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)
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 questions l Three Mock Tests to develop examination temperament l Solved NEET 2018 Chemistry paper 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 About Maestro Series l Idea: World-class content developed by “Master teachers” adapted to the needs of medical aspirants l Process: Developed by best-in class teachers in the field after in depth study of syllabus and relative weightage of topics l Approach: Designing to meet the objective of: Knowledge: Conceptual strength provided by authoritative yet precise content as per syllabus requirement v Comprehension: Supported with illustrations and effective pedagogy v Application: Assessment as per medical entrance through thought-provoking exercises as per the pattern of NEET (previously AIPMT) papers v Scan the QR code with your smart phone to access free resources Wiley India Pvt Ltd Customer Care +91 120 6291100 csupport@wiley.com www.wileyindia.com www.wiley.com ISBN 978-81-265-9833-5 Objective CHEMISTRY NEET SINGH MEHTA ASOKAN 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 788126 598335 2020 FOR NEET Visit us at https://www.wileyindia.com/resources/ Objective CHEMISTRY l ABOUT THE BOOK 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 FOR HIGHLIGHTS OF THE BOOK K Singh Vipul Mehta Asokan K K 2020 Objective CHEMISTRY FOR NEET K Singh Vipul Mehta Asokan K K Objective Chemistry for NEET 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 John Wiley & Sons Australia Ltd, 42 McDougall Street, Milton, Queensland 4064, Australia John Wiley & Sons (Asia) Pte Ltd, Fusionpolis Walk #07-01 Solaris, South Tower Singapore 138628 John Wiley & Sons Canada Ltd, 22 Worcester Road, Etobicoke, Ontario, Canada, M9W 1L1 First Edition: 2018 Reprint: 2019 ISBN: 978-81-265-9833-5 ISBN: 978-81-265-8875-6 (ebk) www.wileyindia.com Printed at: Preface 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 accordance 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 problems 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 ksingh.visiondirector@gmail.com Vipul Mehta marvellousmehta@gmail.com Asokan K.K asokankk007@gmail.com FM_TOC.indd 1/5/2018 5:25:24 PM FM_TOC.indd 1/5/2018 5:25:24 PM About the Authors 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 competitive 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 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 Facilitator for NIOS (National Institute of Open Schooling) in Kochi Centre His earlier publications include Chemistry books for Vidyarthimithram Publications, Kottayam FM_TOC.indd 1/5/2018 5:25:24 PM FM_TOC.indd 1/5/2018 5:25:24 PM Directions to Use the Book 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 2 Then, go to the corresponding chapters in this book and quickly go through Chapter at a Glance provided at the 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 6 Check answers with those given in the Answer Key 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 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! FM_TOC.indd 1/5/2018 5:25:24 PM FM_TOC.indd 1/5/2018 5:25:24 PM CONTENTS Prefaceiii About the Authors v Directions to Use the Book vii PHYSICAL CHEMISTRY 1. Some Basic Concepts of Chemistry Chapter at a Glance Solved Examples Practice Exercises Level I Laws of Chemical Combinations Calculations of Moles, Empirical and Molecular Formula Stoichiometry, Limiting Reagent, and POAC Concentration Terms Equivalent Weight Concept Level II Calculations of Moles, Empirical and Molecular Formula Stoichiometry, Limiting Reagent, and POAC Concentration Terms Equivalent Weight Concept Previous Years’ NEET Questions Answer Key Hints and Explanations 2. Structure of Atom Chapter at a Glance Solved Examples Practice Exercises Level I Early Models, Electromagnetic Waves and Planck’s Theory Bohr’s Model, Calculation of Radius, Velocity and Energy Hydrogen Spectrum and Rydberg Equation Heisenberg Uncertainty Principle and de Broglie Equation Photoelectric Effect Schrödinger’s Wave Equation Aufbau Principle, Hund’s Rule and Pauli’s Exclusion Principle Level II Early Models, Electromagnetic Waves, Planck’s Theory Bohr’s Model, Calculation of Radius, Velocity and Energy Hydrogen Spectrum and Rydberg Equation FM_TOC.indd 3 13 13 13 13 13 14 15 15 15 16 16 17 17 19 19 29 29 34 36 36 36 37 37 38 38 39 39 40 40 40 41 Heisenberg Uncertainty Principle and de Broglie Equation Photoelectric Effect Schrödinger’s Wave Equation Aufbau Principle, Hund’s Rule and Pauli’s Exclusion Principle Previous Years’ NEET Questions Answer Key Hints and Explanations 3. States of Matter Chapter at a Glance Gaseous State Liquid State Solved Examples Practice Exercises Level I Gas Laws Ideal Gas Law Dalton’s Law Graham’s Law of Effusion and Diffusion Kinetic Theory of Gases and Maxwell Boltzmann Distribution Curve Compressibility Factor (Z), Liquefaction of Gas, van der Waals Equation Liquid State Level II Gas Laws Ideal Gas Law Dalton’s Law Graham’s Law of Effusion and Diffusion Kinetic Theory of Gases and Maxwell Boltzmann Distribution Curve Compressibility Factor (Z), Liquefaction of Gas, van der Waals Equation Liquid State Previous Years’ NEET Questions Answer Key Hints and Explanations 4. Thermodynamics Chapter at a Glance Solved Examples Practice Exercises Level I Heat, Work, Internal Energy, Enthalpy, Heat Capacity and First Law of Thermodynamics Entropy and Second Law of Thermodynamics Gibbs Free Energy and Spontaneity of a Reaction 42 42 42 43 43 45 46 55 55 55 60 