JEE Main
CHAPTERWISE

SOLUTIONS
2019-2002

Chemistry
All the 16 Question Papers
of JEE Main Online 2019 (Jan & Apr Attempt)

ARIHANT PRAKASHAN (Series), MEERUT


Arihant Prakashan (Series), Meerut
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PREFACE
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).
To make the students well-versed with the pattern as well as the

level of the questions asked in the exam, this book contains
Chapterwise Solutions of Questions asked in Last 18 Years’
Examinations of JEE Main (formerly known as AIEEE). Solutions to
all the questions have been kept very detailed and accurate. Along
with the indication of level of the exam, this book will also teach you
to solve the questions objectively in the examination.
To give the students a complete practice, along with Chapterwise
Solutions, this book contains 5 Practice Sets, based exactly on JEE
Main Syllabus and Pattern. By practicing these sets, students can
attain efficiency in Time Management during the examination.
We hope this book would be highly beneficial for the students. We
would be grateful if any discrepancy or mistake in the questions or
answers is brought to our notice so that these could be rectified in
subsequent editions.

Publisher


CONTENTS
1. Some Basic Concepts of Chemistry

1-8

2. States of Matter

9-20

3. Atomic Structure

21-28


4. Chemical Bonding

29-40

5. Thermodynamics

41-56

6. Solutions

57-72

7. Equilibrium

73-90

8. Redox Reactions and Electrochemistry
9. Chemical Kinetics and Surface Chemistry

91-107
108-128

10. Periodicity of Elements

129-134

11. Principles and Processes of Metallurgy

135-140


12. Hydrogen, s-and p-Block Elements

141-165

13. d-and f-Block Elements and Coordination Chemistry

166-191

14. Environmental Chemistry

192-198

15. General Organic Chemistry

199-214

16. Hydrocarbons and their Halogen Derivatives

215-238

17. Organic Compounds Containing Oxygen (Alcohols, Ethers,
Aldehydes, Ketones, Carboxylic Acids and their Derivatives)

239-276

18. Organic Compounds Containing Nitrogen
(Amines and Diazonium Salts)

277-293


19. Polymers and Biomolecules

294-307

20. Analytical Chemistry and Chemistry in Everyday Life

308-314

PRACTICE SETS for JEE MAIN
Practice Set 1
Practice Set 2
Practice Set 3
Practice Set 4
Practice Set 5

315-320
321-326
327-331
332-336
337-342


SYLLABUS
SECTION- A (Physical Chemistry)
UNIT 1 Some Basic Concepts in hemistry
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.



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.

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 10 Surface Chemistry
Adsorption Physisorption and chemisorption and
their characteristics, factors affecting adsorption of
gases on solids- Freundlich 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 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).

SECTION- B (Inorganic Chemistry)
UNIT 11 Classification of Elements and
Periodicity in Properties

UNIT 14 s - Block Elements

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.

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 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.

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.

(Alkali and Alkaline Earth Metals)

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.

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.

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 mono-substituted
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, NH and its derivatives), Grignard
3

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.

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+
— Cations — Pb , Cu , Al , Fe , Zn , Ni , Ca , Ba ,
2+
4+
222Mg 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.


1
Some Basic Concepts
of Chemistry
1.

Exp. (d)

In order to oxidise a mixture of one mole of

each of FeC 2O4 , Fe 2 (C 2O4 )3 , FeSO4 and
Fe 2 (SO4 )3 in acidic medium, the number of
moles of KMnO4 required is

Given, WCa(HCO 3 )2 = 0.81 g
WMg(HCO ) = 0.73 g
3 2
MCa(HCO 3 )2 = 162 g mol −1,
MMg(HCO 3 )2 = 146 mol −1

[JEE Main 2019, 8 April Shift-I]

(a) 2

(b) 1

(c) 3

(d) 1.5

MH 2O = 100 mL
Now, neq (CaCO 3)= neq [Ca(HCO 3 )2 ] + neq [Mg(HCO 3 )2 ]

Exp. (a)

W
0.81
073
.
×2 =

×2 +
×2
100
162
146
W
= 0.005 + 0.005
∴
100
W = 0.01 × 100 = 1
1
Thus, hardness of water sample =
× 10 6
100
= 10,000 ppm

The oxidation of a mixture of one mole of each of
FeC 2O 4 , Fe 2 (C 2O 4 )3 FeSO 4 and
Fe 2 (SO 4 )3 in acidic medium with KMnO 4 is as follows :
FeC 2O 4 + KMnO 4 → Fe 3+ + CO 2 + Mn2 + …(i)
Fe 2(C 2O 4 )3 + KMnO 4 → Fe 3+ + CO 2 + Mn2 + …(ii)
FeSO 4 + KMnO 4 → Fe 3+ + SO 24− + Mn2 + …(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.
neqKMnO 4 = neq [FeC 2O 4 + Fe 2 (C 2O 4 )3 + FeSO 4 ]
n × 5 = 1× 3 + 1× 6 + 1× 1
∴
n=2


2.

3.

0.27 g of a long chain fatty acid was dissolved in
100 cm3 of hexane. 10 mL of this solution was
added dropwise to the 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?

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 )
[JEE Main 2019, 8 April Shift-I]

Exp. (a)

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

100 mL (cm3) of hexane contains 0.27 g of fatty acid.
In 10 mL solution, mass of the fatty acid,

0.27
m=
× 10 = 0.027 g
100

5,000 ppm
1,000 ppm
100 ppm
10,000 ppm

[Density of fatty acid = 0.9 g cm −3 ; π = 3]
[JEE Main 2019, 8 April Shift-II]
−6

(a) 10

−4

m (b) 10

m (c) 10−8 m (d) 10−2 m


2

JEE Main Chapterwise Chemistry
Density of fatty acid, d = 0.9 g cm−3

Here, H 2 gas does not act as limiting reagent since
7.5 g of H 2 gas is required for 35 g of N 2 and 8 g of H 2

is present in reaction mixture. Mass of H 2 left unreacted
= 8 − 7.5 g of H 2. = 0.5 g of H 2.
Similarly, in option (c) and (d), H 2 does not act as
limiting reagent.
For 14 g of N 2 + 4 g of H 2.
As we know 28 g of N 2 reacts with 6 g of H 2.
6
14 g of N2 reacts with
× 14 g of H 2 ⇒ 3 g of H 2.
28
For 28 g of N 2 + 6 g of H 2, i.e. 28 g of N 2 reacts with 6 g
of H 2 (by equation I).

