sodium, phosphorus halides, ZnCl2/concentrated HCl,
conversion of alcohols into aldehydes and ketones, Ethers,
Preparation by Williamson's Synthesis, Aldehydes and
Ketones, oxidation, reduction, oxime and hydrazone
formation, aldol condensation, Perkin reaction, Cannizzaro
reaction, haloform reaction and nucleophilic addition
reactions (Grignard addition), Carboxylic acids, formation
of esters, acid chlorides and amides, ester hydrolysis.
Amines, basicity of substituted anilines and aliphatic
amines, preparation from nitro compounds, reaction with
nitrous acid, azo coupling reaction of diazonium salts of
aromatic amines, Sandmeyer and related reactions of
diazonium salts, carbylamine reaction, Haloarenes,
nucleophilic aromatic substitution in haloarenes and
substituted haloarenes (excluding Benzyne mechanism
and Cine substitution).
Carbohydrates Classification, mono and disaccharides
(glucose and sucrose), Oxidation, reduction, glycoside
formation and hydrolysis of sucrose.
Amino Acids and Peptides General structure (only
primary structure for peptides) and physical properties.
Properties and Uses of Some Important Polymers

Natural rubber, cellulose, nylon, teflon and PVC.
Practical Organic Chemistry Detection of elements (N, S,
halogens), Detection and identification of the following
functional groups, hydroxyl (alcoholic and phenolic),
carbonyl (aldehyde and ketone), carboxyl, amino and
nitro, Chemical methods of separation of mono-functional
organic compounds from binary mixtures.


1
Some Basic Concepts
of Chemistry
6. The percentage composition of carbon by mole in methane is

Topic 1 Mole Concept

(2019 Main, 8 April II)

Objective Questions I (Only one correct option)
1. 5 moles of AB2 weight 125 × 10−3 kg and 10 moles of A2 B2

weight 300 × 10−3 kg. The molar mass of A ( M A ) and molar
mass of B ( M B ) in kg mol −1 are
(2019 Main, 12 April I)

(a) M A = 10 × 10−3 and M B = 5 × 10−3
(b) M A = 50 × 10−3 and M B = 25 × 10−3
(d) M A = 5 × 10

and M B = 10 × 10


reactant is for the reaction (Given atomic mass : Fe = 56,
O = 16, Mg = 24, P = 31, C = 12, H = 1) (2019 Main, 10 April II)
C3 H8 ( g ) + 5O2 ( g ) → 3CO2 ( g ) + 4H2 O( l )
P4 ( s ) + 5O2 ( g ) → P4 O10 ( s )
4Fe( s ) + 3O2 ( g ) → 2Fe2 O3 ( s )
2Mg ( s ) + O2 ( g ) → 2MgO( s )

10 mL of a hydrocarbon required 55 mL of O2 for complete
combustion and 40 mL of CO2 is formed. The formula of the
hydrocarbon is
(2019 Main, 10 April I)
(c) C4H10
(d) C4H 8
(a) C4H7Cl (b) C4H 6

4. 10 mL of 1 mM surfactant solution forms a monolayer
covering 0.24 cm 2 on a polar substrate. If the polar head is
approximated as a cube, what is its edge length?
(2019 Main, 9 April II)

(b) 0.1 nm

(c) 1.0 pm

(d) 80%

7. 8 g of NaOH is dissolved in 18 g of H2 O. Mole fraction of
NaOH in solution and molality (in mol kg− 1 ) of the solution
respectively are

(2019 Main, 12 Jan II)
(a) 0.2, 11.11
(b) 0.167, 22.20
(c) 0.2, 22.20
(d) 0.167, 11.11

(d) 2.0 nm

5. For a reaction,
N 2 ( g ) + 3H 2 ( g ) → 2NH 3 ( g ), identify dihydrogen (H 2 )
as a limiting reagent in the following reaction mixtures.
(2019 Main, 9 April I)

(a) 56 g of N 2 + 10 g of H 2 (b) 35 g of N 2 + 8 g of H 2
(c) 14 g of N 2 + 4 g of H 2 (d) 28 g of N 2 + 6 g of H 2

(2019 Main, 12 Jan II)

(d) 5.6

9. The amount of sugar (C12 H22 O11 ) required to prepare 2 L of
its 0.1 M aqueous solution is
(2019 Main, 10 Jan II)
(a) 17.1 g
(b) 68.4 g
(c) 136.8 g
(d) 34.2 g

10. For the following reaction, the mass of water produced from
445 g of C57 H110 O6 is :

2C57 H110 O6 ( s ) + 163O2 ( g ) → 114CO2 ( g ) + 110 H2 O ( l )
(2019 Main, 9 Jan II)

(a) 490 g

3. At 300 K and 1 atmospheric pressure,

(a) 2.0 pm

(c) 25%

(Molar mass of H2 O2 = 34 g mol −1 )
(a) 16.8
(b) 22.4
(c) 11.35

−3

2. The minimum amount of O2 ( g ) consumed per gram of

(a)
(b)
(c)
(d)

(b) 20%

8. The volume strength of 1 M H2 O2 is

(c) M A = 25 × 10−3 and M B = 50 × 10−3

−3

(a) 75%

(b) 495 g

(c) 445 g

(d) 890 g

11. A solution of sodium sulphate contains 92 g of Na + ions per

kilogram of water. The molality of Na + ions in that solution
in mol kg−1 is
(2019 Main, 9 Jan I)
(a) 16
(b) 4
(c) 132
(d) 8

12. The most abundant elements by mass in the body of a healthy
human adult are oxygen (61.4%), carbon (22.9%), hydrogen
(10.0 %), and nitrogen (2.6%). The weight which a 75 kg
person would gain if all 1 Hatoms are replaced by 2 Hatoms is
(2017 JEE Main)

(a) 15 kg
(c) 7.5 kg

(b) 37.5 kg

(d) 10 kg

13. 1 g of a carbonate (M 2 CO3 ) on treatment with excess HCl
produces 0.01186 mole of CO2 . The molar mass of M 2 CO3
in g mol −1 is
(2017 JEE Main)
(a) 1186
(b) 84.3
(c) 118.6
(d) 11.86


2 Some Basic Concepts of Chemistry
14. At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon
requires 375 mL air containing 20% O2 by volume for
complete combustion. After combustion, the gases occupy
330 mL. Assuming that the water formed is in liquid form
and the volumes were measured at the same temperature and
pressure, the formula of the hydrocarbon is
(2016 Main)
(b) C4 H8
(c) C4 H10
(d) C3 H6
(a) C3 H8

15. The molecular formula of a commercial resin used for

24. The normality of 0.3 M phosphorus acid (H3PO3) is
(1999, 2M)


(a) 0.1

(b) 0.9

(c) 0.3

(d) 0.6

25. In which mode of expression, the concentration of a solution
remains independent of temperature?
(a) Molarity (b) Normality (c) Formality

(1988, 1M)

(d) Molality

26. A molal solution is one that contains one mole of solute in
(1986, 1M)

exchanging ions in water softening is C8 H7 SO3 Na
(molecular weight = 206). What would be the maximum
uptake of Ca 2+ ions by the resin when expressed in mole per
gram resin?
(2015 Main)
1
1
2
1
(b)
(c)

(d)
(a)
103
206
309
412

(a) 1000 g of solvent
(b) 1.0 L of solvent
(c) 1.0 L of solution
(d) 22.4 L of solution
27. If 0.50 mole of BaCl 2 is mixed with 0.20 mole of Na 3 PO4 ,
the maximum number of moles of Ba 3 (PO4 )2 that can be
formed is
(1981, 1M)
(a) 0.70
(b) 0.50
(c) 0.20
(d) 0.10

16. 3 g of activated charcoal was added to 50 mL of acetic acid

28. 2.76 g of silver carbonate on being strongly heated yields a

solution (0.06 N) in a flask. After an hour it was filtered and
the strength of the filtrate was found to be 0.042 N. The
amount of acetic acid adsorbed (per gram of charcoal) is
(2015 Main)

(a) 18 mg


(b) 36 mg

(c) 42 mg

(d) 54 mg

17. The ratio mass of oxygen and nitrogen of a particular gaseous
mixture is 1 : 4. The ratio of number of their molecule is
(2014 Main)

(a) 1 : 4

(b) 7 : 32

(c) 1 : 8

(d) 3 : 16

18. The molarity of a solution obtained by mixing 750 mL of
0.5 M HCl with 250 mL of 2 M HCl will be
(2013 Main)
(a) 0.875 M (b) 1.00 M
(c) 1.75 M
(d) 0.0975M

19. Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water
gave a solution of density 1.15 g/mL. The molarity of the
solution is
(2011)

(a) 1.78 M (b) 2.00 M (c) 2.05 M
(d) 2.22 M

20. Given that the abundances of isotopes

54 Fe, 56 Fe and 57 Fe
are 5%, 90% and 5%, respectively, the atomic mass of Fe is

residue weighing
(a) 2.16 g
(b) 2.48 g

(1979, 1M)

(c) 2.32 g

(d) 2.64 g

29. When the same amount of zinc is treated separately with
excess of sulphuric acid and excess of sodium hydroxide, the
ratio of volumes of hydrogen evolved is
(1979, 1M)
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 9 : 4

30. The largest number of molecules is in

(1979, 1M)


(a) 36 g of water
(b) 28 g of CO
(c) 46 g of ethyl alcohol
(d) 54 g of nitrogen pentaoxide (N2 O5 )
31. The total number of electrons in one molecule of carbon
dioxide is
(1979, 1M)
(a) 22
(b) 44
(c) 66
(d) 88
32. A gaseous mixture contains oxygen and nitrogen in the ratio
of 1:4 by weight. Therefore, the ratio of their number of
molecules is
(1979, 1M)
(a) 1 : 4
(b) 1 : 8
(c) 7 : 32
(d) 3 : 16

(2009)

