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

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

SANJAY SHARMA

ARIHANT PRAKASHAN (Series), MEERUT

Complete Theory 2 Levels Exercises Exams Questions

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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).

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 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.

— 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.

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.

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.

— 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.

2 States of Matter : Gaseous and Liquid States 42-79

8 Classification of Elements and Periodicity in Properties 294-328

12 Purification and Characterisation of Organic Compounds 440-464

CONTENTS

PART I

Chapters from Class 11th Syllabus

3 Electrochemistry 745-788

6 General Principles and Processes of Isolation of Metals 872-896

Online JEE Main 2019 Solved Papers

(April & January Attempt) 1-32

PART II

Chapters from Class 12th Syllabus

JEE Main Solved Papers

MASTER

RES URCE

UNIT 1 Some Basic Concepts in Chemistry

UNIT 2 States of Matter

Section- A (Physical 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.

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.

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.

Classification of matter into solid, liquid and gaseous states.

Liquid State Properties of liquids - vapour pressure, viscosity and surface tension and effect of temperature on them

(qualitative treatment only).

SYLLABUS

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.

UNIT 6 Solutions

Elementary idea of metallic bonding Hydrogen bonding and its applications.

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.

UNIT 7 Equilibrium

Second law of thermodynamics Spontaneity of processes; ΔS of the universe and ΔG of the system as criteria for

o

spontaneity, ΔG (Standard Gibb's energy change) and equilibrium constant.

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.

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

Fundamentals of thermodynamics System and surroundings, extensive and intensive properties, state functions,

types of processes.

UNIT 5 Chemical Thermodynamics

Quantum mechanical approach to covalent bonding Valence bond theory - Its important features, concept of

hybridization involving s, p and d orbitals; Resonance.

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.

Meaning of equilibrium, concept of dynamic equilibrium.

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

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

o

(K and K) and their significance, significance of ΔG and ΔG in chemical equilibria, factors affecting equilibrium

concentration, pressure, temperature, effect of catalyst; Le -Chatelier’s principle.

MASTER

RES URCE

Book for JEE Main

UNIT 8 Redox Reactions and Electrochemistry

UNIT 9 Chemical Kinetics

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

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.

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 solids- Freundlich and Langmuir adsorption isotherms, adsorption from solutions.

UNIT 11 Classification of Elements and Periodicity in Properties

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.

Catalysis Homogeneous and heterogeneous, activity and selectivity of solid catalysts, enzyme catalysis and its

mechanism.

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.

Electronic concepts of oxidation and reduction, redox reactions, oxidation number, rules for assigning oxidation number, balancing of redox reactions.

Section- B (Inorganic Chemistry)

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

Book for JEE Main

MASTER

RES URCE

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,(PCl , PCl ); Structures of oxides and 3 5

oxoacids of nitrogen and phosphorus.

UNIT 14 s - Block Elements

(Alkali and Alkaline Earth Metals)

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 13 to Group 18 Elements

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

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 14 Tendency for catenation; Structure, properties and uses of allotropes and oxides of carbon, silicon

tetrachloride, silicates, zeolites and silicones.

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.

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.

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 K Cr O and KMnO 2 2 7 4

Inner Transition Elements Lanthanoids Electronic configuration, oxidation states, chemical reactivity and lanthanoid contraction

Actinoids Electronic configuration and oxidation states.

Unit 18 Environmental Chemistry

Environmental pollution Atmospheric, water and soil Atmospheric pollution Tropospheric and stratospheric.

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).

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.

Strategies to control environmental pollution.

UNIT 19 Purification & Characterisation of Organic Compounds

Soil pollution Major pollutants such as: Pesticides (insecticides, herbicides and fungicides), their harmful effects and

prevention.

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)

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;

Section- C (Organic Chemistry)

Purification Crystallization, sublimation, distillation, differential extraction and chromatography principles and their

applications.

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

Alkenes Geometrical isomerism; Mechanism of electrophilic addition: addition of hydrogen, halogens, water, hydrogen halides (Markownikoff's and peroxide effect); Ozonolysis, oxidation, and 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.

Alkynes acidic character; addition of hydrogen, halogens, water and hydrogen halides; polymerization

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

Alkanes Conformations: Sawhorse and Newman projections (of ethane); Mechanism of halogenation of alkanes.

UNIT 23 Organic Compounds Containing Oxygen

General methods of preparation, properties, reactions and uses

Classification, isomerism, IUPAC nomenclature, general methods of preparation, properties and reactions.

Alcohols, Phenols and Ethers

UNIT 24 Organic Compounds Containing Nitrogen

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.

Phenols Acidic nature, electrophilic substitution reactions: halogenation, nitration and sulphonation, Reimer -

Tiemann reaction.

Ethers Structure.

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 reagent; oxidation; reduction (Wolff 3Kishner and Clemmensen) acidity of α-hydrogen, aldol condensation, Cannizzaro reaction, Haloform reaction; Chemical tests to distinguish between aldehydes and Ketones.

General methods of preparation, properties,

reactions and uses.

Carboxylic Acids Acidic strength and factors affecting it.

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

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.

Aldehyde and Ketones Nature of carbonyl group;

MASTER

RES URCE

— Chemistry involved in the preparation of the following

1 Enthalpy of solution of CuSO4

Proteins Elementary Idea of α-amino acids, peptide bond, polypeptides; proteins: primary, secondary, tertiary and

quaternary structure (qualitative idea only), denaturation of proteins, enzymes.

— Chemical principles involved in the qualitative salt analysis

Nucleic Acids Chemical constitution of DNA and RNA Biological functions of Nucleic acids.

Vitamins Classification and functions.

— 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.

2 Enthalpy of neutralization of strong acid and strong base.

General introduction and importance of biomolecules.

3 Preparation of lyophilic and lyophobic sols.

— Organic compounds Acetanilide,

p-nitroacetan ilide, aniline yellow, iodoform.

UNIT 27 Chemistry in Everyday Life

4 Kinetic study of reaction of iodide ion with hydrogen peroxide at room temperature.

Chemicals in medicines Analgesics, tranquilizers, antiseptics, disinfectants, antimicrobials, antifertility drugs, antibiotics, antacids, antihistamins - their meaning and common examples.