∴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

h cm

5.

10 cm


⇒

V = πr × h

⇒

h=

2

V
πr

2

=

= 1 × 10

4.

0.03 cm3
3 × (10)2 cm2

−4

[JEE Main 2019, 10 April Shift-I]

cm = 1 × 10 −6 m


(a) C4 H7Cl
(c) C4 H10

For a reaction,
N2 ( g ) + 3 H2 ( g ) → 2 NH3 ( g ), identify
dihydrogen ( H2 ) as a limiting reagent in the
following reaction mixtures.

Exp. (b)

[JEE Main 2019, 9 April Shift-I]

1 mol
1 mL
10 mL

Exp. (a)

When 56 g of N 2 + 10 g of H 2 is taken as a
combination then dihydrogen (H 2 ) act as a limiting
reagent in the reaction.
…(I)
N 2( g ) + 3H 2( g ) → 2NH 3( g )
2( 14 + 3)
34g

12 g of H 2 gas is required for 56 g of N 2 gas but
only 10 g of H 2 gas is present in option (a).
Hence, H 2 gas is the limiting reagent.
In option (b), i.e. 35 g of N 2 + 8 g of H 2.

As 28 g N 2 requires 6 g of H 2.
6g
35 g N 2 requires
× 35 g H 2 ⇒ 7.5 g of H 2.
28 g

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

x mol
x mL
10x mL

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 (C xH y ) is C 4H 6.

amount, i.e. which limits the amount of product formed
is called limiting reagent.

28 g N 2 requires 6 g H 2 gas.
6g
56 g of N 2 requires
× 56 g = 12 g of H 2
28 g


x+



Given, (i) VCO2 = 10 x = 40 mL ⇒ x = 4

Key Idea The reactant which is present in the lesser

3 × 2g
6g

(b) C4 H6
(d) C4 H8

In eudiometry,
300 K
y
y
C xH y +  x +  O 2 → x CO 2 + H 2O
1 atm

4
2

(a) 56 g of N2 + 10 g of H2
(b) 35 g of N2 + 8 g of H2
(c) 14 g of N2 + 4 g of H2
(d) 28 g of N2 + 6 g of H2

2 × 14g
28g

At 300 K and 1 atmospheric pressure,

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

6.

[Q x = 4]

The minimum amount of O2 ( g ) consumed per
gram of reactant is for the reaction (Given
atomic mass : Fe = 56, O =16, Mg = 24, P = 31,
[JEE Main 2019, 10 April Shift-II]
C =12, H =1)
(a) C 3H 8 ( g ) + 5O 2( g ) → 3CO 2( g ) + 4H 2O(l )
(b) P4 ( s ) + 5O 2( g ) → P4O 10 ( s )
(c) 4Fe( s ) + 3O 2( g ) → 2Fe 2O 3( s )

(d) 2Mg( s ) + O 2 ( g ) → 2MgO( s )


3

Some Basic Concepts of Chemistry
From equation (i) and (ii)
1 (M A + 2 M B )
125 
= 

2 (2 M A + 2 M B )  300 


Exp. (b)
(a) C 3H8 (g ) + 5O 2 (g ) → 3CO 2 (g ) + 4H2O(l )
44g

160g

On solving the equation, we obtain

160
g of O 2 consumed
44
= 3.64 g
(b) P4 (s ) + 5O 2 (g ) → P4O10 (s )
⇒ 1g of reactant =

124g

M A = 5 × 10 −3

So, the molar mass of A( M A ) is
5 × 10 −3 kgmol −1 and B( M B )is10 × 10 −3 kgmol −1.

160g

160
g of O 2 consumed = 129
. g
124
(c) 4Fe(s ) + 3O 2 (g ) → 2Fe 2O 3 (s )
⇒ 1g of reactant

244g

8.

96g

= 0.43 g
(d) 2Mg(s ) +O 2 (g ) → 2MgO(s )
48 g

32 g

⇒ 1 g of reactant
= 0.67 g

(d) 18 g of carbon and 7 g of hydrogen

32
g of O 2 consumed
48

Exp. (c)
Hydrocarbon containing C and H upon burning
produces CO 2 and water vapour respectively. The
equation is represented as
C xH y + ( x + y / 4)O 2 → xCO 2 + ( y / 2 ) H 2O
12
Mass of carbon =
× mass of CO 2
44

12
=
× 88 g = 24 g
44
2
Mass of hydrogen =
× mass of H 2O
18
2
=
× 9 = 1g
18
So, the unknown hydrocarbon contains 24 g of carbon
and 1g of hydrogen.

So, minimum amount of O 2 is consumed per gram of
reactant (Fe) in reaction (c).

7.

5 moles of AB 2 weight125 × 10 −3 kg and 10 moles
of A 2 B 2 weight 300 × 10 −3 kg. The molar mass of
A ( M A )and molar mass of B ( M B )in kg mol −1 are

[JEE Main 2019, 12 April Shift-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
(c) M A = 25 × 10 −3 and M B = 50 × 10 −3


(d) M A = 5 × 10−3 and M B = 10 × 10−3

Exp. (d)
Key Idea To find the mass of A and B in the given

9.

question, mole concept is used.
Number of moles ( n) =
Compound

given mass ( w )
molecular mass ( M )

Mass of A (g)
MA

2MB

A 2B2

2MA

2MB

2C57H110O6(s ) + 163O2( g ) → 114CO2( g )
+ 110 H2O (l )
(a) 490 g

given mass ( w )

molecular mass ( M )

n × M = w …(A)
Using equation (A), it can be concluded that
5( M A + 2 M B ) = 125 × 10 −3 kg
−3

kg

(b) 495 g

(c) 445 g

(d) 890 g

Exp. (b)

We know that,

10(2 M A + 2 M B ) = 300 × 10

For the following reaction, the mass of water
produced from 445 g of C57H110O6 is :

Mass of B (g)

