Preview Master Resource Book in Chemistry for JEE Main 2020 by Sanjay Sharma (2019)

MASTER
RES URCE

Book for

JEE Main

Chemistry
Specially Prepared Questions for JEE Main with

Complete Theory 2 Levels Exercises Exams Questions

SANJAY SHARMA

ARIHANT PRAKASHAN (Series), MEERUT


MASTER
RES URCE

JEE Main

Book for

Arihant Prakashan (Series), Meerut
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MASTER
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Book for

JEE Main

PREFACE
In sync with the recent changes in the test pattern and format of JEE Main (Joint Engineering Entrance), it
is my pleasure to introduce Master Resource Book in Chemistry for JEE Main, for the Students aspiring
a seat in a reputed Engineering College. JEE Main is a gateway examination for candidates expecting to
seek admission in Bachelor in Engineering (BE), Bachelor of Technology (B.Tech) and Bachelor of
Architecture (B.Arch) at Indian Institutes of Information Technology (IIITs), National Institutes of
Technology (NITs), Delhi Technological University and other Centrally Funded Technical Institutes (CFTIs).
JEE Main is also an examination which is like screening examination for JEE Advanced
(The gateway examination to India's most reputed Technical Institutes, Indian Institutes of Technology
IITs). Only the top 2.2 lacs students passed in JEE Main will be able to attempt JEE Advanced.
Gradually, the number of students aspiring for the seat in the Engineering College has increased rapidly in
the last 5 Years or so. This year nearly 10 lacs students appeared for JEE Main and only a few were able to
reserve a seat in the college of their choice, so there is a cut throat competition among the aspirants.
Thus, it calls for a systematic mastery of all the subjects of the test with paramount importance to
problem-solving. Most of the books now in the market have become repetitive with scant respect to the
needs of true and effective learning. This book has been designed to fulfill the perceived needs of the
students as such.
—

This book comprehensively covers all the topics of JEE Main Chemistry syllabus. The chapters have
been sequenced according to the syllabus of class 11th & 12th. Each chapter has essential theoretical

discussion of the related concepts with sufficient number of solved examples, practice problems and
other solved problems. In each chapter previous years' questions of AIEEE and JEE Main have been
included to help students know the difficulty levels and nature of questions asked in competitive
exams at this level.

—

All types of questions have been included in this book: Single Correct Answer Types, Multiple
Correct Answer Types, Reasoning Types, Matches, Passage-based Questions etc.

—

This is the only book which has its subject matter divided as per class 11th & 12th syllabus. It covers
almost all questions of NCERT Textbook & NCERT Exemplar problems.

It is hoped this new effort will immensely benefit the students in their goal to secure a seat in the
prestigious engineering college, and would be convenient to teachers in planning their teaching
programmes. Suggestions for further improvement are welcome from the students and teachers.

Sanjay Sharma


MASTER
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Book for

JEE Main

CONTENTS

PART I
Chapters from Class 11th Syllabus
1. Some Basic Concepts in Chemistry
2. States of Matter : Gaseous and Liquid States
3. Atomic Structure

3-41
42-79
80-128

4. Chemical Bonding and Molecular Structure

129-170

5. Chemical Thermodynamics

171-212

6. Equilibrium

213-267

7. Redox Reactions

268-293

8. Classification of Elements and Periodicity in Properties

294-328

9. Hydrogen

329-356

10. s-Block Elements-I

357-390

11. p-Block Elements-I

391-439

12. Purification and Characterisation of Organic Compounds

440-464

13. Some Basic Principles of Organic Chemistry

465-533

14. Isomerism in Organic Compounds

534-580

15. Hydrocarbons

581-634

16. Environmental Chemistry

635-660


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PART II
Chapters from Class 12th Syllabus
1. States of Matter : Solid State

663-701

2. Solutions

702-744

3. Electrochemistry

745-788

4. Chemical Kinetics

789-832

5. Surface Chemistry

833-871

6. General Principles and Processes of Isolation of Metals

872-896

7. p-Block Elements-II

897-960

8. The d and f- Block Elements-II

961-988

9. Coordination Compounds

989-1032

10. Organic Compounds containing Halogens

1033-1078

11. Organic Compounds containing Oxygen

1079-1149

12. Organic Compounds containing Nitrogen

1150-1200

13. Polymers

1201-1232

14. Biomolecules

1233-1276

15. Chemistry in Everyday Life

1277-1300

16. Principles Related to Practical Chemistry

1301-1335

JEE Main Solved Papers
Solved Papers 2013 (Online & Offline)

1-30

Solved Papers 2014

31-37

Solved Papers 2015

38-44

Solved Papers 2016

45-50

Solved Papers 2017

1-7

Solved Papers 2018

1-8

Online JEE Main 2019 Solved Papers
(April & January Attempt)

1-32


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SYLLABUS
Section- A (Physical Chemistry)
UNIT 1 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 2 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 3 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.


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UNIT 4 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 5 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 6 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 7 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 ΔG o 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.