61 66 66 66 66 67 68 68 69 70 70 70 70 71 71 72 73 74 74 75 76 83 83 88 91 91 91 92 92 1/5/2018 5:25:24 PM 68 OBJECTIVE CHEMISTRY FOR NEET (1) 157.5 mm Hg (2) 175.5 mm Hg (3) 315.0 mm Hg (4) none 32 Same masses of CH4 and H2 is taken in a container The partial pressure caused by H2 is (1) Kinetic Theory of Gases and Maxwell Boltzmann Distribution Curve 41 A sample of gas is at 0°C The temperature at which rms speed of the molecule will be doubled is (2) 9 (1) 103°C (2) 273°C (3) 723°C (4) 819°C (4) 1 Aqueous tension of water depends upon 42 According to kinetic theory of gases (3) (1) the pressure exerted by given mass of a gas is proportional to mean square velocity of the molecules at constant volume (2) the pressure exerted by the gas is proportional to the root mean square velocity of the molecules (3) the root mean square velocity is inversely proportional to the temperature (4) the mean K.E of the molecule is directly proportional to the temperature in degree centigrade (1) the amount of water taken (2) volume of container in which water is present (3) temperature (4) some external other factors Graham’s Law of Effusion and Diffusion 34 Which of the following pairs will diffuse at the same rate? (1) CO2 and N2O (2) CO2 and NO (3) CO2 and CO (4) N2O and NO 35 According to Graham’s law at a given temperature, the ratio of the rates of diffusion rA/rB of gases A and B is given by p M (1) A A p M 1/2 (3) p A M B p M B A 1/2 B B M A pA (2) M p 1/2 (4) M A pB M p B A 1/2 B 43 At the same temperature and pressure, which of the following gases will have the highest kinetic energy per mole? (1) Hydrogen (2) Oxygen (3) Methane (4) All are the same 44 The kinetic energy of a mole of ideal gas in calories is approximately equal to (1) (2) (3) (4) B 36 A balloon having g Ne gas and radius 20 cm is pricked and g of Ne gas effused from it What will be the radius of balloon under similar conditions? times its absolute temperature times its absolute temperature times its absolute temperature 2/3 times its absolute temperature 45 (1) cm (2) 10 cm (3) 15 cm (4) 20 cm 37 Molecular weight of a gas that diffuses twice as rapidly as the gas with molecular weight 64 is Fraction of molecules (1) 16 (2) 8 (3) 64 (4) 6.4 38 Helium gas diffuses four times faster than gas X Molar mass of gas X is (1) 8 (2) 72 (3) 16 (4) 64 39 The ratio of rates diffusion of SO2, O2 and CH4 is (1) 1: 2:2 (2) 1:2:4 (3) 2: 2:1 (4) 1:2: 40 The pair of gases which can be most easily separated from effusion technique is (1) D2 and H2 (2) CH4 and CD4 (3) C12H4 and C14H4 (4) U235F6 and U238F6 Chapter 3_States of matter.indd 68 C1 Velocity The velocity C1 in the above figure is given by the relation (1) C1 = 3RT (2) C1 = 2RT M M (3) C1 = 8RT (4) C1 = 3RT pM M 46 In a closed vessel, a gas is heated from 300 K to 600 K, the kinetic energy becomes/remains (1) half (2) double (3) same (4) four times 1/4/2018 5:10:29 PM States of Matter 47 A closed flask contains water in all its three states at 0°C In this situation average kinetic energy of water molecule will be (1) (2) (3) (4) same in all the three states the greatest in vapor state the greatest in liquid state the greatest in solid state 48 If p is the pressure of gas, then the kinetic energy per unit volume of the gas is (1) p/2 (2) p (3) 3p/2 (4) 2p 49 Which of the following gases would have the highest rms speed at 0°C? 56 The root mean square speed of CH4 molecules at 25°C is about 0.56 km s-1 What is the root mean square speed of a H2 molecule at 25°C? (1) 0.070 km s-1 (2) 0.20 km s-1 (3) 1.1 km s-1 (4) 1.6 km s-1 Compressibility Factor (Z), Liquefaction of Gas, van der Waals Equation 57 The van der Waals equation explains the behavior of (1) ideal gases (2) real gases (3) vapor (4) non-real gases 58 The compressibility factor for an ideal gas is (1) 1.5 (2) 1 (3) 2 (4) ¥ (1) O2 (2) CO2 (3) SO3 (4) CO 50 Assume that the container is filled with the mixture of SO3 and Ne The molecular weight of SO3 is 80 g mol-1 and the atomic weight of Ne is 20 g mol-1 The average velocity of SO3 molecules is (1) (2) (3) (4) one fourth that of a Ne atom one half that of a Ne atom the same as a Ne atom two times that of a Ne atom 51 The ratio of rms velocity to average velocity of gas molecules at a particular temperature is (1) 1.086:1 (2) 1:1.086 (3) 2:1.086 (4) 1.086:2 52 The temperature at which H2 has the same rms speed (at atm) as that of O2 at STP is (1) 37K (2) 17K (3) 512K (4) 27K 53 Identify a postulate of kinetic theory among the following (1) (2) (3) (4) An atom is indivisible Gases combine in simple ratio There is no influence of gravity on gas molecules None of the above 54 An ideal gas molecule is present at 27°C By how many degree centigrade its temperature should be raised so that its urms, ump and uav all may double? (1) 900°C (2) 108°C (3) 927°C (4) 81°C 55 A certain gas is at a temperature of 350 K If the temperature is raised to 700 K, the average translational kinetic energy of the gas will (1) (2) (3) (4) remain constant increase by a factor of increase by a factor of square root of decrease by a factor of square root of Chapter 3_States of matter.indd 69 69 59 The value of van der Waals constant a is maximum for (1) He (2) N2 (3) CH4 (4) NH3 60 Table gives values of a for different gases O2 H2 NH3 CH4 1.310 1.390 4.17 2.253 Therefore which can most easily liquefied is (1) O2 (2) NH3 (3) H2 (4) CH4 61 The compressibility factor for gas obeying van der Waals’ equation of state is given by (where V is molar volume) (1) V a (2) a V − − V − b RTV RTV V − b (3) V − b − RTV (4) RTV − V − b V V a a 62 Out of the following gases, which one has least value of van der Waals constant a? (1) CO2 (2) NH3 (3) CH4 (4) H2 63 A given gas cannot be liquefied if its temperature is (1) (2) (3) (4) equal to its critical temperature greater than its critical temperature smaller than its critical temperature equal to its inversion temperature 64 Which of the following is incorrect statement? (1) van der Waals’ constant a is a measure of attractive force (2) van der Waals’ constant b is also called co-volume or excluded volume per mole (3) b is expressed in L mol–1 (4) b is one-fourth of critical volume 1/4/2018 5:10:30 PM 70 OBJECTIVE CHEMISTRY FOR NEET 65 a /V given in van der Waals equation is for (1) (2) (3) (4) internal pressure intermolecular attraction both (1) and (2) temperature correction 66 The temperature at which real gases obey the ideal gas laws over a wide range of pressure is called (1) (2) (3) (4) critical temperature Boyle temperature inversion temperature reduced temperature 67 Which of the following statement is not true? (1) The pressure of the real gas is equal to the pressure calculated for an ideal gas (2) The van der Waals’ equation helps to calculate the pressure and volume of real gases (3) Real gas molecules occupy a finite, but small, volume (4) None of these 68 Which of the following gases is the least likely to behave ideally? (1) He (2) N2 (3) HCl (4) H2 Liquid State 69 Which of the following is not true? (1) Density of solids is greater than gases (2) Molecules of solid possess vibratory motion (3) Molecules of gases possess greater kinetic energy than molecules of solid and liquid (4) Gases like liquids possess definite volume 70 With the increasing molecular mass of a liquid, the viscosity (1) decreases (2) increases (3) no effect (4) all wrong 71 Which of the following expression regarding the unit of coefficient of viscosity is not true? (1) dyne cm−2 s (2) dyne cm−2 s−1 (3) Nm−2 s (4) poise = 10−1 Nm−2 s Level II Gas Laws The volume of a large irregularly shaped tank is determined as follows The tank is first evacuated, and then it is connected to a 50 L cylinder of compressed helium gas The gas pressure in the cylinder, originally at 21 atm, falls to 7.0 atm without a change in temperature What is the volume of the tank? Chapter 3_States of matter.indd 70 (1) 100 L (2) 150 L (3) 200 L (4) 300 L A spherical air bubble is rising from the depth of a lake where pressure is p atm and temperature is T Kelvin The percentage increase in its radius when it comes to the free surface of lake will be (Assume temperature and pressure at the surface be respectively p/4 and 2T Kelvin.) (1) 100% (2) 50% (3) 40% (4) 200% Which of the following graphs does not represent isobar given by Charles’ law? V V (1) V V (2) T(ºC) T(ºC) T(K) T(K) (3) Both (1) and (2) (4) None of these A gas cylinder containing cooking gas can withstand a pressure of 14.9 atm The pressure gauge of cylinder indicates 12 atm at 27°C Due to sudden fire in building the temperature starts rising The temperature at which the cylinder will explode is (1) 42.5°C (2) 67.8°C (3) 99.5°C (4) 25.7°C Ideal Gas Law 3.2 g S on heating occupies a volume of 780 mL at 450°C and 723 mm Hg pressure Formula of sulphur is (1) S2 (2) S (3) S4 (4) S8 mol of an ideal gas A with 300 mm of Hg is separated from mol of another ideal gas B with 300 mm of Hg in a closed container at the same temperature If the separation is removed then the total pressure is (1) 200 mm of Hg (2) 300 mm of Hg (3) 500 mm of Hg (4) 600 mm of Hg The volume of balloon filled with g He at 22°C and 720 mm of Hg is (1) 25.565 L (2) 20 L (3) 15 L (4) 30 L 0.2 mol sample of hydrocarbon CxHy yields after complete combustion with excess O2 gas, 0.8 mol of CO2, 1.0 mol of H2O Hence hydrocarbon is (1) C4H10 (2) C4H8 (3) C4H5 (4) C8H16 Excess F2(g) reacts at 150°C and 1.0 atm pressure with Br2(g) to give a compound BrFn If 423 mL of Br2(g) at the 1/4/2018 5:10:30 PM States of Matter 71 same temperature and pressure produced 4.2 g of BrFn what is n ? (At.wt.Br = 80, F = 19) (1) 3 (2) 1 (3) 5 (4) 7 CO2 2.13 atm 1.5 L 10 Assume center of Sun to consist of gases whose average molecular weight is The density and pressure of the gas are 1.3 g cm-3 and 1.12 × 109 atm respectively The temperature of Sun is (1) 2 × 103 K (2) 2 × 105 K (3) 2 × 107 K (4) 2 × 109 K 11 5.40 g of an unknown gas at 27°C occupies the same volume as 0.14 g of hydrogen at 17°C and same pressure The molecular weight of unknown gas is (1) 79.8 (2) 81 (3) 79.2 (4) 83 12 In the figure below mercury columns of 10 cm each are trapped between gas column of 10 cm each If p atm = 75 cm of Hg then the gas pressure in the topmost column will be p gas = ? Hg Ar 1.15 atm 2.0 L H2 1.0 L 0.861 atm (1) 1.41 atm (2) 2.41 atm (3) 3.41 atm (4) 1.12 atm 16 When g gas A is introduced into an evacuated flask kept at 25°C, the pressure is found to be atm If g of another gas B is further added to same flask, the total pressure becomes 1.5 atm The ratio of molecular weights is (1) 1:1 (2) 1:2 (3) 1:3 (4) 1:4 Graham’s Law of Effusion and Diffusion 17 The valves X and Y in the below figure are opened simultaneously The white fumes of NH4Cl will first formed at 10 cm 10 cm Hg NH3 10 cm 10 cm X Hg (1) 55 cm of Hg (2) 35 cm of Hg (3) 65 cm of Hg (4) 45 cm of Hg 13 An open vessel containing air is heated from 300 K to 400 K The fraction of air originally present which goes out of it is (1) 3/4 (2) 1/4 (3) 2/3 (4) 1/8 Dalton’s Law 14 A mixture consisting of 0.10 mol of N2, 0.05 mol of O2 and 