AB2

Number of moles ( n) =


25 g of an unknown hydrocarbon upon
burning produces 88 g of CO2 and 9 g of H 2O.
This unknown hydrocarbon contains
[JEE Main 2019, 12 April Shift-II]
(a) 20 g of carbon and 5 g of hydrogen
(b) 22 g of carbon and 3 g of hydrogen
(c) 24 g of carbon and 1 g of hydrogen

96
g of O 2 consumed
224

⇒ 1 g of reactant

and M B = 10 × 10 −3

…(i)
…(ii)

2 C 57H110O 6( s )+ 163 O 2( g ) →
110 H 2O( l ) + 114 CO 2( g )
Molecular mass of C 57H110O 6
= 2 × (12 × 57 + 1 × 110 + 16 × 6) g = 1780 g
Molecular mass of 110 H 2O = 110 (2 + 16) = 1980 g
1780 g of C 57H110O 6 produced = 1980 g of H 2O.
1980
445g of C 57H110O 6 produced =
× 445 g of H 2O
1780
= 495 of H 2O



4
10.

JEE Main Chapterwise Chemistry
In the reaction of oxalate with permanganate
in acidic medium, the number of electrons
involved in producing one molecule of CO2 is
(a) 2
(c) 1

12.

The hardness of a water sample (in terms of
equivalents of CaCO3 ) containing
10 −3 M CaSO4 is
(Molar mass of CaSO4 = 136 g mol−1 )

(b) 5
(d) 10

(a) 100 ppm
(c) 50 ppm

[JEE Main 2019, 10 Jan Shift-II]

[ JEE Main 2019, 12 Jan Shift-I]

Exp. (c)

Reaction of oxalate with permanganate in acidic
medium.

Exp. (a)
Hardness of water sample can be calculated in terms
of ppm concentration of CaCO 3.
Given, molarity = 10 −3M

5C 2O 42− + 2MnO −4 → 10CO 2 + 2Mn2 + + 8H 2O
( 4 − 3) × 2 = 2

n-factor :
No. of mole

(7 − 2 ) = 5

5

2

i.e. 1000 mL of solution contains10 −3 mole ofCaCO 3.

10

⇒ 5C 2O 42– ions transfer 10 e − to produce
10 molecules of CO 2.
So, number of electrons involved in producing
10 molecules of CO 2 is 10. Thus, number of electrons
involved in producing 1 molecule of CO 2 is 1.


11.

∴Hardness of water = ppm of CaCO 3
=

(a) 8.4
(c) 16.8

13.

Exp. (d)
Given, rate constant (k) = 0.05 µg/year
Thus, from the unit of k, it is clear that the reaction is
zero order. Now, we know that
a
half-life (t 1/ 2 ) for zero order reaction = o
2k
where, ao = initial concentration,
k = rate constant
5 µg
t 1/ 2 =
2 ì 0.05 àg / year

Exp. (a)
2NaHCO 3 + H 2C 2O 4 → 2CO 2 + Na 2C 4O 4 + H 2O
2 mol

⇒ In the reaction, number of mole of CO 2
produced.
pV

1 bar × 0.25 × 10 −3 L
n=
=
RT 0.082 L atm K −1mol −1 × 298.15 K
Number of mole of NaHCO 3
Weight of NaHCO 3
=
Molecular mass of NaHCO 3
∴

wNaHCO 3 = 1.02 × 10 −5 × 84 × 10 3 mg

= 0.856 mg
0.856
× 100
⇒ NaHCO 3 % =
10
= 8.56%

(b) 25
(d) 50
[ JEE Main 2019, 12 Jan (Shift-I)]

[ JEE Main 2019, 11 Jan Shift-I]

= 1.02 × 10 −5 mol

Decomposition of X exhibits a rate constant of
0.05 µg/year. How many years are required for
the decomposition of 5 µg of X into 2.5 µg?

(a) 20
(c) 40

(b) 0.84
(d) 33.6

1 mol

10 −3 × 1000
× 10 6
1000

= 100 ppm

A 10 mg effervescent tablet containing sodium
bicarbonate and oxalic acid releases 0.25 mL of
CO2 at T = 298.15 K and p =1 bar. If molar
volume of CO2 is 25.0 L under such condition,
what is the percentage of sodium bicarbonate
in each tablet?
[Molar mass of NaHCO 3 = 84 g mol −1 ]

2 mol

(b) 10 ppm
(d) 90 ppm

= 50 years
Thus, 50 years are required for the decomposition of
5 µg of X into 2.5 µg.


14.

The ratio of mass per cent of C and H of an
organic compound (C x H yOz ) is 6 : 1. If one
molecule of the above compound (C x H yOz )
contains half as much oxygen as required to burn
one molecule of compound C x H y completely
to CO2 and H 2O. The empirical formula of
compound C x H yOz is
[ JEE Main 2018]
(a) C 3H 6O 3
(c) C 3H 4O 2

(b) C 2H 4O
(d) C 2H 4O 3


5

Some Basic Concepts of Chemistry
Exp. (d)

Exp. (c)

We can calculate the simplest whole number ratio of C
and H from the data given, as
Element

C

H

Relative
mass
6
1

Given, abundance of elements by mass
oxygen = 614
. %, carbon = 22.9%, hydrogen = 10% and
nitrogen = 2.6%

Molar Relative Simplest
mass
mole whole no.
ratio
12
1

6
= 0.5
12

0.5
=1
0.5

1
=1
1


1
=2
0.5

Alternatively this ratio can also be calculated directly
in the terms of x and y as
12 x 6
= (given and molar mass of C = 12, H = 1)
y
1
Now, after calculating this ratio look for condition 2
given in the question i.e. quantity of oxygen is half
of the quantity required to burn one molecule of
compound C xHy completely to CO 2 and H2O. We
can calculate number of oxygen atoms from this as
consider the equation.

Total weight of person = 75 kg
Mass due to 1H =
1

∴Mass gain by person = 7.5 kg

16.