UNIT 26 Biomolecules

Carbohydrates Classification: aldoses and ketoses; monosaccharides (glucose and fructose), constituent

monosaccharides of oligosacchorides (sucrose, lactose, maltose) and polysaccharides (starch, cellulose, glycogen).

Cleansing agents Soaps and detergents, cleansing action.

UNIT 28 Principles Related to Practical Chemistry

— Inorganic compounds Mohr's salt, potash alum.

Chemicals in food Preservatives, artificial sweetening agents - common examples.

— Chemistry involved in the titrimetric excercises - Acids bases and the use of indicators, oxali acid vs KMnO , Mohr's 4salt vs KMnO 4

— Cations — Pb , Cu , Al , Fe , Zn , Ni , Ca , Ba , Mg NH Anions – CO , S , SO , NO , NO , Cl , Br , I (Insoluble 2+ 2+ 3+ 3+ 2+ 2+ 2+ 2+ 2+ 4+ 32- 2- 42- 2 3 - -

-salts excluded).

— Chemical principles involved in the following experiments

Part - I

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

Plasmais 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 sayplasmasaslow

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

Einsteinbut due to lack of equipments, it was only created in 1935 byCornell,

Ketterle and Weimann. Its concept and existence is totally opposite to

plasmas The state is conceptualised at supercold conditions

Matter and Its Nature

Physical Quantities and Their Measurements

in Chemistry

Atomic and Molecular Masses

Equivalent Mass or Equivalent WeightEmpirical and Molecular FormulaeStoichiometric Problems of DifferentKinds

The supercold above means only a few billionth of a

degree above absolute zero Cornell and Weimann

developed BEC at such temperature withrubidium.

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, calledatoms.

(ii) all atoms of an element have identical mass and

similar chemical properties (Atoms of different

elements have different masses and different

chemical properties)

(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.’’

Both atoms and molecules are basic constituents of matterwith the condition that atoms combine to form themolecules

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 , O2 2etc

(iii) triatomic, i.e., contain 3 atoms, e.g., O3etc

(iv) polyatomic, i.e., contain more than 3 atoms e.g., P , S4 8etc

Chemical Classification of Matter

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

(a) Mixtures

These have variable composition and variable propertiesdue to the fact that components retain their characteristicproperties These may be separated into pure components

by applying physical methods

These can be of two types

(i) Thehomogeneous mixtures, have same compositionthroughout and their components are

indistinguishable, e.g.,a liquid solution of sugar and

water etc

(ii) Aheterogeneous mixture,on the other hand, do nothave the same composition throughout and the

components here are distinguishable, e.g., a mixture

of grains of sand and salt Here particles of eachcomponent maintain their own identity

(b) Pure Substances

These have fixed composition and non-variableproperties These cannot be separated into simplersubstances by physical methods

Anelementis a substance that contains only one type ofatoms whereas a compound is formed when atoms ofdifferent elements combine in a fixed ratio

Compounds and elements can be differentiated as theformer can be decomposed into simple substances bychemical methods while later cannot be decomposed intosimpler substances by chemical methods

Mixtures

Homogeneousmixture

Heterogeneousmixture Compounds Elements

substances

Matter

Pure

SampleProblem 1 Which one of the following is not an

element?

(c) Diamond (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 (SiO )2 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]

(i) (ii)

(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

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 calleddigit term.It is a number between 1.000and 9.999

nis called anexponent.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 FiguresThe digits in a properly recorded measurement are known as

significant figuresor in other words, we can say that significantfigures are the meaningful digits in a measured or calculatedquantity

A significant figure includes all those digits that are known withcertainty 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 twosignificant 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 foursignificant figures Such numbers are better represented inscientific notation We can express the number 100 as 1 10´ 2for one significant figure, 10 ´102for two significant figuresand 100 10 ´ 2for three significant figures

5 Counting numbers of objects, for example, 2 balls or 20 eggs,have infinite significant figures as these are exact numbersand 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 ´ 2has three significant figures

Some Basic Concepts in Chemistry 5

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 2 significant figures

1.5 2 significant figures

20.911 5 significant figures

22.541

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

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

e.g.,

0.012080.0236 =0.512

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

3.12g mL-1, the mass of 1.5 mL solution in significant figures is

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 3.52.01

´ ´

(a) 1 (b) 2

(c) 3 (d) 4

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

2.5 or 3.5, has two significant figures, so answer must also bereported in two significant figures

Note Here is no need to do complete calculation

Various Systems of Measurement

It is also called Gaussian system and is based oncentimetre (cm), gram (g) and second (s) as the units oflength, mass and time respectively

(b) FPS System

It is a British system which used foot (ft), pound (lb) andsecond (s) as the fundamental units of length, mass andtime

It is called MKSA system later on It is the system, whichuses metre (m), kilogram (kg) and second (s) respectivelyfor length, mass and time; Ampere (A) was added later onfor electric current

(d) SI System

It is internationally accepted system in 1960s, hencecalled International system of units and containsfollowing 7 basic and 2 supplementary units

(i)Basic unitsincludes metre (m) for length, kilogram (kg)for mass, second (s) for time, ampere (A) for electriccurrent, kelvin (K) for thermodynamic temperature,mole (mol) for amount of substance and candela (Cd)for luminous intensity

(ii)Supplementary units includes radian (rad) for angleand steradian (sr) for solid angle

Sometimes submultiples and multiples are used toreduce or enlarge the size of the different units The namesand symbols of sub-multiples and multiples are listed inthe table given below

Some Basic Concepts in Chemistry 7

Derived Units

The units of all other quantities, which are derived from

the above mentioned units, called thefundamental units,

are called the derived units e.g,

Units of volume = length ´ breadth ´ height =m m m´ ´ = m3

Table 1.2 Some Derived Properties and their Units

Quantity Definition of quantity Expression in terms of

SI base units

Area Length squared m2

Volume Length cubed m3

Density Mass per unit volume kg / m or kg m3 –3

Velocity Distance travelled per

(pascal, Pa)Energy (work,

heat)

Force times distancetravelled

kg m / s or kg m s2 2 2 –2(joule, J)

Electric charge Ampere times second A-s (coulomb, C)

Electric potential Energy per unit

charge

J/(A-s) potentialdifference (volt, V)

Dimensional Analysis

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 calledfactor label methodorunit factor methodor

dimensional analysis.