(a) 55.85
(c) 55.75

(b) 55.95
(d) 56.05

Numerical Answer Type Questions


21. Mixture X = 0.02 mole of [Co(NH3 )5 SO4 ]Br and 0.02 mole
of [Co(NH3 )5 Br]SO4 was prepared in 2 L solution.
1 L of mixture X + excess of AgNO3 solution → Y
1 L of mixture X + excess of BaCl 2 solution → Z
Number of moles of Y and Z are
(a) 0.01, 0.01
(b) 0.02, 0.01
(c) 0.01, 0.02
(d) 0.02, 0.02

22. Which has maximum number of atoms?
(a) 24 g of C (12)
(c) 27 g of Al (27)

(2003, 1M)

(2003, 1M)

(b) 56 g of Fe (56)
(d) 108 g of Ag (108)

23. How many moles of electron weighs 1 kg?
23

(a) 6.023 × 10
(c)

6.023
× 1054

9.108

1
(2002, 3M)
(b)
× 1031
9.108
1
(d)
× 108
9.108 × 6.023

33. A 100 mL solution was made by adding 1.43 g of

Na 2CO3 ⋅ xH 2O. The normality of the solution is 0.1 N. The
value of x is ……… .
(The atomic mass of Na is 23 g/mol)

(2020 Main, 4 Sep II)

34. Galena (an ore) is partially oxidised by passing air through it
at high temperature. After some time, the passage of air is
stopped, but the heating is continued in a closed furnace such
that the content undergo self-reduction. The weight (in kg) of
Pb produced per kg of O 2 consumed is ……… .
(Atomic weights in g mol −1 : O = 16, S = 32, Pb = 207)
(2018 Adv.)

35. To measure the quantity of MnCl 2 dissolved in an aqueous
solution, it was completely converted to KMnO4 using the

reaction,
MnCl 2 + K 2 S2 O8 + H2 O → KMnO4 + H2 SO4 + HCl
(equation not balanced).


Some Basic Concepts of Chemistry 3
Few drops of concentrated HCl were added to this solution
and gently warmed. Further, oxalic acid (225 mg) was
added in portions till the colour of the permanganate ion
disappeared. The quantity of MnCl 2 (in mg) present in the
initial solution is ……… .
(Atomic weights in g mol −1 : Mn = 55, Cl = 35.5)
(2018 Adv.)

36. In the following reaction sequence, the amount of D (in
gram) formed from 10 moles of acetophenone is …….
(Atomic weights in g mol

−1

: H = 1, C = 12, N = 14,
O = 16, Br = 80. The yield (%) corresponding to the
product in each step is given in the parenthesis)
O
NaOBr
H3O+

A

NH3, ∆


(60%)

B

(2005, 3M)

45. In a solution of 100 mL 0.5 M acetic acid, one gram of active
charcoal is added, which adsorbs acetic acid. It is found that the
concentration of acetic acid becomes 0.49 M. If surface area of
charcoal is 3.01 × 102 m2 , calculate the area occupied by single
acetic acid molecule on surface of charcoal.
(2003)

46. Find the molarity of water. Given: ρ = 1000 kg/m3

(2003)

47. 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 cm 3 /g. If the virus is considered to be a single
particle, find its molar mass.
(1999, 3M)
obtain 1 dm 3 of solution of density 1077.2 kg m −3 . Calculate
the molality, molarity and mole fraction of Na 2 SO4 in solution.

C
(50%)

Br2(3 equivalent )

AcOH

number of surface sites occupied per molecule of N2 .

48. 8.0575 × 10−2 kg of Glauber’s salt is dissolved in water to

Br2/KOH

(50%)

K in a container of volume is 2.46 cm 3 . Density of surface sites
is 6.023 × 1014 /cm 2 and surface area is 1000 cm 2 , find out the

(1994, 3M)

D

(100%)

(2018 Adv.)

Fill in the Blanks
37. The weight of 1 × 1022 molecules of CuSO4 ⋅ 5H2 O is
…………. .

(1991, 1M)

49. A is a binary compound of a univalent metal. 1.422 g of A reacts
completely with 0.321 g of sulphur in an evacuated and sealed
tube to give 1.743 g of a white crystalline solid B, that forms a

hydrated double salt, C with Al 2 (SO4 )3 . Identify A, B and C.
(1994, 2M)

50. Upon mixing 45.0 mL 0.25 M lead nitrate solution with

(1980, 1M)

25.0 mL of a 0.10 M chromic sulphate solution, precipitation of
lead sulphate takes place. How many moles of lead sulphate are
formed? Also calculate the molar concentrations of species left
behind in the final solution. Assume that lead sulphate is
completely insoluble.
(1993, 3M)

40. The modern atomic mass unit is based on the mass of

51. Calculate the molality of 1.0 L solution of 93% H2 SO4 ,

38. 3.0 g of a salt of molecular weight 30 is dissolved in 250 g
water. The molarity of the solution is ……….

(1983, 1M)

39. The total number of electrons present in 18 mL of water is
…………. .
…………. .

(1980, 1M)

(weight/volume). The density of the solution is 1.84 g/mL.

(1990, 1M)

Integer Answer Type Questions

52. A solid mixture (5.0 g) consisting of lead nitrate and sodium

41. The mole fraction of a solute in a solution is 0.1. At 298 K,
molarity of this solution is the same as its molality. Density
of this solution at 298 K is 2.0 g cm−3 . The ratio of the

m
molecular weights of the solute and solvent,  solute  is ...
 msolvent 
.

(2016 Adv.)

42. A compound H2 X with molar weight of 80 g is dissolved
in a solvent having density of 0.4 g mL−1 . Assuming no
change in volume upon dissolution, the molality of a 3.2
molar solution is
(2014 Adv.)

43. 29.2% (w/W ) HCl stock solution has density of 1.25g mL
−1

. The molecular weight of HCl is 36.5 g mol − 1 . The
volume (mL) of stock solution required to prepare a 200
mL solution 0.4 M HCl is
(2012)


Subjective Questions

nitrate was heated below 600°C until the weight of the residue
was constant. If the loss in weight is 28.0 per cent, find the
amount of lead nitrate and sodium nitrate in the mixture.
(1990, 4M)

53. n-butane is produced by monobromination of ethane followed
by Wurtz’s reaction.Calculate volume of ethane at NTP
required to produce 55 g n-butane, if the bromination takes
place with 90% yield and the Wurtz’s reaction with 85% yield.
(1989, 3M)

54. A sugar syrup of weight 214.2 g contains 34.2 g of sugar
(C12 H22 O11 ). Calculate (i) molal concentration and (ii) mole
fraction of sugar in syrup.
(1988, 2M)

55. An unknown compound of carbon, hydrogen and oxygen
contains 69.77% C and 11.63% H and has a molecular weight
of 86. It does not reduces Fehling’s solution but forms a
bisulphate addition compound and gives a positive iodoform
test. What is the possible structure(s) of unknown compound?
(1987, 3M)

44. 20% surface sites have adsorbed N2 . On heating N2 gas

56. The density of a 3 M sodium thiosulphate solution ( Na 2 S2 O3 )


evolved from sites and were collected at 0.001 atm and 298

is 1.25 g per mL. Calculate (i) the percentage by weight of


4 Some Basic Concepts of Chemistry
sodium thiosulphate (ii) the mole fraction of sodium
thiosulphate and (iii) the molalities of Na + and S2 O2−
3 ions.

oxygen. All volumes have been reduced to NTP. Calculate
the molecular formula of the hydrocarbon gas. (1979, 3M)

(1983, 5M)

59. In the analysis of 0.5 g sample of feldspar, a mixture of

57. (a) 1.0 L of a mixture of CO and CO2 is taken. This mixture

chlorides of sodium and potassium is obtained, which weighs
0.1180 g. Subsequent treatment of the mixed chlorides with
silver nitrate gives 0.2451 g of silver chloride. What is the
percentage of sodium oxide and potassium oxide in the
sample?
(1979, 5M)

is passed through a tube containing red hot charcoal. The
volume now becomes 1.6 L. The volumes are measured
under the same conditions. Find the composition of
mixture by volume.

(b) A compound contains 28 per cent of nitrogen and
72 per cent of a metal by weight. 3 atoms of metal
combine with 2 atoms of nitrogen. Find the atomic
weight of metal.
(1980, 5M)

58. 5.00 mL of a gas containing only carbon and hydrogen were
mixed with an excess of oxygen (30 mL) and the mixture
exploded by means of electric spark. After explosion, the
volume of the mixed gases remaining was 25 mL.
On adding a concentrated solution of KOH, the volume
further diminished to 15 mL, the residual gas being pure

60. The vapour density (hydrogen = 1) of a mixture consisting of
NO2 and N2 O4 is 38.3 at 26.7°C. Calculate the number of
moles of NO2 in 100 g of the mixture.
(1979, 5M)

61. Accounts for the following. Limit your answer to two
sentences, “Atomic weights of most of the elements are
fractional”.
(1979, 1M)

62. Naturally occurring boron consists of two isotopes whose
atomic weights are 10.01 and 11.01. The atomic weight of
natural boron is 10.81. Calculate the percentage of each
isotope in natural boron.
(1978, 2M)

Topic 2 Equivalent Concept, Neutralisation and Redox Titration

Objective Questions I (Only one correct option)
1. An example of a disproportionation reaction is
(2019 Main, 12 April I)

(a)

2MnO−4

−

+

+ 10I + 16H → 2Mn2 + +5I2 + 8H 2O

(b) 2NaBr + Cl2 → 2NaCl + Br2
(c) 2KMnO4 → K 2MnO4 + MnO2 + O2
(d) 2CuBr → CuBr2 + Cu

2. In an acid-base titration, 0.1 M HCl solution was added to
the NaOH solution of unknown strength. Which of the
following correctly shows the change of pH of the titration
mixture in this experiment?
(2019 Main, 9 April II)

pH

surface of water in a round watch glass. Hexane evaporates
and a monolayer is formed. The distance from edge to centre
of the watch glass is 10 cm. What is the height of the
monolayer? [Density of fatty acid = 0.9 g cm −3 ; π = 3]

