Preview Objective Chemistry for NEET 2020 by K. Singh, Vipul Mehta, Asokan K. K. (2019)

Preview Objective Chemistry for NEET 2020 by K. Singh, Vipul Mehta, Asokan K. K. (2019) Preview Objective Chemistry for NEET 2020 by K. Singh, Vipul Mehta, Asokan K. K. (2019) Preview Objective Chemistry for NEET 2020 by K. Singh, Vipul Mehta, Asokan K. K. (2019) Preview Objective Chemistry for NEET 2020 by K. Singh, Vipul Mehta, Asokan K. K. (2019) Preview Objective Chemistry for NEET 2020 by K. Singh, Vipul Mehta, Asokan K. K. (2019)

Features of the book include:

Focused on NEET examination Chapters summarized in a systematic flow for quick revision

Selected topic-wise Practice Questions to cover all important concepts

Previous Years' NEET Questions (2007-2017) with solutions covered chapter-wise

Three NEET Mock Tests for self-evaluation Includes solved NEET 2018 Chemistry paper

ABOUT THE BOOK

Scan the QR code with your smart phone to access free resources

HIGHLIGHTS OF THE BOOK

About Maestro Series

2020

K Singh Vipul Mehta Asokan K K.

SINGH MEHTA ASOKAN

The book Objective Chemistry for NEET is an endeavour to help students to

prepare for NEET (National Eligibility cum Entrance Test) and other medical entrance examinations NEET-UG has replaced All India Pre-Medical Test (AIPMT), and all individual MBBS and BDS exams conducted by states or colleges themselves for admissions into medical and dental courses

Chapter at a Glance, an attractive feature with concepts summarized in a systematic flow, has been incorporated to enhance the quick-learning The enormous number of practice exercises based on latest NEET pattern has been extensively developed based on NCERT Chemistry books of Class 11 and 12 These are arranged topic-wise under two levels of difficulty and include Previous Years’ NEET Questions The designing of questions is strictly

in accordance with the topic that aids students in approaching the corresponding problems of the topic under study Apart from the Answer Key, the distinctive feature, Hints and Explanations of tricky and difficult questions have also been included to simplify learning At the end of the book, three Mock tests have been provided to give the students an experience of attempting the actual examination.

the needs of medical aspirants.

study of syllabus and relative weightage of topics.

Knowledge: Conceptual strength provided by authoritative yet

v

precise content as per syllabus requirement.

Comprehension: Supported with illustrations and effective

l Chapter at a Glance with flowcharts and tables

for quick revision of all important concepts.

l Solved Examples placed in the same order as the

topics in the feature Chapter at a Glance.

l Practice Exercises based on latest NEET pattern

in big number, arranged topic-wise.

l Previous Years' NEET Questions (2007-2017)

with solutions covered chapter-wise.

l Hints and Explanations for tricky and difficult

Copyright © 2019 by Wiley India Pvt Ltd., 4436/7, Ansari Road, Daryaganj, New Delhi-110002.Cover Image: Fabrizio Denna/Getty Images

All rights reserved No part of this book may be reproduced, stored in a retrieval system, or transmitted

in any form or by any means, electronic, mechanical, photocopying, recording or scanning without the written permission of the publisher

Limits of Liability: While the publisher and the author have used their best efforts in preparing this

book, Wiley and the author make no representation or warranties with respect to the accuracy or completeness of the contents of this book, and specifically disclaim any implied warranties of merchantability or fitness for any particular purpose There are no warranties which extend beyond the descriptions contained in this paragraph No warranty may be created or extended by sales representatives or written sales materials

Disclaimer: The contents of this book have been checked for accuracy Since deviations cannot be

precluded entirely, Wiley or its author cannot guarantee full agreement As the book is intended for educational purpose, Wiley or its author shall not be responsible for any errors, omissions or damages arising out of the use of the information contained in the book This publication is designed to provide accurate and authoritative information with regard to the subject matter covered It is sold on the understanding that the Publisher is not engaged in rendering professional services

Other Wiley Editorial Offices:

John Wiley & Sons, Inc 111 River Street, Hoboken, NJ 07030, USA

Wiley-VCH Verlag GmbH, Pappellaee 3, D-69469 Weinheim, Germany

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John Wiley & Sons (Asia) Pte Ltd, 1 Fusionpolis Walk #07-01 Solaris, South Tower Singapore 138628John Wiley & Sons Canada Ltd, 22 Worcester Road, Etobicoke, Ontario, Canada, M9W 1L1

Chemistry plays a central role among all sciences as well as in our day-to-day life in general The general chemistry course of

CBSE for classes 11 and 12 is designed to give the student an appreciation of this fact and provide a sound foundation of basic

facts and concepts The chemistry syllabus for NEET (earlier AIPMT) is closely aligned with this and the entrance examination

aims to test the competency of the medical aspirants in all three branches of Chemistry – Physical, Inorganic and Organic

The current pattern of examination aims to foster problem-solving skills in students based on the strength or core concepts

and their application The book Objective Chemistry for NEET is designed to prepare the medical aspirants for NEET and

other medical entrance examination along these lines

The book is aimed at students who want to ace their subjects by rigorous practice The coverage of the book is in

accord-ance with the latest NEET syllabus and covers all question-types asked in the target examination It is the endeavor of the

authors of the book to offer it as a must-have resource for these students It is our advice to all the students going through this

book is to solve the problems in a timed manner Here is a brief overview of this book:

1 Chapter at a GlanCe: A summarized, relevant, lucid and up to date coverage of the concepts at the beginning of every

chapter as per the NEET curriculum These can be used as the eleventh hour revision notes

2 Solved exampleS: Ample in number, these are placed in the same order as the topics in the feature Chapter at a Glance

and give students first-hand experience of how to apply concepts to problem-solving

3 praCtiCe exerCiSeS: This section examines the skills of the students to attempt objective problems as per the latest NEET

pattern They are further divided into two levels LEVEL I is the basic level, and LEVEL II is the advanced level The

prob-lems are arranged section-wise to build proper context with the concept

4 previouS YearS’ neet QueStionS: These are solved objective questions from previous years’ NEET (AIPMT) papers This

section will help the student to get acquainted with the actual NEET questions and to test the skills learnt in chapter

5 hintS and explanationS: These are provided for most of the questions in Practice Exercises and Previous Years’ NEET

