Preview Complete Companion For Jee Main 2020 Chemistry Vol 1 by A.K. Singhal, U.K. Singhal (2019)

Preview Complete Companion For Jee Main 2020 Chemistry Vol 1 by A.K. Singhal, U.K. Singhal (2019) Preview Complete Companion For Jee Main 2020 Chemistry Vol 1 by A.K. Singhal, U.K. Singhal (2019) Preview Complete Companion For Jee Main 2020 Chemistry Vol 1 by A.K. Singhal, U.K. Singhal (2019) Preview Complete Companion For Jee Main 2020 Chemistry Vol 1 by A.K. Singhal, U.K. Singhal (2019) Preview Complete Companion For Jee Main 2020 Chemistry Vol 1 by A.K. Singhal, U.K. Singhal (2019)

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Complete Companion for

Also useful for other Engineering Entrance Examinations

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Complete Companion for

Also useful for other Engineering Entrance Examinations

ISBN 978-93-534-3505-9

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Preface xi

Acknowledgements xii

Chapter 1 Some Basic Concepts of

Chemistry 1.1

Physical Quantities and their Measurements 1.2

De Broglie Equation and Dual Nature Theory 2.8

Heisenberg’s Uncertainty Principle 2.9

Hund’s Rule of Maximum Multiplicity 2.14

Chapter 3 Classification of Elements and

Periodicity in Properties 3.1

Trends in Periodic Properties of Elements 3.4

Critical Phenomenon and Liquefaction of Gases 5.7

Contents

Chapter 5B States of Matter: Solid State

Mathematical Analysis of Cubic System 5.37

Packing of Constituents in Crystals 5.40

Magnetic Properties of Solids 5.44

Electrical Properties of Solids 5.45

Resemblance of Hydrogen with Alkali Metals (IA) 9.1

Methods of Preparation of Dihydrogen 9.3

Heavy Water or Deuterium Oxide D 2 O 9.12 Hydrogen Peroxide (Auxochrome) H 2 O 2 9.13

Caustic Soda or Sodium Hydroxide (NaOH) 10.9

Sodium Carbonate or Washing Soda

Sodium Bicarbonate or Baking Soda (NaHCO3) 10.13 Micro Cosmic Salt (Na(NH 4 ) HPO 4 4H 2 O) 10.14

Potassium Super Oxide (KO 2 ) 10.15

Potassium Carbonate (K 2 CO 3 ) 10.16

Potassium Sulphate (K 2 SO 4 ) 10.17 Potassium Bicarbonates (KHCO 3 ) 10.17 Biological Role of Sodium (Na) and

Alkaline Earth Metals and their Compounds

Magnesium Hydroxide Mg(OH)2 10.27

Magnesium Carbonate (MgCO 3 ) 10.27

Magnesium Bicarbonate Mg(HCO 3 ) 2 10.28

Calcium Oxide or Quick–Lime (Cao) 10.29

Calcium Hydroxide or Slaked Lime Ca(OH) 2 10.29

Calcium Oxide or Marble or Lime Stone

Aluminium Chloride AlCl3 or Al2Cl6 11.16

Aluminium Oxide or Alumina Al2O3 11.17

Potash Alum (K2SO4Al2(SO4)3 24H2O) 11.18

Ultra Marine (Na5Al3Si3S3O12) 11.19

Compounds of Tin Stannous Oxide (SnO) 11.54

Red Lead or Tri Lead Tetra-Oxide (Pb 3 O 4 ) 11.60

Lead (II) Halides or Plumbous Halides (PbX 2 ) 11.60 Lead Chloride or Plumbous Chloride (PbCl 2 ) 11.60 Lead (IV) Halides or Plumbic Halides (PbX 4 ) 11.61 Lead Tetrach loride or Plumbic Chloride (PbCl 4 ) 11.61 Lead Acetate or Sugar of Lead (CH 3 COO) 2 Pb 11.61 Basic Lead Carbonate or White Lead

Organic and Inorganic Compounds 12.1

Functional Group Preference Table 12.9

Electrophile or Electrophillic Species 12.47

Nucleophile or nucleophilic species 12.48

Detection of Elements or Qualitative Analysis 12.95

Estimation of Elements or Quantitative Analysis 12.97

Molecular Weight Determination 12.98 Empirical and Molecular Formulas 12.99 Modern Methods of Structure Elucidation 12.99

About the Series

A Complete Resource Book for JEE Main series is a must−have resource for students preparing for JEE Main examination

There are three separate books on Physics, Chemistry, and Mathematics; the main objective of this series is to strengthen the fundamental concepts and prepare students for various engineering entrance examinations It provides class−tested course material and numerical applications that will supplement any ready material available as student resource

To ensure high level of accuracy and practicality, this series has been authored by highly qualified and experienced faculties for all three titles

About the Book

This book, Complete Companion for JEE Main 2020 Chemistry, Volume 1 is an authentic book for all the aspirants

prepar-ing for the joint entrance examination (JEE) This title is designed as per the latest JEE Main syllabus, where the important topics are covered in 34 chapters It has been structured in user friendly approach such that each chapter begins with topic−wise theory, followed by sufficient solved examples and then practice questions

The chapter end exercises are structured in line with JEE questions; with ample number of questions on single choice correct question (SCQ), multiple−type correct questions (MCQ), assertion and reasoning, column matching, passage based and integer type questions are included for extensive practice Previous 15 years’ questions of JEE Main and AIEEE are also added in every chapter Hints and Solutions at the end of every chapter will help the students to evaluate their concepts and numerical applications Because of its comprehensive and in−depth approach, it will be especially helpful for those students who prefers self−study than going for any classroom teaching

Series Features

• Complete coverage of topics along with ample number of solved examples.

• Large variety of practice problems with complete solutions.

• Chapter−wise Previous 17 years’ AIEEE/JEE Main questions.

• Fully solved JEE Main 2019 questions added in opening section of the book

• 5 Mock Tests based on JEE Main pattern in the book.

• 5 Free Online Mock Tests as per the recent JEE Main pattern.

Despite of our best efforts, some errors may have crept into the book Constructive comments and suggestions to further improve the book are welcome and shall be acknowledged gratefully Kindly share your feedback with me at

singhal.atul1974@gmail.com.

A.K Singhal

The contentment and ecstasy that accompany the successful completion of any work would remain essentially incomplete

if I fail to mention the people whose constant guidance and support has encouraged me

I am grateful to all my reverend teachers, especially, late J.K Mishra, D.K Rastogi, late A.K Rastogi and my able guide, S.K Agarwala Their knowledge and wisdom has continued to assist me to present in this work

honour-I am thankful to my colleagues and friends, Deepak Bhatia, Vikas Kaushik, A.R Khan, Vipul Agarwal, Ankit Arora (ASO Motion, Kota), Manoj Singhal, Yogesh Sharma, (Director, AVI), Rajneesh Shukla (Allen, Kota), Anupam Srivastav, Sandeep Singhal, Chandan Kumar (Mentor, Patna), P.S Rana (Vidya Mandir, Faridabad)

I am indebted to my father, B.K Singhal, mother Usha Singhal, brothers, Amit Singhal and Katar Singh, and sister, Ambika, who have been my motivation at every step Their never−ending affection has provided me with moral support and encouragement while writing this book

Last but not the least, I wish to express my deepest gratitude to my wife Urmila and my little, —but witty beyond her years, daughters Khushi and Shanvi who always supported me during my work