(2019 Main, 8 April II)

(a) 10−6 m
(c) 10−8 m

(b) 10−4 m
(d) 10−2 m

4. In order to oxidise a mixture of one mole of each of FeC2 O4 ,
Fe2 (C2 O4 )3 , FeSO4 and Fe2 (SO4 )3 in acidic medium, the
number of moles of KMnO4 required is (2019 Main, 8 April I)
(a) 2
(b) 1
(c) 3
(d) 1.5

5. 100 mL of a water sample contains 0.81 g of calcium
bicarbonate and 0.73 g of magnesium bicarbonate. The
hardness of this water sample expressed in terms of
equivalents of CaCO3 is (molar mass of calcium bicarbonate
is 162 g mol−1 and magnesium bicarbonate is 146 g mol−1 )

pH

V(mL)

V(mL)

(A)


(B)

(2019 Main, 8 April I)

(a) 5,000 ppm
(c) 100 ppm

(b) 1,000 ppm
(d) 10,000 ppm

6. 50 mL of 0.5 M oxalic acid is needed to neutralise 25 mL of
pH

sodium hydroxide solution. The amount of NaOH in 50 mL
of the given sodium hydroxide solution is

pH

(2019 Main, 12 Jan I)

(a) (D)
(c) (B)

V(mL)

V(mL)

(C)

(D)


(a) 40 g
(2019 Main, 9 April II)

(b) (A)
(d) (C)

3. 0.27 g of a long chain fatty acid was dissolved in 100 cm 3 of
hexane. 10 mL of this solution was added dropwise to the

(b) 80 g

(c) 20 g

(d) 10 g

7. 25 mL of the given HCl solution requires 30 mL of 0.1 M
sodium carbonate solution. What is the volume of this HCl
solution required to titrate 30 mL of 0.2 M aqueous NaOH
solution?
(2019 Main, 11 Jan II)
(a) 75 mL
(b) 25 mL
(c) 12.5 mL
(d) 50 mL


Some Basic Concepts of Chemistry 5
8. In the reaction of oxalate with permanganate in acidic
medium, the number of electrons involved in producing one

(2019 Main, 10 Jan II)
molecule of CO2 is
(a) 2
(b) 5
(c) 1
(d) 10

9. The ratio of mass per cent of C and H of an organic
compound (Cx H y O z ) is 6 : 1. If one molecule of the above
compound (Cx H y O z ) contains half as much oxygen as
required to burn one molecule of compound Cx H y
completely to CO2 and H2 O. The empirical formula of
compound Cx H y O z is
(2018 Main)
(c) C3 H4 O2
(d) C2 H4 O3
(a) C3 H6 O3 (b) C2 H4 O

10. An alkali is titrated against an acid with methyl orange as
indicator, which of the following is a correct combination?
(2018 Main)

Base

Acid

End point

(a)


Weak

Strong

Colourless to pink

(b)

Strong

Strong

Pinkish red to yellow

(c)

Weak

Strong

Yellow to pinkish red

(d)

Strong

Strong

Pink to colourless


17. The oxidation number of sulphur in S 8 , S 2 F 2 , H 2 S
respectively, are
(a) 0, +1 and –2
(c) 0, +1 and +2

18. The number of moles of KMnO 4 that will be needed to react
completely with one mole of ferrous oxalate in acidic
medium is
(1997)
2
3
4
(b)
(c)
(d) 1
(a)
5
5
5

19. The number of moles of KMnO 4 that will be needed to react
with one mole of sulphite ion in acidic solution is
(1997)

2
(a)
5

incorrect statement.
(2015 Main)

(a) It can act only as an oxidising agent
(b) It decomposed on exposure to light
(c) It has to be stored in plastic or wax lined glass bottles in
dark
(d) It has to be kept away from dust

12. Consider a titration of potassium dichromate solution with
acidified Mohr’s salt solution using diphenylamine as
indicator. The number of moles of Mohr's salt required per
mole of dichromate is
(2007, 3M)
(a) 3
(b) 4
(c) 5
(d) 6

13. In the standardisation of Na 2 S2 O3 using K 2 Cr2 O7 by
iodometry, the equivalent weight of K 2 Cr2 O7 is (2001, 1M)
(a) (molecular weight)/2
(b) (molecular weight)/6
(c) (molecular weight)/3
(d) same as molecular weight

14. The reaction, 3ClO − (aq) → ClO3– (aq) + 2Cl − (aq) is an
(2001)

15. An aqueous solution of 6.3 g oxalic acid dihydrate is made up
to 250 mL. The volume of 0.1 N NaOH required to
completely neutralise 10 mL of this solution is (2001, 1M)
(a) 40 mL

(b) 20 mL
(c) 10 mL
(d) 4 mL

16. Among the following, the species in which the oxidation
number of an element is + 6
(a) MnO−4
(b) Cr(CN)3−
6
(c) NiF62−

(d) CrO2 Cl 2

3
(b)
5

4
(c)
5

(d) 1

20. For the redox reaction
MnO4− + C2 O24 − + H+ → Mn 2+ + CO2 + H2 O
The correct coefficients of the reactants for the balanced
reaction are

11. From the following statements regarding H2 O2 choose the


example of
(a) oxidation reaction
(b) reduction reaction
(c) disproportionation reaction
(d) decomposition reaction

(1999)

(b) +2, +1 and –2
(d) –2, +1 and –2

(2000)

(a)
(b)
(c)
(d)

MnO−4
2
16
5
2

C2 O42 −
5
5
16
16


H+
16
2
2
5

21. The volume strength of 1.5 N H2 O2 is
(a) 4.8

(b) 8.4

(c) 3.0

(1992)

(1990, 1M)

(d) 8.0

22. The oxidation number of phosphorus in Ba(H2 PO2 )2 is
(a) +3
(c) +1

(b) +2
(d) –1

(1988)

23. The equivalent weight of MnSO 4 is half of its molecular
weight, when it converts to

(a) Mn 2 O3
(b) MnO2
(c) MnO−4

(1988, 1M)

(d) MnO2−
4

Objective Question II (More than one correct option)
24. For the reaction, I− + ClO3− + H2 SO4 → Cl − + HSO4− + I2
the correct statement(s) in the balanced equation is/are
(a) stoichiometric coefficient of HSO−4 is 6
(2014 Adv.)
(b) iodide is oxidised
(c) sulphur is reduced
(d) H2 O is one of the products

Numerical Answer Type Questions
25. 5.00 mL of 0.10 M oxalic acid solution taken in a conical
flask is titrated against NaOH from a burette using
phenolphthalein indicator. The volume of NaOH required for
the appearance of permanent faint pink color is tabulated
below for five experiments. What is the concentration, in
molarity, of the NaOH solution?


6 Some Basic Concepts of Chemistry
Exp. No.
1

2
3
4
5

Integer Answer Type Questions

Vol. of NaOH (mL)
12.5
10.5
9.0
9.0
9.0

31. The difference in the oxidation numbers of the two types of
sulphur atoms in Na 2 S4 O6 is

32. Among the following, the number of elements showing only
one non-zero oxidation state is O, Cl, F, N, P, Sn, Tl, Na, Ti
(2010)
(2020 Adv.)

26. Aluminium reacts with sulphuric acid to form aluminium
sulphate and hydrogen. What is the volume of hydrogen gas
in litre (L) produced at 300 K and 1.0 atm pressure, when
5.4 g of aluminium and 50.0 mL of 5.0 M sulphuric acid are
combined for the reaction?
(Use molar mass of aluminium as 27.0 g mol −1 , R = 0.082
atm L mol −1 K −1 )


(2020 Adv.)

27. A 20.0 mL solution containing 0.2 g impure H 2O2 reacts
completely with 0.316 g of KMnO4 in acid solution. The
purity of H 2O2 (in%) is ............. (molecular weight of
H 2O2 = 34; molecular weight of KMnO4 = 158 ).
(2020 Main, 4 Sep I)

28. The ammonia prepared by treating ammonium sulphate with

calcium hydroxide is completely used by NiCl 2 ⋅ 6H2 O to
form a stable coordination compound. Assume that both the
reactions are 100% complete. If 1584 g of ammonium
sulphate and 952 g of NiCl 2 ⋅ 6H2 O are used in the
preparation, the combined weight (in grams) of gypsum
and the nickel-ammonia coordination compound thus
produced is____
(Atomic weights in g mol −1 : H = 1, N = 14, O = 16, S = 32,
(2018 Adv.)
Cl = 35.5, Ca = 40, Ni = 59)

Assertion and Reason
Read the following questions and answer as per the direction
given below :
(a) Statement I is true; Statement II is true; Statement II is the
correct explanation of Statement I.
(b) Statement I is true; Statement II is true; Statement II is not
the correct explanation of Statement I.
(c) Statement I is true; Statement II is false.
(d) Statement I is false; Statement II is true.


29. Statement I In the titration of Na 2 CO3 with HCl using
methyl orange indicator, the volume required at the
equivalence point is twice that of the acid required using
phenolphthalein indicator.
Statement II Two moles of HCl are required for the
complete neutralisation of one mole of Na 2 CO3 . (1991, 2M)

Fill in the Blanks
30. The

(2011)

compound YBa 2 Cu 3 O7 , which shows super
conductivity, has copper in oxidation state ………. Assume
that the rare earth element yttrium is in its usual + 3 oxidation
state.
(1994, 1M)

33. A student performs a titration with different burettes and
finds titrate values of 25.2 mL, 25.25 mL, and 25.0 mL. The
number of significant figures in the average titrate value is
(2010)

Subjective Questions
34. Calculate the amount of calcium oxide required when it
reacts with 852 g of P4 O10 .

