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Preview A Complete Resource Book in Chemistry for JEE Main 2019 by Atul Singhal (2018) Preview A Complete Resource Book in Chemistry for JEE Main 2019 by Atul Singhal (2018) Preview A Complete Resource Book in Chemistry for JEE Main 2019 by Atul Singhal (2018) Preview A Complete Resource Book in Chemistry for JEE Main 2019 by Atul Singhal (2018) Preview A Complete Resource Book in Chemistry for JEE Main 2019 by Atul Singhal (2018)
About Pearson Pearson is the world’s learning company, with presence across 70 countries worldwide Our unique insights and world-class expertise comes from a long history of working closely with renowned teachers, authors and thought leaders, as a result of which, we have emerged as the preferred choice for millions of teachers and learners across the world We believe learning opens up opportunities, creates fulfilling careers and hence better lives We hence collaborate with the best of minds to deliver you class-leading products, spread across the Higher Education and K12 spectrum Superior learning experience and improved outcomes are at the heart of everything we This product is the result of one such effort Your feedback plays a critical role in the evolution of our products and you can contact us at reachus@pearson.com We look forward to it A01_KUMAR_0283_01_SE_PREL.indd 5/10/2017 11:19:37 AM This page is intentionally left blank A01_KUMAR_0283_01_SE_PREL.indd 5/10/2017 11:19:37 AM A Complete Resource Book in chemistry for JEE Main 2019 A.K Singhal U.K Singhal Dedicated to my grandparents, parents and teachers Copyright © 2018 Pearson India Education Services Pvt Ltd Published by Pearson India Education Services Pvt Ltd, CIN: U72200TN2005PTC057128 No part of this eBook may be used or reproduced in any manner whatsoever without the publisher’s prior written consent This eBook may or may not include all assets that were part of the print version The publisher reserves the right to remove any material in this eBook at any time ISBN: 9789353062156 eISBN: 9789353063412 Head Office:15th Floor, Tower-B, World Trade Tower, Plot No 1, Block-C, Sector-16, Noida 201 301, Uttar Pradesh, India Registered Office: 4th Floor, Software Block, Elnet Software City, TS-140, Block & 9, Rajiv Gandhi Salai, Taramani, Chennai 600 113, Tamil Nadu, India Fax: 080-30461003, Phone: 080-30461060 Website: in.pearson.com, Email: companysecretary.india@pearson.com A01_KUMAR_0283_01_SE_PREL.indd 5/10/2017 11:19:38 AM Contents Preface vii Acknowledgements viii JEE Mains 2018 Paper xi JEE Mains 2017 Paper xviii Chapter Basics of Chemistry 1.1–1.32 Chapter Solid State 2.1–2.34 Chapter Gaseous State 3.1–3.36 Chapter Atomic Structure 4.1–4.38 Chapter Solutions 5.1–5.38 Chapter Energetics 6.1–6.40 Chapter Chemical Equilibrium 7.1–7.42 Chapter Ionic Equilibrium 8.1–8.54 Chapter Redox Reactions and Electrochemistry 9.1–9.50 Chapter 10 Chemical Kinetics 10.1–10.46 Chapter 11 Surface Chemistry 11.1–11.34 Chapter 12 Periodic Properties 12.1–12.26 Chapter 13 Chemical Bonding 13.1–13.38 Chapter 14 Chemistry of Representive Elements 14.1–14.26 Chapter 15 Chemistry of Non-Metals I 15.1–15.30 Chapter 16 Chemistry of Non-Metals II 16.1–16.60 Chapter 17 Chemistry of Lighter Elements 17.1–17.28 Chapter 18 Chemistry of Heavier Elements (Metallurgy) 18.1–18.34 Chapter 19 Transition Metals Including Lanthanides and Actinides 19.1–19.26 Chapter 20 Coordination Compounds 20.1–20.40 Chapter 21 Nuclear Chemistry 21.1–21.28 Chapter 22 Purification and Characterization of Carbon Compounds 22.1–22.26 A01_KUMAR_0283_01_SE_PREL.indd 5/10/2017 11:19:38 AM vi Contents Chapter 23 General Organic Chemistry I�������������������������������������������������������������������������������������������������������� 23.1–23.54 Chapter 24 General Organic Chemistry II������������������������������������������������������������������������������������������������������ 24.1–24.52 Chapter 25 Hydrocarbons and Petroleum ����������������������������������������������������������������������������������������������������� 25.1–25.58 Chapter 26 Organic Compounds with Functional Groups Containing Halogens (X)�������������������������������������� 26.1–26.36 Chapter 27 Alcohol Phenol Ether�������������������������������������������������������������������������������������������������������������������� 27.1–27.56 Chapter 28 Organic Compounds Containing Oxygen-II�������������������������������������������������������������������������������� 28.1–28.70 Chapter 29 Organic Compounds with Functional Groups Containing Nitrogen�������������������������������������������� 29.1–29.44 Chapter 30 Polymers�������������������������������������������������������������������������������������������������������������������������������������� 30.1–30.20 Chapter 31 Biomolecules and Biological Processes�������������������������������������������������������������������������������������� 31.1–31.40 Chapter 32 Chemistry in Everyday Life ��������������������������������������������������������������������������������������������������������� 32.1–32.22 Chapter 33 Environment Chemistry��������������������������������������������������������������������������������������������������������������� 33.1–33.16 Chapter 34 Practical Chemistry��������������������������������������������������������������������������������������������������������������������� 34.1–34.34 Appendix �������������������������������������������������������������������������������������������������������������������������������������������������������������� 35.1–35.20 Mock Test �����������������������������������������������������������������������������������������������������������������������������������������������������M1.1–M1.6 Mock Test �����������������������������������������������������������������������������������������������������������������������������������������������������M2.1–M2.6 Mock Test �����������������������������������������������������������������������������������������������������������������������������������������������������M3.1–M3.6 Mock Test �����������������������������������������������������������������������������������������������������������������������������������������������������M4.1–M4.6 Mock Test �����������������������������������������������������������������������������������������������������������������������������������������������������M5.1–M5.6 A01_KUMAR_0283_01_SE_PREL.indd 5/10/2017 11:19:38 AM Preface About the Series A Complete Resource Book for JEE Main series is a must-have resource for students preparing for JEE Main examination There are three separate books on Physics, Chemistry, and Mathematics; the main objective of this series is to strengthen the fundamental concepts and prepare students for various engineering entrance examinations It provides class-tested course material and numerical applications that will supplement any ready material available as student resource To ensure high level of accuracy and practicality, this series has been authored by highly qualified and experienced faculties for all three titles About the Book A Complete Resource Book in Chemistry for JEE Main 2018 is an authentic book for all the aspirants preparing for the joint entrance examination (JEE) This title is designed as per the latest JEE Main syllabus, where the important topics are covered in 34 chapters It has been structured in user friendly approach such that each chapter begins with topic-wise theory, followed by sufficient solved examples and then practice questions The chapter end exercises are structured in line with JEE questions; with ample number of questions on single choice correct question (SCQ), multiple-type correct questions (MCQ), assertion and reasoning, column matching, passage based and integer type questions are included for extensive practice Previous 15 years’ questions of JEE Main and AIEEE are also added in every chapter Hints and Solutions at the end of every chapter