0.20 mol of CH4 and an unknown amount of CO2 occupied a volume of 9.6 L at 27°C and 1.0 atm pressure How many moles of CO2 are there in this sample? (1) 0.04 mol (2) 0.39 mol (3) 0.05 mol (4) 0.10 mol 15 In the following figure, when the two stop corks are opened, the total pressure inside the flask will be Chapter 3_States of matter.indd 71 HCl B A C Y (1) A (2) B (3) C (4) A, B and C simultaneously 18 A balloon filled with moist air has developed a pinhole It is quickly plunged into a tank of dry air at the same pressure In a short while (1) it will collapse (2) it will enlarge (3) no change will take place (4) can’t be predicted 19 The mole fraction He is 0.4 in gaseous mixture He and CH4 If both the gaseous are effusing through the constant area of the orifice of the container, then what will be % composition by volume of CH4 gas effusing out initially? (1) 50% (2) 40% (3) 43% (4) 75% 20 A mixture of H2 and O2 in 2:1 volume is allowed to diffuse through a porous partition what is the composition of gas coming out initially (1) 1:2 (2) 4:1 (3) 8:1 (4) 1:4 1/4/2018 5:10:31 PM 72 OBJECTIVE CHEMISTRY FOR NEET 21 The rate of diffusion of methane at a given temperature is twice that of gas X The molecular weight of X is (1) 64.0 (2) 32.0 (3) 4.0 (4) 8.0 28 According to the kinetic theory of gases 22 Vegetables are canned, while they are steaming hot because (1) (2) (3) (4) the heat inside will seal the jars the heat increases the atmospheric pressure the heat creates more pressure inside the jars when the jars cool, a vacuum inside will help to seal the jars 23 Which of the following mixture of gases cannot be separated by diffusion method? (1) NO + C2H6 (2) NO + NO2 (3) CO + CO2 (4) C2H4 + C2H6 24 Some moles of SO2 diffuse through a small opening in 20 s Same number of moles of an unknown gas diffuses through the same opening in 60 s Molecular mass of the unknown gas is 60 (1) (64)2 × 20 20 (2) (64)2 × 60 (3) (64) × 60 20 (4) (64) × 20 60 Kinetic Theory of Gases and Maxwell Boltzmann Distribution Curve 25 If EK is the kinetic energy per mole of a gas, then (1) pV = E K (2) p = VE K 2 (3) pV = E K (4) 3pV = E K 26 On increasing temperature, the fraction of total gas molecule which has acquired most probable velocity will (1) increase (2) decrease (3) remains constant (4) cannot say without knowing pressure 27 Distribution of molecules with velocity is represented by the following curve Point A in the curve shifts to the higher value of velocity if A Number of molecules u Chapter 3_States of matter.indd 72 (1) T is increased (2) V is increased (3) p is increased (4) All of the above (1) (2) (3) (4) there are intermolecular attractions molecules have considerable volume no intermolecular attractions the velocity of molecules decreases after each collision 29 Let the most probable velocity of hydrogen molecules at a temperature T °C is V0 Suppose all the molecules dissociate into atoms when temperature is raised to (2 T + 273)°C then the new rms velocity is (1) (2) V0 (3) V0 (4) 3( + 273) V0 T V0 30 Which of the following statements concerning the kinetic theory of gases is (are) correct? (I) Molecules make elastic collisions with each other and with the walls of their container (II) The average kinetic energy of a large number of molecules of mass, M, is proportional to M1/2 at a given temperature (III) The molecules of a gas are in constant random motion (IV) All the molecules of a gas have the same kinetic energy at a given temperature (1) I, II, III, IV (2) I, II, III (3) I, III, IV (4) I, III 31 Compare the root mean square speed of an O2 molecule with that of CH4 molecule at the same temperature and pressure (1) The speed is the same, since the weight of O2 and of CH4 are both 16 g mol-1 (2) The speed is the same because at the same temperature all gas molecules have the same mean square speed (3) CH4 is times faster, because the molecular weight of O2 is times greater than the molecular weight of CH4 (4) CH4 is 1.41 times faster, since at equal temperature of all gas molecules have the same kinetic energy 32 Consider two L flasks, one containing O2, the other containing He, each at STP Which of the following statement is NOT true regarding these gases? (1) Each flask contains the same number of atoms or molecules (2) The gases in each flask have the same average kinetic energy (3) The gases in each flask have the same density (4) The pressure in each flask is the same 1/4/2018 5:10:32 PM States of Matter 33 Molecular velocities of two gases at the same temperature are u1 and u2 and their molecular masses are m1 and m2 respectively Which of the following expression is correct? (1) m1 = m2 u12 u22 (2) m1u1 = m2u2 (3) m1 = m2 (4) m1u12 = m2u22 u1 u2 34 The kinetic molecular theory of gases predicts pressure to rise as the temperature of a gas increases because (1) the average kinetic energy of the gas molecules decreases (2) gas molecules collide more frequently with the container walls (3) gas molecules collide less frequently with the container walls (4) gas molecules collide less energetically with the container walls 35 If the absolute temperature of a sample of gas in a fixed volume container is quadrupled, then the root mean square speed in the initial state ui and that in the final stage uf would be related as (1) uf = ui (2) uf = ui (3) uf = 2ui (4) uf = 4ui Compressibility Factor (Z), Liquefaction of Gas, van der Waals Equation 36 The compressibility factor of He as a real gas at room temperature is (1) unity (2) 1− a RTV RTV (3) 1+ pb (4) 1− a RT 37 At the critical