A.K Singhal

Acknowledgements

19 Organic Compounds containing Halogens 0 2 6 2 2 2 2 2 1

20 Organic compund containing Nitrogen 0 0 0 2 0 0 0 1 2

21 Organic Compounds Containing Oxygen 1 0 3 1 1 1 1 1 2

22 Transition elements and Coordination

Any species having mass and occupying space is known

as matter It can exist in the three physical states, namely,

solid, liquid and gas

FIGURE 1.1 Classification of Matter

Pencil, air, water, justify the physical states and are all

composed of matter

• At the bulk level or macroscopic level, we can further

classify matter as mixtures or pure substances

Mixture

A mixture is composed of two or more substances which

are known as its components or constituents (in any ratio)

The components of the mixture can be separated with the help of physical separation methods like filtration, crystal-lization, distillation

• A mixture is further classified into two categories— homogeneous and heterogeneous

• In a homogeneous mixture all the components under-

go complete mixing forming a uniform composition as, air or sugar solution

• In a heterogeneous mixture the composition formed due to the mixing of components is not entirely uni-form like in the case of grains mixed with dust etc

Pure Substance

Pure substances have fixed compositions and their uents cannot be separated by using simple physical meth-ods of separation

constit-• A pure substance can be further classified into an ment or a compound

ele-• An element is composed of one type of particle which

could either be atoms or molecules Na, Cu, Ag have only one type of atoms

• A compound is formed by the combination of two or

more atoms or different elements For example, H2O,

After reading this chapter, you will be able to:

Explain the basic characteristics of three states of matter

Differentiate between elements, compounds and mixtures

Use scientific notations and determine significant figures

Define SI base units and convert physical quantities from

one system of units to another

Explain various laws of chemical combination

Explain the significance of atomic mass, average atomic

mass, molecular mass and formula mass

Describe the terms – mole and molar mass Calculate the mass per cent of component elements con- stituting a compound

Determine empirical formula and molecular formula for

a compound from the given experimental data Learn to do the stoichiometric calculations

BASIC CONCEPTS

Dalton’s Atomic Theory

An atom is the smallest particle of an element which is

neutral in nature, retains all the properties of the element

and takes part in a chemical reaction The word atom was

introduced by Dalton (atomos means undivided)

The Dalton’s atomic theory was proposed by Dalton on

the basis of laws of chemical combination

Main assumptions

• Matter (of any type) is composed of atoms

• An atom is the smallest, fundamental, undivided particle

(Building block material)

• An atom can neither be created nor destroyed

• Atoms of an element have similar size, energy and

prop-erties while atoms of different element differ in these

aspects

• Atoms combine in whole number ratios to form a

mol-ecule, therefore, a molecule is the smallest identity that

exists individually

Modern view about atom

According to the modem view:

• An atom is divisible into other smaller particles which

are known as subatomic particles It can also

com-bine in non-whole number ratio as in the case of non-

stoichiometric compounds (Berthollide compounds)

like Fe0.93O

• Atoms of same element also differ in mass and mass

related properties as in the case of isotopes

• Chemical reactions involve rearrangement of atoms

Molecule

The term molecule was introduced by Avogadro It is the

smallest particle (identity) of matter that can exist

inde-pendently and retains all the properties of the substance

Normally the diameter of the molecules is in the range of

4–20 Å and the molecular mass is between 2–1000

• In case of macromolecules, the diameter is in the range

of 50–250 Å and the molecular weight may be in lakhs

Berzelius Hypothesis

According to the Berzelius hypothesis, “Equal volumes

of all the gases contain same number of atoms under the

similar conditions of temperature and pressure.”

This hypothesis on application to law of combining

volume confirms that atoms are divisible which is in

con-trary to Dalton’s theory

PHYSICAL QUANTITIES AND THEIR MEASUREMENTS

In order to describe and interpret the behaviour of ical species, we require not only chemical properties but also few physical properties Physical properties are mass, length, temperature time, electric current etc

Further, to express the measurement of any physical quantity we require its numerical value as well as its unit Hence, the magnitude of a physical quantity can be given as Magnitude of physical quantity 5 Its numerical value 3 Unit

Precision and Accuracy

• The measurements are considered accurate when the average value of different measurements is closer to the actual value

An individual measurement is considered more accurate when

it differs slightly from the actual value

• When the values of different measurements are close

to each other as well as to the average value, such measurements are called precise

• In fact, precision is simply the measurement of tability of an experiment

reproduc-Significant Figures

These are some uncertainties in values during ment of matter In order to make accurate measurements

measure-we use these figures

The total number of digits in a number including the last digit with uncertain value is known as the number of significant figures, for example, 14.3256 6 0.0001 has six significant figures

Rules to determine significant numbers

• All non-zero digits as well as the zeros present between the non-zero digits are significant, for example, 6003 has four significant figures

• Zeros to the LHS of the first non-zero digit in a given number are not significant figures, for example, 0.00336 has only three significant figures

• In a number ending with zeros, if the zeros are present at right of the decimal point then these zeros are also sig-nificant figures, for example, 33.600 has five significant figures

• Zeros at the end of a number without a decimal are not counted as significant figures, for example, 12600 has just three significant figures

• The result of division or multiplication must be reported

to the same number of significant figures as possessed

by the least precise term, for example, 3.331 3 0.011 5 0.036641 ≈ 0.037

• The result of subtraction or addition must be reported to

the same number of significant figures as possessed by

the least precise term, for example, 5.1 1 7.21 1 8.008

5 20.318 ≈ 20.3

Rounding-off non-significant figures

Rounding-off non-significant figures means dropping of

the uncertain or non-significant digits in a number It is

possible as follows:

• If the rightmost digit to be rounded-off is 5, then the

preceding number is increased by one, for example,

3.17 is rounded off to 3.2

• If the rightmost digit to be rounded-off is, 5, then the

preceding number is kept unchanged, for example,

5.12 is rounded off to 5.1

• If the rightmost digit to be rounded-off is equal to 5,

the preceding number is kept as such in case of an even

value However, in case of an odd value it is increased

by one, for example, 4.45 is rounded-off to 4.4; 5.35 is

rounded off to 5.4

Exponential notation or scientific notation

In case a number ends in zeros that are not to the right of

decimal point it is not essential that zeros are significant

For example, 290 has 2 or 3 significant figures and 19500

has 3, 4 or 5 significant figures

This confusion can be removed when the values are

expressed in terms of scientific notations, for example, 19500

can be written as 1.95 3 104 (3 significant figures), 1.950 3

104 (4 significant figures), 1.9500 3 104 (5 significant

fig-ures) In this kind of notation, every number can be written as

Amount of Substance Mole (mol)

N 5 Number with non-zero digit

to the left of the decimal point

For example, 0.00069 can be expressed as 6.9 3 1024

2 2

5 kg m 2 1 sec 2 2 5 Pascal (Pa) Work (W) 5 Force 3 Displacement 5 F 3 d

5 kg m 2 sec 2 2 5 Joule

 Plane angle (Radian, that is, ‘rad’)

 Solid angle (Steradian, that is, ‘str’)

LAWS OF CHEMICAL COMBINATIONS

Law of Conservation of Mass

• Law of conservation of mass was proposed by Lavoisier

in 1774

• It was verified by Landolt

• According to this law, “In a chemical change the total

mass of the products is equal to the total mass of the

reactants, that is, mass is neither created nor destroyed.”