will help the students to evaluate their concepts and numerical applications Because of its comprehensive and in-depth approach, it will be especially helpful for those students who prefers self-study than going for any classroom teaching Series Features • • • • • • Complete coverage of topics along with ample number of solved examples Large variety of practice problems with complete solutions Chapter-wise Previous 15 years’ AIEEE/JEE Main questions Fully solved JEE Main 2017 questions Mock Tests based on JEE Main pattern in the book Free Online Mock Tests as per the recent JEE Main pattern Despite of our best efforts, some errors may have crept into the book Constructive comments and suggestions to further improve the book are welcome and shall be acknowledged gratefully Kindly share your feedback with me at singhal.atul50@yahoo.com or singhal.atul1974@gmail.com A.K Singhal A01_KUMAR_0283_01_SE_PREL.indd 5/10/2017 11:19:38 AM Acknowledgements The contentment and ecstasy that accompany the successful completion of any work would remain essentially incomplete if I fail to mention the people whose constant guidance and support has encouraged me I am grateful to all my reverend teachers, especially, late J.K Mishra, D.K Rastogi, late A.K Rastogi and my honourable guide, S.K Agarwala Their knowledge and wisdom has continued to assist me to present in this work I am thankful to my colleagues and friends, Deepak Bhatia, Vikas Kaushik, A.R Khan, Vipul Agarwal, Ankit Arora (ASO Motion, Kota), Manoj Singhal, Yogesh Sharma, (Director, AVI), Vijay Arora, (Director, Dronacharya), Rajneesh Shukla (Allen, Kota), Anupam Srivastav, Rajeev Jain (M V N), Sandeep Singhal, Chandan Kumar (Mentor, Patna), P.S Rana (Vidya Mandir, Faridabad) I am indebted to my father, B.K Singhal, mother Usha Singhal, brothers, Amit Singhal and Katar Singh, and sister, Ambika, who have been my motivation at every step Their never-ending affection has provided me with moral support and encouragement while writing this book Last but not the least, I wish to express my deepest gratitude to my wife Urmila and my little, —but witty beyond her years, daughters Khushi and Shanvi who always supported me during my work A.K Singhal A01_KUMAR_0283_01_SE_PREL.indd 5/10/2017 11:19:38 AM Chemistry Trend Analysis (2007 to 2018) S No Chapters 07 08 09 10 11 12 13 14 15 16 17 18 Basics of Chemistry – – – – 1 – – 2 Solid State – 1 1 1 – Gaseous State – – – 1 1 – Atomic Structure 2 1 1 1 – Solutions 2 2 1 1 1 Energetics 3 1 1 2 Chemical Equilibrium – – 1 1 2 Ionic Equilibrium – – 1 – – – Redox Reactions and Electrochemistry 1 1 1 1 10 Chemical Kinetics – 1 – – 1 11 Surface Chemistry – 1 – 1 – 2 – 12 Periodic Properties – – 1 1 1 1 – 13 Chemical Bonding 1 – – 1 14 Chemistry of Representive Elements – – 1 – – – – – 15 Chemistry of Non-Metals I – – – – – – – – – 1 16 Chemistry of Non-Metals II – 2 – – – 17 Chemistry of Lighter Elements – – – – – – – 1 18 Chemistry of Heavier Elements (Metallurgy) – – – – – 1 – – 19 Transition Metals Including Lanthanides and Actinides – – – – 1 – – 20 Coordination Compounds 2 2 1 – 21 Nuclear Chemistry – – – – – – 1 – – – (Continued ) A01_KUMAR_0283_01_SE_PREL.indd 5/18/2018 5:14:23 PM 3.22 Chapter More than One Option Correct Type 116 Pick out the correct statements of the following about liquids? (a) The intermolecular forces of attraction in a liquid are high (b) All liquids suffer cooling on evaporation (c) Lower the boiling point of a liquid, greater the vapour pressure at room temperature (d) At higher altitudes water boils at a higher temperature than at the sea level temperature ‘T’ is the same as that of gas of molar mass ‘2M’ at temperature ‘T/2’ (d) The product of pressure and volume of a fixed amount of a gas is independent of temperature dV (a) = k dT P dT = k (b) dV P 117 A mole mixture of Ne(g), H2(g) and O2(g) are placed in a closed container at a pressure equal to 50 atm An electric spark is passed and pressure noted is 12.5 atm after cooling Oxygen gas is