(2005, 2M)


35. Hydrogen peroxide solution (20 mL) reacts quantitatively
with a solution of KMnO4 (20 mL) acidified with dilute
H2 SO4 . The same volume of the KMnO4 solution is just
decolourised by 10 mL of MnSO4 in neutral medium
simultaneously forming a dark brown precipitate of hydrated
MnO2 . The brown precipitate is dissolved in 10 mL of 0.2 M
sodium oxalate under boiling condition in the presence of
dilute H2 SO4 . Write the balanced equations involved in the
reactions and calculate the molarity of H2 O2 .
(2001)

36. How many millilitres of 0.5 M H2 SO4 are needed to dissolve
0.5 g of copper (II) carbonate?

(1999, 3M)

37. An aqueous solution containing 0.10 g KIO3

(formula weight = 214.0) was treated with an excess of KI
solution. The solution was acidified with HCl. The liberated
I2 consumed 45.0 mL of thiosulphate solution decolourise
the blue starch-iodine complex. Calculate the molarity of the
sodium thiosulphate solution.
(1998, 5M)

38. To a 25 mL H2 O2 solution, excess of acidified solution of
potassium iodide was added. The iodine liberated required
20 mL of 0.3 N sodium thiosulphate solution. Calculate the
volume strength of H2 O2 solution.
(1997, 5M)


39. A 3.00 g sample containing Fe3 O4 , Fe2 O3 and an inert
impure substance, is treated with excess of KI solution in
presence of dilute H2 SO4 . The entire iron is converted into
Fe2+ along with the liberation of iodine. The resulting
solution is diluted to 100 mL . A 20 mL of the diluted
solution requires 11.0 mL of 0.5 M Na 2 S2 O3 solution to
reduce the iodine present. A 50 mL of the dilute solution,
after complete extraction of the iodine required 12.80 mL of
0.25 M KMnO4 solution in dilute H2 SO4 medium for the
oxidation of Fe2+ . Calculate the percentage of Fe2 O3 and
(1996, 5M)
Fe3 O4 in the original sample.
40. A 20.0 cm 3 mixture of CO, CH4 and He gases is exploded by
an electric discharge at room temperature with excess of
oxygen. The volume contraction is found to be 13.0 cm 3 .
A further contraction of 14.0 cm 3 occurs when the residual
gas is treated with KOH solution. Find out the composition
of the gaseous mixture in terms of volume percentage.
(1995, 4M)


Some Basic Concepts of Chemistry 7
41. A 5.0 cm 3 solution of H2 O2 liberates 0.508 g of iodine from
an acidified KI solution. Calculate the strength of H2 O2
solution in terms of volume strength at STP.
(1995, 3M)

Calculate the amount of H2 C2 O4 and NaHC2 O4 in the
mixture.

(1990, 5M)

47. An organic compound X on analysis gives 24.24 per cent
carbon and 4.04 per cent hydrogen. Further, sodium extract
of 1.0 g of X gives 2.90 g of silver chloride with acidified
silver nitrate solution. The compound X may be represented
by two isomeric structures Y and Z. Y on treatment with
aqueous potassium hydroxide solution gives a dihydroxy
compound while Z on similar treatment gives ethanal. Find
out the molecular formula of X and gives the structure
of Y and Z.
(1989, 5M)

42. One gram of commercial AgNO3 is dissolved in 50 mL of
water. It is treated with 50 mL of a KI solution. The silver
iodide thus precipitated is filtered off. Excess of KI in the
filtrate is titrated with (M/10) KIO 3 solution in presence of
6 M HCl till all I− ions are converted into ICl. It requires
50 mL of (M/10) KIO 3 solution, 20 mL of the same stock
solution of KI requires 30 mL of (M/10) KIO3 under similar
conditions. Calculate the percentage of AgNO3 in the
sample.
Reaction KIO3 + 2KI + 6HCl → 3ICl + 3KCl + 3H2 O
(1992, 4M)

43. A 2.0 g sample of a mixture containing sodium carbonate,
sodium bicarbonate and sodium sulphate is gently heated till
the evolution of CO2 ceases. The volume of CO2 at 750 mm
Hg pressure and at 298 K is measured to be 123.9 mL. A 1.5 g
of the same sample requires 150 mL of (M/10) HCl for

complete neutralisation. Calculate the percentage
composition of the components of the mixture. (1992, 5M)

44. A 1.0 g sample of Fe2 O3 solid of 55.2% purity is dissolved in
acid and reduced by heating the solution with zinc dust. The
resultant solution is cooled and made up to 100.0 mL. An
aliquot of 25.0 mL of this solution requires for titration.
Calculate the number of electrons taken up by the oxidant in
the reaction of the above titration.
(1991, 4M)

48. An equal volume of a reducing agent is titrated separately
with 1 M KMnO4 in acid, neutral and alkaline medium. The
volumes of KMnO4 required are 20 mL in acid, 33.3 mL in
neutral and 100 mL in alkaline media. Find out the oxidation
state of manganese in each reduction product. Give the
balanced equations for all the three half reaction. Find out the
volume of 1M K 2 Cr2 O7 consumed, if the same volume of the
reducing agent is titrated in acid medium.
(1989, 5M)
49. A sample of hydrazine sulphate ( N2 H 6 SO4 ) was dissolved in
100 mL of water, 10 mL of this solution was reacted with
excess of ferric chloride solution and warmed to complete
the reaction. Ferrous ion formed was estimated and it,
required 20 mL of M/50 potassium permanganate solution.
Estimate the amount of hydrazine sulphate in one litre of the
solution.
Reaction 4Fe3+ + N2 H4 → N2 + 4Fe2+ + 4H+
MnO−4 + 5Fe2+ + 8H+ → Mn 2+ + 5Fe3+ + 4H2 O


45. A solution of 0.2 g of a compound containing Cu 2+ and
C2 O2−
4 ions on titration with 0.02 M KMnO4 in presence of
H2 SO4 consumes 22.6 mL of the oxidant. The resultant
solution is neutralised with Na 2 CO3 , acidified with dilute
acetic acid and treated with excess KI. The liberated iodine
requires 11.3 mL of 0.05 M Na 2 S2 O3 solution for complete
reduction. Find out the mole ratio of Cu 2+ to C2 O2−
4 in the
compound. Write down the balanced redox reactions
involved in the above titrations.
(1991, 5M)

(1988, 3M)

50. 5 mL of 8 N nitric acid, 4.8 mL of 5 N hydrochloric acid and
a certain volume of 17 M sulphuric acid are mixed together
and made up to 2 L. 30 mL of this acid mixture exactly
neutralise 42.9 mL of sodium carbonate solution containing
one gram of Na 2 CO3 ⋅ 10H2 O in 100 mL of water. Calculate
the amount in gram of the sulphate ions in solution.
(1985, 4M)

51. 2.68 × 10−3 moles of a solution containing an ion A n+ require
1.61 × 10−3 moles of MnO−4 for the oxidation of A n+ to A O−3
in acidic medium. What is the value of n ?
(1984, 2M)

46. A mixture of H2 C2 O4 (oxalic acid) and NaHC2 O4 weighing
2.02 g was dissolved in water and the solution made up to one

litre. Ten millilitres of the solution required 3.0 mL of 0.1 N
sodium hydroxide solution for complete neutralisation. In
another experiment, 10.0 mL of the same solution, in hot
dilute sulphuric acid medium, required 4.0 mL of 0.1 N
potassium permanganate solution for complete reaction.

52. 4.08 g of a mixture of BaO and unknown carbonate MCO3
was heated strongly. The residue weighed 3.64 g. This was
dissolved in 100 mL of 1 N HCl. The excess acid required
16 mL of 2.5 N NaOH solution for complete neutralisation.
Identify the metal M.
(1983, 4M)

Answers
Topic 1
1. (d)
5. (d)
9. (b)

2. (c)
6. (b)
10. (b)

3. (b)
7. (d)
11. (b)

4. (a)
8. (c)
12. (c)


13.
17.
21.
25.
29.

(b)
(b)
(a)
(d)
(a)

14.
18.
22.
26.
30.

(*)
(a)
(a)
(a)
(a)

15.
19.
23.
27.
31.


(d)
(c)
(d)
(d)
(a)

16.
20.
24.
28.
32.

(d)
(b)
(d)
(a)
(c)


8 Some Basic Concepts of Chemistry
33. (10.00)
34. (6.47kg)
35.
37. (4.14 g)
38. (0.4)
39.
isotope
41. (9)
42. (8)

43.
45. (5 × 10 −19 m 2 ) 46. (55.56 mol
L −1) 47. (70.91 × 10 6g) 48. (4.3 × 10 −3)
51. (10.42)
52. (1.7 g)
53.
56. (i) (37.92), (ii) (0.065), (iii) (7.73m)
59. (i) (0.0179 g), (ii) (10.6 %)
60.

(126 mg) 36. (495 g)
(6.023×10 24 ) 40. C-12
(8 mL)

44. (2)

(55.55 L) 54. (9.9 × 10 −3)
57. (a) (0.6), (b) (24)
(0.437)
62. (20 %)

Topic 2
1. (d)

2. (b)

3. (a)

4. (a)


5.
9.
13.
17.
21.
25.
28.
32.
37.
45.
51.

(d)
(d)
(b)
(a)
(b)
(0.11)
(2992)
(2)
(0.062 M)
(1:2)
(2)

6.
10.
14.
18.
22.
26.

29.
33.
38.
48.
52.

(*)
(c)
(c)
(b)
(c)
(6.15)
(b)
(3)
(1.334 V)
(16.67 mL)
(Ca)

7.
11.
15.
19.
23.
27.
30.
34.
42.
49.

(b)

(a)
(a)
(a)
(b)
(85)
7/3
(1008 g)
(85%)
(6.5gL −1)

8.
12.
16.
20.
24.

(c)
(d)
(d)
(a)
(a,b,d)

31.
36.
44.
50.

(5)
(8.096 mL)
(1.04)

(6.5376 g)

Hints & Solutions
Topic 1 Mole Concept
1.