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UNIT 8 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.
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 9 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 temperature on rate of
reactions - Arrhenius theory, activation energy and its calculation, collision theory of bimolecular gaseous reactions (no
derivation).

UNIT 10 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.

Section- B (Inorganic Chemistry)
UNIT 11 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 12 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.

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UNIT 13 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 14 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 15 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 16 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.


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UNIT 17 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 18 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 19 Purification & Characterisation of Organic Compounds
Purification Crystallization, 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 20 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.


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UNIT 21 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.
Alkynes 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 monosubstituted benzene.

UNIT 22 Organic Compounds Containing Halogens
General methods of preparation, properties and reactions; Nature of C—X bond; Mechanisms of substitution
reactions. Uses/environmental effects of chloroform, iodoform

UNIT 23 Organic Compounds Containing Oxygen
General methods of preparation, properties, reactions and uses.
Alcohols, Phenols and Ethers
Alcohols Identification of primary, secondary and tertiary alcohols; mechanism of dehydration.
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 and factors affecting it.

UNIT 24 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 25 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.


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UNIT 26 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 27 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 28 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.
— Chemistry involved in the preparation of the following
— Inorganic compounds Mohr's salt, potash alum.
— 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.
— Chemical principles involved in the qualitative salt analysis
2+
2+
3+
3+
2+
2+
2+
2+

2+
4+
222- — Cations — Pb , Cu , Al , Fe , Zn , Ni , Ca , Ba , Mg NH . Anions – CO3 , S , SO4 , NO2, NO3, Cl , Br , I (Insoluble
salts excluded).
— 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.


Part - I
Chapters from Class 11

th

Syllabus



Some

1 Basic Concepts
in Chemistry
JEE Main MILESTONE
Matter and Its Nature
Physical Quantities and Their Measurements
in Chemistry
Atomic and Molecular Masses

Equivalent Mass or Equivalent Weight
Empirical and Molecular Formulae
Stoichiometric Problems of Different
Kinds

1.1 Matter and Its Nature
Matter is anything which occupies space and has mass. All the things around
us e.g., water, air, book, table etc., are matter.
There are five states of matter namely solid, liquid, gases, plasma and
Bose-Einstein condensate. Out of these, three states i.e., solid, liquid and gas
are general states and taught in our schools. These three states provide a basis
for the physical classification of matter.
Solids have a definite volume and shape; liquids have a definite volume but not
definite shape; gases have neither a definite volume nor a definite shape.
These three states of matter are the result of competition between
intermolecular interactions (attractive force between molecules) and thermal
energy (responsible for repulsion between molecules).
On heating, a solid usually changes to a liquid and the liquid on further
heating changes to the gaseous (or vapour) state. In the reverse process, a gas
on cooling liquifies to the liquid and the liquid on further cooling freezes to
the solid.

Plasma is seen as a state containing gaseous ions and free electrons and exists
when gaseous state is taken to very high temperatures (say 1000 to
1,000,000,000°C). Here, it is necessary that the entire gas as a whole have no
charge and is not of too much density. So, in short we can say plasmas as low
density ionised gases at very high temperatures. Plasmas can be seen in
northern lights or ball lightenings, flames, lightenings, neon lights, stars in
particular sun, clouds of gas and dust around stars.

BE condensate was predicted in 1924 by Satyendra Nath Bose and Albert
Einstein but due to lack of equipments, it was only created in 1935 by Cornell,
Ketterle and Weimann. Its concept and existence is totally opposite to
plasmas. The state is conceptualised at supercold conditions.

Chemistry is the branch of science
which deals with the composition,
properties and interaction of all
kinds of matter such as air, water,
rocks, plants, earth, etc.


4 JEE Main Chemistry
The supercold above means only a few billionth of a
degree above absolute zero. Cornell and Weimann
developed BEC at such temperature with rubidium.

Dalton’s Atomic Theory
J. Dalton in 1803, proposed the atomic theory of matter on
the basis of laws of chemical combinations. (which are
given later in this chapter)
According to which
(i) matter is made up of indivisible and indestructible
particles, called atoms.
(ii) all atoms of an element have identical mass and
similar chemical properties. (Atoms of different
elements have different masses and different
chemical properties).

Both atoms and molecules are basic constituents of matter

with the condition that atoms combine to form the
molecules.
The molecules may be
(i) monoatomic, i.e., contain 1 atom only, e.g., Na, K etc.
(ii) diatomic, i.e., contain 2 atoms, e.g., N 2 , O 2 etc.
(iii) triatomic, i.e., contain 3 atoms, e.g., O3 etc.
(iv) polyatomic, i.e., contain more than 3 atoms e.g., P4 , S 8 etc.

Chemical Classification of Matter
On the basis of chemical composition and properties,
matter can be classified as

Matter
Pure
substances

Mixtures

(iii) when atoms combine, they do so in the ratio of small
whole numbers to form compound atoms or simply

compounds or molecules. Compounds formed by
such combinations are alike in every respect.
(iv) chemical reactions involve only combination,
separation or rearrangement of atoms.
(v) atoms are neither created nor destroyed in the course
of an ordinary chemical reaction.