Questions Special care has been taken to answer questions in the simplest and shortest possible way so that the NEET aspirants develop an attitude for solving problems in the least time

6 moCk teStS: This section, is provided at the end of this book, helps in a holistic revision of all chapters of Chemistry

after the completion of the syllabus These tests are designed after the analysis of the past NEET papers, and the difficulty level of the questions in these tests will be very close to the actual NEET test

We hope that the students would benefit from the experience of using the book It would be the privilege and reward for the

authors if the book will help any student improve his/ her score even by a small percentage We will be grateful for all the readers

(particularly students) for their evaluation, of this book Their comments, criticism and suggestions will be incorporated in

the subsequent editions

Dr K Singh

Dr K Singh has done his post-graduation in Chemistry with special paper in Organic Chemistry and his Ph.D in Water

Pollution from Veer Kunwar Singh University, Ara, Bihar He is an accomplished faculty in Chemistry, imparting quality education

to mainly engineering and also medical aspirants across north India He has a vast teaching experience of more than 21 years

and has taught more than sixty thousand students He has churned out thousands of students successful in competitive

examinations, many of whom attained top hundred ranks and studied in institutes of their choice Many of these students are

doing exceedingly well in the field of research and in corporate sector

Dr Singh is one of the very few faculties from Bihar who had the privilege of teaching at Kota in early days He then went

on to establish the first Coaching Institute at Patna, Bihar, which is based on Kota pattern As a leading educationist of Patna,

he very successfully stopped the migration of the students from Bihar to Kota

Vipul Mehta

With over 10 years of experience in the education industry, Vipul Mehta, an IIT alumnus has excelled in many roles in the past

decade, from being a teacher, a mentor to an entrepreneur He has taught over 20,000 students for various Indian

competi-tive examinations and international Olympiads and has produced numerous all-India toppers and Olympiad medalists in

Chemistry

Vipul has addressed over 1000 educational seminars and has also written a series of articles for leading newspapers to

help students excel in competitive examination In his current role, he is working on his venture Edhola (www.edhola.com),

an initiative that provides a one-stop solution to students from class 8 to 12 for education abroad

Asokan K K.

Mr Asokan K K started his career as a PGT for Chemistry in Kendriya Vidyalaya and has served a long tenure in different

capacities in Jawahar Navodaya Vidyalaya across Kerala For the past 15 years teaching, he has concentrated on teaching

Inorganic Chemistry for Engineering and Medical entrance examinations in Brilliant, Pala He is also the Academic

Facilita-tor for NIOS (National Institute of Open Schooling) in Kochi Centre His earlier publications include Chemistry books for

Vidyarthimithram Publications, Kottayam

1 To start with, read the chapter thoroughly from your text books, notes and NCERT books You must have a comprehensive

knowledge of the entire topic whose questions you are going to crack

beginning of each chapter

3 Now approach the exhaustive topic-wise question bank, to self asses your understanding of the topic Before you start

practicing the questions, set an alarm of 40 minutes in your watch

4 It is recommended that you to aim at attempting at least 45 questions in a single sitting.

5 As the alarm rings, stop solving the paper.

7 If you have doubts, then check the explanations given at the end of the chapter.

8 If you are not satisfied, then ask the doubts from your respective teachers at school or institute.

9 If you do not have access to a trainer, you can email the undersigned.

10 After completing your complete syllabus, you can start solving previous years’ papers to understand the weightage of

each chapter from the examination point of view

11 As a last step in your preparation, start with Mock Tests based on the pattern of NEET and check your answer from the

provided Answer Key list

ALL THE BEST and Enjoy Solving the MCQs!

Preface iii

Directions to Use the Book vii

Laws of Chemical Combinations 13

Calculations of Moles, Empirical and

Hydrogen Spectrum and Rydberg Equation 37

Heisenberg Uncertainty Principle and

Schrödinger’s Wave Equation 39

Aufbau Principle, Hund’s Rule and Pauli’s Exclusion Principle 39

Early Models, Electromagnetic Waves,

Bohr’s Model, Calculation of Radius,

Hydrogen Spectrum and Rydberg Equation 41

Heisenberg Uncertainty Principle and

Schrödinger’s Wave Equation 42

Aufbau Principle, Hund’s Rule and Pauli’s Exclusion Principle 43Previous Years’ NEET Questions 43

Graham’s Law of Effusion and Diffusion 68

Kinetic Theory of Gases and Maxwell Boltzmann Distribution Curve 68

Compressibility Factor (Z), Liquefaction

of Gas, van der Waals Equation 69

Graham’s Law of Effusion and Diffusion 71

Kinetic Theory of Gases and Maxwell Boltzmann

Entropy and Second Law of Thermodynamics 92

Gibbs Free Energy and Spontaneity of a Reaction 92

Thermochemistry, Hess’ Law and

Heat, Work, Internal Energy, Enthalpy, Heat Capacity and First Law of Thermodynamics 94

Entropy and Second Law of Thermodynamics 95

Gibbs Free Energy and Spontaneity of a Reaction 96

Thermochemistry, Hess’ Law and

Relation between K p and K C , Characteristics of Equilibrium Constant K 119

Reaction Quotient (Q), Extent of Reaction 120

Calculation of K in Different Reactions and Degree of Dissociation 120

Thermodynamics of Equilibrium 121

Le Chatelier’s Principle 121

Relation between K p and K C and Characteristics of Equilibrium Constant K 122

Reaction Quotient (Q) and Extent of Reaction 123

Calculation of K in Different Reactions and Degree of Dissociation 123

Thermodynamics of Equilibrium 125

Le Chatelier’s Principle 125Previous Years’ NEET Questions 127

Applications of Activity Series 170

Balancing of Redox Reactions 170

n-Factor and Equivalent Weight 170

Oxidation Number Concept, Oxidizing and

Applications of Activity Series 173

Balancing of Redox Reactions 173

n-Factor and Equivalent Weight 173

Tetrahedral and Octahedral Voids 193

Types of Ionic Crystals, Radius Ratio 193

Tetrahedral and Octahedral Voids 195

Types of Ionic Crystals, Radius Ratio 195

Imperfections in Solids 196

Electrical Properties 197Previous Years’ NEET Questions 197

Construction of Cells and Reference Electrodes 240

Electrochemical Series, Calculation of E o for the Cell and Feasibility of Cell 241