Information sought = information given ´ CF

Some conversion factors [CCF] are as follows

1m 39.37= inch

1 inch = 2.54 cm1L 1000 mL= = 1000 cm3

=10- 3m3=1dm3

1lb=453 59237 g1J 1N m = 1 kgm s= - 2 - 2

1cal =4.184 J 2.613 10 eV= ´ 19

1eV=1 602 10 ´ -19J1eV/atom 96 485 kJ mol= -1

1 D (Debye) = ´1 10- 18esu-cm

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

Caution PointRemember that CF must always have the numerator

and denominator representing equivalent quantities.

SampleProblem 5 If the speed of light is 3.0´10 ms8 -1, calculate the distance covered by light in 2.00 ns.

Sample Problem 7 Pressure is determined as force per

unit area of the surface The SI unit of pressure, pascal is as

=1034 kg´100´100´9.8 ms

-1000 m

2 2

=101332.0 Nm-2 (1 N=kg ms )-2

=1.01332´10 Pa5

Laws of Chemical Combinations

The combination of elements to form compounds is

governed by following basic laws.

Law of Conservation of Mass

(by Lavoisier)

According to this law the matter can neither be created

nor destroyed in a chemical reaction

Law of Definite Proportions (by J Proust)

“A sample of a pure compound always consists of the

same elements combined in same proportions by mass,

whatever be its source.”

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 NH ,314 g

of nitrogen requires 3 g of hydrogen and in hydrazine

(N H )2 4 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”

These lines are considered as the law of definiteproportions 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 H2combines with

1 vol of O2to form 2 vol of H O2 (steam) This simply meansthat 100 mL of H2 gas combines with 50 mL of O2 toproduce exactly 100 mL of steam (if volume of all the gasesare measured at same temperature and pressure)

Avogadro’s Law

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

According to this law

V µn

where, n = number of moles of gas

In simpler words, the law can also be stated as “equalvolumes of all gases, under the same conditions oftemperature and pressure contain equal number ofmolecules’’ which is infact 6.023 10´ 23or 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 different isotopes of the same element, e.g.,H O2 andD O2 or H O216and

H O218 Moreover, law of conservation of mass does not hold good for nuclear reactions.

SampleProblem 8 Which of the following reactions is not correct according to the law of conservation of mass?

Some Basic Concepts in Chemistry 9

Sample Problem 9 The following data are obtained

when dinitrogen and dioxygen react together to form different

Which law of chemical combination is obeyed by the above

(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, 12C is chosen as standard and is

arbitrarily assigned the mass of 12 atomic mass unit

(amu)

\ Atomic mass

= 1 mass of one atom of the element

12th part of the mass of one atom of C 12

-Therefore, one amu or u (unified mass) is equal to exactly

the 1/12th of the mass of12Catom

1 u 112

12 g6.022 1023

= ´

´

=1.66 10´ - 24g

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, m2and m3in the ratio, x, y and z, the average

Be, B, C and Si, is related to specific heat as

Average atomic mass = 6.4

specific heatThis is calledDulong and Petit’s method.

Mass spectrometer is used to determine the atomic massexperimentally

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 (C H O )6 12 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 atomicmasses of all atoms in the formula unit of the compound

It is normally calculated for ionic compounds e.g.,

formula mass of NH3is 14 3 17+ = amu or 17u

or formula mass of NaCl is

23 35.5 58.5+ = amu or 58.5u

1. How many millimetres are there in 14.0 cm?

2. Explain why in calculations involving more than onearithmetic operation, rounding off to the proper number

of significant figures may be done once at the end if allthe operations are multiplications or divisions or if they areall additions and subtractions, but not if they arecombinations of additions or subtractions with divisions ormultiplications?

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

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

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

If A is metal and B is H (or O or Cl) then

Eq wt of metal mass of metal

mass of hydrogen displaced 1.008

molecular weight of baseacidity (number of repl

=

aceable OH )

-e.g.,Equivalent weight of NaOH =40=

1 40

Eq wt of salt molecular weight of salt

total positive valency

e.g.,When KMnO4reacts under acidic conditions, change

in oxidation number (from +7 to +2) is 5, hence;

Equivalent weight of KMnO4in acidic medium

=158=

5 31.6For volatile metal chlorides, eq wt and atomic weight arerelated as

Atomic wt = eq wt ´ valency

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 ofCr, SeandSinK CrO ,2 4

K SeO2 4andK SO2 4are same.

SampleProblem 10 0.5g of a metal on oxidation gave

0.79g of its oxide The equivalent weight of the metal is

Sample Problem 11 The equivalent weight of iron in

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

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

(ii) Molesµ molecules

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.

One mole of object will always mean 6.023´1023of those objects

The number of object per mole, 6.023´1023 mol-1 is called

Thus, one mole of any substance is defined as

(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

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´1023formula units

One mole of NaCl = 6.023´1023NaClunits6.023 1023

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

-= 58.5 g NaCl

SampleProblem 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]

Interpret (d) For comparing number of atoms, first we

calculate the moles as all are monoatomic and hence,moles´N A= number of atoms

Moles of 4g He= =4

4 1 mol46g Na 46

Volume ofgas at STP

Mola

r ma ss

Divi

deby

Mu ltiplied by

Flow chart showing conversion of mole in other units

Mole Concept

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 H SO2 4

means a solution which contains 0.12 g-equivalent of

H SO2 4 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 Na CO 2 3

Thus,

Normality N = g- equivalent of solute

volume of solution (in L)

or g - equivalent of solute

volume of solution (in

=

mL)´1000

If specific gravity is known, normality is calculated as

Normality = specific gravity % strength 10

equi

valent 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

mg-molecules per millilitre of solution The molarity is

usually designated by M, e.g., if the molarity of H PO3 4is0.18, it means a concentration corresponding to 0.18 mol

of H PO3 4per litre of solution

Thus, molarity is given as

Molarity, M = moles of solute

volume of solution (in L)