introduced for pressure to become 25 atm Again electrical spark is passed and pressure drops to 10 atm (all measurements are at same T and P) (a) Moles fraction of O2 in original mixture is 0.25 (b) Moles of O2 added after first spark is 0.5 (c) Mole fraction of H2 in original mixture is 0.7 (d) Mole fraction of Ne in original mixture is 0.005 (c) (d) V α 118 Kinetic energy per mole of an ideal gas is (a) Zero at zero Kelvin temperature (b) Independent of temperature (c) Proportional to the absolute temperature of the gas (d) Proportional to pressure at constant temperature 119 Mark the correct statements (a) At low pressure, the van der Waal’s equation is written as ( P+ a V ) × V = RT (b) When Z > 1, at STP, Vreal > Videal (c) Mean free path of O2 is greater than that of H2 (d) At 273 K, the total kinetic energy of O2 will be eight times that of one mole of He 120 Which of the following statement(s) is/are incorrect? (a) A gas can be liquefied at a temperature ‘T’ such that T < TC and p = PC TC and PC are critical temperature and pressure (b) Rise in the compressibility factor with increasing pressure is due to equal contribution of both a and b (Van der Waal’s parameter) (c) The fraction of molecules having speeds in the range of u to u + du of a gas of molar mass ‘M’ at Objective_PCM_Chapter 03.indd 22 121 According to Charle’s Law V T − T = P T 122 At low pressures, the Van der Waal's gas equation for mole of a gas may be written as a (a) PV = RT − V (b) PV = RT (c) PV = RT + Pb (d) P + a = RT V2 V 123 Identify the correct statements (a) Unit of gas constant, R = 0.082 kPa dm3 K-1 mole-1 (b) Inversion temperature, Ti = (c) Boyle’s temperature, TB = 2a Rb 2a Rb (d) Critical temperature, T = 8a C 27Rb 124 If a real gas follows equation P(V – nb) = RT at low pressure, then for a graph between d/P vs P, (where d is the density of gas) (a) Intercept is MR T (b) Intercept is (c) Slope is – b M ( RT )2 (d) Slope is – M RT Mb ( RT )2 125 The Maxwell–Boltzmann distribution for molecular speeds is shown in the following figure In the figure, H is the height of the peak, L is the location of the 13/05/16 1:05 PM Gaseous State 3.23 maximum and W is the width at half height As the temperature is increased (a) L increase (d) H decreases (c) H increases (d) W increases Passage Based Questions Passage-1 A gas, which obeys Boyle’s law, Charle’s law, Avogadro’s law etc., or ideal gas equation PV = nRT under all conditions of temperature and pressure, is called ideal gas No gas is ideal All gases are real gases The real gas obeys these gas laws only when the temperature is high or pressure is low The extent of derivations of a real gas form ideal behaviour is expressed in terms of compressibility factor Z defined PV as Z = nRT Real gases have characteristic temperatures like critical temperature, inversion temperature and Boyle temperature These temperatures can be calculated using van der Waal constants 126 The compressibility factor of a gas is less than unity at STP Therefore (a) Vm > 22.4 lit (b) Vm < 22.4 lit (c) Vm = 22.4 lit (d) Vm = 11.2 lit 127 Identify the correct relationship 27 27 (b) Tb : Ti : Tc = : : (a) Tb : Ti : Tc = : : (c) Tb : Ti : Tc = : : (a) 2 : (c) : (b) : (d) : Passage II In the given sample of a gas all molecules not possess same speed Due to frequent molecular collisions, the molecules move with ever changing speeds and also with changing direction There are three types of velocities (i) root mean square velocity (ii) average velocity and (iii) most probable velocity 129 Relation between the three types of velocities, i.e., most probable velocity : average velocity : root mean square velocity is (a) : : π (b) 3 : : (c) : (8 / π ) : (d) : : 130 Oxygen has a density of 1.429 gm /L at STP The RMS velocity of O2 molecules in cms–1 (a) 4.61 × 103 (b) 4.16 × 103 (c) 46.1 × 103 (d) 6.41 × 103 131 By how may folds the temperature of the gas would increase when the RMS velocity of gas molecules in a container of fixed volume is increased from × 104 cm sec–1 to 10 × 104 cm sec-1? (a) times (b) times (c) times (d) times 27 (d) Tb : Ti : Tc = : 27 : 16 128 Density of two gases of same molecular weight are in the ratio : and their temperatures are in the ratio : The ratio of respective pressures is Objective_PCM_Chapter 03.indd 23 13/05/16 1:05 PM 3.24 Chapter Match the Column Type 132 Match the following Column-I (A) Ur.m.s (B) Um.p (C) Uav (D) √P Column-II (p) log P = – log V + constant (q) r = K.P M (r) d = PM RT Column-II (p) dRT (s) M (q) 2.5RT M 3P (r) d (s) 8P πd 2RT (t) M 133 Match the following Column-I (a) Boyle’s law (b) Charles’ law (c) Graham’s law (d) Ideal gas (t) 134 Match the following Column-I (a) Compressibility factor, Z = (b) Compressibility factor, Z > (c) Compressibility factor, Z < (d) Boyle temperature Column-II (p) Attractive forces dominate (q) PV = nRT (r) Repulsive forces dominate (s) Attractive force and repulsive forces cancel each other (t) Gas is less compressible Integer Type 135 A man weigh 72.15 kg and want to fly in the sky with the aid of balloons itself weighing 20 kg and each containing 50 moles of H2 gas at 0.05 atm and 27oC If the density of air at the given conditions is 1.25 g/L, how many such types of balloons he is needed to fly in the sky 137 60 g of gaseous C2H6 are mixed with 28 g of gaseous CO The pressure of the resulting gaseous mixture is atm The partial pressure of C2H6 in the mixture is _ atm 136 When 10 ml of a strong acid is added to 10 ml of an alkali, the temperature rises by 5oC If 100 ml of each liquid is mixed, the temperature rise would be _ 139 At 27oC, 0.5 L of H2 at 0.8 bar and 2.0 L of oxygen at 0.7 bar are introduced in a 1L vessel The total pressure of the gas mixture is 1.8 × 10n bar where n is _ Objective_PCM_Chapter 03.indd 24 138 The ration between the r.m.s velocity of H2 at 50 K ad that of O2 at 800 K is 13/05/16 1:05 PM Gaseous State 3.25 140 An LPG cylinder contains 20 kg of butane gas at 27oC and 10 atm pressure It was leaking and the pressure came down to atmosphere after 20 hours The gas lost is _ kg 142 Ar and He are both gases at room temperature The average molecular velocity of He atoms is x times of the average molecular velocity of Ar atoms at this temperature The numerical value of x is _ 141 A gas at 350 K 15 atm has a molar volume 12 percent smaller than that calculated from the perfect gas law Compressibility factor under these conditions can be expressed in scientific notation as 88 × 10-x The value of x is 143 4gm of oxygen diffuses through a very narrow hole in 30 sec What mass of hydrogen in gm will diffuse in the same time under identical conditions ? 144 If a mixture of mole of H2 and mole of N2 is completely converted into NH3, what would be the ratio of the initial and final volume at same temperature and pressure? Previous Years' Questions 145 For an ideal gas, number of mole per litre in terms of its pressure P, temperature T and gas constant R is [2002] (a) PT/R (b) PRT (c) P/RT (d) RT/P 146 Based on kinetic theory of gases following laws can be proved: [2002] (a) Boyle’s law (b) Charles law (c) Avogadro’s law (d) All of these 147 According to the kinetic theory of gases, in an ideal gas, between two successive collisions the gas molecules travels [2003] (a) in a circular path (b) in a wavy path (c) in a straight line path (d) with an accelerated velocity 148 As the temperature is raised from 20°C to 40°C, the average kinetic energy of neon atoms changes by a factor of which of the following? [2004] (a) 1/2 (b) 2 (c) √313/293 (d) 313/293 149 In van der Waals equation of state of the gas law, the constant ‘b’ is a measure of [2004] (a) intermolecular attraction (b) intermolecular repulsions (c) intermolecular collision per unit volume (d) volume occupied by the molecules Objective_PCM_Chapter 03.indd 25 150 For which of the following parameters the structural isomers C2H5OH and CH3OCH3 would be expected to have the same vaules? (assume ideal behaviour) [2004] (a) gaseous densities at the