Key Idea To find the mass of A and B in the given question,
mole concept is used.
given mass (w)
Number of moles( n) =
molecular mass (M )

(c) 4Fe(s) + 3O2(g ) → 2Fe2O3 (s)
244g
96g
96
g of O2 consumed = 0.43 g
⇒ 1 g of reactant =
224
(d) 2Mg(s) + O2(g ) → 2MgO(s)
48 g

32 g

32
g of O2 consumed = 0.67 g
48
So, minimum amount of O2 is consumed per gram of reactant
(Fe) in reaction (c).
⇒ 1 g of reactant =


Compound

Mass of A (g)

Mass of B (g)

AB2

MA

2 MB

A 2B 2

2 MA

2 MB

3. In eudiometry,
300 K
y
y

CxH y +  x +  O2 → x CO2 + H2O

1 atm
4
2


We know that,
Number of moles (n) =

given mass (w)
molecular mass (M )

n×M = w
Using equation (A), it can be concluded that

1 mol

…(A)

5(M A + 2M B ) = 125 × 10−3 kg

…(i)

10(2M A + 2M B ) = 300 × 10−3 kg

…(ii)

From equation (i) and (ii)
1 (M A + 2M B )  125 
=

2 (2M A + 2M B )  300
On solving the equation, we obtain
M A = 5 × 10−3
M B = 10 × 10−3


and

So, the molar mass of A (M A ) is
5 × 10−3 kg mol −1 and B (M B ) is 10 × 10−3 kg mol −1.

2. (a) C3H8 (g ) + 5O2 (g ) → 3CO2 (g ) + 4H2O(l )
44g

160g

⇒ 1g of reactant =

160
g of O2 consumed = 3.64 g
44

(b) P4 (s) + 5O2(g ) → P4O10 (s)
124g

160g

⇒1 g of reactant =

160
g of O2 consumed = 129
. g
124

1 mL
10 mL


y

 x +  mol

4
y

 x +  mL

4
y

 x +  × 10 mL

4

x mol
x mL
10x mL

Given, (i) VCO2 = 10x = 40 mL ⇒ x = 4
y

(ii) VO2 = 10  x +  mL = 55 mL

4
y

10  4 +  = 55


4
y × 10
⇒
40 +
= 55
4
10
4
⇒
y×
= 15 ⇒ y = 15 ×
=6
4
10
So, the hydrocarbon (CxH y ) is C4H6.
⇒

[Qx = 4]

4. Given, volume = 10 mL
Molarity = 1mM = 10−3 M
∴Number of millimoles = 10 mL × 10−3 M = 10−2
Number of moles = 10−5
Now, number of molecules
= Number of moles × Avogadro’s number
= 10−5 × 6 × 1023 = 6 × 1018


Some Basic Concepts of Chemistry 9

Surface area occupied by 6 × 1018 molecules = 0.24 cm2
∴Surface area occupied by 1 molecule
0.24
=
= 0.04 × 10−18 cm2
6 × 1018
As it is given that polar head is approximated as cube. Thus,
surface area of cube = a2, where
a =edge length
2
∴
a = 4 × 10−20 cm2
a = 2 × 10−10 cm = 2 pm

5. Key Idea The reactant which is present in the lesser amount,
i.e. which limits the amount of product formed is called
limiting reagent.
When 56 g of N2 + 10 g of H2 is taken as a combination then
dihydrogen (H2 ) act as a limiting reagent in the reaction.
…(I)
N2 (g ) + 3H2 (g ) → 2NH3 (g )
2 × 14 g
28g

3 × 2g
6g

2(14 + 3) g
34g


28g N2 requires 6g H2 gas.
6g
× 56 g = 12g of H2
56g of N2 requires
28 g
12g of H2 gas is required for 56g of N2 gas but
only 10 g of H2 gas is present in option (a).
Hence, H2 gas is the limiting reagent.
In option (b), i.e. 35g of N2 + 8 g of H2.
As 28 g N2 requires 6g of H2.
6g
× 35 g H2 ⇒ 7.5 g of H2.
35g N2 requires
28 g
Here, H2 gas does not act as limiting reagent since 7.5 g of H2
gas is required for 35g of N2 and 8g of H2 is present in reaction
mixture. Mass of H2 left unreacted = 8 − 7.5 g of H2.
= 0.5 gof H2.
Similarly, in option (c) and (d), H2 does not act as limiting
reagent.
For 14 g of N2 + 4 g of H2.
As we know 28g of N2 reacts with 6g of H2.
6
14 g of N2 reacts with
× 14 g of H2 ⇒ 3g of H2.
28
For 28g of N2 + 6 g of H2, i.e. 28g of N2 reacts with 6g of H2
(by equation I).

6. Key Idea The percentage composition of a compound is given

by the formula.
% composition = [Composition of a substance in a compound /
Total composition total of compound] ×100
In CH4 ,
mole of carbon = 1
mole of hydrogen = 4
∴ % of carbon by mole in CH4 =

1
× 100 = 20%
1+ 4

7. Mole fraction of solute
=

number of moles of solute + number of moles solvent
number of moles of solute

χ Solute

Given,

w Solute
n Solute
Mw Solute
=
=
w Solute
w
n Solute + nSolvent

+ Solvent
Mw Solute MwSolvent
wSolute = wNaOH = 8 g

Mw Solute = MwNaOH = 40g mol − 1
w Solvent = w H2 O = 18g

Mw Solvent = 18 g mol− 1
0.2
0.2
8 / 40
=
=
= 0167
.
∴ χ Solute = χ NaOH =
8 18 0.2 + 1
12
.
+
40 18
Moles of solute
Now, molality (m) =
Mass of solvent (in kg)
w Solute
8
Mw Solute
40
=
× 1000 =

× 1000
wSolvent (in g )
18
0.2
=
× 1000 = 11.11mol kg − 1
18
Thus, mole fraction of NaOH in solution and molality of the
solution respectively are 0.167 and 11.11 mol kg − 1 .

8. Concentration of H2O2 is expressed in terms of volume strength,
i.e. “volume of O2 liberated by H2O2 at NTP”. Molarity is
connected to volume strength as:
x
or x = Molarity × 11.2
Molarity (M) =
112
.
where, x = volume strength
So, for 1 M H2O2, x = 1 × 112
. = 112
.
Among the given options, 11.35 is nearest to 11.2.
Number of moles of solute (n)
9. Molarity =
Volume of solution (in L)
wB (g)
Also, n =
M B (gmol−1 )
w / MB

Molarity = B
∴
V
Given, wB = mass of solute ( B ) in g
M B = Gram molar mass of B (C12H22O11 ) = 342 g mol −1
Molarity = 01
. M
Volume (V ) = 2 L
w / 342
01
. = B
⇒ wB = 01
. × 342 × 2 g = 68.4 g
⇒
2

10. 2 C57 H110O6 (s) + 163 O2 (g ) → 110H2O(l ) + 114 CO2 (g )
Molecular mass of C57H110O6
= 2 × (12 × 57 + 1 × 110 + 16 × 6) g = 1780 g
Molecular mass of 110 H2O = 110 (2 + 16) = 1980 g
1780 g of C57H110O6 produced = 1980 g of H2O.
1980
445g of C57H110O6 produced =
× 445 g of H2O
1780
= 495 of H2O
Number of moles of solute
11. Molality (m) =
× 1000
Mass of solvent (in g)



10 Some Basic Concepts of Chemistry
Mass of solute (in g) × 1000
Molecular weight of solute 
× mass of solvent (in g)

wNa + × 1000 92 × 1000
=
=
= 4 mol kg − 1
M Na + × wH 2 O 23 × 1000
=

16. Given, initial strength of acetic acid = 0.06 N
Final strength = 0.042 N;

∴Initial millimoles of CH3COOH = 0.06 × 50 = 3
Final millimoles of CH3COOH = 0.042 × 50 = 2.1
∴ Millimoles of CH3COOH adsorbed = 3 − 2.1= 0.9 mmol

12. Given, abundance of elements by mass

oxygen = 614
. %, carbon = 22.9%, hydrogen = 10% and
nitrogen = 2.6%
Total weight of person = 75 kg
75 × 10
Mass due to 1 H =
= 7.5 kg

100
1
2
H atoms are replaced by H atoms,
Mass due to 2 H = (7.5 × 2) kg
∴ Mass gain by person = 7.5 kg

13. M 2CO3 + 2HCl → 2M Cl + H2O + CO2
1g

0.01186
mole

Number of moles of M 2CO3 reacted = Number of moles of
CO2 evolved
1
[M = molar mass of M 2CO3]
= 0.01186
M
1
M =
= 84.3 g mol − 1
0.01186
y
y
14. CxH y (g ) +  x +  O2 (g ) → xCO2 (g ) + H2O(l )


4


75 mL

30 mL

2

= 20% of 375 = 75 mL
O 2 used
Inert part of air = 80% of 375 = 300 mL
Total volume of gases = CO2 + Inert part of air
= 30 + 300 = 330 mL
x 30
=
⇒x=2
1 15
y
x+
4 = 75 ⇒ x + y = 5
1
15
4
⇒
x = 2, y = 12 ⇒ C2 H12

= 0.9 × 60 mg = 54 mg

(mO 2 )

17.


nO 2
nN 2

=

(M O 2 )
(mN 2 )
(M N 2 )

where, mO 2 = given mass of O2 , mN 2 = given mass of N2 ,
M O 2 = molecular mass of O2 , M N 2 = molecular mass of N2 , nO 2 =
number of moles of O2 , nN 2 = number of moles of N2
 mO  28 1 28 7
= 2
= ×
=
 mN 2  32 4 32 32

18. From the formula, M f =

M 1V1 + M 2V2
V1 + V2

Given, V1 = 750 mL, M 1 = 0.5 M
V2 = 250 mL, M 2 = 2 M
750 × 0.5 + 250 × 2 875
=
=
= 0.875 M
750 + 250

1000

19. Molarity =

Moles of solute
Volume of solution (L)

120
=2
60
Weight of solution = Weight of solvent + Weight of solute
= 1000 + 120 = 1120 g
1120 g
1
⇒
Volume =
×
= 0.973 L
1.15 g / mL 1000 mL / L
2.000
Molarity =
= 2.05M
⇒
0.973
Moles of urea =

20. From the given relative abundance, the average weight of Fe can be
calculated as

15. We know the molecular weight of C8 H7 SO3Na

= 12 × 8 + 1 × 7 + 32 + 16 × 3 + 23 = 206
we have to find, mole per gram of resin.