If specific gravity is given,

Molarity =specific gravity % strength 10

molec

ular weight

(c) Formality (F)

It is practically same as molarity

Formality =gram formula weight

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 themass which does not vary with temperature

Molality, m= moles of solute ´

weight of solvent (in g) 1000

Formulae Used to Calculate the Number of Moles

1.Number of millimoles = molarity ´volume in mL

or Moles =millimoles

1000

2. Number of equivalents of a substance

weight (g)equivalent weight

4. Normality=molarity´balance factor( )y

where, y= acidity/basicity/number of replaceable

H-atoms/change in oxidation number

5. Number of equivalents= ´y number of moles

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=

+

n

n N

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

For very dilute solutions,Mole fraction of solute in solution= Mm

ừ÷

9.Per cent by weight of solute in solution

=weight of solute (g) ´100weight of solution (g)

10.Percent by volume of solute in solution

=weight of solute (g) ´100volume of solution (mL)

Some Basic Concepts in Chemistry 13

Sample Problem 13 If the concentration of glucose

(C H O )6 12 6 in blood is 0.9 g L-1 what will be the molarity of

glucose in blood? [ NCERT Exemplar]

SampleProblem 14 Calculate the mass of sodium acetate,

CH COONa3 required to make 500 mL of 0.375 molar aqueous

solution Molar mass of sodium acetate is 82.0245 g mol-1

1000volume of solution(mL)where, w= mass of solute and m = molar mass of solute

Given, molarity of the solution= 0 375 M

Molar mass of solute, M=82.0245g mol-1

Volume of solution=500 mL

\ Mass of solute, w =0.375´82.0245´500

1000

=15.379 g»15.38g

Sample Problem 15 Calculate the concentration of nitric

acid in mol per litre in a sample which has a density,

1.41 g mL-1and the mass per cent of nitric acid in it being 69%.

[ NCERT]

(a) 15.4 M (b) 10.9 M (c) 5.43 M (d) 18.21 M

Interpret (a)

Given, d=1.14 g mL-1, mass % of HNO3=69%

69% HNO3 means 100g of its solution contains 69 g HNO3

(nitric acid)

Hence, mass of HNO3(solute)= 69 g

Molar mass of nitric acid,

-11.4 mLMolarity= ´

´

w m

1000volume of solution (mL)

= ´ ´

´ =

69 1000 1.4163.0146 100 15.439M

Note Concentration of a substance in mol per liter is known as molarity

Alternative method

Molarity=density´mass per cent´10

molar mass of solute

=1.14´69 ´1063.0146 = 15.439 M

Sample Problem 16 Sulphuric acid reacts with sodium hydroxide as follows.

H SO2 4+ 2NaOH¾®Na SO2 4+2H O2

When 1 L of 0.1 M sulphuric acid solution is allowed to react with 1 L of 0.1 M sodium hydroxide solution, the amount of sodium sulphate formed and its molarity in the solution obtained are [NCERT Exemplar]

(a) 0.1 mol L , 3.55 g-1 (b) 0.025 mol L , 7.10 g-1

(c) 0.01 mol L , 5.33 g- 1 (d) 0.25 mol L , 5.33 g- 1

Interpret (b) Moles of H SO2 4=1L´0.1M 0.1 mol=

moles of NaOH =1L´0.1M=0.1 mol

H SO2 4+2NaOH¾®Na SO2 4+ 2H O2

0.1mol 0.1mol 0.05 mol

0.0

2 4 5

1 1= 0.025 mol L-1

SampleProblem 17 If 500 mL of a 5 M solution is diluted to

1500mL, what will be the molarity of the solution obtained?

X1+X2=1Hence, XH O2 = -1 XC H OH2 5

0.96 55.5555.55 C H OH

Sample Problem 19 One mole of any substance contains

6.022´1023 atoms/molecules Number of molecules of H SO2 4

present in 100 mL 0.02M H SO2 4solution is [ NCERT Exemplar]

(a) 12.044´1020molecules (b) 6.022´1023molecules

(c) 1 10´ 23molecules (d) 12.044´1023molecules

Interpret (a) Number of millimoles of H SO2 4 = molarity

´ volume in mL

=0.02´100=2 millimol = ´2 10-3molNumber of molecules =Number of moles´N A

ê ùû

ê ùû

Molar Mass and Molar Volume

Molar mass of an element is defined as mass of 1 mole of

that element, i.e., mass of 6.023 10´ 23 entities or particles

of that element e.g., molar mass of oxygen = 32 g/mol,

that means 6.023 10´ 23molecules of oxygen weigh 32 g,

or molar mass of Na = 23 g/mol, that means 6.023 10´ 23

monoatomic molecules of Na weigh 23 g

When molar mass is divided by density, molar volume is

obtained It is the volume of one mole of a substance

Since molar volumes of solids and liquids do not vary

much with temperature and pressure, these can be

calculated easily by the following relation :

Molar volume molar mass

density

=The molar volumes of gases, change considerably with

temperature and pressure For an ideal gas, the molar

volume at 0°C and 1 atm pressure is 22.4 L

Percentage Composition of CompoundsPercentage composition of the compounds is the relative mass

of the each of the constituent elements in 100 parts of it It isreadily calculated from the formula of the compound Masspercentage of an element

=mass of that element in the compound´100

molar mass of the compoundSampleProblem 21 What is the mass per cent of carbon

in carbon dioxide? [NCERT Exemplar]

(a) 0.034% (b) 3.4%

(c) 27.27% (d) 28.7%

Interpret (c) In order to solve such problem, first write the

formula of molecule/compound

The formula of carbon dioxide= CO2

Molecular mass of CO2=12.0+ 16.0 ´ =2 44 g mol-1

Atomic mass of C=12.0 g atom-1

the empirical formula, i.e., Molecular formula = (empirical formula) ´ n where, n = 1,2,3,Ketc

The molecular formula conveys two informations mainly :

1 The relative number of each type of atoms in a

molecule

2 The total of atoms of each element in the molecule

Deriving Molecular & Empirical

Formula

Use the following steps to determine the empirical formula of the

compound.