Limitations of Dalton’s Atomic Theory
(i) It failed to explain how atoms of different elements

differ from each other.
(ii) It failed to explain how and why atoms of elements
combine with each other to form compound atoms or
molecules.
(iii) It failed to explain the nature of forces that bind
together different atoms in a molecule.
(iv) It did not make any distinction between ultimate
particle of an element that takes part in reaction
(atoms) and the ultimate particle that has independent
existence (molecules).

The hypothesis of Dalton is even accepted today by the
scientific community with two modifications only.
1. Atom is divisible and destructible.
2. All atoms of an element are not identical in mass.

Atoms and Molecules
An atom is defined as “the smallest particle of matter which
may or may not exist independently but can take part in a
chemical reaction.’’
A molecule is defined as “the smallest particle of matter which
can exist independently but cannot take part in a chemical
reaction.’’

Homogeneous
mixture

Heterogeneous
mixture

Compounds

Elements

(a) Mixtures
These have variable composition and variable properties
due to the fact that components retain their characteristic
properties. These may be separated into pure components
by applying physical methods.
These can be of two types
(i) The homogeneous mixtures, have same composition
throughout
and
their
components
are
indistinguishable, e.g.,a liquid solution of sugar and
water etc.
(ii) A heterogeneous mixture, on the other hand, do not
have the same composition throughout and the
components here are distinguishable, e.g., a mixture
of grains of sand and salt. Here particles of each
component maintain their own identity.

(b) Pure Substances
These have fixed composition and non-variable
properties. These cannot be separated into simpler
substances by physical methods.
An element is a substance that contains only one type of
atoms whereas a compound is formed when atoms of

different elements combine in a fixed ratio.
Compounds and elements can be differentiated as the
former can be decomposed into simple substances by
chemical methods while later cannot be decomposed into
simpler substances by chemical methods.


Some Basic Concepts in Chemistry
Sample Problem 1 Which one of the following is not an
element?
(a) Graphite
(c) Diamond

(b) Silica
(d) Plasma sulphur

Interpret (b) Elements contain only one type of atoms.
Graphite and diamond both contain only C (carbon), plastic
sulphur contains only S but silica (SiO2) contains two different
atoms, i.e., Si and O, so it is not an element.

1.2 Physical Quantities and
Their Measurements in
Chemistry
The description, interpretation and prediction of the
behaviour of chemical substances can be done on the
basis of the knowledge of their physical and chemical
properties determined from careful experimental
measurements. The properties like mass, length, time,
temperature etc., are physical quantities and their

measurement does not involve any chemical reaction.
These properties are expressed in numerals with suitable
units.
The measurement of any physical quantity is represented
by a number followed by units in which it is measured.
For example length of a room can be represented as 12 m;
here 12 is the number and m denotes metre-the unit in
which the length is measured.

Precision and Accuracy
Precision is the measure of reproducibility of an experiment
while accuracy is the measurement of closeness of a result
to its true value. Good accuracy means good precision but
reverse is not always true.
i.e., precision = individual value - arithmetic mean value
Accuracy = mean value - true value

Sample Problem 2 Two students performed the same
experiment separately and each one of them recorded two
readings of mass which are given below. Correct reading of
mass is 3.0 g. On the basis of given data mark the correct option
out of the following statements.
[NCERT Exemplar]
Student

Readings
(i)
(ii)
A
3.01

2.99
B
3.05
2.95
(a) Results of both the students are neither accurate nor precise.
(b) Results of student A are both precise and accurate.
(c) Results of student B are neither precise nor accurate.
(d) Results of student B are both precise and accurate.

Interpret (b) Results of student A are close to true value as well
as to each other, that’s why these are precise as well as accurate.

5

Scientific Notation
In scientific notation, all numbers (however large or small)
are expressed as a number between 1.000 and 9.999
multiplied or divided by 10, i.e., here a number is
generally expressed in the form
N ´ 10n
Here, N is called digit term. It is a number between 1.000
and 9.999.
n is called an exponent. If decimal is shifted towards left,
the value of n is positive (and is equal to the number of
places by which decimal is shifted) and if decimal is
shifted towards right, the value of n is negative.
e.g., 138.42 can be written as 1.3842 ´ 102
or 0.013842 can be written as 1.3842 ´ 10-2

Significant Figures

The digits in a properly recorded measurement are known as
significant figures or in other words, we can say that significant
figures are the meaningful digits in a measured or calculated
quantity.
A significant figure includes all those digits that are known with
certainty plus one more which is uncertain or estimated.
Always remember that greater the number of significant
figures in a reported result, smaller the uncertainty.
There are certain rules for determining the number of significant
figures. These are as follows
1. All non-zero digits are significant. e.g., in 852 cm, there are
three significant figures and in 0.25 L there are two
significant figures.
2. Zeros preceding to first non-zero digit are not significant.
Such zero indicates the position of decimal point. e.g., 0.03 has
one significant figure and 0.0052 has one significant figure.
3. Zeros between two non-zero digits are significant e.g., 3.007
has four significant figures.
4. Zeros at the end or right of a number are significant provided
they are on the right side of the decimal point. e.g., 0.200 g has
three significant figures.
But, if otherwise, the terminal zeros are not significant if there
is no decimal point. e.g., 100 has only one significant figure,
but 10.0 has three significant figures and 100.0 has four
significant figures. Such numbers are better represented in
scientific notation. We can express the number 100 as1´ 10 2
for one significant figure, 10
. ´ 10 2 for two significant figures
2
and 100