1 g of a carbonate (M 2CO3 ) on treatment with
excess HCl produces 0.01186 mole of CO2 . The
molar mass of M 2CO3 in g mol − 1 is
(a) 1186

(c) 118.6

[ JEE Main 2017 (Offline)]
(b) 84.3
(d) 11.86

Exp. (b)
M2CO 3 + 2HCl → 2M Cl + H 2O +
1g

CO 2
0.01186 mole

Number of moles of M2CO 3 reacted = Number of moles
of CO 2 evolved
1
= 0.01186 [M = molar mass of M2CO 3]
M
1
M =
= 84.3 g mol − 1
0.01186

17.

Thus, the simplest ratio figures for x, y and z are
x = 1, y = 2 and z = 15
.

At 300 K and 1 atm, 15 mL of a gaseous

hydrocarbon requires 375 mL air containing
20% O 2 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 [ JEE Main 2016 (Offline)]

Now, put these values in the formula given i.e.
C xH yO z = C 1H 2O1.5

(a) C3 H8
(c) C4 H10

So, empirical formula will be

Exp. (None)

Here we consider x and y as simplest ratios for C
and H so now putting the values of x and y in the
above equation.
2
y
z =  x +  = 1 +  = 1.5



4 
4 


[C 1H 2O1.5 ] × 2 = C 2H 4O 3

15.

H atoms are replaced by 2H atoms,
Mass due to 2H = (7.5 × 2) kg

y
y
C xH y +  x +  O 2 → xCO 2 + H 2O

4 
2

Number of oxygen atoms required
y
y
= 2 ×  x +  = 2 x + 

4  
2 
1
y
y
Now given, z = 2 x +  =  x + 
2 
2  
4 

75 × 10

= 7.5 kg
100

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 H atoms are replaced
by 2 H atoms is
[ JEE Main 2017 (Offline)]
(a) 15 kg
(c) 7.5 kg

(b) 37.5 kg
(d) 10 kg

(b) C4 H8
(d) C3 H6

y
y
C xH y ( g ) +  x +  O 2( g ) → xCO 2 ( g ) + H 2O ( l )

4
75 mL 2 30 mL
O 2 used = 20% of 375 = 75 mL
Inert part of air = 80% of 375 = 300 mL
Total volume of gases = CO 2 + Inert part of air
= 30 + 300 = 330 mL
y

x+
y
x 30
75
4
=
⇒ x=2 ⇒
=
⇒ x+ =5
4
1 15
1
15
⇒
x = 2, y = 12 ⇒ C 2 H12


6

JEE Main Chapterwise Chemistry

18.

The molecular formula of a commercial resin
used for exchanging ions in water softening is
C 8H 7 SO3 Na (mol. wt. = 206). What would be
the maximum uptake of Ca 2+ ions by the resin
when expressed in mole per gram resin?

20.


(a) 1 : 4
(c) 1 : 8

[ JEE Main 2015]

1
(a)
103

1
(b)
206

2
(c)
309

(d)

(b) 7 : 32
(d) 3 :16

Exp. (b)

1
412

Exp. (d)


The number of moles is given by
weight ( w )
n=
molecular weight ( M )

We know the molecular weight of C 8H 7SO 3Na

Thus, ratio of moles of O 2 and N 2 is given by

= 12 × 8 + 1 × 7 + 32 + 16 × 3 + 23 = 206

nO 2

We have to find, mole per gram of resin.
∴ 1 g of C 8H 7SO 3Na has number of mole
Weight of given resin
1
mol
=
=
Molecular weight of resin 206
Now, reaction looks like
2C 8H 7SO 3Na + Ca 2 + → (C 8H 7SO 3 )2Ca + 2Na
∴ 1 mole of C 8H 7SO 3Na will combine with

n N2

21.
2+


1
mol Ca 2 +
2

3 g of activated charcoal was added to 50 mL of
acetic acid 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
(c) 42 mg

 MN 2

M
 O2






Exp. (a)
Given, initial strength of acetic acid = 0.06N
Final strength = 0.042 N
Volume given = 50 mL
∴ Initial m moles of CH 3COOH = 0.06 × 50 = 3
Final m moles of CH 3COOH = 0.042 × 50 = 2.1
∴ m moles of CH 3COOH adsorbed
= 3 − 2.1 = 0.9 m mol
Hence, mass of CH 3COOH absorbed per gram of

charcoal
0.9 × 60
=
3

(b) 1.00 M
(d) 0.0975 M

Exp. (a)
The molarity of a resulting solution is given by
M V + M 2V2 750 × 0.5 + 250 × 2
875
M = 11
=
=
V1 + V2
750 + 250
1000
= 0.875 M

22.

A gaseous hydrocarbon gives upon combustion,
0.72 g of water and 3.08 g of CO2 . The empirical
formula of the hydrocarbon is [ JEE Main 2013]
(a) C2 H4

(d) 54 mg

(Q molar mass of CH 3COOH = 60 gmol −1)

54
=
= 18 mg.
3






The molarity of a solution obtained by mixing
750 mL of 0.5 M HCl with 250 mL of 2M HCl
will be
[JEE Main 2013]
(a) 0.875 M
(c) 1.75 M

[JEE Main 2015]

(b) 36 mg

  WO 2
=
 W
  N2

Hence, ratio of nO 2 and n N 2 is 7 : 32 .

1
mole of C 8H 7SO 3Na will combine with

∴
206
1
1
1
mol Ca 2 +
mol Ca 2 + =
×
412
2 206

(a) 18 mg

 WO 2 / MO 2
= 
W / M
 N2
N2

1
28
7
=   ×   =
 4   32  32

+

Q 2 moles of C 8H 7SO 3Na combines with 1 mol Ca

19.


The ratio of masses of oxygen and nitrogen in a
particular gaseous mixture is 1 : 4. The ratio of
number of their molecule is
[ JEE Main 2014]

(b) C3 H4

(c) C6 H5

(d) C7 H8

Exp. (d)
18 g H 2O contain 2g H.
∴ 0.72 g H 2O contain 0.08 g H
44 g CO 2 contain 12 g C.
∴ 3.08 g CO 2 contain 0.84 g C.
0.84 0.08
C :H =
:
∴
12
1
= 0.07 : 0.08 = 7 : 8
∴ Empirical formula = C 7 H 8

23.