1 Calculate the amount of elements and their percentage composition.

2 Divide the percentage of each element by its atomic mass It gives

atomic ratio of the elements present in the compound

3 Divide the atomic ratio of each element by the minimum value of

atomic ratio as to get simplest ratio of the atoms of elements present

in the compound

4 If the simplest ratio is fractional, then multiply the values of simplest

ratio of each element by a smallest integer to get a simplest whole

number for each of the element

5 To get the empirical formula, write symbols of various elements

present side by side with their respective whole number ratio as a

subscript to the lower right hand corner of the symbol

Deriving Molecular Formula

The molecular formula of a substance may be determined from the

empirical formula if the molecular mass of the substance is known The

molecular formula is always a simple multiple of empirical formula and

the value of simple multiple is obtained by dividing molecular mass with

empirical formula mass

SampleProblem 23 A welding fuel gas contains carbon

and hydrogen only Burning a small sample of it in oxygen gives

3.38g carbon dioxide, 0.690 g of water and no other products.

A volume of 10.0 L (measured at STP) of this welding gas is

found to weigh 11.6 g The molecular formula of the compound

Total mass of compound=0.9218+ 0.0767=0.9985g

(because compound contains only carbon and hydrogen)

Percentange of C in the compound=0.9218´ =

0.9985 100 92.32Percentange of H in the compound=0.0767´ =

0.9985 100 7.68

Calculation for Empirical Formula

Element Per cent

by mass

Atomic mass

Relative number

of moles of elements

Simplest molar ratio

12 7 69

7 68 1

»

1 7 68

7 68 1

8 =

Hence, empirical formula= CH(ii) Calculation for molar masss of the gas10.0 L of the given gas at STP weigh=11.6 g

\ 22.4 L of the given gas at STP will weigh

=11.6´22.4=

10 25.984 gMolar mass=25.984»26 g mol-1

(iii) Empirical formula mass (CH)=12 1 13+ =

\ n= molecular mass

empirical formula mass

=26=

13 2Hence, molecular formula= ´n CH= ´2 CH=C H2 2

Sample Problem 24 A compound having empirical formula (C H O)3 4 n has vapour density 84 The molecular formula of this compound is

= C H O9 12 3

Some Basic Concepts in Chemistry 15

Stoichiometry is the most important topic of this chapter Generally questions are seen from this topic The level of question is from moderate to typical.

Calculations based on the quantitative relationship between the

reactants and products are referred as stoichiometry.

(Stoichiometry from Greek words stoichion = element; metron =

measure) Solving of stoichiometric problems require a firm grasp of

mole concept, balancing chemical equations and care in consistent

use of units

The numerals used to balance a chemical equation are called

stoichiometric coefficients.The goal of these calculations is to

predict the relationship between the amounts of the reactants and

products of chemical reaction

Always remember, never lose your goal while solving such problems

Use mole concept carefully and then also find type of stoichiometric

problem before solving

To solve out stoichiometric problems, follow a right approach that is based on a few relatively easy ground rules, which can be understood

by following sample problem.

SampleProblem 25 Predict the amount of oxygen that must be inhaled to digest (burn) 100 of sugar The sugar burns according to the following equation.

Limiting Reactant or Limiting Reagent

If a reaction involves two or more reactants, the reactant that is

consumed first is calledlimiting reagentorlimiting reactantas

it limits the amount of product formed e.g., A limiting reactant is like

a part in an automobile factory, if there are 1000 head lights and 600

car bodies, then the maximum number of automobile will be limited

by the number of headlights Because each body requires two

headlights and the headlights are available only for 500 cars So the

headlights play the role of limiting reagents

Calculation of Limiting Reagent

This can be best understood by considering the following sample

problem

Sample Problem 26 Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation,

Interpret (d) To find the limiting reagent, we have to follow

the following sequence of steps

Chemical Equations

and Stoichiometry

Step I Identify the goal of the problem e.g., in the

above problem

Find out how many grams of O2are consumed when 100 g of sugar is burned?

Step II Write down the key elements of the problem, or

draw a simple picture that summarizes the key

information in the problem

We have 100 g of sugar

Sugar has formula C H O12 22 11.

We know molar equation

Step III Try to do what can be done We can convert the mass (100 g) in moles, as we know molar mass of

C H O12 22 11 i.e.,100

342=0.29 mol

Step IV Do not try to do impossible There is no way to get moles of oxygen directly form moles of sugar in one step

Step V Never lose sight of what you have

accomplished

What else we can do?

We know the molar relationship of O2and sugar in balance chemical equation

We can calculate the number of moles of oxygen needed to burn 0.29 mole ofsugar as, 0.29 12

1= 3.48 mole of O2

Step VI Don’t try to work the problem in your head Write

down all the intermediate steps

Now we can calculate the weight of O as 3.482 ´32=112.28 g O2

Step I Consider the possibility that there may be a limiting amount

of one of the reactants e.g., in the above problem,

dihydrogen appears as a limiting reagent because its amount

is less than dinitrogen

Step II Assume that one of the reactants is the limiting reagent Let

dihydrogen is the limiting reagent

Step III See if you have enough of the other reactant to consume the

material you have assumed to be the limiting reactant If you

do, your original assumption was correct

g

( )g

Q 6g dihydrogen (H )2 reacts with dinitrogen (N )2 = 28 g

Q 1 10´ 3g H2will react with N 28 1 10

3

2= ´ ´

=4.66´10 g3

But the available amount of N is2 = 2.00´10 g3 which is

less than 4.66´10 g.3 That means, our assumption is not

correct

Note If your assumption is true, left the step IV.

Step IV If you don’t, assume that another reagent is the limiting

reagent and test this assumption

the amount of ammonia produced

Amount of H2remain unreacted

Sample Problem 27 The reactant which is entirely consumed in reaction is known as limiting reagent In the reaction 2 A + 4B ®3C + 4D , whein 5 moles of A react with

6 moles of B, then the amount of C formed is [NCERT Exemplar]

Interpret (d) 2 A+ 4B®3C+ 4D

According to the above equation, 2 moles of ‘A’ require 4 moles of

‘B’ for the reaction.