For example, when a solution with calculated weight of

AgNO3 and NaCl is mixed, white precipitates of AgCl

are formed while NaNO3 remains in solution The weight

of the solution remains the same before and after this

experiment

• It is not applicable to nuclear reactions

Law of Constant Composition or Law of Definite

Proportion

• Law of constant composition was proposed by Proust in

1779

• It was verified by Star and Richards

• According to this law, “A chemical compound always

contains same elements combined together in same

pro-portion by mass.” For example, NaCl extracted from sea

water or achieved from deposits will have 23 g Na and

35.5 g of chlorine in its one mole

• It is not applicable to non-stoichiometric compounds like

Fe0.93 O

Law of Multiple Proportion

• Law of multiple proportion was proposed by Dalton

in 1804

• It was verified by Berzilius

• According to this law, “Different weights of an ment that combine with a fixed weight of another element bear a simple whole number ratio.” For example,

ele-in case of CO, and CO2 weight of oxygen which bines with 12 g of carbon is in 1 : 2 ratio

com-• It is applicable when same compound is prepared from different isotopes of an element For example, H2O, D2O

Law of Reciprocal Proportion

• Law of reciprocal proportion was proposed by Richter in

1792

• It was verified by Star

• According to this law, “When two different elements undergo combination with same weight of a third element, the ratio in which they combine will either be same or some simple multiple of the ratio in which they combine with each other.”

• It is also known as Law of equivalent proportion which

states “Elements always combine in terms of their alent weight.”

equiv-Law of Combining Volume

• Law of combining volume was proposed by Gay-Lussac

• Avogadro’s law explains law of combining volumes

• According to this law “Under similar conditions

of temperature and pressure equal volume of gases contain equal number of molecules.”

• It is used in:

1 Deriving molecular formula of a gas

2 Determining atomicity of a gas

3 Deriving a relation Molecular mass 5 2 3 Vapour Density (M 5 2 3 V.D.)

4 Deriving the gram molecular volume

• Avogadro number (N0 or NA) 5 6.023 3 1023.

• Avogadro number of gas molecules occupies

22.4 litre or 22400 mL or cm3 volume at STP

• The number of molecules in 1 cm3 of a gas at STP is

equal to Loschmidt number that is, 2.68 3 1019

• Reciprocal of Avogadro number is known as avogram

MOLE

• Mole is a unit which represents 6.023 3 1023

parti-cles, atoms, molecules or ions etc., irrespective of their

nature

• Mole is related to the mass of substance, the volume of

gaseous substance and the number of particles

• Volume of one mole of any gas is equal to 22.4 litres or

22.4 dm3 at STP It is known as molar volume

• Mole 5WM

5 Molar mass of substance (g.m.m.)Wt of substance in g _

Here, g.m.m 5 Gram molecular mass

Mole 5 Vol of substance in litre 22.4 L

Mole Concept: An Example

A mole of any substance (like N2) stands for:

• 6.023 3 1023 molecules of N2

• 2 3 6.023 3 1023 atoms of nitrogen

• 28 g of nitrogen

• 22.4 litre of N2 at STP

To find total number of identities

⇒ Total Number of Molecules = mole (n) × NA

⇒ Total Number of Atoms in = mole (n) × NA ×

No of atoms present in one molecule

⇒ Total Number of Electrons = mole (n) × NA ×

No of electrons in one molecule

⇒ Total Charge on Any Ion = mole (n) × NA × charge on

• The atomic mass of any element expressed in grams is called g.a.m (gram atomic mass) or gram atom

• A gram atom has number of atoms of the element Atomic mass 5 E 3 V

Here, E 5 Equivalent weight

Here 1 99 × 10–23 g is wt of one C–12 atom

Average atomic mass

Determination of molecular mass

Vapour density method

Mol mass 5 2 3 V.D

V.D 5 Volume at STP (in mL) W 3 22400 _

Here, W 5 Weight of substance in g V.D = Vapour Density

Graham’s diffusion method

r

r =

MM

1 2

2 1









Here r1, r2 are rates of diffusion/effusion for two species

while M1, M2 are their molecular masses respectively

Colligative properties method

pV 5 W _ m RT Here, p 5 Osmotic pressure in atm

• Equivalent weight is the weight of an element or a

com-pound which will combine with or displace 1.008 part

by weight of H2 or 8 part by weight of O2 or 35.5 part by

weight of Cl2

• Equivalent weight is a number and when it is denoted in

grams, it is called gram equivalent.

• It depends upon the nature of chemical reaction in which

substances take part

Methods to Find Equivalent Weight

For acids E = protocity or basicity of acid Molecular weight _

For example, for H3PO4, E = M 3

For H2SO4, E = M 2 (Also for H2C2O4 · 2H2O)

For bases E = Acidity or number of OHMolecular weight 2 ions

For example, for Ca(OH)2, E = M 2

For Al(OH)3, E = M3

or

Fe(OH)3

For ions E = Molecular weightCharge on ion _

For example, for SO 42 2, E = M 2

For P O 43 2

, E = M3

proticity or basicity of acid

For compounds

E = Valency of cation or anion Molecular weight

For example, for CaCO3, E = M 2 For AlCl3, E = M 3

Na3PO4, E = M 3

For redox reactions

=

Totalchange in oxidation number

Let’s take KMnO4 as an example

2KMnO4 1 H2O 2KOH 1 2MnO2 1 3[O]

3 units change in oxidation number

For acidic salts

E 5 Molecular weight of AcidNumber of relaceable H-atoms For H3PO4, for example,

E 5 Wt of metal Wt of oxygen 3 8 Weight of oxygen 5 Weight of metal oxide

2 Weight of metal

(c) Chloride formation method

E 5 Wt of metal Wt of chloride 3 35.5

Weight of chloride 5 Weight of metal chloride

2 Weight of metal (d) Double decomposition method

Eq wt of salt taken

5 _ WW1 2

MOLE FRACTION

• Mole fraction is t he ratio of moles of one component to

the total number of moles present in the solution It is

expressed by X, for example, for a binary solution of two

solute respectively

• Mole fraction does not depend upon temperature as both

the solute and the solvent are expressed by weight

CHEMICAL EQUATION AND STOICHIOMETRY

OF CHEMICAL REACTIONS

1 A balanced chemical reaction represents a

stoichio-metric equation

2 In a stoichiometric equation, the coefficient of reactants

and products represents their stoichiometric amounts

3 The reactant which is completely used up during an

irre-versible reaction is called the limiting reagent while the

reactant left is called the excess reagent, for example, 20

g of calcium is burnt in 32 g of O2, then Ca is the limiting reagent while O2 is the excess reagent

4 Stoichiometric calculations help in finding whether

the production of a particular substance is cally feasible or not

5 These stoichiometric calculations are of following

four types:

(b) Calculations based on weight–volume relationships

(c) Calculations based on volume–volume relationships

(d) Calculations based on weight–volume–energy relationships

6 If the amount of the reactant in a particular reaction is

known, then the amount of the other substance needed

in the reaction or the amount of the product formed in the reaction can be calculated

7 For stoichiometric calculations the following steps

must be considered:

(a) A balanced chemical equation using cal formulas of reactants and products must be written

(b) Here, the coefficients of balanced chemical tion provide the mole ratio of the reactants and products

(c) This mole ratio is convertible into weight–weight (w/w) ratio, weight–volume (w/v) ratio

or volume–volume (v/v) ratio These are called percentage by weight, percentage by volume and percentage by strength respectively

SOLVED EXAMPLES

Mole Concept

1 If 1 Faraday was to be 60230 coulombs instead of 96500

coulombs, what will be the charge on an electron?