Nernst Equation and Concentration Cells 241

Electrolysis and Faraday’s Laws of Electrolysis 242

Conductance, Kohlrausch’s Law, Batteries and

Construction of Cells and Reference Electrodes 244

Electrochemical Series, Calculation of E o for the Cell and Feasibility of Cell 244

Nernst Equation and Concentration Cells 245

Electrolysis and Faraday’s Laws of Electrolysis 246

Conductance, Kohlrausch’s Law, Batteries and

Order and Molecularity 269

Integrated Rate Law and Half-Life (t 1/2 ) 269

Methods to Find Order of the Reaction 270

Replacement of Concentration Terms

Arrhenius Equation and Collision Theory 271

Mechanism of Complex Reactions 271

Order and Molecularity 272

Integrated Rate Law and Half-Life (t 1/2 ) 272

Methods to Find Order of the Reaction 273

Replacement of Concentration Terms

Arrhenius Equation and Collision Theory 273

Mechanism of Complex Reactions 274Previous Years’ NEET Questions 274

Shielding Effect and Effective Nuclear Charge 316

Atomic and Ionic Radius 316

Electron Gain Enthalpy and Electron Affinity 317

Development of Periodic Table and Modern

Shielding Effect and Effective Nuclear Charge 318

Atomic and Ionic Radius 318

Hints and Explanations 322

14 Chemical Bonding and

Kossel–Lewis Approach to Chemical Bonding 337

Ionic or Electrovalent Bond 337

Molecular Orbital Theory 340

Bonding in Some Homonuclear

Kossel–Lewis Approach to Chemical Bonding 340

Ionic or Electrovalent Bond 340

Hints and Explanations 373

Anomalous Properties of the First Element

of Each Group and Diagonal Relationships 386

Preparation and Properties of Some Important Compounds of Sodium 386

Anomalous Properties of the First Element

of Each Group and Diagonal Relationships 388

Preparation and Properties of Some Important Compounds of Sodium 388

Compounds of Calcium 388

Biological Significance of Na, K, Mg and Ca 389Previous Years’ NEET Questions 389

Hints and Explanations 391

Group 17 Elements – Halogens 416 Group 18 Elements – Noble Gases 420

Hints and Explanations 442

18 General Principles and Processes of

Thermodynamic Principles of Metallurgy 465

Electrochemical Principles of Metallurgy 466

Thermodynamic Principles of Metallurgy 468

Electrochemical Principles of Metallurgy 468

Previous Years’ NEET Questions 469

Hints and Explanations 471

Werner’s Theory of Coordination Compounds 513

Definition of Some Important Terms Pertaining to Coordination Compounds 513

Nomenclature of Coordination Compounds 514

Isomerism in Coordination Compounds 515

Bonding in Coordination Compounds 515

Bonding in Metal Carbonyls 516

Importance and Applications of Coordination

Werner’s Theory of Coordination Compounds 517

Definition of Some Important Terms Pertaining

to Coordination Compounds 517

Nomenclature of Coordination Compounds 518

Isomerism in Coordination Compounds 518

Bonding in Coordination Compounds 518

Bonding in Metal Carbonyls 519

Stability of Coordination Compounds 520Previous Years’ NEET Questions 520

Hints and Explanations 543

ORGANIC CHEMISTRY

22 Organic Chemistry – Some Basic

Isomerism and Dipole Moment 568

Qualitative Analysis of Organic Compounds 571

Quantitative Estimation of Elements 571

Reaction Intermediates 571

Isomerism and Dipole Moment 572

Electrophiles, Nucleophiles, Inductive Effect, Resonance and Hyperconjugation 572

Purification and Characterization of

Qualitative Analysis of Organic Compounds 573

Quantitative Estimation of Elements 573Previous Years’ NEET Questions 573

Hints and Explanations 659

Hints and Explanations 692

Hints and Explanations 716

Acidic Character of Carboxylic Acids 734

Methods of Preparation of Carboxylic Acids 735

Chemical Reactions of Carboxylic Acids 736

Chemical Reactions of Carboxylic

Methods of Preparation of Carboxylic Acids 737

Chemical Reactions of Carboxylic Acids 737

Physical Properties and Chemical Reactions

of Carboxylic Acids Derivatives 738Previous Years’ NEET Questions 739

Hints and Explanations 815

PHYSICAL CHEMISTRY

Some Basic Concepts of Chemistry

1

Chapter at a Glance

1 Matter: It is anything that occupies space.

(a) It exists in three physical states viz.

(i) Gas (ii) Liquid, and (iii) Solid.

(b) Chemically it can be classified into:

(i) Elements, (ii) Compounds, and (iii) Mixture

2 Elements

These cannot be decomposed further into simpler products by any physical and chemical means Majority of them are metals, few are metalloids and non-metals Oxygen is the most abundant element, whereas Si, Al, Fe are the second most, third most and fourth most abundant element of earth They can be further classified into metals and non-metals.

3 Compounds

Two or more different elements in a fixed ratio or proportion of their mass are known as compounds.

4 Mixture

It consists of two or more substances in any proportion in which compounds do not lose their identity They may

be homogeneous or heterogeneous Majority of heterogeneous mixture can be easily separated by physical methods

Homogeneous composition mixtures are difficult to separate by physical methods, (e.g., alloys).

5 Precision and Accuracy

(a) Precision refers to closeness of various measurements for the same quantity to each other and to the average.

(b) Accuracy refers to an agreement of a particular value to the true value of the result.

6 Significant Figures

It is the total number of digits in a number that results from a measurement which is known with certainty.

(a) The rule for determining the number of significant figures is as follows:

(i) All non-zero digits are significant.

(ii) Zero proceeding to first non-zero digit is not significant.

(iii) Zero between two non-zero digits is significant.

(iv) Zero at the end or right of a number is significant provided it is on the right side at the decimal point.

12 inches in a foot.

(vi) If number is written in scientific notation, all digits are significant For example 5.026 × 1023 has 4 cant figures.

the original numbers For example, 16.24 + 13.0 + 1.221 = 30.461 But 13.0 has only one digit after decimal

point hence result will be 30.4 or around to 30.5.

number of significant figures in the least precise measurement For example, 2.4 × 1.21 = 2.904, as 2.4 has two

significant figures, therefore, the result should be 2.9.