The density of a solution prepared by
dissolving 120 g of urea (mol. mass = 60 u) in

1000 g of water is 1.15 g/mL. The molarity of
this solution is
[AIEEE 2012]
(a) 0.50 M
(c) 1.02 M

(b) 1.78 M
(d) 2.05 M


7

Some Basic Concepts of Chemistry
Exp. (d)

= 300 mL ×

Total mass of solution = 1000 g water + 120 g urea
= 1120 g
Density of solution = 1.15 g / mL
mass
1120 g
Thus, volume of solution =
=
density 1.15 g / mL

120
= 2 mol
60
2 mol

moles of solute
Molarity =
=
volume (L) of solution 0.974 L

= 300 g
= 299.9 g = 0.2999 kg
0.0100
Molality =
= 5.55 × 10 −4 mol kg −1
60 × 0.2999

26.

= 2.05 mol L−1

24.

(b) furnishes H + ions in addition to those from
oxalic acid
(c) reduces permanganate to Mn 2 +
(d) oxidises oxalic acid to carbon dioxide and water

(b) 0.45 g
(d) 2.2 g

Exp. (c)

Exp. (c)


Titration of oxalic acid by KMnO 4 in the presence of HCl
gives unsatisfactory result because HCl is a better
reducing agent than oxalic acid and HCl reduces
preferably MnO –4 to Mn2+ .

Mohr’s salt is FeSO4 ⋅ (NH 4 )2SO 4 ⋅ 6H 2O.

Only oxidisable part Fe 2+ is
[Fe 2+ → Fe 3+ + e − ] × 6
Cr2O 27 − +14H+ + 6e − → 2Cr 3+ +7H2O

27.

6Fe 3+ +2Cr 3 + +7H2O
Millimoles of Fe 2+ = 750 × 0.6 = 450

(a) 6L HCl (aq ) is consumed for every 3L H2 ( g )
produced
(b) 33.6 L H2 ( g ) is produced regardless of
temperature and pressure for every mole Al
that reacts
(c) 67.2 L H2 ( g ) at STP is produced for every mole
Al that reacts
(d)11.2 L H2 ( g ) at STP is produced for every mole
HCl (aq ) consumed

450
= 0.450 mol
1000
≡ 1 mol Cr2O 27 −


Moles of Fe 2+ =

∴

0.450 mol Fe 2+ ≡

0.450
= 0.075 mol Cr2O 2−
7
6

= 0.075 × 294 g = 22.05 g

25.

The molality of a urea solution in which
0.0100 g of urea, [( NH 2 )2 CO] is added to
0.3000 dm 3 of water at STP is
[AIEEE 2011]
(a) 5.55 × 10 − 4 M
(c) 3.33 × 10 −2 M

Exp. (d)
2Al( s ) + 6HCl( aq ) → 2Al 3+ ( aq ) + 6Cl − ( aq ) + 3H 2( g )
From the equation, it is clear that,
6 mol of HCl produces 3 mol of H 2

(b) 33.3M
(d) 0.555 M


Exp. (a)
moles of the solute
mass of the solvent in kg
0.0100 g
Moles of urea ( nurea ) =
60 g mol −1
Molality =

Mass of 1000 mL of solution = volume × density

In the reaction,

2Al (s ) + 6HCl (aq ) → 2Al 3+ (aq )
+ 6Cl − (aq ) + 3H 2( g ) [AIEEE 2007]

6Fe 2+ +Cr2O 27 − +14H+ →

6 mol Fe 2 +

Amount of oxalic acid present in a solution can
be determined by its titration with KMnO4
solution in the presence of H 2 SO4 . The titration
gives unsatisfactory result when carried out in
the presence of HCl because HCl [AIEEE 2008]
(a) gets oxidised by oxalic acid to chlorine

The mass of potassium dichromate crystals
required to oxidise 750 cm3 of 0.6 M Mohr’s salt
[AIEEE 2011]

solution is (molar mass = 392)
(a) 0.49 g
(c) 22.05 g

[Q1 dm3 = 1000 mL]

Mass of solvent = 300 g − 0.0100 g

= 973.91 mL = 0.974 L
Moles of solute =

1g
mL

or 1 mole of HCl =

28.

3 × 22.4
L of H 2 = 11.2 L of H 2
6

How many moles of magnesium phosphate,
Mg 3 (PO4 )2 will contain 0.25 mole of oxygen
atoms?
[AIEEE 2006]
(a) 0.02
(c) 1.25 × 10 −2

(b) 3.125 × 10 −2

(d) 2.5 × 10 −2

Exp. (b)
8 moles of O-atom are contained by 1 mole Mg 3(PO 4 )2.


8

JEE Main Chapterwise Chemistry
Hence, 0.25 moles of O-atom are contained by
1
= × 0.25 mol Mg 3 ( PO 4 )2 = 3.125 × 10 −2
8

29.

(c) Both (a) and (b)

Exp. (c)
560 g of Fe

If we consider that 1/6, in place of 1/12, mass of
carbon atom is taken to be the relative atomic
mass unit, the mass of one mole of a substance
will
[AIEEE 2005]

Number of moles =

30.


For 20 g of H
1 g H = 1 mol of H -atom
20 g H ≡ 20 mol of H -atom

32.

What volume of hydrogen gas, at 273 K and
1 atm pressure will be consumed in obtaining
21.6 g of elemental boron (atomic mass = 10.8)
from the reduction of boron trichloride by
[AIEEE 2003]
hydrogen?
(a) 89.6 L
(c) 44.8 L

(b) 67.2 L
(d) 22.4 L

Exp. (b)
2BCl 3 + 3H 2 → 2B + 6HCl
2 mol
2 mol 3 mol
21.6 g
21.6 g B = 2 mol B ≡ 3 mol H 2

pV = nRT
∴

V =


nRT
p

3 × 0.0821 × 273
=
= 67.2 L
1

31.

Number of atoms in 560 g of Fe
(atomic mass = 56 g mol −1 ) is
(a) twice that of 70 g N

[AIEEE 2002]
(b) half that of 20 g H

= 10 mol

14 g N = 1 mol of N -atom
70 g N = 5 mol of N -atom

(b) remain unchanged
(c) increase two fold
(d) decrease twice
The mass of one mole of a substance will remain
unchanged.