3 Calculate the number of atoms of oxygen present in

88 g of CO2 What would be the mass of CO having the

same number of oxygen atoms?

Solution

Number of moles of CO2 5 44 g mol88 g 21

5 2 moles

1 mole of CO2 contains 2 moles of oxygen atoms,

2 moles of CO2 will contain 4 moles of oxygen atoms

Number of oxygen atoms 5 4 3 6.023 3 1023

5 2.4092 3 1024

1 mole oxygen atom is present in 1 mole of CO,

4 moles oxygen atoms are present in 4 moles of CO

As one molecule of CH4 contains (6 1 4) 5 10

ele-ctrons, 6.02 3 1022 molecules of CH4 will have

10 3 6.02 3 1022 5 6.02 3 1023 electrons

5 How many atoms of carbon has a young man given

to his bride-to-be if the engagement ring contains 0.5 carat diamond? (1 carat 5 200 mg)

6 A mixture of aluminium and zinc weighing 1.67 grams

was completely dissolved in acid and the evolved 1.69 litres of hydrogen gas was measured at 273 K and one atmosphere pressure What was the mass of alumin-ium in the original mixture?

5 (1.67 2 A) 22.465.4 L (ii)From (i) and (ii)

8 How many years would it take to spend Avogadro

num-ber of rupees at the rate of 10 lac rupees per second?

Solution

Avogadro number 5 6.023 3 1023

Total rupees 5 6.023 3 1023 Rs

Rate of spending 5 10 lac rupees/s 5 106 Rs/s

Number of years to spend all the rupees

5 6.023 3 1023 Rs

106 3 60 3 60 3 24 3 365 Rs/year

5 1.90988 3 1010 years

9 Oxygen is present in a one litre flask at a pressure of

7.6 3 10210 mm of Hg Calculate the number of

oxy-gen molecules in the flask at 0 °C

required 1.61 3 1023 moles of Mn O 42 for the oxidation

of An1 to A O 32 in an acidic medium What is the value

of 5 N HNO3 are mixed and the volume of the mixture

is made 1000 mL by adding water Find the normality

of the resulting solution

dissolve 0.5 g of copper (II) carbonate?

con-verted into soft water required 22.2 g Ca(OH)2 Calculate the amount of Ca(HCO3)2 present per litre

of hard water

Solution

Reaction Ca(HCO3)2 1 Ca(OH)2 2CaCO3 1 2H2O

14 The formula weight of an acid is 82 In a titration,

100 cm3 of a solution of this acid containing 39.0 g

of the acid per litre were completely neutralized

by 95.0 cm3 of aqueous NaOH containing 40.0 g of

NaOH per litre What is the basicity of the acid?

the solution is made up to 250 mL To 50 mL of this

made up solution, 50 mL of 0.1 N HCl is added and the

mixture after shaking well required 10 mL of 0.16 N

sodium hydroxide solution for complete neutralization

Calculate the per cent purity of the sample of Na2CO3

Solution

Strength of the Na2CO3 solution 5 4 g L21

Suppose the normality of Na2CO3 solution 5 Nx

As after mixing Na2CO3 and HCl solution, NaOH

solution is added, so, according to the normality

V1 5 5 3 100 _ 0.01 5 50000 mL

So, volume of water to be added 5 50000 2 100

525 mL of 0.1 N HCl and no acid is left in the end After converting all calcium chloride to CaSO4, how much plaster of paris can be obtained?

Solution

525 mL of 0.1 N HCl 5 525 mL of 0.1 N CaCl2

5 525 mL of 0.1 N plaster of parisMolecular mass of plaster of paris 5 145 Equivalent mass of plaster of paris 5 145 2

Calculations Based on Reactions

18 Metallic tin in the presence of HCl is oxidized by

K2Cr2O7 solution to stannic chloride What volume of decinorrnal dichromate solution would be reduced by

The reaction is as follows:

6CaO 1 P4O10 2Ca3(PO4)2

852 g P4O10 3 mol P4O10

1 mole of P4O10 neutralizes 6 moles of CaO

3 moles of P4O10 will neutralize 18 moles of CaO Mass of CaO 5 18 3 56 5 1008 g

20 Find the weight of iron which will be converted into its

oxide by the action of 18 g of steam

Solution

The reaction is 3Fe 1 4H2O Fe3O4 1 4H2

4 moles steam reacts with 3 moles Fe

1 mole (18 g) steam reacts with 3/4 moles Fe

5 3 4 mole 3 56 g mol21

5 42 g Fe

21 The mineral haematite is Fe2O3 Haematite ore

con-tains unwanted material called gangue in addition of

Fe2O3 If 5.0 kg of ore contains 2.78 kg of Fe, what per

of an aqueous solution so that it contains 70 mg Na

= 12935 mg

= 12.935 g

CONCEPTS AT A GLANCE

 Giorgi introduced MKS system

 π has infinite number of significant numbers

 20 carat gold is a mixture having 20 parts by weight of

gold and 4 parts by weight of copper

 Some substances like CuSO4.5H2O, Na2CO3.10H2O have a

tendency to lose water in air These are called efflorescent

substances and this tendency is called efflorescence.

 Some solid substances like NaOH, KOH, which have a

tendency to absorb moisture greatly from air and to get

wet are called deliquescent and this tendency is called

deliquescence

 Hygroscopic substances like quicklime (CaO) anhydrous

P2O5 etc., absorb moisture from air

 Compounds having similar chemical composition in the

same crystalline form are called isomorphs

For example, all alums [M2SO4 M2 (SO4)3 24H2O]

Here, M = Monovalent (K)

M = Trivalent (Al) FeSO4.7H2O (Green vitriol) and ZnSO4.7H2O

 Different crystalline forms of a substance are called

polymorphs and this phenomenon is called polymorphism.

For example, ZnS → Zinc blende

Here rl, r2 are rates of diffusion for two species while

M1, M2 are their molecular masses respectively

1 2 1 2

 Specific gravity = Mass of Liquid

NCERT EXEMPLARS

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

(d) Results of student B are both precise and accurate

2 A measured temperature on Fahrenheit scale is

200 °F What will this reading be on Celsius scale?

(a) 40 °C (b) 94 °C

(c) 93.3 °C (d) 30 °C

3 What will be the molarity of a solution, which

contains 5.85 g of NaCl(s) per 500 mL?

(a) 4 mol L–1 (b) 20 mol L–1

5 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?

(a) 4g He (b) 46g Na

(c) 0.40g Ca (d) 12g He

is 0.9 g L–1, what will be the molarity of glucose in

atoms/molecules Number of molecules of H2SO4 present in 100 mL of 0.02M H2SO4 solution is

(a) 12.044 × 1020 molecules (b) 6.022 × 1023 molecules (c) 1 × 1023 molecules (d) 12.044 × 1023 molecules

9 What is the mass percent of carbon in carbon dioxide?

(a) 0.034% (b) 27.27%

(c) 3.4% (d) 28.7%

10 The empirical formula and molecular mass of a

compound are CH2O and 180 g respectively What will be the molecular formula of the compound?