7 Measurement of Properties

Properties of matter like length, area, volume, etc., are quantitative in nature and they must be suffixed by certain units.

(a) The international system of units (SI) is a well-accepted system to express.

(b) The SI system allows the use of prefix to indicate the multiples or sub-multiples of a unit.

Multiple Prefix Symbol

8 Laws of Chemical Combination

These were basic laws based on the experimental observations of many scientists in the 18th and early 19th centuries that governed the combination of atoms in the formation of compounds.

It states that no detectable gain or loss of mass occurs in chemical reactions, that is, mass is conserved.

are always combined in the same proportions by mass It is also known as law of definite composition.

compound, the different masses of one element that combine with the same mass of the other element are in the ratio of small whole numbers.

ratios of the volumes of reacting gases are small whole numbers.

(e) Avogadro’s law: It states that equal volumes of different gases at the same temperature and pressure contain the

same number of molecules This constant number of molecules was later determined as Avogardro’s constant and defined as mole The law could be stated as: one mole of any gas contains 6.023 × 1023 molecules and it occupies a volume of 22.4 L at STP.

9 Dalton’s Atomic Theory

(a) Every element is composed of small indivisible particles called atoms.

(b) Atoms of the same element are alike, but different elements have different types of atoms.

(c) Atoms of different elements can combine in simple ratio to form compounds.

(d) The relative number and kind of atoms are always the same in a given compound.

10 Atomic and Molecular Masses

(a) Atomic masses: It is the ratio of mass of one atom of an element to 1/12th part mass of carbon = 12 (12C), which

is an isotope of carbon The unit in which atomic mass measurements are reported is the atomic mass unit (amu)

One atomic mass unit is equal to 1/12th of the mass of one 12C atom Thus,

By definition, the mass of a single atom of the 12C isotope is exactly 12 atomic mass units, or 12 amu.

1 amu = 1.66 × 10-24 g

So, the actual mass of an element is (amu) × 1.66 × 10-24g.

Nowadays, amu has been replaced by u which stands for unified mass.

(b) Average atomic mass: Most of the elements exist as isotopes and since there is a large difference in the natural abundance of these isotopes, the average mass of an element is a weighted average of the masses of the different isotopes For example, chlorine has two isotopes, 1735

weight of the element.

(i) Dulong-Petit law: It is applicable to metals and states that

Atomic weight × specific heat = 6.4 where specific heat is in g cal-1.

(d) Molecular mass: It is defined as the sum of the atomic masses of the elements in the molecule For example, the molecular mass of water (H2O) is determined from the atomic weights of hydrogen and oxygen as 18.02 amu.

Molecular mass = 2 × vapor density

(e) Formula mass: It is the sum of the atomic masses of the elements in the formula It is used for substances which

do not contain discrete molecules as their constituent units For example, in case of sodium chloride, formula mass is used instead of molecular mass as in the solid state it does not exist as a single entity; instead positive and negative entities are arranged in three dimensional structures So, it has a formula (NaCl) mass of 58.44 u, deter- mined by adding the atomic weights of sodium and chlorine For more complex ionic compounds, the formula mass is calculated the same way For example, the formula mass of Cu(NO3)2, is 187.52 u.

11 Mole Concept and Molar Masses

of the 12C isotope of carbon It is the SI unit for the amount of substance The number of particles in a mole is known as Avogadro’s constant (previously known as Avogadro’s number) and is equal to 6.023 × 1023.

element It is the absolute mass in grams of 6.023 × 1023 atoms of an element.

expressed in grams of 6.023 × 1023 molecules of a substance Thus,

Molar Mass

formula of the compound.

12 Percentage Composition: In general, the percentage composition of an element in a compound can be

calcu-lated using the relation

13 Empirical Mass and Molecular Mass

molecule of a substance.

substance.

Molecular formula Empirical formula

14 Equivalent Weight (Eq wt.)

Valenceof anelementt

Totalcharge onionicc part

Number of eleectrons participatinginthe reaction

15 Stoichiometry and Stoichiometric Calculations

which each kind of atom in the reactant and product side is balanced When reactants are mixed in exactly the

mass ratio as determined from the balanced equation, the mixture is said to be stoichiometric The coefficients used with reactants and products to balance these are called stoichiometric coefficients A balanced equation

also mentions the physical state of reactants and products The three main steps in stoichiometric calculations for any chemical reaction are:

(i) Determine the number of moles of starting substance.

(ii) Determine the mole ratio of the desired substance to the starting substance.

(iii) Calculate the desired substance in the units specified in the problem (i.e., in terms of mass, mole, volume or density).

16 Limiting Reagent

(a) In the case of a chemical reaction, if specific amounts of each reactant are mixed, the reactant that produces the least amount of product is called the limiting reagent In other words, the amount of product formed is limited

by the reactant that is completely consumed.

(i) Calculate the amount of product (moles or grams, as needed) formed from each reactant.

determined by the limiting reactant.

reactant required to react with the limiting reactant, and then subtract this amount from the starting quantity

of the reactant.

17 Principle of Atom Conservation (POAC)

The principle states that moles of atoms of an element are conserved throughout the reaction (provided the reactants are converted to products completely).

Advantages of using POAC method over stoichiometry:

(a) No need to balance the equation.

(b) Complete reactions are not required.

18 Methods for Expressing Concentration in Solutions

(b) Mass percent: This expresses the mass of solute per 100 g of solution.

×

×

100 ( )

where n is the number of electron participating or acidity or basicity.

ppm is used whenever concentration is very low.

Solved Examples

and 0.112 L of CH4 at STP What is average molecular

weight of the gas mixture?

Solution

fraction of components of gas mixture and MAvg is the average molecular weight

Number of moles can be calculated as

Moles fractions can be calculated as

2

1 25 2 0 005

=+ +( ),

0.8 g mL-1, 1.425 L of petrol on complete combustion will

⇒Number of moles of O2 = 125 mol

3 149 g of potassium chloride (KCl) is dissolved in 10 L of

an aqueous solution Determine the molarity of the

bub-bled through a solution of 0.205 mol Ba(OH)2 is

5 There are 10 g of mixture of NaCl and NaBr If the amount

of sodium is 25% of the weight of total mixture, calculate the amount of NaCl and NaBr present in the mixture

(Given, atomic weights of Na, Cl and Br are 23, 35.5 and

80, respectively)

(1) NaCl = 1.5734 g, NaBr = 1.5734 g (2) NaCl = 8.4266 g, NaBr = 1.5734 g (3) NaCl = 8.4266 g, NaBr = 8.4266 g (4) NaCl = 1.5734 g, NaBr = 8.4266 g Solution

(4) Let the weight of NaCl be a.