560 g

56 g mol −1

For 70 g of N

(a) be a function of the molecular mass of the
substance

Exp. (b)

(d) None of these

In an organic compound of molar mass
108 g mol –1 C, H and N atoms are present in
9 : 1 : 3.5 by weight. Molecular formula can be
(a) C 6 H 8 N 2
(c) C 5H 6 N 3

[AIEEE 2002]
(b) C 7 H 10 N
(d) C 4 H 18 N 3

Exp. (a)
Molar mass 108 g mol −1
Total part by weight = 9 + 1 + 3.5 = 13.5
9
Weight of carbon =
× 108 = 72 g
13.5
72
Number of carbon atoms =

=6
12
1
Weight of hydrogen =
× 108 = 8 g
13.5
8
Number of hydrogen atoms = = 8
1
Weight of nitrogen =
Number of nitrogen atom =
Hence,

3.5
× 108 = 28 g
13.5

28
=2
14

molecular formula = C 6H 8N 2.


2
States of Matter
1.

Element ‘B ’ forms ccp structure and ‘A ’
occupies half of the octahedral voids, while

oxygen atoms occupy all the tetrahedral voids.
The structure of bimetallic oxide is
[JEE Main 2019, 8 April Shift-I]

(b) AB2O 4
(d) A 4 B2O

(a) A 2 BO 4
(c) A 2B2O

The number of element ‘B ’ in the crystal
structure = 4 N
Number of tetrahedral voids = 2N
Number of octahedral voids = N
N 4
∴Number of ‘A’ in the crystal = = = 2
2 2
Number of oxygen (O) atoms = 2 N = 2 × 4 = 8
∴The structure of bimetallic oxide = A2 B4 O 8
= AB2 O 4

Consider the bcc unit cells of the solids 1 and 2
with the position of atoms as shown below.
The radius of atom B is twice that of atom A.
The unit cell edge length is 50% more is solid 2
than in 1. What is the approximate packing
efficiency in solid 2?
A
A


B

A

A
Solid 1

(a) 65%

A
A

A

A
A

A

A

A

A

A
A

A


Solid 2

(b) 90%

(c) 75%

Key Idea Packing efficiency
Volume occupied by sphere
=
× 100
Volume of cube
Given, rB = 2 rA
a2 = a1 +

Exp. (b)

2.

Exp. (b)

A

(d) 45%

[JEE Main 2019, 8 April Shift-II]

For bcc lattice
4rA =

50

. a1
a1 = 15
100

3 a1

3 a1
4
4r
a1 = A
3
4r
3 4r
.  A  =  A 
a2 = 15
∴
 3 2  3
a2 = 2 3 rA
4 3
4
πrA × zA + πrB3 × zB
3
3
Packing efficiency =
a23
rA =

[As the atoms A are present at the edges only
1
zA = × 8 = 1, atom B is present only at the body

8
centre zB = 1]
 4 πr 3 × 1 +  4 πr 3 × 1

 

A

 3 B

PE 2 = 3
∴
3
a2
4 3 4
πrA + π (2 rA )3
3
= 3
(2 3 rA )3
4 3
πrA × 9
π
=
= 3
8 × 3 3 rA3 2 3
= 90.72% ≈ 90%


10
3.


JEE Main Chapterwise Chemistry
pV − pb = RT
pV = RT + pb
pV
pb
= 1+
RT
RT
pV
As,
Z=
RT
pb
so,
Z = 1+
⇒ y = c + mx
RT
The plot of z vs p is found to be

Consider the van der Waals’ constants,
a and b, for the following gases.
Gas

⇒

Ar

Ne


Kr

Xe

a/(atm dm mol )

1.3

0.2

5.1

4.1

b/(10 −2 dm 3 mol −1)

3.2

1.7

1.0

5.0

6

−2

Which gas is expected to have the highest
critical temperature ?

[JEE Main 2019, 9 April Shift-I]

(a) Kr

(b) Xe

(c) Ar

slope =

(d) Ne
Z

Exp. (a)
Critical temperature is the temperature of a
gas above which it cannot be liquefied what
ever high the pressure may be. The kinetic
energy of gas molecules above this
temperature is sufficient enough to overcome
the attractive forces. It is represented as Tc .
8a
Tc =
27 Rb
8 × 1.3
For Ar, Tc =
= 0.0144
27 × 8.314 × 3.2
8 × 0.2
For Ne, Tc =
= 0.0041

27 × 8.314 × 1.7
8 × 51
.
For Kr, Tc =
= 0.18
27 × 8.314 × 1.0
8 × 4.1
For Xe, Tc =
= 0.02
27 × 8.314 × 5.0
The value of Tc is highest for Kr (Krypton).

4.

At a given temperature T , gases Ne, Ar, Xe
and Kr are found to deviate from ideal gas
behaviour. Their equation of state is given
RT
atT .
as, p =
V −b
Here, b is the van der Waals’ constant.
Which gas will exhibit steepest increase in
the plot of Z (compression factor)vs p?
[JEE Main 2019, 9 April Shift-II]

(a) Xe

(b) Ar


(c) Kr

(d) Ne

Exp. (a)
Noble gases such as Ne, Ar, Xe and Kr found
to deviate from ideal gas behaviour.
Xe gas will exhibit steepest increase in plot of
Z vs p. Equation of state is given as:
RT
p=
(V − b)
⇒

b
RT

p(V − b ) = RT

p

The gas with high value of b will be steepest as
slope is directly proportional to b. b is the van der
Waals’ constant and is equal to four times the actual
volume of the gas molecules. Xe gas possess the
largest atomic volume among the given noble gases
(Ne, Kr, Ar). Hence, it gives the steepest increase in
the plot of Z (compression factor) vsp.

5.


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
[JEE Main 2019, 9 April Shift-II]
is its edge length?
(a) 2.0 pm (b) 0.1 nm (c) 1.0 pm (d) 2.0 nm

Exp. (a)
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
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
∴
a2 = 4 × 10−20 cm2
a = 2 × 10−10 cm = 2 pm



11

States of Matter
7.