(a) C9H18O9 (b) CH2O (c) C6H12O6 (d) C2H4O2

11 If the density of a solution is 3.12 g mL–1, the mass of 1.5 mL solution in significant figures is _

(a) 4.7g (b) 4680 × 10–3g (c) 4.680g (d) 46.80g

12 Which of the following statements about a compound

constit-(d) The ratio of atoms of different elements in a pound is fixed

com-13 Which of the following statements is correct about

the reaction given below:

4Fe(s) + 3O2(g) → 2Fe2O3(g)(a) Total mass of iron and oxygen in reactants = total mass of iron and oxygen in product therefore it follows law of conservation of mass

(b) Total mass of reactants = total mass of product; therefore, law of multiple proportions is followed NCER

HINTS AND EXPLANATIONS

1 Average of both students is close to the correct

reading But readings of student A are close to each

other and above to the average value

12 A compound does not retain the physical or chemical

properties of its constituent elements

14 In non-balanced equation mass of reactant and

products are different which is against the law of conservation of mass

ANSWER KEYS

1 (b) 2 (c) 3 (c) 4 (b) 5 (d) 6 (c) 7 (d) 8 (a) 9 (b) 10 (c)

11 (a) 12 (c) 13 (d) 14 (b) 15 (b)

(c) Amount of Fe2O3 can be increased by taking any

one of the reactants (iron or oxygen) in excess

(d) Amount of Fe2O3 produced will decrease if the

amount of any one of the reactants (iron or

oxy-gen) is taken in excess

14 Which of the following reactions is not correct

according to the law of conservation of mass

(a) 2Mg(s) + O2(g) → 2MgO(s)

(b) C3H8(g) + O2(g) → CO2(g) + H2O(g)

(c) P4(s) + 5O2(g) → P4O10(s)

(d) CH4(g) + 2O2(g) → CO2(g) + 2H2O (g)

15 Which of the following statements indicates that law

of multiple proportion is being followed

(a) Sample of carbon dioxide taken from any source will always have carbon and oxygen in the ratio 1:2

(b) Carbon forms two oxides namely CO2 and CO, where masses of oxygen which combine with fixed mass of carbon are in the simple ratio 2:1 (c) When magnesium burns in oxygen, the amount

of magnesium taken for the reaction is equal to the amount of magnesium in magnesium oxide formed

(d) At constant temperature and pressure 200 mL of hydrogen will combine with 100 mL oxygen to produce 200 mL of water vapour

PRACTICE EXERCISES Single Option Correct Type

1 The number of significant figures for the three

(d) All have same no.of O-atoms

4 A mole of any substance is related to

(a) number of particles

(b) volume of gaseous substances

(c) mass of a substance

(d) all of these

5 What is the weight of oxygen required for the

com-plete combustion of 2.8 kg of ethylene?

(18%) Find the average atomic weight of boron

11 The correct relationship between molecular mass and

vapour density is (a) V.D 5 2M (b) V.D 5 M 2 (c) M 5 (V.D ) 1 2 (d) V.D 5 M2

12 A bivalent metal has an equivalent mass of 32, the

molecular mass of metal nitrate is?

(a) 168 (b) 182 (c) 184 (d) 188

13 How many moles of potassium chlorate should be

decomposed completely to obtain 67.2 litres of gen at STP?

(a) 1 (b) 2 (c) 3 (d) 4

14 How many grams of phosphoric acid is required to

complete neutralize 120 g of sodium hydroxide?

(a) 0.98 (b) 98 (c) 89 (d) 49

in mass on heating and becomes anhydrous The value

of n is (a) 4 (b) 6 (c) 8 (d) 10

16 The vapour density of a mixture having NO2 and N2O4

is 27.6 The mole fraction NO2 in the mixture is (a) 1.6 (b) 0.8

(c) 2.4 (d) 0.6

17 Among the following pairs of compounds, the one that

illustrates the law of multiple proportions is (a) Cu and CuSO4 (b) CuO and Cu2O (c) H2S and SO2 (d) NH3 and NCl3

18 How many grams of KCl must be added to 75 g of

water to produce a solution with a molality of 2.25 (a) 1.257 g (b) 125.7 g

(c) 12.57 g (d) 25.14 g

19 Normality of 0.04 M H2SO4 is

(a) 0.02 N (b) 0.01 N

(c) 0.04 N (d) 0.08 N

20 Which among the following is the heaviest?

(a) one mole of oxygen

(b) one molecule of sulphur trioxide

(c) 100 amu of uranium

(d) 44 g of carbon dioxide

21 The empirical formula of a commercial ion exchange

resin is C8H7SO3Na The resin can be used to soften

water according to the reaction Ca+2 + 2C8H7SO3Na →

(C8H7SO3)2 Ca + 2Na+ What would be the maximum

uptake of Ca+2 by the resin expressed in mole/g resin?

(a) 0.0024 (b) 0.0246

(c) 0.246 (d) 24.6

22 A boy drinks 500 mL of 9% glucose solution The

number of glucose molecules he has consumed are

[mol wt of glucose 5 180]

(a) 0.5 3 1023 (b) 1.0 3 1023

(c) 1.5 3 1023 (d) 2.0 3 1023

volume of SO2 per litre of air is

25 The number of grams of a dibasic acid (molecular

weight 200) present is 100 mL of its aqueous solution

to give decinormal strength is

64 respectively At 15 8C and 150 mm Hg

pres-sure, one litre of O2 contains ‘N’ molecules The

number of molecules in two litres of SO2 under

the same conditions of temperature and pressure

will be

(a) N (b) N 5 (c) 4N (d) 2N

29 7.5 gram of a gas occupies 5.6 litres as STP The gas is

(a) CO (b) NO (c) CO2 (d) N2O

30 50 gram of calcium carbonate was completely burnt in

air What is the weight (in grams) of the residue? (a) 28 (b) 2.8

(c) 46 (d) 4.8

31 At STP the density of a gas (mol wt 5 45) in g/L is

(a) 11.2 (b) 1000 (c) 2 (d) 224

of N2 and 2 L of H2 is?

(a) 2 L (b) 1 L (c) 0.66 L (d) 1.33 L

oxidized by one mole of KMnO4? (a) 20 (b) 10 (c) 5 (d) 0.5

34 A compound possess 8% sulphur by mass The least

molecular mass is (a) 200 (b) 400 (c) 155 (d) 355

35 The vapour density of ozone is

(a) 24 (b) 16 (c) 48 (d) 72

36 The weight of one molecule of a compound C60H122 is (a) 1.3 3 10220 g (b) 5.01 3 10221 g

(c) 3.72 3 1023 g (d) 1.4 3 10221 g

37 1000 g calcium carbonate solution contains 10 g

car-bonate The concentration of solution is (a) 10 ppm (b) 100 ppm (c) 1000 ppm (d) 10,000 ppm

(a) 4.0 g atoms of hydrogen (b) 3.0 g atom of carbon (c) 6.02 3 1023 atoms of hydrogen (d) 1.81 3 1023 molecules of CH4

combustion of 1 kg coal is?

(a) 22.4 L (b) 2240 L (c) 1866 L (d) 100 L

40 The maximum number of molecules is present in

42 The incorrect statement for 14 g of CO is

(a) it occupies 2.24 litre at NTP

44 Which of the following statement is correct?

(a) 1 mole of electrons weighs 5.4 mg

(b) 1 mole of electrons weighs 5.4 kg

(c) 1 mole of electrons weighs 0.54 mg

(d) 1 mole of electrons has 1.6 3 10219 C of charge

45 Which of the following pairs of gases contain equal

number of molecules?