. ´ 10 for three significant figures.
5. Counting numbers of objects, for example, 2 balls or 20 eggs,
have infinite significant figures as these are exact numbers
and can be represented by writing infinite number of zeros
after placing a decimal i.e., 2 = 2.000000 or 20 = 20.000000
6. In numbers written in scientific notation, all digits are
significant e.g., 4.01 ´ 10 2 has three significant figures.


6 JEE Main Chemistry
(a) 1
(c) 3

Significant Figures in Calculations
1. When adding or subtracting, the number of decimal
places in the answer should not exceed the number
of decimal places in either of the numbers. e.g.,

0.13
1.5
20.911

2 significant figures
2 significant figures
5 significant figures

2. In multiplication and division, the significant figures
in the answer should be the same as that in the
quantity with the least number of significant figures.

(a) CGS System
It is also called Gaussian system and is based on
centimetre (cm), gram (g) and second (s) as the units of
length, mass and time respectively.

(b) FPS System
It is a British system which used foot (ft), pound (lb) and
second (s) as the fundamental units of length, mass and
time.

0.01208
= 0.512
0.0236

The number 0.0236 has only three significant figures
that’s why the answer must also be limited to three
significant figures. Similarly, the product

132.07 ´ 0.12 = 15.8484
The answer 15.8484 should be reported as 15
because 0.12 has only two significant figures.
3. When a number is rounded off, the number of
significant figures is reduced. The last digit retained is
increased by 1 only if the following digit is > 5 and is
left as such if the following digit is £ 4. e.g.,

12.696 can be written as 12.7
13.93 can be written as 13.9
If the following digit is 5, left the number as such if it is
even or add 1 if it is odd. e. g .,

18.35 can be written as 18.4

Caution Point While calculating the significant figures of numbers,
it is better to convert them into scientific notation because exponential
term does not contribute to the significant figures.

Sample Problem 3 If the density of a solution is
-1

3.12 g mL , the mass of 1.5 mL solution in significant figures is
[NCERT Exemplar]

(b) 4680 ´ 10 -3 g
(d) 46.80 g

(c) MKS System
It is called MKSA system later on. It is the system, which
uses metre (m), kilogram (kg) and second (s) respectively
for length, mass and time; Ampere (A) was added later on
for electric current.

(d) SI System
It is internationally accepted system in 1960s, hence
called International system of units and contains
following 7 basic and 2 supplementary units.
(i) Basic units includes metre (m) for length, kilogram (kg)
for mass, second (s) for time, ampere (A) for electric
current, kelvin (K) for thermodynamic temperature,
mole (mol) for amount of substance and candela (Cd)

for luminous intensity.
(ii) Supplementary units includes radian (rad) for angle
and steradian (sr) for solid angle.

Sometimes submultiples and multiples are used to
reduce or enlarge the size of the different units. The names
and symbols of sub-multiples and multiples are listed in
the table given below.

Table 1.1 SI Prefixes
-1

Interpret (a) Mass = volume ´ density = 1.5 mL ´ 3.12 g mL

= 4.68 g
The digit 1.5 has only two significant figures, so the answer must
also be limited to two significant figures. Hence, it is rounded off to
reduce the number of significant figures. Hence, the answer is
reported as 4.7g.

Sample Problem 4 How many significant figures
should be present in the answer of the following
calculations?
[NCERT Exemplar]
2.5 ´ 1.25 ´ 3.5
2.01

2.5 or 3.5, has two significant figures, so answer must also be
reported in two significant figures.

Various Systems of Measurement

1.5 has only one digit after the decimal point and the
result should be reported only up to one digit after the
decimal point which is 22.5.

(a) 4.7 g
(c) 4.680 g

Interpret (b) Since the number with least significant figure i.e.,

Note Here is no need to do complete calculation.