Hence, for 5 moles of ‘A’ the moles of ‘B’ required

But we have only 6 moles of ‘B’ hence, ‘B’ is the limiting reagent So

amount of ‘C’ formed is determined by amount of ‘B’.

Since 4 moles of ‘B' give 3 moles of ‘C’.

Hence, 6 moles of ‘B’ will give

Depending on the units, stoichiometric problems may be

classified into the following relationship

(a) mass-mass relationship-gravimetric analysis

(b) mass-volume relationship

(c) volume-volume relationship

Such relations further be classified as

(i) Gas-gas analysis or eudiometry

(ii) Solutions-volumetric analysis or titration

I Calculations Based on Mass-mass Relationship

In making necessary calculations, following methods can

be used

It involves the following steps

Step I Write the complete and balanced chemical reaction

concerned in the problem

Step II The stoichiometric coefficients in the balanced

chemical reactions represent the relative number ofmoles of the different reaction components Relatethe amounts of the reaction components concerned,with the help of mole

Step III Calculate the unknown amount of substance using

unitary method

Some Basic Concepts in Chemistry 17

Sample Problem 28 Calculate the weight of carbon

dioxide formed by complete combustion of 1.5 g ethane.

This is another approach of solving problems by mole

method, without balancing the reaction As the reactions

are balanced by conserving the atoms of each element,

the mole of atoms of each element in the reactant side

should be equal to that in the product side On applying

the conservation of atoms of each element with the help of

mole, we may get relations needed to solve the problem

For example, the sample problem 28 may be solved by

In this method, the required amount is determined directly

by first converting the given amount of substance into its

moles, then relating the moles of given substance with the

moles of required substance as per balanced chemical

equation or atomic conservation and finally converting the

moles into the amount of the required substance

For example, the sample problem 28 may be solved by

factor label method as

ừ

´ỉè

ừ

÷ ´ỉè

(Based on equivalent concept)

The number of gram-equivalents of each reactantsreacted will be same and the same number ofgram-equivalents of each products will form The sampleproblem 28 may be solved by equivalent concept as

no of g-equivalents of C H2 6= no of g-equivalents of CO2

or 1.5

3014

447

ỉ

èç ưø÷

=ỉ

(a) 122.5 g (b) 196 g (c) 245 g (d) 98 g

Interpret (b) Decomposition of KClO3takes place as,

2KClO ( )3s ¾®2KCl( )s+ 3O ( )2 g

2 moles of KClO3 º3moles of O2

Q 3 moles of O2 are formed by 2 moles of KClO3

\ 2.4 moles of O will be formed by 2

3 2.4

2 ỉ ´

èç ưø÷ moles of KClO3

= 1.6 mol KClO3Mass of KClO3=Number of moles ´ molar mass

´

25 32

2 114gThus, for burning 570 g of octane, oxygen required

=25 mol=25´32 gFor burning 5 moles octane, oxygen required

=25´32´ =2.0 5.0g 2000 g

Proportion method Let x g of oxygen be required for

burning 570.0 g of octane It is known that 2 114´ g of the

octane require 25 32´ g of oxygen; then, the proportion,

Calculation Involving Per cent Yield

In general, when a reaction is carried out in the laboratory

we do not obtain actually the theoretical amount of the

product The amount of the product that is actually

obtained is called the actual yield Knowing the actual

yield and theoretical yield the per cent yield can be

calculated by using the formula,

Per cent yield Actual yield

Theoretical yield 1

SampleProblem 31 For the reaction,

CaO+ 2HCl¾®CaCl2+ H O2

1.23g of CaO is reacted with excess of hydrochloric acid and

1.85g of CaCl2is formed What is the per cent yield?

+ ¾® +

56 g of CaO produces CaCl2=111 g

1.23 g of CaO will produce CaCl 111

56 1.23

2= ´ = 2.43 gThus, theoretical yield= 2.43 g

Actual yield= 1.85 gPer cent yield=1.85´ =

2.43 100 76.1

Calculations Involving Per cent Purity

Depending upon the mass of the product, the equivalent

amount of the reactant present can be determined with

the help of a chemical equation Knowing the actual

amount of the reactant taken and the amount calculated

with the help of a chemical equation, the percentage

purity can be determined as

% purity =wt of reactant required

wt of reactant taken ´100

SampleProblem 32 Chlorine evolved by the reaction of

45.31 g of pyrolusite (impure) and excess of HCl is found to

combine completely with the hydrogen produced by the

reaction of 10 g of magnesium and excess of dilute hydrochloric

acid Find the percentage purity of MnO2 in the given

the mass of pure MnO2required is = 87´ =

24 10 36.25 g

So, 45.31 g of pyrolusite contain MnO2 (pure)= 36.25 g

\ 100 g of pyrolusite contain MnO2(pure)

=36.25´ =45.31 100 80.004 g

\ Percentage of purity= 80.00

Analysis of Mixtures

In such problems, one of the components is supposed to be x g and the other

will be the difference from the total Balanced chemical equations for thereactions of both the components are now written and the total amount ofthe common product produced by the components of the mixture iscalculated It is equated with the data given and the unknown factors are,thus, worked out

SampleProblem 33 A solid mixture (5.0 g) consisting of lead nitrate and sodium nitrate was heated below 600 ° C until

the mass of the residue was constant If the loss in mass is

28.0per cent, find the amount of lead nitrate in the mixture.

(a) 1.67 (b) 3.32 (c) 1.40 (d) 5.00

Interpret (b) Let the amount of NaNO3in the mixture be xg.

\The amount of Pb (NO )3 2in the mixture=(5.0-x)g

170 g of NaNO3evolve oxygen=32 g

x g of NaNO3evolve oxygen= 32

Reactions in Succession

In such problems, the amount of any one of the reaction component

belonging from a reaction is to be determined from the given amount of

some other reaction component belonging from some other reaction with

the help of some common components

SampleProblem 34 How many gram of ethylene can be

burnt completely by the oxygen gas produced from complete

decomposition of 49 g KClO3?