Consider the following table.
Gas

a/(k Pa dm6 mol−1)

b/(dm3 mol−1 )

A

642.32

0.05196

B

155.21

0.04136

C

431.91

0.05196


D

155.21

0.4382

Points I, II and III in the following plot
respectively correspond to (v mp : most
probable velocity)
[JEE Main 2019, 10 April Shift-II]
Distribution function f( v)

6.

a and b are van der Waals’ constants. The
correct statement about the gases is
[JEE Main 2019, 10 April Shift-I]

(a) gas C will occupy lesser volume than gas A; gas
B will be lesser compressible than gas D
(b) gas C will occupy more volume than gas A; gas
B will be more compressible than gas D
(c) gas C will occupy more volume than gas A; gas
B will be lesser compressible than gas D
(d) gas C will occupy more volume than gas A; gas
B will be lesser compressible than gas D

Exp. (b)
For 1 mole of a real gas, the van der Waals’
equation is

 p + a  (V − b ) = RT



V2 
The constant ‘a’ measures the intermolecular
force of attraction of gas molecules and the
constant ‘b’ measures the volume correction by
gas molecules after a perfectly inelastic binary
collision of gas molecules.
For gas A and gas C given value of ‘b’ is
0.05196 dm3 mol −1. Here,
a ∝ intermolecular force of attraction
∝ compressibility ∝ real nature
1
∝
volume occupied
6

−1

b/(dm mol ) for gas B (0.04136) < Gas D
(0.4382)
So, gas B will be more compressible than gas D.

II

III

Speed, v


(a) v mp of H2 (300 K); v mp of N 2 (300 K); v mp
of O 2 (400 K)
(b)v mp of O 2 (400 K); v mp of N 2 (300 K); v mp
of H2 (300 K)
(c) v mp of N 2 (300 K); v mp of O 2 ( 400 K ); v mp
of H2 (300 K)
(d)v mp of N 2 (300 K); v mp of H2 (300 K); v mp
of O 2 (400 K)

Exp. (c)
Key Idea From kinetic gas equation,
2 RT
Most probable velocity (v mp ) =
M
where, R = gas constant, T = temperature,
M = molecular mass
2 RT
M
T
∝
M

vmp =
i.e.

vmp

−1


Value of a/(kPa dm mol ) for gas A (642.32) >
gas C (431.91) So, gas C will occupy more
volume than gas A. Similarly, for a given value of
a say 155.21 kPa dm 6 mol −1 for gas B and gas D
1
∝ intermolecular force of attraction
b
∝ compressibility ∝ real nature
1
∝
volume accupied
3

I

Gas

M

T(K)

T /M

H2

2

300

300 / 2 = 150 …III (Highest)


N2

28

300

300 / 28 = 1071
. …I (Lowest)

O2

32

400

400 / 32 = 12.5 … II

So,
I corresponds to vmp of N2 (300 K)
II corresponds to vmp of O 2 (400 K)
III corresponds to vmp of H2 (300 K)


12
8.

JEE Main Chapterwise Chemistry
An element has a face-centred cubic (fcc)
structure with a cell edge of a. The distance

between the centres of two nearest tetrahedral
voids in the lattice is

10.

[JEE Main 2019, 12 April Shift-I]

(a)

(b) a
3
(d) a
2

2a

a
(c)
2

(a)
(c)

Exp. (c)

1
—(AB)
2

B′

√3 a
—
4

B

Position of 2 TVs
on one of the
body diagonal
of the unit cell

1
—(AB)
4

11.

The ratio of number of atoms present in simple
cubic, body centred cubic and face centered
cubic structure are 1 : 2 : 4 respectively.

x3

(b) Monoclinic
(d) Triclinic
[JEE Main 2019, 10 Jan Shift-I]

Exp. (d)
Triclinic primitive unit cell has dimensions as,
a ≠ b ≠ c and α ≠ β ≠ γ ≠ 90° .

Among the seven basic or primitive crystalline
systems, the triclinic system is most
unsymmetrical. In other cases, edge length and
axial angles are given as follows :
Hexagonal : a = b ≠ c and α = β = 90°, γ = 120°
Monoclinic : a ≠ b ≠ c and α = γ = 90°, β ≠ 90°
Tetragonal : a ≠ b ≠ c and α = β = γ = 90°

[JEE Main 2019, 12 April Shift-II]

Exp. (b)

x3
422

Which primitive unit cell has unequal edge
lengths (a ≠ b ≠ c ) and all axial angles different
from 90°?
(a) Hexagonal
(c) Tetragonal

The ratio of number of atoms present in a
simple cubic, body centered cubic and face
centered cubic structure are, respectively.
(a) 8 : 1 : 6
(b) 1 : 2 : 4
(c) 4 : 2 : 1
(d) 4 : 2 : 3

(d)


205

Edge length, a = x Å = x × 10−8 cm
Z×M
density (d ) =
NA × a3
4 × 63.55
=
6.023 × 1023 × ( x × 10−8 )3
422.048
g cm−3
=
x3

The angle between body diagonal and an edge
is cos −1 (1/ 3 ). So, the projection of the line on
an edge is a/ 4.
Similarly, other tetrahedral void also will be a/4
away.
So, the distance between these two is
a − a  − a = a .

4  4 2

9.

x3

(b)


For fcc, rank of the unit cell (Z ) = 4
Mass of one Cu-atom, M = 63.55 u
Avogadro’s number, NA = 6.023 × 1023 atom

A

A′

x3
105

Exp. (d)

One of the body diagonal
of cubic unit cell

√3 a
—
4

211

[JEE Main 2019, 9 Jan Shift-II]

In fcc unit cell, two tetrahedral voids are formed on
each of the four non-parallel body diagonals of the
cube at a distance of 3a / 4 from every corner
along the body diagonal.


1
— AB
4

At 100°C, copper (Cu) has FCC unit cell
structure with cell edge length of x Å. What is
the approximate density of Cu (in g cm −3 ) at
this temperature?
[Atomic mass of Cu = 63.55 u]

12.

A compound of formula A 2B3 has the hcp
lattice. Which atom forms the hcp lattice and
what fraction of tetrahedral voids is occupied
by the other atoms ?
2
tetrahedral voids-B
3
1
(b) hcp lattice-A, tetrahedral voids-B
3
(a) hcp lattice- A,


13

States of Matter
1
tetrahedral voids-A

3
2
(d) hcp lattice-B, tetrahedral voids-A
3

(c) hcp lattice-B,

14.