22.541

e.g.,

(b) 2
(d) 4

Multiple
24

10
1021
1018
1015
1012
109
106

103
102
10

Prefix
yotta
zetta
exa
peta
tera
giga
mega
kilo
hecto
deca

Symbol
Y
Z
E
P
T
G
M
k
h
da

Multiple
-1

10
10-2
10-3
10-6
10-9
10-12
10-15
10-18
10-21
10-24

Prefix
deci
centi
milli
micro
nano
pico
femto
atto
zepto
yocto

Symbol
d
c
m
m
n

p
f
a
z
y


Some Basic Concepts in Chemistry
1 J = 107 erg

Derived Units

1 dyne = 10-5 N

The units of all other quantities, which are derived from
the above mentioned units, called the fundamental units,
are called the derived units. e.g,

1 atm = 101325 Nm-2 = 101325 Pa
1 bar = 1 ´ 105 Nm-2 = 1 ´ 105 Pa

3

Units of volume = length ´ breadth ´ height =m ´ m ´ m = m

1 L-atm = 101.3 J = 24.21 cal
1 mol (gas) = 22 . 4 L at STP

Table 1.2 Some Derived Properties and their Units
Definition of quantity

Expression in terms of
SI base units

Area

Length squared

m2

Volume

Length cubed

m3

Density

Mass per unit volume

kg / m3 or kg m–3

Velocity

Distance travelled per
unit time

m/s or ms -1

Acceleration

Velocity changed per
unit time

m / s 2 or ms –2

Force

Mass times
acceleration of object

kg m / s 2 or kg ms –2
(newton, N)

Quantity

Pressure

7

Force per unit area

1 mol (substance) = N A molecules
1 g -atom = N A atoms
1 K = t ° C + 273.15
t (° F) =

2

1 D (Debye) = 1 ´ 10-18 esu-cm

1 g-cm-3 = 1000 kg cm-3

–1 –2

kg / (ms ) or kg m s
(pascal, Pa)
2

9
t ° C +32
5

2

Caution Point Remember that CF must always have the numerator
and denominator representing equivalent quantities.

Sample Problem 5 If the speed of light is 3.0 ´ 10 8ms-1,
calculate the distance covered by light in 2.00 ns.
[NCERT Exemplar]

(a) 6.0 ´10
(c) 0.06 ´10

2 –2

Energy (work,
heat)

Force times distance

travelled

kg m / s or kg m s
(joule, J)

Electric charge

Ampere times second

A-s (coulomb, C)

Electric potential

Energy per unit
charge

J/(A-s) potential
difference (volt, V)

(b) 0.6 ´ 10
(d) 60.0 ´ 10 -1

Interpret (c) 1ns = 10 -9s
Conversion factor =

Dimensional Analysis

10 -9 s
1ns

2.00ns = 2.00ns ´

In calculations, generally there is a need to convert units
from one system to other. This can be done with the help
of conversion factor, so the method used to accomplish
this is called factor label method or unit factor method or

10 -9 s
1ns

= 2.00 ´ 10 -9 s
Distance covered = speed x time
= 3.0 ´ 10 8 ms-1 ´ 2 . 0 ´ 10 -9 s
= 6 . 0 ´ 10 -1 m = 0 . 6 m

dimensional analysis.
Information sought = information given ´ CF
Some conversion factors [CCF] are as follows.

Sample Problem 6 The relationship between picometer
(pm) and nanometer (nm) is

1m = 39.37 inch

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

1 inch = 2.54 cm

1L = 1000 mL = 1000 cm3
= 10-3m3= 1dm3
1 lb = 453 . 59237g

Interpret (a) 1 pm = 10 –12 m,1 nm = 10 -9 m

1J = 1N - m = 1 kgm 2 s-2

Conversion factor =

1cal = 4.184 J = 2.613 ´ 1019 eV
1eV = 1. 602 ´ 10-19 J
1eV /atom = 96. 485 kJ mol-1
1 u = 1.66 ´ 10-27 kg
= 931.5 MeV

1 nm = 100.0 pm
1 nm = 10 pm
1 pm = 10 nm
1 pm = 100 nm
10 -12m
1nm
and -9
1pm
10 m

1pm = 10 -12m ´
Þ

1 pm = 10 -3nm

or

1 nm = 1000 pm

1nm
10 -9m


8 JEE Main Chemistry
Sample Problem 7 Pressure is determined as force per
unit area of the surface. The SI unit of pressure, pascal is as
shown below.

Interpret (a) Pressure is the force or weight per unit area.
Pressure =

1 Pa = 1 Nm-2

=

If the mass of air at sea level is1034gcm-2 , calculate the pressure
[NCERT]
in pascal.
(a) 1.01 ´ 10

5

(c) 2 . 32 ´ 10

(b) 1.01 ´ 10
5

1034 kg ´ 100 ´ 100 ´ 9.8 ms-2
1000 m2

= 101332.0 Nm-2

4

(d) 1.03 ´ 10

1034g ´ 9.8ms-2
cm2

(1 N = kg ms-2)

= 1.01332 ´ 10 5 Pa

5

Laws of Chemical Combinations

According to this law the matter can neither be created
nor destroyed in a chemical reaction.

These lines are considered as the law of definite
proportions by volume given by Gay-Lussac, a French
chemist. e.g., in the reaction of hydrogen with oxygen to
produce water, it was found that 2 vol of H2 combines with

1 vol of O 2 to form 2 vol of H2O (steam). This simply means
that 100 mL of H2 gas combines with 50 mL of O 2 to
produce exactly 100 mL of steam (if volume of all the gases
are measured at same temperature and pressure).

Law of Definite Proportions (by J. Proust)

Avogadro’s Law

“A sample of a pure compound always consists of the
same elements combined in same proportions by mass,
whatever be its source.”