Volume = 50 mL

A=

54 × 5 + 56 × 90 + 57 × 5
= 55.95
100

21. 1.0 L of mixture X contain 0.01 mole of each [Co(NH3 )5 SO4 ]Br

∴ 1g of C8 H7 SO3Na has number of mole
weight of given resin
1
=
=
mol
Molecular, weight of resin 206

and [Co(NH3 )5 Br]SO4. Also, with AgNO3, only
[Co(NH3 )5 SO4 ]Br reacts to give AgBr precipitate as

Now, reaction looks like

With BaCl 2, only [Co(NH3 )5 Br]SO4 reacts giving BaSO4
precipitate as
[Co(NH3 )5 Br]SO4 + BaCl 2 → [Co(NH3 )5 Br]Cl 2 + BaSO4


2C8 H7 SO3Na + Ca 2+ → (C8 H7 SO3 )2 Ca + 2Na
Q 2 moles of C8 H7 SO3Na combines with 1 mol Ca 2+
1
∴1 mole of C8 H7 SO3Na will combine with mol Ca 2+
2
1
mole of C8 H7 SO3 Na will combine with
∴
206
1
1
1
mol Ca 2+ =
mol Ca 2+
×
412
2 206

[Co(NH3 )5 SO4 ]Br + AgNO3 → [Co(NH3 )5 SO4 ]NO3 + AgBr
1.0 mol

1.0 mol

1.0 mol

Excess

1 mol

Excess


Hence, moles of Y and Z are 0.01 each.

22. Number of atoms = Number of moles
Number of atoms in 24 g C =

× Avogadro’s number (N A)
24
× NA = 2NA
12


Some Basic Concepts of Chemistry 11
56
NA = NA
56
27
Number of atoms in 27 g of Al =
NA = NA
27
108
Number of atoms in 108 g of Ag =
NA = NA
108
Hence, 24 g of carbon has the maximum number of atoms.

Number of atoms in 56 g of Fe =

30. Number of molecules present in 36 g of water
36

× N A = 2N A
18
28
Number of molecules present in 28 g of CO =
× NA = NA
28
46
Number of molecules present in 46 g of C2H5OH =
× NA = NA
46
54
Number of molecules present in 54 g of N2O5 =
× N A = 0.5 N A
108
Here, NA is Avogadro’s number. Hence, 36 g of water contain
the largest (2NA ) number of molecules.
=

23. Mass of an electron = 9.108 × 10−31 kg
Q 9.108 × 10−31 kg = 1.0 electron
1
1031
1
electrons
=
×
∴ 1 kg =
−31
9.108 6.023 × 1023
9.108 × 10

1
=
× 108 mole of electrons
9.108 × 6.023

31. In a neutral atom, atomic number represents the number of
protons inside the nucleus and equal number of electrons around
it. Therefore, the number of total electrons in molecule of CO2
= electrons present in one carbon atom
+ 2 × electrons present in one oxygen atom
= 6 + 2 × 8 = 22.

24. Phosphorus acid is a dibasic acid as :
O

H—P — OH only two replaceable hydrogens

OH
Therefore, normality = molarity × basicity = 0.3 × 2 = 0.60

25. Molality is defined in terms of weight, hence independent of
temperature. Remaining three concentration units are defined in
terms of volume of solution, they depends on temperature.

26. Molality of a solution is defined as number of moles of solute
present in 1.0 kg (1000 g) of solvent.

27. The balanced chemical reaction is
3BaCl 2 + 2Na 3PO4 → Ba 3 (PO4 )2 + 6NaCl
In this reaction, 3 moles of BaCl 2 combines with 2 moles of

Na 3PO4. Hence, 0.5 mole of of BaCl 2 require
2
× 0.5 = 0.33 mole of Na 3PO4.
3
Since, available Na 3PO4 (0.2 mole) is less than required mole
(0.33), it is the limiting reactant and would determine the
amount of product Ba 3 (PO4 )2.
Q 2 moles of Na 3PO4 gives 1 mole Ba 3 (PO4 )2
1
∴0.2 mole of Na 3PO4 would give × 0.2 = 0.1 mole Ba 3 (PO4 )2
2

28. Unlike other metal carbonates that usually decomposes into
metal oxides liberating carbon dioxide, silver carbonate on
heating decomposes into elemental silver liberating mixture of
carbon dioxide and oxygen gas as :
Heat

Ag2CO3 (s) → 2Ag (s) + CO2 (g ) +

1
O (g )
2 2

MW = 276 g
2 × 108 = 216 g
Hence, 2.76 g of Ag2CO3 on heating will give
216
× 2.76 = 2.16g Ag as residue.
276


29. The balanced chemical reaction of zinc with sulphuric acid and
NaOH are
Zn + H2SO4 → ZnSO4 + H2 (g ) ↑
Zn + 2NaOH + 2H2O → Na 2[ Zn(OH)4 ] + H2 (g ) ↑
Since, one mole of H2 (g ) is produced per mole of zinc with both
sulphuric acid and NaOH respectively, hydrogen gas is
produced in the molar ratio of 1:1 in the above reactions.

32.

Weight of a compound in gram (w)
= Number of moles (n)
Molar mass (M )
Number of molecules (N )
=
Avogadro number (NA )
w (O2 ) N (O2 )
…(i)
⇒
=
32
NA
w (N2 ) N (N2 )
And
…(ii)
=
28
NA
Dividing Eq. (i) by Eq. (ii) gives

N (O2 ) w (O2 ) 28 1 28 7
= ×
=
=
×
N (N2 ) w (N2 ) 32 4 32 32

33. Molar mass of Na 2CO3⋅ xH2O .
(Atomic mass of Na = 23, C = 12, O = 16)
= 23 × 2 + 12 + 48 + 18x
= 46 + 12 + 48 + 18x
= (106 + 18x )
Equivalent weight of Na 2CO3⋅ xH2O
Molar mass M
=
=
= (53 + 9x )
2
n factor
[ Here, m = molar mass and n factor = 2]
Weight
Gram equivalent =
Equivalent weight
[Given, weight of Na 2CO3 ⋅ xH2O = 143
. g]
Hence, gram equivalent of
1.43
Na 2CO3 ⋅ xH2O =
53 + 9x
Gmeq

Normality =
Vlitre
143
.
01
. =
53 + 9x
.
01
As, volume = 100 mL = 0.1 L


12 Some Basic Concepts of Chemistry
So,

10−2 =

143
.
53 + 9x

36. Given,
O

53 + 9x = 143
9x = 90
x = 10.00

NH3/∆
Br2/KOH

Br2(3-eqiv.)
NaOBr
C
D
A
B
AcOH
H3O+ (60%)
(50%)
(100%)
(50%)

34. The equations of chemical reactions occurring during the process
are
In the presence of oxygen

HO

O

2PbS + 3O2 → 2PbO + 2SO2
By self reduction

207 g

So, 32 g of O 2 gives 207 g of Pb
207
g of Pb
1 g of O 2 will give
32

207
1000g of O 2 will give
× 1000 = 6468.75 g
32
= 6.46875 kg ≈ 6.47kg

Benzoic acid
(60%)
i.e.,
6 mol
(A)

Acetophenone
10 mol

(1) 2MnCl 2 + 5K 2S2O 8 + 8H2O →
2KMnO 4 + 4 K 2SO 4 + 6H2SO 4 + 4HCl
(2) 2KMnO 4 + 5H2C 2O 4 + 3H2SO 4 →
K 2SO 4 + 2MnSO 4 + 8H2O +10CO 2
Given, mass of oxalic acid added = 225mg
225
So, millimoles of oxalic acid added =
= 2.5
90
Now from equation 2
Millimoles of KMnO 4 used to react with oxalic acid=1 and
Millimoles of MnCl 2 required initially=1
∴ Mass of MnCl 2 required initially = 1 × (55 + 71) = 126mg
Alternative Method


O

NH3/∆

NaOBr
H3O+

Benzamide
(50%)
i.e.,
3 mol
(B)

NH2

Br2/KOH

NH2

Br

Br
Br2(3-eqiv.)
AcOH

Br
2,4,6-tribromo aniline
(100%)
i.e.,
1.5 mol

(D)

Aniline
(50%)
i.e.,
1.5 mol
(C)

NH2
Br

Br

So, 1.5 mol of

are produced from

35. The balanced equations are

m moles of MnCl 2 = m moles of KMnO 4 = x (let)
and M eq of KMnO 4 = M eq of oxalic acid
225
So,
x×5=
×2
90
Hence, x = 1
∴ m moles of MnCl 2 = 1
Hence mass of MnCl 2 = (55 + 71) × 1 = 126 mg.