(a) CO2 and NO2 (b) CO and (CN)2

(c) NO and CO (d) N2O and CO2

46 The samples of NaCl are produced when Na combines

separately with two isotopes of chlorine Cl35 and Cl37

Which law is illustrated?

(a) Law of constant volume

(b) Law of multiple proportions

(c) Law of reciprocal proportions

48 A breakfast cereal in advertized to contain 110 mg of

sodium per 100 g of the cereal The per cent of sodium

in the cereal is (a) 0.110 % (b) 0.01 10 % (c) 11.0 % (d) 0.22 %

49 Express 145.6 L of chlorine in terms of gram moles.

(a) 6.5 g moles (b) 4.5 g moles (c) 0.65 g moles (d) 9.5 g moles

50 The number of significant figures in 306.45 and 40440

are respectively (a) 4, 5 (b) 5, 5 (c) 5, 4 (d) 4, 6

51 The quantity of PV K

BT represents the (a) molar mass of a gas

(b) number of molecules in a gas (c) mass of gas

(d) number of moles of a gas

52 Which is the correct order of micro, nano, femto and

pico here?

(a) micro < nano < pico < femto (b) pico < femto < nano < micro (c) femto < pico < nano < micro (d) femto < nano < micro < pico

53 Find the number of atoms present in 0.016 g of methane.

(a) 0.5 N0 (b) 0.05 N0 (c) N0 (d) 1.6 N0

54 15 litre atmosphere is equal to

(a) 1.515 3 108 erg (b) 15.15 3 109 erg (c) 1.515 3 1010 erg (d) 15.15 3 1012 erg

55 If equal moles of water and urea are taken in a

ves-sel what will be the mass percentage of urea in the solution?

(a) 22.086 (b) 11 536 (c) 46.146 (d) 23.076

[Co(NH3)5Br]SO4 was prepared in 2 litre of solution

1 litre of mixture X 1 excess AgNO3 Y

1 litre of mixture X 1 excess BaCl2 Z Number of moles of Y and Z are

(a) 0.02, 0.01 (b) 0.01, 0.01 (c) 0.01, 0.02 (d) 0.02, 0.02

nearest to (a) 8.67 (b) 6.87 (c) 5.67 (d) 4.26

58 The amount of zinc required to produce 224 mL of

H2 at STP on treatment with dilute H2SO4 will be

(Zn 5 65)

(a) 65.0 g (b) 0.65 g

(c) 6.35 g (d) 0.065 g

59 Ratio of Cp and Cv of a gas ‘X’ is 1.4 The number of

atoms of the gas ‘X’ present in 11.2 litres of it at NTP

will be

(a) 6.02 3 1021 (b) 60.2 3 1023

(c) 6.02 3 1023 (d) 1.02 3 1023

60 When 18 g of glucose is dissolved in 180 g of water

then the mole fraction of glucose is

electrons in 4.2 g of nitride ions (N3-) is

64 What is the molarity of H2SO4 solution that has a density

1.84 g/cc at 358C and contains 98% H2SO4 by weight?

(a) 1.84 M (b) 81.4 M

(c) 18.4 M (d) 184 M

65 The number of moles of oxygen in one litre of air

containing 21% oxygen by volume, in standard

68 The percentage weight of Zn in white vitriol

[ZnSO47H2O] is approximately equal to (Zn 5 65,

S 5 32, O 5 16 and H 5 1) (a) 21.56% (b) 32.58%

(c) 22.65% (d) 26.55%

1 : 4 by weight Then the ratio of their number of ecules in the mixture is

(a) 3 : 32 (b) 7 : 32 (c) 1 : 4 (d) 3 : 16

70 1 cc N2O at NTP contains (a) 1.32 _ 224 3 1023 electrons (b) 6.02 22400 3 1023 molecules (c) 1.8 224 3 1022 atoms (d) all of these

at STP in the Victor Meyer’s method The molecular mass of the liquid is

(a) 54.44 g (b) 34.64 g (c) 64.76 g (d) 74.66 g

72 A 5 molar solution of H2SO4 is diluted from 1 litre to

a volume of 10 litres, the normality of the solution will be

(a) 0.5 N (b) 1 N (c) 2.5 N (d) 5 N

Na3PO4, the maximum number of moles of Ba3(PO4)2that can be formed is

(a) 0.10 (b) 0.20 (c) 0.30 (d) 0.40

weight when it is converted to (a) MnO (b) MnO 422 (c) MnO2 (d) MnO 42

75 An aqueous solution of 6.3 g oxalic acid dehydrate

is made up to 250 ml The volume of 0.1 N NaOH required to completely neutralize 10 mL of this solu-tion is

(a) 4 mL (b) 20 mL (c) 40 mL (d) 60 mL

76 In the standardization of Na2S2O3 using K2Cr2O7 by

iodometry, the equivalent weight of K2Cr2O7 is

(a) same as mol wt (b) mol wt2

(c) mol wt

mol wt6

77 The number of molecules in 4.25 g of ammonia is

(a) 1.5 3 1023 (b) 2.5 3 1023

(c) 3.5 3 1023 (d) 15 3 1023

of 90% pure limestone is heated completely is

STP the volume of CO 2 obtained is

(a) 2.24 L (b) 4.48 L

(c) 44.8 L (d) 48.4 L

Suppose there is no reaction between them If the

density of ethanol is 0.789 g/mL then the molality of

resulting solution is

(a) 0.0256 (b) 0.1056

(c) 1.1288 (d) 0.2076

82 800 g of a 40% solution by weight was cooled 100 g

of solute precipitated The percentage composition of

to water to prepare 150 mL of a solution that is 2.0 M

CH3OH?

(a) 9.6 g (b) 906 g

(c) 4.3 3 102 g (d) 9.6 3 103 g

85 The oxide of an element contains 67.67% of oxygen

and the vapour density of its volatile chloride is 79

Equivalent weight of the element is

(a) 2.46 (b) 3.82

(c) 4.36 (d) 4.96

86 The molar concentration of 20 g of NaOH present in

5 litre of solution is (a) 0.1 mol/L (b) 0.2 mol/L (c) v1.0 mol/L (d) 2.0 mol/L

of molecules in it is (a) 3.01 3 1012 (b) 3.01 3 1018 (c) 3.01 3 1024 (d) 3.01 3 1030

88 Maximum number of molecules will be in

(a) 1 g of H2 (b) 10 g of H2 (c) 22 g of O2 (d) 44 g of CO2

89 Haemoglobin contains 0.33% of iron by weight The

molecular weight of haemoglobin is approximately

67200 The number of iron atom (at wt of Fe is 56) present in one molecule of haemoglobin are

(a) 1 (b) 6 (c) 4 (d) 2

the reaction:

NaOH 1 H3PO4 NaH2PO4 1 H2O is (a) 89 (b) 98

(c) 59 (d) 29

91 2.76 g of silver carbonate (At mass of Ag 5 108) on

being heated strongly yields a reduce weighing (a) 2.32 g (b) 3.32 g

(c) 1.36 g (d) 0.32

92 4 g caustic soda is dissolved in 100 cc of solution The

normality of solution is (a) 1 (b) 0.8 (c) 0.6 (d) 0.10

at 423K 112 ml of CO2 was formed only What is the

% of Na2 CO3 here in the mixture?

(a) 84% (b) 16%

(c) 32% (d) 68%

passed through a solution having 0.205 mole of Ba (OH)2 is?