“The volume of a gas (at constant pressure and
temperature) is proportional to the number of moles
(or molecules) of gas present”.

The combination of elements to form compounds is
governed by following basic laws.

Law of Conservation of Mass
(by Lavoisier)

e.g., ammonia always has the formula NH3 i.e., one
molecule of NH3 always contains one atom of nitrogen
and three atoms of hydrogen or 17.0 g of NH3 always
contains 14 g of nitrogen and 3 g of hydrogen. These
findings always remain the same for NH3.

Law of Multiple Proportions

(by John Dalton)
An element may form more than one compound with
another element. For a given mass of an element, the
masses of other elements (in two or more compounds)
come in the ratio of small integers.”
This is called law of multiple proportions. e.g., in NH3, 14 g
of nitrogen requires 3 g of hydrogen and in hydrazine
(N2H4 ) 14 g of nitrogen requires 2 g of hydrogen. Hence,
fixed mass of nitrogen requires hydrogen in the ratio 3 : 2
in two different compounds (3 : 2 is a simple ratio). Thus,
this is in agreement with “law of multiple proportions’’.

Gay Lussac’s Law of Gaseous Volumes
The volume of reactants and products in a large number
of chemical reactions are related to each other by small
integers, provided the volumes are measured at same
temperature and pressure”.

According to this law
V µn
where, n = number of moles of gas
In simpler words, the law can also be stated as “equal
volumes of all gases, under the same conditions of
temperature and pressure contain equal number of
molecules’’ which is infact 6.023 ´ 1023 or in multiples of it.

Caution Point Law of definite proportions and law of multiple
proportions do not hold good when same compound is prepared by
16

different isotopes of the same element, e.g., H2 O and D2 O or H2 O and
18

H2 O. Moreover, law of conservation of mass does not hold good for
nuclear reactions.

Sample Problem 8 Which of the following reactions is not
correct according to the law of conservation of mass?
[NCERT Exemplar]

(a) 2Mg( s) + O2( g ) ắắđ 2MgO( s)
(b) C3H8( g ) + O2( g ) ắắđ CO2( g ) + H2O( g )
(c) P4( s) + 5O2( g ) ắđ P4 O10( s)
(d) CH4( g ) + 2O2( g ) ắđ CO2( g ) + 2H2O( g )

Interpret (b) In equation,
C3H8( g ) + O2( g ) ắđ CO2( g ) + H2O( g )
44g + 32g

44 + 18

i.e., mass of reactants ¹ mass of products. Hence, law of
conservation of mass is not followed.


Some Basic Concepts in Chemistry
Sample Problem 9 The following data are obtained
when dinitrogen and dioxygen react together to form different
compounds.
S. No.

Mass of dinitrogen

Mass of dioxygen

(i)

14g

16g

(ii)

14g

32g

(iii)

28g

32g

(iv)

28g

80g

9

If in place of ratio, relative abundance is given, the
m ´ r + m2 ´ r2 + m3 ´ r3
Mav = 1 1
r1 + r2 + r3
Here, r1, r2 and r3 = relative abundances of the isotopes.
The approximate atomic mass of solid elements except
Be, B, C and Si, is related to specific heat as
Average atomic mass =

6.4
specific heat

This is called Dulong and Petit’s method.
Which law of chemical combination is obeyed by the above
experimental data?
[NCERT]
(a) Law of multiple proportions
(b) Law of conservation of mass
(c) Law of definite proportions
(d) All of the above

Interpret (a) On fixing the mass of dinitrogen as 28g, the
masses of dioxygen combined are 32, 64, 32 and 80 in the given
four oxides. These are in the simple whole number ratio i.e.,
2 : 4 : 2 : 5. Hence, the given data obey the law of multiple
proportions.

1.3 Atomic and Molecular Masses
Atomic Mass

Dalton gave the idea of atomic masses in relative terms,
i.e., the average mass of one atom relative to the average
mass of the other. We can make accurate measurement of
mass by comparing mass of an atom with the mass of a
particular atom chosen as standard. On the present
atomic mass scale, 12 C is chosen as standard and is
arbitrarily assigned the mass of 12 atomic mass unit
(amu).

Mass spectrometer is used to determine the atomic mass
experimentally.

Molecular Mass
Molecular mass is the sum of atomic masses
of the elements present in a molecule. It is obtained
by multiplying the atomic mass of each element
by the number of its atoms and adding them together e.g.,
Molecular mass of glucose (C6H12O 6 )
= 6(12.011u) + 12(1.008u) + 6(16.00u)
= (72.066u) + (12 . 096u) + (96.00u)
= 180.162u

Formula Mass
The formula mass of a substance is the sum of the atomic
masses of all atoms in the formula unit of the compound.
It is normally calculated for ionic compounds. e.g.,
formula mass of NH3 is 14 + 3 = 17 amu or 17u
or formula mass of NaCl is
23 + 35.5 = 58.5 amu or 58.5u.