temperature (1) (2) (3) (4) liquid and vapor exist in equilibrium the meniscus between liquid and vapor disappears vapor state does not exist at all the vapor condense into solid 38 The compressibility factor Z = pV nRT a will be T= Rb of a gas above (1) always less than unity (2) always equal to unity (3) always greater than unity (4) depends on pressure 39 A gas deviates from ideal behavior at a high pressure because its molecules Chapter 3_States of matter.indd 73 (1) (2) (3) (4) 73 have kinetic energy are bound by covalent bonds have different shapes attract one another 40 At critical temperature, which kind of forces is dominating? (1) Attractive forces (2) Repulsive forces (3) Both are the same (4) Cannot be determined 41 For a non-zero volume of molecules having no force of attraction, the variation of compressibility factor, Z vs p is best represent by Z II 1.0 I III p (atm) (1) I (2) II (3) III (4) All of the above 42 A gas is easily liquefied (1) above critical temperature and below critical pressure (2) below critical temperature and above critical pressure (3) below critical temperature and critical pressure (4) above critical temperature and critical pressure 43 van der Waals constant a and b are related with (1) attractive force and bond energy of molecules respectively (2) volume and repulsive forces of molecules respectively (3) shape and repulsive forces of molecules respectively (4) attractive force and volume of the molecules respectively 44 For non-zero value of force of attraction between gas molecules, gas equation will be (1) pV = nRT - n a (2) pV = nRT + nbp V nRT (3) pV = nRT (4) p = V −b 45 Which of the following is an incorrect statement? (1) Ideal gases have Z = and cannot be liquefied (2) When Z > 1, real gases are difficult to compress (3) When Z > 1, real gases are easier to compress (4) When Z < 1, real gases are easier to compress 1/4/2018 5:10:34 PM 74 OBJECTIVE CHEMISTRY FOR NEET 46 One mole of SO2 gas occupies a volume of 900 mL at 24 atm pressure at 27°C Then the gas will show (1) (2) (3) (4) negative deviation from ideal behavior positive deviation from ideal behavior no deviation from ideal behavior Cannot be determined 47 At low pressure van der Waals equation for mol of a real gas will have its simplified form as pV (1) =4 RT + pb (3) pV pV =1 RT + 4pb =4 (2) RT − 4a V pV =4 (4) RT − a V 48 Which of the following statement is true for van der Waals gas constant a and b? (1) a depends on size and shape, b depends only on size of molecule (2) b depends on size and shape, a depends only on size of molecule (3) Both a and b depends on shape and size of molecule (4) Both a and b depends only on size of molecule 49 Dominance of strong repulsive forces among the molecules of real gas (Z = compressibility factor) (1) (2) (3) (4) depends on Z and indicated by Z = depends on Z and indicated by Z < depends on Z and indicated by Z > is independent of Z 50 If for a gas, critical parameters are pC = atm, TC = 47°C, then approximate value of VC will be (1) 10 L (2) L (3) 20 L (4) L 51 When you expect a real gas to behave like an ideal gas? (1) (2) (3) (4) When both the temperature and pressure are low When both the temperature and pressure are high When the temperature is high and pressure is low When the temperature is low and pressure is high Liquid State 52 Viscosity of a liquid is increased by (1) (2) (3) (4) increase in temperature decrease in molecular size increase in molecular size none of the above 53 If n1 and n2 are the coefficients of viscosity of two liquids, r1 and r2 their densities and t1 and t2 the flow times in Ostwald viscometer, then Chapter 3_States of matter.indd 74 (1) n1 r1t = n2 r2t1 n rt (3) = 1 n2 r2t (2) n1 r2t = n2 r1t1 (4) n1 r2t1 = n2 r1t Previous Years’ NEET Questions If an ideal gas expands at a constant temperature, it indicates that (1) (2) (3) (4) the number of the molecules of gas increases kinetic energy of molecule decreases pressure of the gas increases kinetic energy of molecules remains the same (AIPMT 2008) What volume of oxygen gas (O2) measured at 0°C and atm, is needed to burn completely L of propane gas (C3H8) measured under the same conditions? (1) 10 L (2) L (3) L (4) L (AIPMT 2008) The pressure exerted by 6.0 g of methane gas in a 0.03 m3 vessel at 129°C is (atomic masses: C = 12.01, H = 1.01 and R = 8.314 JK−1 mol−1) (1) 13409 Pa (2) 41648 Pa (3) 31684 Pa (4) 215216 Pa (AIPMT MAINS 2010) By what factor does the average velocity of a gaseous molecule increase when the temperature (in Kelvin) is doubled? (1) 1.4 (2) 2.0 (3) 2.8 (4) 4.0 (AIPMT PRE 2011) Two gases A and B having the same volume diffuse through a porous partition in 20 and 10 seconds respectively The molecular mass of A is 49 u Molecular mass of B will be (1) 25.00 u (2) 50.00 u (3) 12.25 u (4) 6.50 u (AIPMT PRE 2011) A gaseous mixture was prepared by taking equal moles of CO and N2 If the total pressure of the mixture was found atm, the partial pressure of the nitrogen (N2) in the mixture is (1) atm (2) 0.5 atm (3) 0.8 atm (4) 0.9 atm (AIPMT PRE 2011) 1/4/2018 5:10:35 PM States of Matter A bubble of air is underwater at temperature 15°C and the pressure 1.5 bar If the bubble rises to the surface where the temperature is 25°C and the pressure is 1.0 bar, what will happen to the volume of the bubble? (1) (2) (3) (4) Volume will become greater by a factor of 2.5 Volume will become greater by a factor of 1.6 Volume will become greater by a factor of 1.1 Volume will become greater by a factor of 0.70 (AIPMT MAINS 2011) 50 mL of each gas A and of