Interpret (c) Method I Solve each concerned reaction

separately First determine the amount of oxygen gas produced by

the complete decomposition of KClO3

Q 245 g KClO3will produce 96 g O2

\ 49 g KClO3will produce O2= 96 ´49

245

= 19.2 g O2

Now, determine the amount of ethane which can be burnt

completely by the oxygen gas produced

Method II Add or subtract the concerned reactions properly such

that the common compound, by which the reactions are related

cancels out The reaction, thus obtained, may be a hypothetical

reaction but it will give the true molar relation

2KClO3 ¾®2KCl+ 3O2

C H + 3O2 4 2¾®2CO2+ 2H O22KClO + C H3 2 4 2KCl 2CO2 2H O2

Method III Relate the moles of the component of given amount

with the component of required amount, with the help of common

substance, with the help of balanced chemical reactions

2 mol KClO3 º3 mol O2º1 mol C H2 4

÷ ´(49 g KClO ) 1 mol KClO

ỉè

ừ

2 4

2 4

´ ỉè

ừ

÷ ´ỉè

ừ

= 5.6 g

Sample Problem 35 0.2415 g sample of a mixture of

MgSO4×7H O2 and MgCl2×6H O2 containing inert impurities was subjected to suitable treatment, as a result of which there were obtained 0.1362 g of BaSO4 and 0.1129 g of Mg P O2 2 7 Calculate the percentage impurity in the original mixture.

Interpret (b) Let the original sample contains a mol

MgSO 7H O4× 2 andb mol MgCl 6H O.2× 2 As BaSO4will form onlyfrom MgSO 7H O4× 2 , from the conservation of SO42–ion,a mole of

BaSO4will form

From question, a´233=0.1362 (i)Now, as Mg P O2 2 7 will form from both MgSO 7H O4× 2 andMgCl 6H O,2× 2 from the conservation of Mg-atom, mole of Mg P O2 2 7formed=a/2+b/2

0.2415 100 59.59%

MgCl 6H O2× 2 =0.000432 mol=0.000432´203=0.0877 gPercentage purity=0.0877´ =

0.2415 100 36.31%

Impurity=100-(59.59+ 36.31)=4.1%

(ii) Calculations Involving Mass-volume Relationship

In such calculations masses of reactants are given and

volume of the product is required and vice-versa 1 mole

of a gas occupies 22.4 L volume at STP thus, mass of a gascan be related to volume according to the following gasequation

Zn+ 2HCl¾®ZnCl2+ H2

Calculate the volume of hydrogen gas liberated at STP when

32.65g of zinc reacts with HCl I mole of a gas occupies 22.7 L volume at STP; atomic mass of Zn = 65.3 u. [NCERT Exemplar]

(a) 22.7 L (b) 22.4 L (c) 11.3L (d) 5.2 L

Interpret (c) From the equation, 65.3 g of zinc liberates 22.7 L

of hydrogen So 32.65 g zinc will liberate

Sample Problem 37 A vessel contains 1.6 g of dioxygen

at STP (273.15 K, 1 atm pressure) The gas is now transferred to

another vessel at constant temperature, where pressure

becomes half of the original pressure The volume of the new

vessel, and number of molecules of dioxygen, are respectively.

[NCERT Exemplar]

(a) 2.24, 6.02´1023 (b) 1.12, 3.01´1022

(c) 2.24 ,3.01´1021 (d) 2.24, 3.01´1022

Interpret (d) (i) p1=1 atm T1=273K, V1=?

32 g of oxygen occupies 22.4 L of volume at STP

Hence, 1.6 g of oxygen will occupy,

These calculations are based on two laws

(i) Avogadro’s law (ii) Gay-Lussac’s law

(See Topic 1.3 for laws.)

Sample Problem 38 What volume of air containing 21%

oxygen by volume is required to completely burn 1 kg of carbon

containing 100% combustible substances?

Q 12 g carbon requires 1 mole of O2for complete combustion

\1000 g carbon will require 1

12´1000 mol O for2combustion,i.e., 83.33 mol O2

(i)Simple titrationsIn a simple titration, a solution of

substance A of unknown concentration is made to react with a solution of B whose concentration is known, so that the concentration of A may be

calculated The two are reacted in such a way that the

volume of B required to completely react with A can be

found out by using some indicators

If the normality of B is N1and its volume used is V1then

the gram-equivalents of B reacted = N V1 1.According to the law of equivalents, the

gram-equivalents of A would be equal to that of B.

\ Gram equivalents of A = N´ V

If the volume of A is V then the normality of A would be

N V V

(ii)Back titrationLet us assume that we have a solid C

which is impure and we want to calculate the

percentage purity We are given two reagents A and B, where the concentration of A is unknown while that of

Some Basic Concepts in Chemistry 21

For back titration to work the following two conditions

should be satisfied.

I A, B and C should be such that A reacts with B, A reacts

with C and the product of A and C cannot react with B.

II The amount of A taken in the beaker before adding C

to it should be such that the gram-equivalents of A in it

should be either greater than or equal to

gram-equivalent of C.

\It can be seen that the gram-equivalents of B that reacted

with A in the first titration = N V1 1which is also equal to the

gram-equivalents of A The gram equivalents of B in the

second titration = N V2 2 which is equal to the

gram-equivalents of A that were left in excess after

reacting with C.

\ Gram-equivalents of A consumed by C = N V1 1- N V2 2

which is also equal to gram-equivalents of C If the ‘n’

factor of C is ´ then the moles of C is N V N V

Sample Problem 40 1.6g of pyrolusite ore was treated

with 50 cc of 1.0 N oxalic acid and some sulphuric acid The

oxalic acid left undecomposed was raised to 250 cc in a flask.

25cc of this solution when titrated with 0.1 N KMnO4required

32cc of the solution Find out the percentage of pure MnO2in

the sample.