\ Atomic mass
=

mass of one atom of the element
1
th part of the mass of one atom of C - 12
12

Therefore, one amu or u (unified mass) is equal to exactly
the 1/12th of the mass of 12 C atom.
1 u=

1
12 g
´
12 6.022 ´ 1023

= 1.66 ´ 10-24 g
Since most of the elements have isotopes, the atomic mass
of an element is, infact, the average of masses of its all the
naturally occurring isotopes, so generally in fractions e.g.,
If an element exists in three isotopic forms having atomic
masses, m1, m2 and m3 in the ratio, x, y and z, the average
atomic mass,
m ´ x + m2 ´ y + m3 ´ z
Mav = 1
x+y+z

Check Point 1
1. How many millimetres are there in 14.0 cm?

2. Explain why in calculations involving more than one
arithmetic operation, rounding off to the proper number
of significant figures may be done once at the end if all
the operations are multiplications or divisions or if they are
all additions and subtractions, but not if they are
combinations of additions or subtractions with divisions or
multiplications?

3. Why do atomic masses of most of the elements in atomic mass
unit be in fractions?

4. On analysis it was found that the black oxide of copper and red
oxide of copper contain 79.9% and 88.8% of copper
respectively. This data is in accordance with which law of
combination?


10 JEE Main Chemistry

1.4 Equivalent Mass or
Equivalent Weight
The number of parts of a substance that combines with or
displaces, directly or indirectly, 1.008 parts by mass of
hydrogen or 35.5 parts by mass of chlorine or 8 parts by
mass of oxygen is called the equivalent mass of the
substance. Mathematically,
Eq. wt. of substance A Wt. of substance A
=
Eq. wt. of substance B Wt. of substance B

Equivalent weight of a substance that undergoes
oxidation/ reduction =

e.g., When KMnO4 reacts under acidic conditions, change
in oxidation number (from +7 to +2) is 5, hence;
Equivalent weight of KMnO4 in acidic medium
158
=
= 31.6
5
For volatile metal chlorides, eq. wt. and atomic weight are
related as
Atomic wt. = eq. wt. ´ valency

If A is metal and B is H (or O or Cl) then
Eq. wt. of metal =

mass of metal
´ 1.008
mass of hydrogen displaced

or

mass of metal
´ 8.0
mass of oxygen combined

=

mass of metal

or
=
´ 35.5
mass of chlorine combined
Equivalent weight of acid
molecular weight of acid
=
basicity (number of replaceable H+ )
e.g., Equivalent weight of H2SO4 =

98
= 49
2

Equivalent weight of base
molecular weight of base
=
acidity (number of replaceable OH- )
40
= 40
1
molecular weight of salt
Eq. wt. of salt =
total positive valency of metal atoms

Caution Points
(i) Atomic and molecular masses of elements and compounds are
always constant but equivalent mass may vary with change of
valency.
(ii) The valencies of elements forming isomorphous compounds

(i.e., the compounds that have similar constitution and chemical
formulae) are same. e.g., valencies of Cr, Se and S in K 2 CrO4 ,
K 2SeO4 and K 2SO4 are same.

Sample Problem 10 0.5 g of a metal on oxidation gave
0.79 g of its oxide. The equivalent weight of the metal is
(a) 10
(c) 20

(b) 14
(d) 40

Interpret (b) Eq. wt. =
=

e.g., Equivalent weight of NaOH =

58.5
e.g.,
Equivalent weight of NaCl =
= 58.5
1
106
Equivalent weight of Na2CO3 =
= 53
2

molecular weight
change in oxidation number

wt. of metal
´8
wt. of oxygen combined
0.5
´ 8 » 14
0.79 – 0.5

Sample Problem 11 The equivalent weight of iron in
Fe2O3 would be
(a) 18.6
(c) 56

(b) 28
(d) 112

Interpret (a) Eq. wt. of metal =

atomic wt. 56
=
= 18.6
valency
3


Hot Spot 1

Mole Concept

Mole concept is an important topic for JEE Main examination and a small practice can help you in solving
problems based on this topic very quickly as the level of questions is easy to average.

While solving problems based on mole concept, always keep in mind

ul

(i) The amount which weighs exactly same as its formula weight
in gram or atomic mass in gram or molecular mass in gram.
(ii) The amount which has same number of entities (atoms,
molecules or other particles) as there are atoms in exactly
0.012 kg (or 12 g) of carbon-12 isotope (i.e., 6.023 ´1023
entities).
(iii) The amount which occupies 22.4 L at STP
(if it is taken for a gas).

The formulae used to convert amount of substance into
moles are
weight in gram
Number of moles of molecule =
molecular weight
weight in gram
Number of moles of atoms =
atomic weight
number of particles
Number of moles =
Avogadro’ s number
or Number of moles of gases =

volume of gas at STP (in L)
22.4

As in atoms and molecules, mole concept is also applicable to ionic

compounds, which do not contain molecules. In such cases, the
formula of any ionic compound is representation of ratio between
constituent ions. One mole of an ionic compound is represented
by 6.023 ´ 1023 formula units.

e

ol
D

iv

id

M

by

ar

m
as
s

by

Thus, one mole of any substance is defined as

d

One mole of object will always mean 6.023 ´ 1023 of those objects.
The number of object per mole, 6.023 ´ 1023 mol-1 is called
Avogadro’s Number/Constant represented by NA .