gas B takes 150 and 200 s respectively for effusing through a pin-hole under the similar condition If molecular mass of gas is 36, the molecular mass of gas A will be (1) 96 (2) 128 (3) 21 (4) 64 11 Maximum deviation from ideal gas is expected from (1) H2(g) (2) N2(g) (3) CH4(g) (4) NH3(g) (NEET 2013) 12 Equal masses of H2, O2 and methane have been taken in a container of volume V at temperature 27°C in identical conditions The ratio of the volumes of gases H2:O2: methane would be (1) 8:16:1 (2) 16:8:1 (3) 16:1:2 (4) 8:1:2 (AIPMT 2014) 13 A gas such as carbon monoxide would be most likely to obey the ideal gas law at (1) (2) (3) (4) (AIPMT PRE 2012) A certain gas takes three times as long to effuse out as helium Its molecular mass will be (1) 27 u (2) 36 u (3) 64 u (4) u (AIPMT MAINS 2012) 10 For real gases, van der Waals equation is written as an p + (V − nb ) = nRT V where a and b are van der Waals constants Two sets of gases are: (I) He < H2 < CO2 < O2 (I) O2 < He < H2 < CO2 (I) H2 < He < O2 < CO2 (I) H2 < O2 < He < CO2 (RE-AIPMT 2015) 14 Equal moles of hydrogen and oxygen gases are placed in a container with a pin-hole through which both can escape What fraction of the oxygen escapes in the time required for one-half of the hydrogen to escape? (NEET-I 2016) The gases given in Set I in increasing order of b and gases given in Set II in decreasing order of a, are arranged below Select the correct order from the following: (1) (2) (3) (4) high temperatures and high pressures low temperatures and low pressures high temperatures and low pressures low temperatures and high pressures (1) 1/8 (2) 1/4 (3) 3/8 (4) 1/2 (I) O2, CO2, H2 and He (II) CH4, O2 and H2 75 (II) CH4 > H2 > O2 (II) H2 > O2 > CH4 (II) CH4 > O2 > H2 (II) O2 > CH4 > H2 15 A 20 L container at 400 K contains CO2(g) at pressure 0.4 atm and an excess of SrO (neglect the volume of solid SrO) The volume of the container is now decreased by moving the movable piston fitted in the container The maximum volume of the container, when pressure of CO2 attains its maximum value, will be (Given that: SrCO3(s) SrO(s) + CO2(g ), Kp = 1.6 atm) (1) 10 L (2) L (3) L (4) L (NEET 2017) (AIPMT MAINS 2012) Answer Key Level I 1. (3) 2. (4) 3. (1) 4. (1) 5. (2) 6. (3) 7. (1) 8. (1) 9. (3) 10. (4) 11. (4) 12. (4) 13. (4) 14. (4) 15. (4) 16. (3) 17. (1) 18. (3) 19. (3) 20. (2) 21. (1) 22. (4) 23. (2) 24. (4) 25. (2) 26. (2) 27. (3) 28. (3) Chapter 3_States of matter.indd 75 1/4/2018 5:10:35 PM 76 OBJECTIVE CHEMISTRY FOR NEET 29. (4) 30. (1) 31. (1) 32. (1) 33. (3) 34. (1) 35. (3) 36. (2) 37. (1) 38. (4) 39. (1) 40. (1) 41. (4) 42. (1) 43. (4) 44. (1) 45. (2) 46. (2) 47. (2) 48. (3) 49. (4) 50. (2) 51. (1) 52. (2) 53. (3) 54. (1) 55. (2) 56. (4) 57. (2) 58. (2) 59. (4) 60. (2) 61. (1) 62. (4) 63. (2) 64. (4) 65. (2) 66. (2) 67. (1) 68. (3) 69. (4) 70. (2) 1. (1) 2. (1) 3. (4) 4. (3) 5. (4) 6. (2) 7. (1) 8. (1) 9. (3) 10. (3) 11. (1) 12. (1) 13. (2) 14. (1) 15. (1) 16. (3) 17. (3) 18. (1) 19. (3) 20. (3) 21. (1) 22. (4) 23. (1) 24. (3) 25. (3) 26 (2) 27. (1) 28. (3) 29. (4) 30. (4) 31. (4) 32. (3) 33. (4) 34. (2) 35. (3) 36. (3) 37. (2) 38. (3) 39. (4) 40. (1) 41. (2) 42. (2) 43. (4) 44. (1) 45. (3) 46. (1) 47. (2) 48. (1) 49. (3) 50. (2) 51. (3) 52. (3) 53. (3) 71. (3) Level II Previous Years’ NEET Questions 1. (4) 2. (4) 3. (2) 4. (1) 5. (3) 6. (2) 7. (2) 8. (3) 9. (2) 10. (4) 11. (4) 12. (3) 13. (3) 14. (1) 15. (4) Hints and Explanations Level I 10 (4) From ideal gas equation, we know V = (3) We know p1V1 = p2V2 p 0.95 Therefore, pV = p2(0.95V ) Þ p2 = Percentage increase in pressure can be calculated as ỉ p - p ÷ ´ 100 ỗ p2 - p 0.95 ố ứ = 100 = = 5.26 % p p (4) pV = constant (2) According to Charle’s law V1 V2 = T1 T2 100 200 = Þ T2 = 746 K or 473°C 373 T2 (1) We have V1 V2 V 1.1 V = Þ = Þ T2 = 1.1T T1 T2 T T2 Hence, percentage of increase in temperature = 10% Chapter 3_States of matter.indd 76 nRT p Therefore, V µ n Since, nHI is minimum, therefore, VHI is minimum 12 (4) The average speed is given by 8RT 8pV = pM pM uavg = Given (uavg )A = 2(uavg )B 8p A ´ pB ´ = Þ p A = pB pM pM 13 (4) From ideal gas equation pV = nRT pV = nB RT (1) On dividing Eq (2) by Eq (1), we get 2p × 2V = nA × 2T (2) 4= nA ´ nB 1/4/2018 5:10:37 PM States of Matter or 2nB = nA x (Mass of A ) or ì ị Mass of A = 2x g = M M r1 T2 = r2 T1 16 T2 = Þ T2 = 312 K or 39°C 14 273 14 (4) We know n1T1 = n2T2 n ´ 300 = 0.5n ´ T2 Þ T2 = 600 K or 327°C 15 (4) Density is given by r = 26 (2) We have r = pM RT pM rRT or p = RT M p A rA ´ M B pA ´ Þ = = = 1 pB rB ´ M A pB 18 (3) nH2 = = 1, therefore, number of molecules of H2 = NA nN = Hence, number of molecules in flask A is greater than that in flask B pV = n (1) RT p If V = L, then Eq (1) becomes n = RT p p 20 (2) We know = T1 T2 19 (3) From ideal gas equation Therefore, 21 (1) 17 = Þ T2 = 150 ´ 17 = 2550 K or 2127°C 300 T2 22 (4) From combined gas law, we have p1V1 p2V2 = T1 T2 where p is the partial pressure and x is the mole fraction pµx Number of moles can be calculated as nH2 = = 0.5 ; nHe = = 2; nN2 = = 0.25; nO2 = = 0.25 2 28 32 As mole fraction of helium would be highest, therefore, it will also have highest partial pressure 28 (3) Given that pN + pO2 = p p Also, pN = pO2 = If N2 is removed, net pressure = pO2 = p 30 (1) Mole fraction of CH4 can be calculated as xCH4 = p1V1 p2V2 = T1 T2 ´ 275 2.5 ´ 200 = Þ T2 = 303 K T2 500 p1 p2 Þ p2 = atm p2 27 (3) Using Dalton’s law p = ptotal x N = , therefore, number of molecules of N2 = A 14 28 14 pM or r µ p RT r1 = r2 = 800 ´ M Air ´ R ´ 300 ( r )Delhi ( r )Delhi Þ = = ( r )Mumbai