Therefore, weight of NaBr = (10 – a)

Number of moles of NaCl

( )

´ + - ´ = ´ Þ = g

Weight of NaBr (10 – a) = 8.4266 g

6 The number of water molecules present in a drop of

water (volume = 0.0018 mL) at room temperature is

(1) 6 023 10 ´ 19 (2) 1 084 10 × 18

(3) 4 84 10 × 17 (4) 5.023 10× 23

Solution (1) We know that

Density = Mass

Volume

Weight of 0.0018 mL = 0.0018 g (as rH2O=1g mL- 1)

Number of moles = Weight

Molecular weight

= 0.0018

18 = 1 10× ––4

Therefore, number of water molecules =

6.023 10´ 23´ ´1 10- 4 = 6.023 10× 19

and 6% C2H4 100 mL of coal gas is mixed with 150 mL of

O2 and the mixture is exploded What will be the volume

and composition of mixture when cooled to original

volume of CO = 14 mL and volume of C2H4 = 6 mL

The reactions involved are

H g O g H OInitial volume mL mL

Final volume

12

50 25 0( ) + ( ) → ( )l

00 0 25 mL

CH O CO g H OInitial volume mL mL

Final vol

30 60 0 0+ → ( ) + ( )lu

ume 0 0 30mL 60mL

CO O COInitial volume mL mL

Final volume

+ 1 →2

14 7 00

0 14 mL

C H O CO H OInitial volume mL mL

Final volum

6 18 0 0+ → + ( )l

Cl KOH KClO H

→ +

→ ++ → + O+ 12KCl

Adding these equations, we get

with 320 mL of HCl solution having density of 1.15 g

mL-1 and containing 30% by weight HCl, what mass of

Cl2 is generated? (Given atomic mass of K and Cr are 39 and 52, respectively)

(1) 43.52 g (2) 39.62 g (3) 46.015 g (4) 35.73 g Solution

(3) The reaction involved is

Since, HCl is the limiting reagent, therefore

Number of moles of HCl Number of moles of Cl

prepared by heating styrene with tribromobenzoyl oxide in the absence of air If it was found to contain

per-10.46% bromine by weight, find the value of n.

(1) 16 (2) 17 (3) 18 (4) 19

(4) Let the weight of polystyrene prepared be 100 g.

Therefore, number of moles of Br in 100 g of polystyrene

= 10 46

80

= 0.1308 mol From the formula of polystyrene, we have,

0 1308 3. = × .=315 1043 100×

+

WeightMol wt n

On solving we get n = 19.

(w/V ) of H2SO4 (specific gravity 1.84) If the specific

gravity of the resulting mixture is 1.4, what is the

molar-ity of the mixture? (Given that the specific gravmolar-ity of 0.6

M H2SO4 is 1.02 and also consider there is no loss of

mass due to mixing.)

We know that

Total volume of solution L

Calculate, the mole fraction of Na2SO3

(1) 0.065 (2) 0.13

(3) 0.26 (4) 0.39

Solution

We know, volume of solution = 1 L = 1000 mL

Thus, weight of water = 1250 – 474 = 776 g

is 38.3 at 27°C Calculate the moles of NO2 in 100 mol mixture

(1) 33.48 (2) 44.66 (3) 76.46 (4) 91.22 Solution

14 In a victor Meyer determination of the relative

molecu-lar mass of benzene, the heating vessel was maintained

at 120°C A mass of 0.1528 g of benzene was used and the volume of displaced air collected over water at 15°C, was 48 cm3 The barometric pressure was 743 mm Hg

Calculate the relative molecular mass of benzene The vapor pressure of water at 15°C is 13 mm Hg

(1) 39.13 (2) 78.26 (3) 117.39 (4) 156.52 Solution

(2) Actual pressure of displaced air = 743 – 13 = 730 mm Hg

760 0 048

= 78.26 g mol-1

constant weight, gains 5% in its weight Find out position of mixture

144

= ´ a

of its solution was prepared by dissolving 1.25 g of the

sample 25 mL of this solution required 17.75 mL of M/10

AgNO3 solution Calculate the composition of the

(2) In 250 mL of solution, let the weight of NaCl be a.

In 25 mL of solution,

10 and weight of NaNO3

1 2510

=( -a)

(250 10000 2116 )=0 8464

-1

./ . g L

17 What volume of oxygen gas at NTP is necessary for

com-plete combustion of 20 L of propane measured at 0°C

[Using Ideal gas equation]

× = ⇒ =

.

V V

18 On heating 60 mL of a mixture of equal volumes of chlorine

and its gaseous oxide and cooling to atmospheric ture, the resulting gas measured is 75 mL Treatment of this resulting gas mixture with caustic soda (absorbs chlorine) solution resulted in contraction to 15 mL Assuming that all measurements were made at the same temperature and pressure, deduce the formula of oxide (Consider that due

tempera-to heating entire chlorine oxide is decomposed)

Solution

19 Naphthalene (compound of C and H) contains 93.71%

carbon If its molar mass is 128 g mol–1, calculate its molecular formula

Solution (1) Calculation of empirical formula

mass

Relative number

of moles

Simple ratio of moles

Simplest whole number ratio

C 93.71 12 7.809 1.3 × 3 4

H 6.29 1 6.29 1 × 3 3

Calculation of molecular formula

Emprical formula gMolecular

Emprical

mass mass

Laws of Chemical Combinations

and 37Cl) respectively to form two samples of KCl Their

formation follows the law of

(1) constant proportions.

(2) multiple proportions.

(3) reciprocal proportions.

(4) None of these.

2 Irrespective of the source, pure sample of water always

yields 88.89% mass of oxygen and 11.11% mass of

hydro-gen This is explained by the law of

3 A sample contains 200 atoms of hydrogen, 0.05 g atom

of nitrogen, 10-20 g atom of oxygen What is the

approxi-mate number of total atoms?