Interpret (a) Meq of excess dil oxalic acid in 25 cc= Meq of

KMnO4=0.1´32

\Meq of oxalic acid in 250 cc dilute solution =32

Meq of MnO2=Meq of oxalic acid added - Meq of excess

NaOH 100% reaction is indicated 100% reaction is

indicated

Na CO2 3 50% of reaction up to

NaHCO3stage is indicated

100% reaction isindicatedNaHCO3 No reaction is indicated 100% reaction is

indicatedNaHCO3+ HCl¾®NaCl+ H O2 +CO2

Suppose, the volume of given standard acid solution (HCl)

is required,for complete reaction of NaOH =V amLfor complete reaction of Na CO =2 3 V bmLfor complete reaction of NaHCO =3 V cmLThere may be different combinations of mixture of basesare possible We will opt the following two methods,

Method I We carry two titration separately with twodifferent indicators

Method II We carry single titration but adding secondindicator after first end point is reached

Table 1.4 Results with Two Indicators

b a

2 +

If mixture contains NaOH, Na CO and NaHCO then these are the following cases with the following cases with the indicators.

Case 1 The titre readings of HCl, using phenolphthalein

from beginning =V a+ V b

2

Case2 If methyl orange is added after the first end point,

then the titre readings of HCl = V b + V

c

2

Case3 If methyl orange is added from the very beginning,

the titre readings of HCl = V a+ V b+ V c

Sample Problem 41 A solution contains a mixture of

Na CO2 3 and NaOH Using phenolphthalein as indicator,

25mL of mixture required 19.5 mL of 0.995 N HCl for the end

point With methyl orange, 25mL of solution required 25 mL of

the same HCl for the end point The grams per litre of Na CO2 3

in the mixture is

(a) 23.2 (b) 18.5 (c) 19.9 (d) 12.8

Interpret (a) Let the moles of Na CO and NaOH2 3 in 25 mL

mixture bex and y respectively.

Case 1 when HPh is used as indicator

On solving Eqs (i) and (ii), we get,

x=13.93´10-3 mol and y=5.4725´10-3mol

Now, weight of NaOH in 25 mL mixture

(v)Iodometry This is an indirect way of doing iodimetry

An oxidizing agent A is made to react with excess of

solid KI The oxidizing agent oxidizes I to I- 2 Thisiodine is then made to react with Na S O2 2 3solution Ascan be seen the gram-equivalents of Na S O2 2 3 would

be equal to I2which in turn will be equal to that ofreacted KI and this would be equal to the number of

Milliequivalents of Cl2in 500mL=2 435 ´20=48.7Meq of Cl2= meq of bleaching powder = Meq of available Cl2inthe bleaching powder

Percentage of chlorine= 48.7´ ´

1000

35.55.7 100

= 30.33%

1. How are 0.50 mole Na CO2 3and 0.50 M Na CO2 3different?

2. Write the empirical formula of the compounds havingmolecular formulae H O ,2 2 B H2 6and Fe O 2 3

3. Which reactant checks the amount of barium phosphateformed, when 2 moles each of barium chloride and sodiumphosphate react?

Some Basic Concepts in Chemistry 23

Example 1 10 g of hydrogen fluoride gas occupies

5.6 L of volume at NTP The molecular formula of the gas is

= ´ = 40 gAmong the given molecular formulae, molecular mass of H F2 2is

40 Thus, the molecular formula of the gas is H F 2 2

Example 2 A sample of pure compound contains 2.04 g of

sodium, 2.65´1022atoms of carbon and 0.132 mol of oxygen

atoms Its empirical formula is

Ratio of number of moles =

0.0887 : 0.0440 : 0.132

\ Empirical formula of the compound is Na CO 2 3

Example 3 One volume hydrogen combines with sulphur

to give one volume of a gas X If the vapour density of X is 17,

the number of sulphur atoms in the gas X is

-=2.5´1032

4

= 781.25

\ Number of moles of N2= ´3 781.25=2343.75Mass of nitrogen in the cylinder=2343.75´28

= 65625 g =6.5625´10 g4Total mass of the gas in the cylinder

WORKED OUT

Examples

Example 6 The density of a gaseous element is 5 times that

of oxygen under similar conditions If the molecule of the

element is triatomic, what will be its atomic mass?

(a) 53.33 (b) 55.84 (c) 43.47 (d) 78.86

Solution (a) Molecular mass of oxygen= 32

Vapour density of oxygen=32=

2 16Thus, vapour density of the gaseous element=16´ =5 80

Molecular mass of the gaseous element=80´ =2 160

As the molecule is triatomic, its atomic mass mol mass

atomicity

=

=160=

3 53.33

Example 7 What is the empirical formula of vanadium

oxide if 2.74 g of metal oxide contains 1.53 g of metal?

Example 8 The sulphate of a metal contains 20% metal.

This sulphate is isomorphous with zinc sulphate hepta hydrate.

The atomic mass of the metal is

Example 9 Hydrated sulphate of a divalent metal of atomic

weight 65.4 loses 43.85% of its weight on dehydration The

number of molecules of water of crystallisation in the salt is

=(161.4+ 18x)

Percentage of water=

+ ´ =

18161.4 18 100 43.85

x x

On solving, x= 7

Example 10 1.020 g of metallic oxide contains 0.540 g

of the metal If the specific heat of the metal, M is 0.216 cal deg-1g-1, the molecular formula of its oxide is

(c) M2O4 (d) M2O

Solution (b) Mass of oxygen in the oxide

=(1.020-0.540)=0.480 gEquivalent mass of the metal 0.540

0.480 8 9.0

= ´ =According to Dulong and Petit’s law,

Approx, atomic mass 6.4

sp heat

6.40.216 29.63

= = =

Valency of the metal at mass

eq mass

29.639.0 3

= = =

Hence, the formula of the oxide is M2O 3

Example 11 The atomicity of a molecule, M, if 10 g of it combine with 0.8 g of oxygen to form an oxide, is (specific heat

of the molecule, M is 0.033 cal deg-1g-1and molecular mass of molecule is 199.87 g)

(a) 1 (b) 2

(c) 3 (d) 8

Solution (a) Equivalent mass of M

= mass of metal ´mass of oxygen 8

=10 ´ =8 1000.8

Approximate atomic mass 6.4

sp heat

6.40.033

= =

=193.93 gValency ofM=193.93=

100 2 (nearest whole number)

So, accurate atomic mass= eq mass ´ valency

=100´ =2 200 gAtomicity mol mass

at mass

=

=199.87=

200 1Some Basic Concepts in Chemistry 25