R ´ 320 ´ 750 ´ M Air ( r )Mumbai 16 (3) We know r = 16 /16 = = 0.25 ỉ 56 44 16 ( + + 1) + + ỗ ữ ố 28 44 16 ø Partial pressure is given by pCH4 = xCH4 × pTotal = 0.25 × 720 = 180 mm 31 (1) We have xO2 = 0.21 pO2 = ptotal × xO2 pV ( p / )V2 = ⇒ V2 = 4V ( 2T ) T = 0.21 × 750 = 157.5 mm Hg 32 (1) Mole fraction of H2 can be calculated as pM p or r µ RT T Option (1): r µ ; Option (2): r µ 273 273 Option (3): r µ ; Option (4): r µ 546 546 Therefore, density of neon is highest at 0°C and atm 23 (2) Density is given by r = 24 (4) We know r = pM or r µ M RT Therefore, density of propane (C3H8) is the maximum 25 (2) We have r = Chapter 3_States of matter.indd 77 pM or r µ RT T 77 x H2 ổxử ỗ ữ ố2ứ = = ổ x ổxử ỗ ữ+ỗ ữ ố 16 ø è ø Therefore, partial pressure = xH2 ´ pTotal = ´ pTotal 34 (1) The rate of diffusion is directly proportional to its molar mass Since, CO2 and N2O have same molar mass, therefore, they will diffuse at the same rate n n 36 (2) We have = V1 V2 1/4/2018 5:10:43 PM 78 OBJECTIVE CHEMISTRY FOR NEET 49 (4) Root mean square speed is given by (8/M ) (1/M ) = 4 3 p ( 20) pr 3 20 r = 1/3 = 10 cm (8 ) 37 (1) From Graham’s law of diffusion urms = r1 M2 = r2 M1 (uavg )SO3 (uavg )Ne 39 (1) We have rSO2 : rO2 = M O2 : M SO2 (uavg )SO3 rSO2 : rO2 = : (1) rO2 : rCH4 = M CH4 : M O2 rO2 : rCH4 = : (2) Therefore, from (1) and (2), we get rSO2 : rO2 : rCH4 = : : (uavg )Ne 51 (1) We have 41 (4) From root mean square speed, we have (urms )1 T = (urms )2 T2 273 = Þ T2 = 1092 K or 819°C T2 43 (4) The average kinetic energy per mole (Ek) is equal to kinetic energy per molecule (E K) multiplied by the Avoga- dro’s number (NA) RT Therefore, (K.E.)1 300 = Þ (K.E.)2 = 2(K.E.)1 (K.E.)2 600 (3) Kinetic energy is given by K.E = nRT Chapter 3_States of matter.indd 78 20 Þ (uavg )SO3 = (uavg )Ne 80 3RT 3R ´ 273 = Þ T @ 17 K 32 54 (1) We know urms , ump , uav g ∝ T Therefore, (urms )2 T = (urms )1 T1 Hence, temperature should be raised by 900°C 55 (2) Average translational kinetic energy = RT ´ n K.E µT (K.E.)2 T2 = (K.E.)1 T1 (K.E.)2 700 = Þ (K.E.)2 = 2(K.E.)2 (K.E.)1 350 2RT M K.E nRT = = p Volume V = 3RT pM urms = ´ 8RT u Avg M ´ ´ T (cal) 46 (2) We know K.E µT M Ne M SO3 52 (2) Given that urms, H2 = urms, O2 K.E./ mol = 3T 45 (2) C1 = ump = = T = T2 T2 = 4T1 ⇒ T2 = × 300 = 1200 K or 927°C As temperature is constant, therefore, kinetic energy per mole is same for all K.E = 8RT or uavg µ (As rest are pM M 3p urms = = 1.086 : u Avg 3 E k = N A ´ kT = RT 2 44 (1) Kinetic energy per mole is given by Molecular weight of CO is least, therefore, (urms)CO is maximum 50 (2) We know uavg = constant) 64 = Þ M gas = 16 M gas 3RT or urms µ M M 56 (4) Root mean square speed is given by urms = Therefore, urms, CH4 urms, H2 = 3RT M 3RT ´ 16 3RT 0.56 = Þ urms, H2 = 1.6 km s -1 urms, H2 1/4/2018 5:10:48 PM States of Matter 60 (2) Higher the value of a, higher is the force of attraction; easier it is to liquefy the gas a ổ 61 (1) We know ỗ p + ÷(V - b ) = RT V ø è Therefore, p + V= CxH y + x + a RT = V (V - b ) V a (V - b ) (VRT ) 62 (4) a is a measure of attractive forces and is proportional to molar mass p1V1 p2V2 = T1 T2 p1 p2 = T1 T2 12 14.9 = Þ T2 = 99.5°C 300 T2 Let the formula of sulphur be Sx Therefore, n = From Eq (1), we get 3.2 32 x 3.2 (723 /760) × (780 × 10 −3 ) = ⇒x=8 32 x 0.0821 × 723 Hence, the formula of sulphur is S8 n Br2 + F2 → BrFn 2 nBr2 nBrFn = 1/ 1 × 423 4.2 ⇒n = = 0.5 × 1000 × 0.0821 × 423 (80 + 19n ) 10 (3) r = pM 1.12 ´ 109 ´ ÞT = = ´ 107 K RT 0.0821 ´ 1.3 ổ 0.14 R 290 5.40 pỗ ´ R ´ 300 ÷= p ø M è M = 79.8 g 12 (1) We have pgas + 10 + 10 = patm n1 ´ 300 = n2 ´ 400 n2 = n1 nRT (1) Volume can be calculated as V = p Therefore, fraction of air left = n1 - n2 = n1 14 (1) pTotal V = nTotal RT ´ 9.6 = (0.1 + 0.05 + 0.2 + x) ´ (0.0821) ´ (300) x = 0.04 mol 15 (1) pTotal V = nTotal RT é 2.13 ´ 1.5 ´ 0.861 1.15 ´ ù pTotal (1.5 + + 2) = ê + + RT RT RT úû ë RT pTotal = 1.41 atm pTotal (V A + VB ) = (1 + 2)RT ổ RT 2RT pTotal ỗ + ữ = 3RT Þ pTotal = 300 mm Hg è 300 300 ø pgas = 75 − 20 = 55 cm of Hg 13 (2) We have n1T1 = n2T2 (2) We have pTotal VTotal = nTotal RT Chapter 3_States of matter.indd 79 nH2 O (3) The reaction is pV (4) According to ideal gas equation n = (1) RT = 11 (1) From ideal gas equation, pV = nRT p p ´ p r13 ´ p r2 = T 2T 8r13 = r23 r2 = 2r1 or 100% increase (3) From combined gas law, we have nCO2 Hence, the hydrocarbon is C4H10 1 (1) Let the volume of the tank be V L We know (1) We know = ( y/ 2) x 0.2 0.8 = ⇒x=4 x 0.2 = ⇒ y = 10 ( y/ 2) Level II p1V1 = p2V2 21 × 50 = 7(V + 50) ⇒ V = 100 L y y O2 → xCO2 + H 2O nCx H y RT a (V - b ) V pV V a = RT (V - b ) VRT Z= ( /4) ´ (0.0821) ´ ( 295) = 25.625 L (720/760 ) (1) The combustion reaction is p= 79 16 (3) We know pV = nRT ´ V = ´ R ´ 298 (1) MA 1/4/2018 5:10:52 PM 80 OBJECTIVE CHEMISTRY FOR NEET ỉ 1.5 ´ V = ỗ + ữ R 298 (2) M M B ø è A On dividing Eq (2) by Eq (1), we get ỉ + ỗ ữ M MB ứ M 1.5 = ố A ị A= MB ổ ỗ ÷ è MA ø rms velocity is given by urms = On dividing Eq (2) by Eq (1), we get Vnew = V0 Vrms, O2 31 (4) 19 (3) From Graham’s law of diffusion rHe p = He rCH4 pCH4 M CH4 Vrms, CH4 rH2 rO2 rH2 rO2 = = pH M O2 pO2 M H2 35 (3) rms velocity is given by urms = rCH4 rgas X = 32 = 1 ui = Þ uf = 2ui uf 36 (3) We have MX 16 runknown = M unknown 60 ỉ 60 = Þ M unknown = ỗ ữ 64 M SO2 20 ố 20 ø 25 (3) Average kinetic energy per mole is given by RT (1) EK = Also, for mol pV = RT(2) From Eq (1) and Eq (2), we get EK = pV Þ pV = E K 2RT (4) Most probable velocity is given by ump = M Chapter 3_States of matter.indd 80 ổ a ỗ p + ữ(Vm - b ) = RT V m ø è