(a) 40.5 gm (b) 20.25 gm (c) 81 gm (d) 4.05 gm

95 120 gram of urea is present in 5 litre of solution The

active mass of urea is (a) 0.06 (b) 0.2

96 The normality of orthophosphoric acid having purity

of 70 % be weight and specific gravity 1.54 is

(a) 11 N (b) 22 N

(c) 33 N (d) 44 N

moles of CO2 left are

(a) 2.88 3 1023 (b) 28.8 3 1023

(c) 288 3 1023 (d) 28.8 3 103

98 What is the volume (in litres) of oxygen at STP

required for complete combustion of 32 g of CH4?

(mol wt of CH4 5 16)

(a) 89.6 (b) 189.6

(c) 98.4 (d) 169.5

99 Two grams of sulphur is completely burnt in oxygen

to form SO2, In this reaction, what is the volume (in

litres) of oxygen consumed at STP? (At wt of sulphur

and oxygen are 32 and 16 respectively)

(a) 22.414 16 (b) 16 _ 22.441

(c) 32.414 18 (d) 42.414 16

100 How many water molecules are there in one drop of

water (volume 5 0.0018 mL) at room temperature?

(a) 4.86 3 1017 (b) 6.023 3 1024

(c) 2.584 3 1019 (d) 6.023 3 1019

101 ‘X’ litres of carbon monoxide is present at STP It

is completely oxidized to CO2 The volume of CO2

formed is 11.207 litres at STP What is the value of ‘X’

103 One mole of fluorine is reacted with two mole of

hot and concentrated KOH The products formed are

KF, H2O and O2 The molar ratio of KF, H2O and O2

105 What is the volume (in litre) of oxygen required at

STP to completely convert 1.5 moles of sulphur to phur dioxide?

(a) 33.6 (b) 43.6 (c) 11.2 (d) 23.6

106 In acidic medium, dichromate ion oxidize ferrous ion

to ferric ion If the gram molecular weight of sium dichromate is 294 g, its equivalent weight is (a) 19 (b) 49

(c) 99 (d) 294

CaO X is passed into an aqueous solution containing one mole of sodium carbonate What is the number of moles of sodium bicarbonate formed?

(Mol wt of CaCO3 5 100, Na2CO3 5 106, NaHCO3 5 84)

(a) 0.010 (b) 0.2 (c) 0.4 (d) 10

in the production of 270 kg of aluminium metal from bauxite by the Hall process is (atomic mass of Al 5 27) (a) 180 kg (b) 270 kg

(c) 145 kg (d) 90 kg

of KI in alkaline medium is (a) 2 (b) 1 (c) 5 (d) 6

110 How many grams of dibasic acid (mol wt 200) should

be present in 100 mL of the aqueous solution to give 0.1 normality?

(a) 1 g (b) 1.5 g (c) 0.5 g (d) 20 g

111 ‘X’ gram of calcium carbonate was completely burnt

in air The weight of solid residue formed is 28 g What

is the value of ‘X’ (in grams)?

(a) 50 (b) 100 (c) 150 (d) 200

112 One mole of acidified K2Cr2O7 on reaction with excess

KI will liberate …… moles (s) of I2 (a) 2 (b) 3

(c) 6 (d) 7

113 0.59 g of the silver salt of an organic acid (molar mass

210) on ignition gave 0.36 g of pure silver The ity of the acid is

(a) 2 (b) 3 (c) 4 (d) 5

114 KMnO4 (mol wt 5 158) oxidizes oxalic acid in acidic

medium to CO2 and water as follows

5C2 O 422 1 2Mn O 42 1 16H1 10CO2 1

2Mn21 1 8H2O What is the equivalent weight of KMnO4?

(a) 158 (b) 31.6

(c) 39.5 (d) 79

115 Sodium bicarbonate on heating decomposes to form

sodium carbonate, CO2 and water If 0.2 moles of

sodium bicarbonate is completely decomposed, how

many moles of sodium carbonate is formed?

(a) 0.1 (b) 0.2

(c) 0.05 (d) 0.025

116 A purified pepsin was subjected to amino acid

anal-ysis The amino acid present in the smallest amount

was lysine, C6H14N2O2 and the amount of lysine was

found to be 0.431 g per 100 g of protein The

mini-mum molecular mass of protein is

118 Given below is the graphical representation of

vol-umes occupied by several gases at STP find out which

gas/gases is/are not placed at the correct position?

11.2 5.6

H2SO4, then the number of moles of H2SO4 left are

(a) 0.1 3 1023 (b) 5 3 1024

(c) 1.2 3 1024 (d) 1.5 3 1023

120 A metal oxide has the formula M2O3 It can be reduced

by hydrogen to give free metal and water 0.1595 g of

the metal oxide requires 6 mg of hydrogen for

com-plete reduction What is the atomic weight of metal?

(a) 54.4 (b) 46.56

(c) 55.8 (d) 58.5

121 A chloride of a metal (M) has 65.5% of chlorine 100

ml of vapour of the chloride of metal at STP weighs 0.72 g The molecular formula of this metal chloride is (a) MCl3 (b) MCl4

(c) M2Cl3 (d) MCl5

molality is nearly when the density of the tion is 1.1 g/mL

(a) 9 3 1023 (b) 1.8 3 1023 (c) 4.5 3 1023 (d) 1.1 3 1023

123 One mole of magnesium in the vapour state absorbed

1200 kJ mol21 of energy If the first and second tion energies of Mg are 750 and 1450 kJ mol21 respec-tively, the final composition of the mixture is

(a) 86% Mg1 1 14% Mg21 (b) 36% Mg1 1 64% Mg21 (c) 69% Mg1 1 31% Mg21 (d) 31% Mg1 1 69% Mg21

124 A metal oxide has the formula Z2O3 It can be reduced

by hydrogen to give free metal and water 0.1596 g of the metal oxide requires 6 mg of hydrogen for com-plete reduction The atomic weight of the metal is (a) 55.8 (b) 65.8

(c) 6.58 (d) 15.9

125 The volume of carbon dioxide gas evolved at STP by

heating 7.3 g of Mg(HCO3)2 will be (a) 1100 mL (b) 1120 mL (c) 2230 mL (d) 3240 mL

126 The amount of Zinc (atomic weight 5 65) necessary

to produce 224 mL of H2 by the reaction with an acid will be

(a) 0.65 g (b) 7.6 g (c) 6.5 g (d) 8.5 g

at STP on heating 9.85 g of BaCO3 (atomic mass,

Ba 5 137) will be (a) 1.12 L (b) 4.84 L (c) 2.12 L (d) 2.06 L

128 Specific volume of cylindrical virus particle is 6.02 3

1022 cc/g Whose radius and length are 7 Å and

10 Å respectively If NA 5 6.02 3 1023, find molecular weight of virus

(a) 15.4 kg/mol (b) 1.54 3 104 kg/mol (c) 4.68 3 104 kg/mol (d) 2.08 3 103 kg/mol PRA

when one mole of ammonia and one mole of oxygen

are made to react to completion, then

(a) 1.0 mol of H2O is produced

(b) all the oxygen is consumed

(c) 1.5 mol of NO is formed

(d) all the ammonia is consumed

131 The number of gram molecules of oxygen in 6.02 3

1024 CO molecules is

(a) 10 g molecules (b) 5 g molecules

(c) 1 g molecules (d) 0.5 g molecules

132 The number of oxalic acid molecules in 100 ml of 0.02

N oxalic acid solution is

(a) 6.023 3 1022 (b) 1023

(c) 6.022 3 1020 (d) none of these

133 In the reaction

4NH3 (g) 1 5O2 (g) 4NO (g) 1 6H2O (l)

when 1 mol of ammonia and 1 mol of O2 are made to

react to completion then

(a) 1.0 mol of H2O is produced

(b) 1.0 mol of NO will be produced

(c) all the ammonia will be consumed

(d) all the oxygen will be consumed

134 Pressure in a mixture of 4 g of O2 and 2g of H2

con-fined in a container of 1 litre capacity at 0 °C is

(a) 25.2 atm (b) 35.6 atm

136 Percentage of Se in peroxidase anhydrous enzyme is

0.5% by weight (at wt 5 78.4) then minimum lar weight of peroxidase anhydrous enzymes is (a) 1.568 3 103 (b) 1.568 3 104 (c) 25.68 (d) 4.316 3 104

137 For the formation of 3.65 gm of HCl, what volume of

H2, and Cl2 are needed at N.T.P?