H2N

O

…(i)

2PbO + PbS → 3Pb + SO2
Thus 3 moles of O2 produces 3 moles of Pb
i.e. 32 × 3 = 96 g of O 2 produces 3 × 207 = 621 g of Pb
So 1000 g (1kg) of oxygen will produce
621
× 1000 = 6468.75 g
96
= 6.4687 kg ≈ 6.47 kg
Alternative Method
From the direct equation,
PbS + O2 → Pb + SO2
32 g

The products formed are

Br

10 moles of acetophenone.
NH2
Br

Br

Molar mass of


= 240 + 14 + 4 + 72 = 330
Br
NH2
Br

Br

Hence, amount of

produced is 330 × 1.5 = 495 g
Br

37. Molar mass of CuSO4 ⋅ 5H2O
= 63.5 + 32 + 4 × 16 + 5 × 18
= 249.5 g
Also, molar mass represents mass of Avogadro number of
molecules in gram unit, therefore
Q 6.023 × 1023 molecules of CuSO4 ⋅ 5H2O weigh 249.5 g
249.5
× 1022 = 4.14 g
∴ 1022 molecules will weigh
6.023 × 1023


Some Basic Concepts of Chemistry 13
Number of moles of solute
Volume of solution in litre
Weight of solute
1000

=
×
Molar mass
Volume in mL
3 1000
=
×
= 0.4 M
30 250

38. Molarity =

39. Considering density of water to be 1.0 g/mL, 18 mL of water is
18 g (1.0 mol) of water and it contain Avogadro number of
molecules. Also one molecule of water contain
2 × (one from each H-atom) + 8 × (from oxygen atom)
= 10 electrons.
⇒ 1.0 mole of H2O contain = 10 × 6.023 × 1023
= 6.023 × 1024 electrons.

40. Carbon-12 isotope. According to modern atomic mass unit, one
atomic mass unit (amu) is defined as one-twelfth of mass of an
atom of C-12 isotope, i.e.
1
1 amu (u) =
× weight of an atom of C-12 isotope.
12
w
w
41. Moles of solute, n1 = 1 ; Moles of solvent, n2 = 2

m1
m2
χ 1 (solute) = 0.1and χ 2 (solvent) = 0.9
χ 1 n1 w1 m2 1
= =
⋅
=
χ 2 n2 m1 w2 9
Solute (moles) w1 × 1000 × 2
Molarity =
=
Volume (L)
m1 (w1 + w2 )

∴

Total mass of solution  w1 + w2 
Note Volume =
=
 mL
 2 
Density
Molality =
Given,
hence,
∴
∴

42.


Solute (moles) w1 × 1000
=
Solvent (kg)
m1 × w2

molarity = molality
2000 w1
1000 w1
=
m1 (w1 + w2 )
m1 w2
w2
1
=
⇒ w1 = w2 = 1
w1+ w2 2
w1 m2 1
m (solute)
= ⇒ 1
=9
m1 w2 9
m2 (solvent)

PLAN This problem can be solved by using concept of conversion of
molarity into molality.

Molarity = 3.2 M
Let volume of solution = 1000 mL = Volume of solvent
Mass of solvent = 1000 × 0.4 = 400 g
Since, molarity of solution is 3.2 molar

∴
n solute = 3.2 mol
3.2
Molality (m) =
=8
400 / 1000
Hence, correct integer is (8).

43. Mass of HCl in 1.0 mL stock solution
= 1.25 ×

29.2
= 0.365 g
100

Mass of HCl required for 200 mL 0.4 M HCl
200
=
× 0.4 × 36.5 = 0.08 × 36.5 g
1000
∴ 0.365 g of HCl is present in 1.0 mL stock solution.
0.08 × 36.5
= 8.0 mL
0.08 × 36.5 g HCl will be present in
0.365

44. Partial pressure of N2 = 0.001 atm,
T = 298 K, V = 2.46 dm 3.
From ideal gas law : pV = nRT
pV 0.001 × 2.46

n(N2 ) =
=
= 10−7
RT
0.082 × 298
⇒ Number of molecules of N2 = 6.023 × 1023 × 10−7
= 6.023 × 1016
Now, total surface sites available
= 6.023 × 1014 × 1000 = 6.023 × 1017
20
Surface sites used in adsorption =
× 6.023 × 1017
100
= 2 × 6.023 × 1016
⇒ Sites occupied per molecules
2 × 6.023 × 1016
Number of sites
=
= 2
=
Number of molecules
6.023 × 1016

45. Initial millimol of CH3COOH = 100 × 0.5 = 50
millimol of CH3COOH remaining after adsorption
= 100 × 0.49 = 49
⇒ millimol of CH3COOH adsorbed = 50 – 49 = 1
⇒ number of molecules of CH3COOH adsorbed
1
=

× 6.023 × 1023 = 6.023 × 1020
1000
3.01 × 102
⇒ Area covered up by one molecule =
6.02 × 1020
= 5 × 10−19 m 2

46. Mass of 1.0 L water = 1000 g
⇒

Molarity =

1000
= 55.56 mol L−1
18

47. Volume of one cylinderical plant virus = πr2l
= 3.14 (75 × 10−8 )2 × 5000 × 10−8 cm 3 = 8.83 × 10−17 cm 3
Volume of a virus
⇒ Mass of one virus =
Specific volume
=

8.83 × 10−17 cm 3
= 1.1773 × 10−16 g
0.75 cm 3 g−1

⇒ Molar mass of virus
= Mass of one virus × Avogadro’s number
= 1.1773 × 10−16 × 6.023 × 1023 g

= 70.91 × 106 g

48. Molar mass of Glauber’s salt (Na 2SO4 ⋅ 10H2O)
= 23 × 2 + 32 + 64 + 10 × 18 = 322g


14 Some Basic Concepts of Chemistry
⇒ Mole of Na 2SO4 ⋅ 10H2O in 1.0 L solution =

80.575
= 0.25
322

⇒ Molarity of solution = 0.25 M
Also, weight of 1.0 L solution = 1077.2 g
weight of Na 2SO4 in 1.0 L solution = 0.25 × 142 = 35.5 g
⇒ Weight of water in 1.0 L solution = 1077.2 – 35.5 = 1041.7 g
0.25
× 1000 = 0.24 m
⇒ Molality =
1041.7
Mole of Na 2SO4
Mole fraction of Na 2SO4 =
Mole of Na 2SO4 + Mole of water
0.25
= 4.3 × 10−3.
=
1041.7
0.25 +
18


49. Compound B forms hydrated crystals with Al 2 (SO4 )3. Also, B is
formed with univalent metal on heating with sulphur. Hence,
compound B must has the molecular formula M 2SO4 and
compound A must be an oxide of M which reacts with sulphur to
give metal sulphate as
A + S → M 2SO4
B

Q 0.321 g sulphur gives 1.743 g of M 2SO4
∴ 32.1 g S (one mole) will give 174.3 g M 2SO4
Therefore, molar mass of M 2SO4 = 174.3 g
⇒ 174.3 = 2 × Atomic weight of M + 32.1 + 64
⇒ Atomic weight of M = 39, metal is potassium (K)
K2SO4 on treatment with aqueous Al 2 (SO4 )3 gives potash-alum.
K2SO4 + Al 2 (SO4)3 + 24H2O → K2SO4Al 2 (SO4)3 ⋅ 24H2O
B

C

If the metal oxide A has molecular formula MOx, two moles of it
combine with one mole of sulphur to give one mole of metal
sulphate as
2KOx + S → K2SO4
⇒
x = 2, i.e. A is KO2.

50. The reaction involved is
3Pb(NO3 )2 + Cr2 (SO4 )3 → 3PbSO4 (s) ↓ + 2Cr(NO3 )3
millimol of Pb(NO3 )2 taken = 45 × 0.25 = 11.25

millimol of Cr2 (SO4 )3 taken = 2.5
Here, chromic sulphate is the limiting reagent, it will determine
the amount of product.
Q 1 mole Cr2 (SO4 )3 produces 3 moles PbSO4.
∴ 2.5 millimol Cr2 (SO4 )3 will produce 7.5 millimol PbSO4.
Hence, mole of PbSO4 precipitate formed = 7.5 × 10−3
Also, millimol of Pb(NO3 )2 remaining unreacted
11.25 – 7.50 = 3.75
⇒ Molarity of Pb(NO3 )2 in final solution
millimol of Pb(NO3 )2 3.75
= 0.054 M
=
=
Total volume
70
Also, millimol of Cr(NO3 )2 formed
= 2 × millimol of Cr2 (SO4 )3 reacted
5
= 0.071 M
⇒ Molarity of Cr(NO3 )2 =
70

51. 93% H2SO4 solution weight by volume indicates that there is
93 g H2SO4 in 100 mL of solution.

If we consider 100 mL solution, weight of solution = 184 g
Weight of H2O in 100 mL solution = 184 – 93 = 91 g
Moles of solute
⇒ Molality =
× 1000

Weight of solvent (g)
93 1000
= 10.42
=
×
98
91

52. Heating below 600°C converts Pb(NO3 )2 into PbO but to
NaNO3 into NaNO2 as

∆

Pb(NO3 )2 → PbO(s) + 2NO2 ↑ +
MW :

330

∆

222

NaNO3 → NaNO2 (s) +
MW :

85

1
O ↑
2 2


1
O ↑
2 2

69

28
Weight loss = 5 ×
= 1.4 g
100
⇒ Weight of residue left = 5 – 1.4 = 3.6 g
Now, let the original mixture contain x g of Pb(NO3 )2.
Q 330 g Pb(NO3 )2 gives 222 g PbO
222 x
g PbO
∴ x g Pb(NO3 )2 will give
330
Similarly, 85 g NaNO3 gives 69 g NaNO2
69 (5 − x )
g NaNO2
⇒ (5 – x) g NaNO3 will give
85
222 x 69 (5 − x )
= 3.6 g
⇒ Residue :
+
330
85
Solving for x gives, x = 3.3 g Pb(NO3 )2 ⇒ NaNO3 = 1.7 g.