X
NA

lie

Amount
of substance
(g )

tip

The word mole was introduced around 1896 by W. Ostwald who
derived it from Latin word moles means a heap or pile. In 1967 this
word was accepted as a unit of chemical substances under SI system. It
is represented by the symbol mol.

M

(i) Convert all in same unit, i. e., mol/atoms/mass and then compare.
(ii) Moles µ molecules

No. of
entities

Mol
X

%
22.4 L

Volume of
gas at STP
Flow chart showing conversion of mole in other units

One mole of NaCl = 6.023 ´ 1023 NaCl units
= 6.023 ´ 1023 units of Na+ + 6.023 ´ 1023 units of ClMass of one mole of NaCl = 23.0g + 35.5g
= 58.5 g NaCl

Sample Problem 12 The number of atoms present in
one mole of an element is equal to Avogadro number.
Which of the following element contains the greatest
number of atoms?
[NCERT Exemplar]
(a) 4 g He
(c) 0.4 g Ca

(b) 46 g Na
(d) 12 g He

Interpret (d) For comparing number of atoms, first we
calculate the moles as all are monoatomic and hence,
moles ´ NA = number of atoms.
Moles of 4g He =
46g Na =
0.40 g Ca =
12g He =

4
= 1 mol
4
46
= 2 mol
23
0.40
= 0.1 mol
40
12
= 3 mol
4

Therefore, number of atoms are highest in 12g He as it has the
maximum number of moles.


12 JEE Main Chemistry
Various Concentration Terms
Different concentration terms are given below:

(a) Normality (N)
It is defined as the number of g-equivalents of solute per
litre of solution or as the number of mg-equivalents of a
substance per millilitre of solution. e.g., 0.12 N H2SO4
means a solution which contains 0.12 g-equivalent of
H2SO4 per litre of solution. This also means that each
millilitre of this solution can react, for example, with
0.12 mg eq. of CaO or with 0.12 mg-eq. of Na2CO3.

Thus, molarity is given as
Molarity, M =

moles of solute
volume of solution (in L)

If specific gravity is given,
specific gravity ´ % strength ´ 10
Molarity =
molecular weight

(c) Formality (F)

Thus,
Normality N =

or

mg-molecules per millilitre of solution. The molarity is
usually designated by M, e.g., if the molarity of H3PO4 is
0.18, it means a concentration corresponding to 0.18 mol
of H3PO4 per litre of solution.

=

g- equivalent of solute
volume of solution (in L)

g - equivalent of solute
´ 1000

volume of solution (in mL)

If specific gravity is known, normality is calculated as
Normality =

specific gravity ´ % strength ´ 10
equivalent weight

(b) Molarity (M)
It is defined as the number of moles of solute per litre of
solution or the same numerically, as the number of

It is practically same as molarity.
gram formula weight
Formality =
volume in litre

(d) Molality (M)
It is defined as the number of moles of solute dissolved in
1000 g of the solvent. It is designated by m. Molality is
independent of temperature, as it depends only upon the
mass which does not vary with temperature.
Molality,

m=

moles of solute
´ 1000
weight of solvent (in g)

Formulae Used to Calculate the Number of Moles
1. Number of millimoles = molarity ´ volume in mL
millimoles
or
Moles =
1000

For very dilute solutions,

2. Number of equivalents of a substance
weight (g)
=
equivalent weight

where, M = molecular weight of solvent, m = molality

Mole fraction of solute in solution =

8. Density of solution
molecular weight of solute ử
ổ 1
= molarity ỗ
+
ữ
ố molality
ứ
1000

3. Number of milli-equivalent
1000 weight (g)

(Meq) =
equivalent weight
= normality ´ volume (mL)

9. Per cent by weight of solute in solution
weight of solute (g) ´ 100
=
weight of solution (g)

4. Normality = molarity ´ balance factor (y).
where, y = acidity/basicity/number of replaceable
H-atoms/change in oxidation number.

10. Percent by volume of solute in solution
weight of solute (g) ´ 100
=
volume of solution (mL)

5. Number of equivalents = y ´ number of moles

11.
12.
13.
14.

6. Number of milli-equivalents = y ´ number of millimoles
where, y is same as for formula (4).
7. Mole fraction of solute in the solution =

Mm

1000

n
n+N

where, n = moles of solute, N = moles of solvent

1 mL, 1 N KMnO 4 º 1 mL, 1 N H 2O 2
1 mL, 1 N Na 2S 2O3 º 1 mL, 1 N I 2 solution
1 volume H 2O 2 = 0.1785 N H 2O 2
If two compounds A and B neutralise completely each other
then
milli-equivalents of A = milli-equivalents of B
15. Molecular weight = 2 ´ vapour density (for gaseous phase).