(a) 1.12 L, 1 12 L (b) 1.12 L, 2.24 L (c) 3.65 L, 1.83 L (d) 1 L, 1 L

138 Study the following table:

Compound (mol wt)

Wt of compound (in g) taken

11.2 dm3 of CO2 gas at STP The mass of KOH required

to completely neutralize the gas is (a) 56 g (b) 28 g (c) 42 g (d) 20 g

Previous Years' Questions

140 Number of atoms in 560 g of Fe (atomic mass 56 g

141 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 5 10.8) from the tion of boron trichloride by hydrogen? [2003]

(a) 89.6 L (b) 67.2 L (c) 44.8 L (d) 22.4 L

142 25 mL of a solution of barium hydroxide on titration

with 0.1 molar solution of hydrochloric acid gave a titre value of 35 mL The molarity of barium hydroxide

(a) 0.07 (b) 0.14

(c) 0.28 (d) 0.35

143 To neutralize completely 20 mL of 0.1 M aqueous

solution of phosphorus acid, the volume of 0.1 M

aqueous KOH solution required is [2004]

(a) 10 mL (b) 40 mL

(c) 60 mL (d) 80 mL

of its solution The concentration of urea solution is

[2004]

(a) 0.02 M (b) 0.001 M

(c) 0.01 M (d) 0.1 M

145 If we consider that 16 , in place of 12 1 , mass of carbon

atom is taken to be the relative atomic mass unit, the

mass of one mole of a substance will [2005]

(a) decrease twice

(b) increase two fold

(c) remain unchanged

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

(PO4)2 will contain 0.25 mole of oxygen atoms?

[2006]

(a) 0.02 (b) 3.125 3 1022

(c) 1.25 3 1022 (d) 2.5 3 1022

147 Density of a 2.05 M solution of acetic acid in water is

1.02 g/mL The molality of the solution is [2006]

(a) 1.14 mol kg21 (b) 3.28 mol kg21

(a) 33.6 L H2 (g) is produced regardless of

tempera-ture and pressure for every mole of Al that reacts

(b) 67.2 L H2 (g) at STP is produced for every mole

solution, that is, 29% H2SO4 (molar mass 5 98 g

mol21) by mass will be [2007]

(a) 1.88 (b) 1.22

(c) 1.45 (d) 1.64

150 A 5.2 molal aqueous solution of methyl alcohol,

CH3OH, is supplied What is the mole fraction of methyl alcohol in the solution? [2007]

(a) 0.86 (b) 0.086 (c) 0.043 (d) 1.0

151 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 solutions is: [2012]

(a) 1.02 M (b) 0.50 M (c) 2.05 M (d) 1.78 M

152 The molarity of a solution obtained by mixing 750

mL of 0.5(M) HCl with 250 mL of 2(M) HCl will be

[2013]

(a) 1.75 M (b) 0.975 M (c) 0.875 M (d) 1.78 M

153 The molecular formula of a commercial resin used

for exchanging ions in water softening is C8H7SO3Na (mol wt 206) What would be the maximum uptake

of Ca2+ ions by the resin when expressed in mole per

154 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 per-son would gain if all 1H atoms are replaced by 2H

(a) 15 kg (b) 37.5kg (c) 7.5 kg (d) 10 kg

excess HCl produces 0.01186 mole of CO2 The molar mass of M2CO3 in g mol–1 is: [2017]

(a) 1186 (b) 84.3 (c) 118.6 (d) 11.86

156 An aqueous solution contains an unknown

concen-tration of Ba2+ When 50 mL of a 1 M solution of

Na2SO4 is added, BaSO4 just begins to precipitate The final volume is 500 mL The solubility product of BaSO4 is 1 × 10–10 What is the original concentration

(a) 1.0 × 10–10 M (b) 5 × 10–9 M (c) 2 × 10–9 M (d) 1.1 × 10–9 M

ANSWER KEYS

Single Option Correct Type

1 (b) 2 (d) 3 (d) 4 (d) 5 (a) 6 (c) 7 (d) 8 (b) 9 (b) 10 (b)

11 (b) 12 (d) 13 (b) 14 (b) 15 (d) 16 (b) 17 (b) 18 (c) 19 (d) 20 (d)

21 (d) 22 (c) 23 (a) 24 (a) 25 (a) 26 (d) 27 (a) 28 (d) 29 (b) 30 (a)

31 (c) 32 (d) 33 (c) 34 (b) 35 (a) 36 (d) 37 (d) 38 (a) 39 (c) 40 (a)

41 (a) 42 (d) 43 (b) 44 (c) 45 (d) 46 (d) 47 (b) 48 (a) 49 (a) 50 (b)

51 (b) 52 (c) 53 (b) 54 (b) 55 (d) 56 (b) 57 (b) 58 (b) 59 (c) 60 (a)

61 (b) 62 (c) 63 (a) 64 (c) 65 (c) 66 (c) 67 (d) 68 (c) 69 (b) 70 (d)

71 (d) 72 (b) 73 (a) 74 (c) 75 (c) 76 (d) 77 (a) 78 (c) 79 (c) 80 (b)

81 (b) 82 (a) 83 (c) 84 (a) 85 (b) 86 (a) 87 (a) 88 (d) 89 (c) 90 (b)

91 (a) 92 (a) 93 (b) 94 (a) 95 (c) 96 (c) 97 (a) 98 (a) 99 (a) 100 (d)

101 (d) 102 (b) 103 (d) 104 (a) 105 (a) 106 (b) 107 (b) 108 (d) 109 (b) 110 (a)

111 (a) 112 (b) 113 (b) 114 (b) 115 (a) 116 (c) 117 (b) 118 (a) 119 (b) 120 (c)

121 (a) 122 (a) 123 (c) 124 (a) 125 (b) 126 (a) 127 (a) 128 (a) 129 (c) 130 (b)

157 The ratio of mass percent of C and H of an organic

compound (CxHyOz) is 6 : 1 If one molecule of the

above compound (CxHyOz) contains half as much

oxy-gen as required to burn one molecule of compound

CxHy completely to CO2 and H2O, then empirical

for-mula of compound CxHyOz is: [2018]

(a) C2H4O3 (b) C3H6O3

(c) C2H4O (d) C3H4O2

158 A solution of sodium sulfate contains 92 g of Na+ ions per kilogram of water The molality of Na+ ions in that solution in mol kg-1 is: [2019]

(a) 16 (b) 8 (c) 4 (d) 12

HINTS AND EXPLANA

This is law of multiple proportion

So 15 litre atm 5 15 3 1.01 3 10 9 erg