5 The element A (At wt = 75) and B (At wt = 32) combine

to form a compound X If 3 mol of B combine with 2 mol

of A to give 1 mol of X, the weight of 5 mol of X is

7 Under similar conditions, oxygen and nitrogen are taken in

the same mass The ratio of their volume will be

num-ber is 6.02 × 1023 Calculate the mass of one molecule (in g) of CO2

11 The number of silver atoms present in a 90% pure silver

wire weighing 10 g is (At wt of Ag = 108)

12 An organic compound on analysis was found to

con-tain 0.014% of nitrogen If its molecule concon-tains two N

atoms, then the molecular mass of the compound is

(1) 200 (2) 2000 (3) 20,000 (4) 200000

per-centage abundance of X20 is 90% and its average atomic mass of the element is 20.11 The percentage abundance

of X21 should be

(1) 9% (2) 8%

(3) 10% (4) 0%

14 The chloride of a metal contains 71% chlorine by weight

and the vapor density of it is 50 The atomic weight of the metal will be

(1) 29 (2) 58 (3) 35.5 (4) 71

Stoichiometry, Limiting Reagent, and POAC

and Fe SO2( 4 3) provide equal number of sulphate ions then, the ratio of Fe2 + and Fe3+ ions in mixture is

(1) 1:2 (2) 2:3 (3) 2:1 (4) 3:2

16 A gas mixture of 3 L of propane and butane on complete

combustion at 25°C produced 10 L of CO2 The propane and butane are respectively

(1) 1, 2 L (2) 2, 1 L

(3) 1.5 L each (4) None of these

ratio of number of C:H atoms in the mixture is

(1) 1/5 (2) 2/3

(3) 4/5 (4) 1/3

18 One mole of potassium chlorate is thermally decomposed

and excess of aluminium is burnt in the gaseous product

How many moles of aluminium oxide are formed?

mixed with x mL of oxygen and electrically sparked The

volume after explosion is (16 + x) mL under the same

conditions What would be the residual volume if 20 mL

of the original mixture is treated with aqueous NaOH ?

(1) 12 mL (2) 10 mL

(3) 9 mL (4) 8 mL

number of moles of D formed starting 4 mol of A, are

(1) 8 (2) 16

(3) 4 (4) 10.67

until all the water is driven off If 5.0 g of a mixture gives

3 g of anhydrous salts, what is the percentage by mass of

CuSO4×5H O2 in the original mixture?

(1) 44% (2) 64%

(3) 74% (4) 94%

23 Element X reacts with oxygen to produce a pure sample

of X O2 3 In an experiment it is found that 1 g of X

pro-duces 1.16 g of X O2 3 Calculate the atomic weight of X

(Given atomic weight of oxygen, 16 0 g mol- 1.)

be needed to neutralize 100 g of magnesium hydroxide

Mg(OH)2 ?

(1) 66.7 g (2) 252

(3) 112.6 g (4) 168 g

25 Calculate the weight of iron which will be converted into

its oxide by the action of 18 g of steam on it from the

reaction 2Fe+3H O2 →Fe O2 3+3H2

(1) 37.3 g (2) 3.73 g (3) 56 g (4) 5.6 g

26 2.4 kg of carbon is made to react with 1.35 kg of

alumin-ium to form Al4C3 The maximum amount (in kg) of minium carbide formed is

alu-(1) 5.4 (2) 3.75 (3) 1.05 (4) 1.8

27 Hydrogen evolved at NTP on complete reaction of 27 g

of Al with excess of aqueous NaOH would be (Chemical reaction: 2Al + 2NaOH + 2H2O → 2NaAlO2 + 3H2)

(1) 22.4 L (2) 44.8 L (3) 67.2 L (4) 33.6 L

28 On reduction with hydrogen, 3.6 g of an oxide of metal

left 3.2 g of the metal If the atomic weight of the metal is

64, the simplest formula of the oxide would be

the moles of CO2 formed and remaining moles of excess reagent

(1) 4, 1 (2) 1, 4 (3) 0, 5 (4) 5, 0

to give CaCl2 according to the reaction,

CaCO s3( )+2HCl(aq)→CaCl (aq) CO (g) H O(l)2 + 2 + 2 The mass of CaCO3 required to react completely with 25 mL

of 0.75 M HCl is

(1) 0.1 g (2) 0.84 g (3) 8.4 g (4) 0.94 g

precipi-tate the metal ions from 20 mL each of 1 M Cd(NO3)2 and 0.5 M CuSO4 is

(1) 1:1 (2) 2:1 (3) 1:2 (4) Indefinite

Concentration Terms

33 The molarity of pure water is almost (1) 5.55 M (2) 55.55 M (3) 2 M (4) 1/18 M

34 3.0 molal NaOH solution has a density of 1.11 g mL-1

The molarity of the solution is

(1) 2 97 M (2) 3.05 M

(3) 3 64 M (4) 3 050 M

0.01 M KMnO4 solution in acid medium? (1 mol KMnO4

requires 5 mol of FeSO4 for complete reaction)

Equivalent Weight Concept

38 0.5 g of metal on oxidation gave 0.79 g of its oxide The

equivalent weight of the metal is

(1) 10 (2) 14

(3) 20 (4) 40

39 74.5 g of the metallic chloride contains 35.5 g of chlorine

The equivalent weight of the metal is

(1) 19.5 (2) 35.5

(3) 39.0 (4) 78.0

40 The sulphate of an element contains 42.2% element The

equivalent weight of the metal would be

(1) 17.0 (2) 35.0

(3) 51.0 (4) 68.0

41 0.1 g of a metal gave on reaction with dilute acid at STP

34.2 mL hydrogen gas The equivalent weight of the

metal is

(1) 32.7 (2) 48.6

(3) 64.2 (4) 16.3

42 The equivalent weight of a metal is 4.5 and the

molecu-lar weight of its chloride is 80 The atomic weight of the

10-3 NA molecules of it is

3 Cortisone is a molecular substance containing 21 atoms

of carbon per molecule The weight percentage of bon in cortisone is 69.98% What is the molecular weight

car-of cortisone?

(1) 176.5 (2) 252.2 (3) 287.6 (4) 360.1

4 A crystalline hydrated salt on being rendered anhydrous,

looses 45.6% of its weight The percentage tion of anhydrous salt is Al = 10.5%, K = 15.1%, S = 24.8%

composi-and O = 49.6% The empirical formula of the line salt is

crystal-(1) KAlS O2 8×12H O2 (2) K Al S O2 2 2 8×12H O2

(3) KAl S O2 2 8×12H O2 (4) None of these

5 The atomic weight of a triatomic gas is a The correct

for-mula for the number of moles of gas in its w g is

electrons in 4.2 g of nitride ion (N3-) is (Given one atom

of Nhas 5 valence electrons.)