[JEE Main 2019, 10 Jan Shift-II]

(a) 0.134 a
(c) 0.047 a

Exp. (c)

Exp. (d)
For body centred cubic bcc structure,
3
radius (R ) =
...(i)
a
4
Where, a = edge length
According to question, the structure of cubic unit
cell can be shown as follows:

...(ii)
∴
a = 2(R + r )

On substituting the value of R from Eq. (i) to Eq. (ii),
we get
a
3
=
a+ r
2
4
2 a − 3a
a
3
r= −
a=
4
2
4
a(2 − 3 )
r=
4
⇒
r = 0.067 a

[JEE Main 2019, 11 Jan Shift-I]

Exp. (a)
⇒ M=

d × NA × a3
Z


Given, d = 9 × 103 kg m −3

15.

M = Molar mass of the solid
Z = 4 (for fcc crystal)
NA = Avogadro’s constant = 6 × 1023 mol −1
a = Edge length of the unit cell
= 200 2 pm
= 200 2 × 10−12 m

=

3

−3

× (6 × 10

= 0.0305 kg mol −1

23

) mol
4

−1

× (200 2 × 10


The volume of gas A is twice than that of gas B.
The compressibility factor of gas A is thrice
than that of gas B at same temperature. The
pressures of the gases for equal number of
moles are
(a) p A = 2p B
(c) p A = 3p B

(b) 2p A = 3p B
(d) 3p A = 2p B
[JEE Main 2019, 12 Jan Shift-I]

On substituting all the given values, we get
(9 × 10 ) kg m

R

a

(b) 0.4320 kg mol −1
(d) 0.0216 kg mol −1

Density of a crystal
M×Z
d=
NA × a3

2r

R


A solid having density of 9 × 10 3 kgm−3 forms
face centred cubic crystals of edge length
200 2 pm. What is the molar mass of the solid?
[Avogadro constant = 6 × 10 23 mol−1 , π = 3]
(a) 0.03050 kg mol −1
(c) 0.0432 kg mol −1

(b) 0.027 a
(d) 0.067 a
[JEE Main 2019, 11 Jan Shift-II]

Total effective number of atoms in hcp unit lattice =
Number of octahedral voids in hcp = 6
∴ Number of tetrahedral voids (TV) in hcp
= 2 × Number of atoms in hcp lattice
= 2 × 6 = 12
As, formula of the lattice is A2 B3 .
Suppose,
A
B
 1 × TV 
(hcp)


3

1
6
× 12

⇒
3
2
1
⇒
3
2
3
⇒
1
So, A = tetrahedral voids, B = hcp lattice
3

13.

The radius of the largest sphere which fits
properly at the centre of the edge of a body
centred cubic unit cell is
(Edge length is represented by ‘a’)

−12 3

) m

3

Exp. (b)
Given, Z A = 3Z B
Compressibility factor (Z) =


pV
[for real gases]
nRT


14

JEE Main Chapterwise Chemistry
On substituting in equation (i), we get
pA VA
3 pB VB
=
nA RTA
nB RTB

Exp. (c)
It is the ‘‘Frenkel defect’’ in which cations leave
their original site and occupy interstitial site as
shown below.

…(ii)

Also, it is given that

+
–
+
–
+
–

+
–

VA = 2 VB , nA = nB and TA = TB
∴ Eq. (ii) becomes
pA × 2 VB 3 pB VB
=
nB RTB
nB RTB
⇒

16.

2 pA = 3 pB

An open vessel at 27ºC is heated until two fifth
of the air (assumed as an ideal gas) in it has
escaped from the vessel. Assuming that the
volume of the vessel remains constant, the
temperature at which the vessel has been
heated is
(a) 750 K
(c) 750ºC

18.

[JEE Main 2019, 12 Jan Shift-II]

Volume of vessel = constant


Which type of ‘defect’ has the presence of
cations in the interstitial sites? [ JEE Main 2018]
(a) Schottky defect
(b) Vacancy defect
(c) Frenkel defect
(d) Metal deficiency defect

–
+
–
+
–
+
–
+

Original
vacant site
of cation
Cation in
interstitial site

(b) 2 2 a

(c) 2 a

(d)

a
2


For fcc arrangement, 4r = 2 a
where, r = radius and a = edge length
2 a
a
=
∴ Closest distance = 2 r =
2
2

Given, temperature(T1 ) = 27 ° C = 273 + 27 = 300K

17.

– + – +
+ – + –
– + – +
– + –
– + – +
+
+ – + –
– + – +
+ – + –

Exp. (d)

Exp. (b)

Let at T1 the volume of air inside the vessel is n so at
3

T2 the volume of air will be n.
5
Now, as p and V are constant, so
3
...(i)
n ⋅T1 = n T2
5
Putting the value of T1 in equation (i) we get,
3
n × 300 = n × T2
5
5
or
T2 = 300 × = 500 K
3

+
–
+
–
+
–
+
–

A metal crystallises in a face centred cubic
structure. If the edge length of its unit cell is ‘a’,
the closest approach between two atoms in
metallic crystal will be [ JEE Main 2017 (Offline)]
(a) 2 a


(b) 500 K
(d) 500ºC

Pressure in vessel = constant
2
so the remaining
Volume of air reduced by
5
3
volume of air is .
5

–
+
–
+
–
+
–
+

19.

Two closed bulbs of equal volume (V )
containing an ideal gas initially at pressure pi
and temperature T1 are connected through a
narrow tube of negligible volume as shown in
the figure below. The temperature of one of the
bulbs is then raised to T2 . The final pressure p f

is
[ JEE Main 2016 (Offline)]
T1
pi, V

T1

T2
pi, V

T2
pf, V

 T1 
(a) 2pi 

 T1 + T2 

 T2 
(b) 2pi 

 T1 + T2 

 TT 
(c) 2pi  1 2 
 T1 + T2 

 TT 
(d) pi  1 2 
 T1 + T2 


pf, V

Exp. (b)
Initially,
Number of moles of gases in each container =

pV
i
RT1

Total number of moles of gases in both containers
pV
=2 i
RT1