For He, a can be neglected Therefore, p(Vm - b ) = RT 23 (1) The molecular weights of NO and C2H6 are the same, thus, they cannot be separated by diffusion rSO or urms µ T ui T = uf T2 MX 2= Þ M X = 64 16 24 (3) 3RT M Therefore, Therefore, composition of gas coming out initially is 8:1 21 (1) We have 16 32 1 m1u12 = m2u22 Þ m1u12 = m2u22 2 Percentage of CH4 coming out = ´ 100 = 43% 20 (3) = 33 (4) For the same temperature, kinetic energy remains same Therefore, M He rHe 0.4 16 = = rCH4 0.6 3R( 2t + 546 ) (2) Vnew = Vnew 3R ´ 2(t + 273) = = V0 R ´ (t + 273) 18 (1) As pmoist gas > pdry gas, therefore, gas will flow from higher pressure to lower pressure and the balloon will collapse 3RT M 2R(t + 273) V0 = (1) Therefore, Z = pVm pb = 1+ RT RT 44 (1) van der Waals equation is given by ỉ a ç p + ÷(Vm - b ) = RT V m ø è Neglecting b in the above equation, we get ổ an ỗ p + ÷ ´ V = nRT V ø è pV = nRT - an V 46 (1) Compressibility factor can be calculated as Z= pVm 24 × (900 /1000 ) = 0.87 = RT 0.0821 × 300 As Z < 1, the gas will show negative deviation from ideal behavior 1/4/2018 5:10:55 PM States of Matter 47 (2) van der Waals equation is given by ổ an ỗ p + ÷(V - nb ) = nRT V ø è Substituting n = in the above equation, we get pV 4a ỉ pV = ỗ RT =4 ữị 4a V ứ ổ ố RT ỗ ữ V ứ ố Dividing Eq (2) by Eq (1), we get TC a æ 27b = ỗ ữ = 160 pC 27 bR è a ø (uavg)2 = 1.44 (uavg)1 (3) The rate of diffusion is related to the molecular mass by the relation rA = rB where r = Substituting the values in Eq (1), we get Volume diffused (V ) Time interval (t ) V/20 = V/10 MB 49 MB 49 Þ MB = = 12.25 u 49 = (2) We know pN = xN ´ pTotal As nN = nCO Þ xN = 0.5 (4) The reaction involved is Substituting the values, we get = 0.5 atm (2) According to combined gas law, we have p1V1 p2V2 = T1 T2 1.5 ´ V1 ´ V2 = Þ V2 = 1.55 V1 » 1.6 V1 288 298 (3) Using the relationship Vol Given that the volume of C3H8 at STP = L, therefore the volume of O2 at STP = L 3 (2) From ideal gas equation, we have pV = nRT pV = w RT (1) M Substituting the values in Eq (1), we get 8.314 ´ 402 ´ 16.05 0.03 = 41647.7 Pa » 41648 Pa p= Chapter 3_States of matter.indd 81 V A / tA M B = VB / t B M A 1/2 Given that VA = VB = 50 mL; tA = 150 s; tB = 200 s; MB = 36 u Substituting in Eq (1), we get C 3H8(g) + O2(g) ® H 2O(g) + CO2(g) Vol pN = 0.5 ´ atm K.E = kT where k is the Boltzmann’s constant Kinetic energy of gaseous molecules depends on temperature only (1) 8b = 160 Þ b = 1.642 R VC = 3b = 4.926 L or L approx (4) We know MB MA Previous Years’ NEET Questions T1 T2 As T2 = 2T1, therefore VC = 3b = a pC = = (1) 27b 8a TC = = 320 (2) 27bR 50 (2) We have (uavg )1 At low pressure, b can be neglected ổ an ỗ p + ÷ V = nRT V ø è nRT an p= - V V an pV = nRT V 8RT pM (1) We have uavg = (uavg )2 81 50 /150 36 = 50 /200 M A 1/2 ⇒ 200 36 = 150 M A (1) 1/2 ⇒ M A = 21 u (2) As per kinetic theory of gases, r µ (1/M)1/2 r2 M = r1 M1 1/2 Vgas t ⇒ × = t VHe M 1/2 t = ⇒ M 1/2 = ⇒ M = 36 u 3t M 1/2 1/4/2018 5:10:58 PM 82 OBJECTIVE CHEMISTRY FOR NEET 10 (4) In the van der Waals equation, constant a is the measure of magnitude of intermolecular forces It can be seen that as the molar mass increases, a increases and also constant b increases with the size of the molecule 11 (4) A real gas, deviates from ideal behavior for two important reasons: (a) The model of an ideal gas assumes that gas molecules are infinitesimally small, but real molecules take up some space (b) In an ideal gas, there would be no attractions between molecules, but in a real gas molecules experience weak attractions towards each other In case of NH3, the force of attraction between the molecules is highest due to hydrogen bonding, hence, it deviates maximum from the ideal gas (b) At high temperature, the molecules are moving so rapidly and are so far apart that the attraction between them is hardly felt 14 (1) Given Let us assume that rate of escape of O2 molecule = rO2 Rate of escape of H2 molecule = rH2 According to Graham’s law rO2 V∝n Let the mass of H2, O2 and methane be x The ratio of the number of moles of gases x/2 : x/32 : x/16 = 16:1:2 (a) At low temperature, the space between the molecules is so large that the volume occupied by the molecules is insignificant Chapter 3_States of matter.indd 82 = M H2 M O2 = 1 = 16 Fraction of H2 molecules escaped in time t = 1/2 Fraction of O2 molecules escaped in time t = x Substituting in Eq (1) we get rO2 rH2 = (1) 1/2t = x /t 1 = x (at constant T and p) 13 (3) The law pV = nRT is called ideal because it follows from the assumptions of kinetic theory, which describes an ideal gas The molecules of an ideal gas have negligible volume and no attraction or repulsion for each other At low temperature and high pressure, Carbon monoxide will show ideal gas behavior due to following two reasons: rH2 12 (3) According to Avogadro’s principle, the volume of a gas is directly proportional to its number of moles, n nO2 = nH2 15 (4) We have SrCO3 (s) SrO(s) + CO2(g) Also, From combined gas law p1V1 = p2V2(1) Substituting p1 = 0.4 atm, V1 = 20 L and p2 = 1.6 atm in Eq (1), we get K p = pCO2 = 1.6 atm 0.4 ´ 20 = 1.6 ´ V2 0.4 ´ 20 V2 = =5L 1.6 1/4/2018 5:10:59 PM .. .2020 Objective CHEMISTRY FOR NEET K Singh Vipul Mehta Asokan K K Objective Chemistry for NEET Copyright © 2019 by Wiley India Pvt Ltd., 4436/7, Ansari... Nowadays, amu has been replaced by u which stands for unified mass Chapter 1_Some Basic Concepts of Chemistry. indd 1/4/2018 5:08:23 PM OBJECTIVE CHEMISTRY FOR NEET (b) Average atomic mass: Most... 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