7 Chlorophyll contains 2.68% of magnesium by mass

Calculate the number of magnesium atoms in 3 g of chlorophyll

8 Two flasks of equal volumes are evacuated, then one is

filled with gas A and other with gas B at the same perature and pressure The weight of B was found to be 0.80 g while the weight of gas A is found to be 1.40 g

tem-What is the weight of one molecule of B in comparison

10 A spherical ball of radius 7 cm contains 56% iron If

density is 1.4 g cm-3, number of moles of Fe present

12 The specific heat of a metal is 0.16 Its approximate

atomic mass would be

(1) 32 (2) 16

(3) 40 (4) 64

13 Hemoglobin contains 0.33% of iron by weight The

molecular mass of hemoglobin is approximately 67200

The number of iron atoms (At mass of Fe = 56) present

in one molecule of hemoglobin is

15 A partially dried clay mineral contains 8% water The

original sample contained 12% water and 45% silica

Percentage of silica in the partially dried sample is nearly

(1) 50% (2) 49%

(3) 55% (4) 47%

Stoichiometry, Limiting Reagent, and POAC

16 510 mg of liquid on vaporization in Victor Meyer’s

appa-ratus displaces 67.2 cm3 of air at (STP) The molecular

weight of the liquid is

combus-tion of 2.5 mol Calculate value of x?

(1) 24 (2) 32

(3) 12 (4) 22

by weight How many grams of this ore would have

to be processed in order to obtain 1 g of pure solid silver Ag?

(1) 74.6 g (2) 85.7 g (3) 134.0 g (4) 171.4 g

Cl2 water The solution is treated with BaCl2 solution The amount of BaSO4 precipitated is

(1) 1 mol (2) 0.5 mol (3) 0.24 mol (4) 0.25 mol

O2 After explosion and cooling, the mixture was treated with KOH, where a reduction of 165 mL was observed

Therefore, the composition of the mixture is

22 An element (X) reacts with hydrogen leading to formation

of a class of compounds that is analogous to bons 5 g of X forms 5.628 g of a mixture of two com-pounds of X XH( 4andX H2 6) in the molar ratio of 2:1

hydrocar-Determine the molar mass of X

(1) 28 (2) 58 (3) 72 (4) 83

Concentration Terms

by weight Its density is 1.2 g mL-1, Its molarity will be

(1) 0.12 (2) 0.06 (3) 1.20 (4) 1.595

24 What volume of a 1.36 M HCl solution should be added

to a 200 mL 2.4 M HCl solution and finally diluted to

500 mL so that molarity of final HCl solution becomes 1.24 M?

(1) 29.2 mL (2) 102.94 mL (3) 46.34 mL (4) 9.4 mL

25 What volume of 0.010 M NaOH (aq) is required to react

completely with 30 g of an aqueous acetic acid solution

in which mole fraction of acetic acid is 0.15?

(1) 108.55 L (2) 18.55 L (3) 34.66 L (4) 42 L

26 An aqueous solution of urea containing 18 g urea in 1500

cm3 of solution has a density of 1.052 g cm-3 If the ular weight of urea is 60, then the molality of solution is

molec-(1) 0.2 (2) 0.192 (3) 0.064 (4) 1.2

time ammonia was found to be 40% by mol The mole

fraction of N2 at that time in the mixture of N2, H2 and

NH3 is

(1) 0.15 (2) 0.3

(3) 0.45 (4) none of these

28 25.4 g of iodine and 14.2 g of chlorine are made to react

completely to yield a mixture of ICl and ICl3 Calculate

the ratio of moles of ICl and ICl3

(1) 1:1 (2) 1:2

(3) 1:3 (4) 2:3

29 25.0 mL of HCl solution gave, on reaction with excess

AgNO3 solution 2 125 g of AgCl The molarity of HCl

solution is

(1) 0.25 (2) 0.6

(3) 1.0 (4) 0.75

Equivalent Weight Concept

(1) 98 (2) 49

(3) 32.66 (4) 24.5

31 0.534 g Mg displaces 1.415 g Cu from the salt solution

of Cu Equivalent weight of Mg is 12 The equivalent

weight of Cu would be

(1) 15.9 (2) 47.7

(3) 31.8 (4) 8.0

If the equivalent weight of the metal is 9, then the atomic

weight of the metal will be

(1) 9 (2) 18

(3) 27 (4) None of these.

from its salt solutions and if the equivalent weights are E1

and E2, respectively, then the equivalent weight of A can

= ´

(1) 18.6 (2) 28

(3) 56 (4) 112.0

35 When a metal is burnt, its weight is increased by 24%

The equivalent weight of the metal will be

(1) 2 (2) 24

(3) 33.3 (4) 76

36 1.5 g of a divalent metal displaced 4 g of copper (At wt

= 64) from a solution of copper sulphate The atomic

weight of the metal is

(1) 12 (2) 24 (3) 48 (4) 6

37 A metallic oxide contains 60% of the metal The

equiva-lent weight of the metal is

(1) 12 (2) 24 (3) 40 (4) 48

38 The weight of a metal of equivalent weight 12, which will

give 0.475 g of its chloride, is

(1) 0.12 g (2) 0.16 g (3) 0.18 g (4) 0.24 g

39 0.84 g of a metal hydride contains 0.042 g of hydrogen Its

equivalent weight is

(1) 80 (2) 40 (3) 60 (4) 20

40 A bivalent metal has the equivalent weight of 12 The

molecular weight of its oxide will be

(1) 24 (2) 34 (3) 36 (4) 40

Previous Years’ NEET Questions

1 An element, X, has the following isotopic composition:

200X : 90%

199X : 8.0%

202X : 2.0%

The weighted average atomic mass of the naturally

occurring element X is closest to

(1) 199 amu (2) 200 amu (3) 201 amu (4) 202 amu

(AIPMT 2007)

mass and has a density of 1.80 g mL-1 The volume of acid required to make 1 L of 0.1 M H2SO4 solution is

(1) 5.55 mL (2) 11.10 mL (3) 16.65 mL (4) 22.20 mL

(AIPMT 2007)

3 How many moles of lead (II) chloride will be formed

from a reaction between 6.5 g of PbO and 3.2 g of HCl?