53. Reactions involved are
C2H6 + Br2 → C2H5Br + HBr
2C2H5Br + 2Na → C4H10 + 2NaBr
Actual yield of C4H10 = 55 g which is 85% of theoretical yield.
55 × 100
= 64.70 g
⇒ Theoretical yield of C4H10 =
85
Also, 2 moles (218 g) C2H5Br gives 58 g of butane.
⇒ 64.70 g of butane would be obtained from
2
× 64.70 = 2.23 moles C2H5Br
58
Also yield of bromination reaction is only 90%, in order to have
2.23 moles of C2H5Br, theoretically
2.23 × 100
= 2.48 moles of C2H5Br required.
90
Therefore, moles of C2H6 required = 2.48
⇒ Volume of C2H6 (NTP) required = 2.48 × 22.4 = 55.55 L.
34.2
54. Moles of sugar =
= 0.1
342
Moles of water in syrup = 214.2 – 34.2 = 180 g
Moles of solute
Therefore, (i) Molality =
× 1000
Weight of Solvent (g)

0.1
=
× 1000 = 0.55
180
Mole of sugar
(ii) Mole fraction of sugar =
Mole of sugar + Mole of water
0.1
=
= 9.9 × 10−3
0.1 + 10


Some Basic Concepts of Chemistry 15
55. From the given elemental composition, empirical formula can
be derived as :
Element
Weight %
Mole %

C
69.77
5.81

H
11.63
11.63

Simple ratio


5

10

O
18.60
1.1625 (obtained by
dividing from M )
1

Hence, empirical formula is C5H10O and empirical formula
weight is 86.
Since, empirical formula weight and molecular weight both are
(86), empirical formula is the molecular formula also.
Also, the compound does not reduce Fehling’s solution,
therefore it is not an aldehyde, but it forms bisulphite, it must be
a ketone.
Also, it gives positive iodoform test, it must be a methyl ketone.
O

C3H7 — C — CH3
Based on the above information, the compound may be one of
the following :
O
CH3 O



CH3CH2CH2— C — CH3 or CH3 — CH— C — CH3
2-pentanone


3-methyl -2-butanone

56. (a) Let us consider 1.0 L solution for all the calculation.
(i) Weight of 1 L solution = 1250 g
Weight of Na 2S2O3 = 3 × 158 = 474 g
474
⇒ Weight percentage of Na 2S2O3=
× 100 = 37.92
1250
(ii) Weight of H2O in 1 L solution = 1250 − 474 = 776 g
3
Mole fraction of Na 2S2O3 =
= 0.065
776
3+
18
3×2
(iii) Molality of Na + =
× 100 = 7.73 m
776
57. (a) After passing through red-hot charcoal, following reaction
occurs
C(s) + CO2 (g ) → 2CO(g )
If the 1.0 L original mixture contain x litre of CO2, after
passing from tube containing red-hot charcoal, the new
volumes would be :
2x (volume of CO obtained from CO2) + 1
– x(original CO) = 1 + x =1.6 (given)
⇒

x = 0.6
Hence, original 1.0 L mixture has 0.4 L CO and 0.6 L of CO2 ,
i.e. 40% CO and 60% CO2 by volume.
(b) According to the given information, molecular formula of
the compound is M 3N2. Also, 1.0 mole of compound has 28 g
of nitrogen. If X is the molar mass of compound, then :
28
X ×
= 28
100
⇒
X = 100 = 3 × Atomic weight of M + 28
72
⇒ Atomic weight of M =
= 24
3

58. In the present case, V ∝ n (Q all the volumes are measured
under identical conditions of temperature and pressure) Hence,
the reaction stoichiometry can be solved using volumes as :
y
y

CxH y (g ) +  x +  O2 (g ) → x CO2 (g ) + H2O (l )

4
2
volume of CO2 (g ) + O2 (g ) (remaining unreacted) = 25
⇒ Volume of CO2 (g ) produced
= 10 mL (15 mL O2 remaining)

Q 1 mL CxH y produces x mL of CO2
∴ 5 mL CxH y will produce 5x mL of CO2 = 10 mL ⇒ x = 2
y

Also, 1 mL CxH y combines with  x +  mL of O2

4
y

5 mL CxH y will combine with 5  x +  mL of O2

4
y

⇒ 5  x +  = 15 (15 mL of O2 out of 30 mL)

4
(remaining unreacted)
⇒ y = 4, hence hydrocarbon is C2H4.

59. Oxides of sodium and potassium are converted into chlorides
according to following reactions :
Na 2O + 2HCl → 2NaCl + H2O
K2O + 2HCl → 2KCl + H2O
Finally all the chlorides of NaCl and KCl are converted into
AgCl, hence
moles of (NaCl + KCl) = moles of AgCl
(one mole of either NaCl or KCl gives one mole of AgCl)
Now, let the chloride mixture contain x g NaCl.
x

0.118 − x 0.2451
⇒
+
=
58.5
74.5
143.5
Solving for x gives x = 0.0338 g (mass of NaCl)
⇒

Mass of KCl = 0.118 – 0.0338 = 0.0842 g
1
Also, moles of Na 2O = × moles of NaCl
2
1 0.0338
× 62 = 0.0179 g
⇒ Mass of Na 2O = ×
2
58.5
1 0.0842
Similarly, mass of K2O = ×
× 94 = 0.053 g
2
74.5
0.0179
Mass % of Na 2O =
⇒
× 100 = 3.58 %
0.5
0.053

Mass % of K2O =
× 100 = 10.6 %
0.5

60. From the vapour density information
Molar mass = Vapour density × 2 (Q Molar mass of H2 = 2)
= 38.3 × 2 = 76.6
Now, let us consider 1.0 mole of mixture and it contains
x mole of NO2.
⇒

46 x + 92 (1 − x ) = 76.6 ⇒ x = 0.3348
100
Also, in 100 g mixture, number of moles =
76.6
100
⇒ Moles of NO2 in mixture =
× 0.3348 = 0.437
76.6


16 Some Basic Concepts of Chemistry
61. Most of the elements found in nature exist as a mixture of
isotopes whose atomic weights are different. The atomic weight
of an element is the average of atomic weights of all its naturally
occurring isotopes.

62. Average atomic weight
Σ Percentage of an isotope × Atomic weight
100

10.01x + 11.01 (100 − x)
⇒ x = 20%
⇒ 10.81 =
100
=

Therefore, natural boron contains 20% (10.01) isotope and 80%
other isotope.

Topic 2 Equivalent Concept, Neutralisation
and Redox Titration
1. In disproportionation reactions, same element undergoes

The difference in the volume of NaOH solution between the end
point and the equivalence point is not significant for most of the
commonly used indicators as there is a large change in the pH
value around the equivalence point. Most of them change their
colour across this pH change.

3. 100 mL (cm3) of hexane contains 0.27 g of fatty acid.
In 10 mL solution, mass of the fatty acid,
0.27
× 10 = 0.027 g
100
Density of fatty acid, d = 0.9 g cm−3
m=

∴Volume of the fatty acid over the watch glass,
m 0.027
V = =

= 0.03 cm3
d
0.9
Let, height of the cylindrical monolayer = h cm
Q Volume of the cylinder = Volume of fatty acid

oxidation as well as reduction.
Reduction
e.g.
+2

+1

2CuBr

0

h cm

CuBr2 +Cu
Oxidation

Here, CuBr get oxidised to CuBr2 and also it get reduced to Cu.
Other given reactions and their types are given below.

10 cm

⇒

V = πr 2 × h


⇒

h=

Reduction
+7

–

2 MnO4

–

+10I + 16 H

+

+2

2Mn

+ 5I2 + 8H2O

= 1 × 10−4cm

Oxidation

In the given reaction, MnO−4 get oxidised to Mn 2+ and I− get
reduced to I2. It is an example of redox reaction. The reaction

takes place in acidic medium.
2KMnO4 → K 2MnO4 + MnO2 + O2
The given reaction is an example of decomposition reaction.
Here, one compound split into two or more simpler compounds,
atleast one of which must be in elemental form.
2NaBr + Cl2 → 2NaCl + Br2
The given reaction is an example of displacement reaction. In
this reaction, an atom (or ion) replaces the ion (or atom) of
another element from a compound.

2. The graph that shows the correct change of pH of the titration
mixture in the experiment is

FeC2O4 , Fe2 (C2O4 )3 FeSO4 and
Fe2 (SO4 )3 in acidic medium with KMnO4 is as follows :
…(i)
FeC2O4 + KMnO4 → Fe3 + + CO2 + Mn 2+
Fe2(C2O4 )3 + KMnO4 → Fe3 + + CO2 + Mn 2+
FeSO4 + KMnO4 → Fe

3+

+ SO24−

+ Mn

2+

…(ii)
…(iii)


Change in oxidation number of Mn is 5. Change in oxidation
number of Fe in (i), (ii) and (iii) are +3, + 6, + 1, respectively.
neq KMnO 4 = neq [ FeC 2O 4 + Fe 2 ( C 2O 4 ) 3 + FeSO 4 ]

∴

n × 5 = 1× 3 + 1× 6 + 1× 1
n=2

5. Given, W Ca (HCO 3) 2 = 0.81 g
W Mg (HCO ) = 0.73 g
3 2
M Ca (HCO ) = 162 g mol −1,
3 2

M Mg(HCO 3 ) 2 = 146 mol −1
V H 2 O = 100mL

V(mL)

Now,
In this case, both the titrants are completely ionised.
⊕

= 1 × 10−6 m

4. The oxidation of a mixture of one mole of each of

pH


HCl + NaOH

0.03 cm3
V
=
πr2 3 × (10)2 cm2

+ −

- N a Cl + H O
2

As H is added to a basic solution, [ OHÈ ] decreases and [ H+ ]
increases. Therefore, pH goes on decreasing. As the equivalence
point is reached,[OHÈ ] is rapidly reduced. After this point [OH È]
decreases rapidly and pH of the solution remains fairly constant.
Thus, there is an inflexion point at the equivalence point.

neq (CaCO3) = neq [Ca(HCO3 )2 ]+ neq [Mg(HCO3 )2 ]

W
0.81
0.73
×2=
×2+
×2
100
162
146

W
= 0.005 + 0.005
∴
100
W = 0.01 × 100 = 1
1
Thus, hardness of water sample =
× 106 = 10,000 ppm
100