(1) 0.029 (2) 0.044

(AIPMT 2008)

4 An organic compound contains carbon, hydrogen and

oxygen Its elemental analysis gave C, 38.71% and H, 9.67% The empirical formula of the compound would be

(AIPMT 2008)

5 10 g of hydrogen and 64 g of oxygen were filled in a steel

vessel and exploded Amount of water produced in this

reaction will be

(1) 1 mol (2) 2 mol

(3) 3 mol (4) 4 mol

(AIPMT 2009)

enough water to make 250 mL of solution If sodium

car-bonate dissociates completely, molar concentration of

sodium ion, Na+ and carbonate ions, CO3- are

respec-tively (Molar mass ofNa CO2 3=106g mol- 1)

9 How many grams of concentrated nitric acid solution

should be used to prepare 250 mL of 2.0 M HNO3? The

concentrated acid is 70% HNO3

(NEET 2013)

solu-tion of dichlorotetraaquachromium(III) chloride The

number of moles of AgCl precipitated would be

(1) 0.001 (2) 0.002

(3) 0.003 (4) 0.01

(NEET 2013)

solution The concentration of solution is

(1) 0.02 M (2) 0.01 M

(3) 0.001 M (4) 0.1 M

(NEET 2013)

at STP, the moles of HCl(g) formed is equal to

(1) 1 mol of HCl(g) (2) 2 mol of HCl(g).

(3) 0.5 mol of HCl(g) (4) 1.5 mol of HCl(g).

(AIPMT 2014)

vessel Which reactant is left in excess and how much?

(At wt Mg = 24; O = 16)

(AIPMT 2014)

of 1:4 (w/w) What is the molar ratio of the two gases in

the mixture?

(1) 4:1 (2) 16:1 (3) 2:1 (4) 1:4

mol-1 to 6.022 × 1020 mol-1, this would change

(1) the ratio of chemical species to each other in a

bal-anced equation

(2) the ratio of elements to each other in a compound.

(3) the definition of mass in units of grams.

(4) the mass of one mole of carbon.

(RE AIPMT 2015)

17 20.0 g of a magnesium carbonate sample decomposes

on heating to give carbon dioxide and 8.0 g magnesium oxide What will be the percentage purity of magnesium carbonate in the sample?

(RE AIPMT 2015)

19 What is the mass of the precipitate formed when 50 mL

of 16.9% solution of AgNO3 is mixed with 50 mL of 5.8%

NaCl solution? (At wt Ag = 107.8, N = 14, O = 16, Na = 23,

Cl = 35.5)

(1) 7 g (2) 14 g

(3) 28 g (4) 3.5 g

(RE AIPMT 2015)

20 Suppose the elements X and Y combine to form two

compounds XY2 and X3Y2 When 0.1 mol of XY2 weighs

10 g and 0.05 mol of X3Y2 weighs 9 g, the atomic weights

of X and Y are

(1) 60, 40 (2) 20, 30 (3) 30, 20 (4) 40, 30

(NEET-II 2016)

21 Which of the following is dependent on temperature?

(1) Molarity (2) Mole fraction (3) Weight percentage (4) Molality

Weight of 1 mol of X = 246 g

18´ NA

18´NA´10 20= NA

Let the mass taken be x

32 and no of moles of N2

= x28

Hence, ratio of volume of O2:N2 = x

x

//

3228

2832

78

g

12 (4) Let the molecular mass of the compound be M

M + 35.5 × x = 50 × 2 (1)

35 5 100

= ´+ ´

.( )

x x

M (2)

On solving Eq (1) and Eq (2), we get x = 2, M = 29

and b respectively.

4

+ → +

-a) ( -a)

Therefore, volume of propane = 2 L and volume of butane

= 1 L

n mol of H2 contains = 2 × n × NA hydrogen atoms

10

15

n n

N N

A A

18 (1) The reactions involved are as follows:

32

At end

12

0 0

0 12

Final volume can be calculated as

èç öø÷´+ = + - Þ =

12

® 2

3C DNo.of moles 8 8

®

´ = =4

4 5 246

´ + - ´ =+ - =

Let the atomic weight of X be a

24 (3) The reaction is

3Mg OH( )2+2H PO3 4®Mg PO3( 4)2+6H O2No.of moles of Mg OH( ) No.of moles of H PO3( / ) ( /

3 2

100 583

23

= ×3 × =2

No of moles of M Ox No of moles of M

x 11

3 2641

´

=

x

29 (3) The balanced reaction is

Fe SO2( 4 3) +3BaCl2→3BaSO4+2FeCl3

12

34

t

t t 88

2

82

8

2 2

1 4 8

0mol mol mol

´

= = =

Therefore, number of moles of CO2 = 4 mol

Number of moles of excess reagent = 1 mol

31 (4) From the reaction, we have

nCaCO nHCl

Mass of CaCO

3

1 2100

25 0 75 102

34 (1) Let the volume of solution be V mL

Weightof solvent(kg)

=3

5×50 0 01× mol KMnO4

.( ) ( )

.( ) ( )

( )

3 (4) Let the molecular weight of cortisone be M.

69 98100

252

360 1

= 0 38

0 38 1

=

39 0 38

= 0 38

0 38 1

=

32 0 775

= 0 775

0 38 2

=

16 = -3 1 3 1

0 38 8

=

Therefore, empirical formula of anhydrous salt is

KAlS2O8

5 (2) As gas is triatomic, therefore, mol wt = 3a

a

3

3-N3 4 2 A A

14 8 2 4

7 (1) Number of Mg atoms =2 68× ×

100

324

13 (3) Let the number of iron atoms be x Therefore,

0 33100

56

67200 4

= x× ⇒ =x

43+

“100 g”

SiO2

H2OImpurity

100100100

8(92 - a)+

“y” g

( ) ( )

( )( )

92

100 43 1

100 45 292

Dividing Eq (2) by Eq (1), we get

( )

( )( )( )

92

100 43 1

100 45 292

-Molecular weight Molecular weight g

28mol

From the reaction, we have

No.of moles of C H2 4 No.of moles of CH CH

Hence, weight of polyethylene = 100 g

18 (3) The reaction involved is

( )