same temperature and pressure (b) heat of vaporization (c) boiling points (d) vapour pressure at the same temperature 151 Which one of the following statement is not true about the effect of an increase in temperature on the distribution molecular speeds in a gas? [2005] (a) the most probable speed increases (b) the fraction of the molecules with the most probable speed increases (c) the distribution becomes broader (d) the area under the distribution curve remains the same as the under the lower temperature 152 An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system If Ti is the initial temperature and Tf is the final temperature, which one of the following statements is correct? [2006] (a) (Tf)irrev (TF)rev (b) Tf Ti for reversible process but Tf = Ti for irreversible process (c) (Tf)irrev = (TF)rev (d) Tf = Ti for both reversible and irreversible processes 153 Equal masses of methane and oxygen are mixed in an empty container at 25°C The fraction of the total pressure exerted by oxygen is [2007] 13/05/16 1:05 PM 3.26 Chapter (a) 1/3 273/298 (c) 1/2 (b) 1/3 (d) 2/3 154 ‘a’ and ‘b’ are van der Wall’s constants for gases Chlorine is more easily liquefied than ethane because [2011] (a) a for Cl2 < a for C2H6 but b for Cl2 > b for C2H6 (b) a and b for Cl2 < a and b for C2H6 (c) a and b for Cl2 > a and b for C2H6 (d) a for Cl2 > a for C2H6 but b for Cl2 < b for C2H6\ 155 The compressibility factor for a real gas at high pressure is – [2012] (a) + Pb/RT (b) + RT/Pb (c) (d) – Pb/RT 156 For gaseous state if most probable speed is denoted – C*, average speed by C and and mean square speed by C, then for a large number of molecules the ratios of these speeds are: [2013] – (a) C* : C : C = : 128 : 1.225 – (b) C* : C = : 1.225 : 1.128 – (c) C* : C : C = 1.225 : 1.128 : – (d) C* : C : C = 1.128 : 1.225 : 157 If Z is a compressibility factor, van der Waals equation at low pressure can be written as: [2014] Pb Pb (a) Z = − (b) Z = 1+ RT RT (c) Z = + Objective_PCM_Chapter 03.indd 26 158 The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1 : 4 The ratio of number of their molecule is: [2014] (a) 1 : 8 (b) 3 : 16 (c) 1 : 4 (d) 7 : 32 159 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is [2015] (a) Ion-ion interaction (b) Ion-dipole interaction (c) London force (d) Hydrogen bond 160 Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure Pl and temperature Tl are connected through a narrow tube of negligible volume as shown in the figure below The temperature of one of the bulbs is then raised to T2 The final pressure pf is [2016] T1 pi, V T1 T1 pi, V pf, V T2 pf, V T2 T1 pi (a) pi (b) T1 + T2 T + T 2 TT TT pi (c) pi (d) T1 + T2 T1 + T2 RT a (d) Z = 1+ Pb VRT 13/05/16 1:05 PM Gaseous State 3.27 ANSWER KEYS Single option correct type 1. (d) 2. (d) 3. (b) 4. (c) 5. (d) 6. (d) 7. (c) 8. (c) 9. (e) 10. (d) 11. (a) 12. (a) 13. (b) 14. (b) 15. (b) 16. (a) 17. (b) 18. (d) 19. (a) 20. (a) 21. (a) 22. (a) 23. (a) 24. (d) 25. (c) 26. (b) 27. (b) 28. (c) 29. (c) 30. (c) 31. (a) 32. (d) 33. (a) 34. (a) 35. (c) 36. (b) 37. (b) 38. (d) 39. (a) 40. (c) 41. (d) 42. (c) 43. (b) 44. (a) 45. (b) 46. (b) 47. (c) 48. (c) 49. (b) 50. (a) 51. (b) 52. (a) 53. (b) 54. (c) 55. (a) 56. (d) 57. (a) 58. (b) 59. (c) 60. (a) 61. (a) 62. (c) 63. (b) 64. (a) 65. (c) 66. (b) 67. (b) 68. (c) 69. (b) 70. (b) 71. (c) 72. (c) 73. (c) 74. (d) 75. (a) 76. (b) 77. (c) 78. (a) 79. (d) 80. (a) 81. (d) 82. (a) 83. (a) 84. (b) 85. (a) 86. (c) 87. (b) 88. (a) 89. (a) 90. (a) 91. (a) 92. (c) 93. (b) 94. (b) 95. (a) 96. (a) 97. (c) 98. (b) 99. (c) 100. (c) 101. (b) 102. (a) 103. (c) 104. (b) 105. (b) 106. (d) 107. (c) 108. (d) 109. (c) 110. (c) 111. (b) 112. (c) 113. (d) 114. (b) 115. (c) More than One Option Correct Type 116. (a, b, c) 121. (a, b, c) 117. (a, b, c) 122. (a, d) 118. (a, c) 123. (b, d) 119. (a, b) 124. (b, d) 120. (b, c, d) 125. (a, b, d) Passage Based Questions 126. (b) 127. (c) 128. (d) 129. (c) 130. (c) 131 (b) Match the Column Type 132. (a) – (r), (b) – (t), (c) - (q, s), (d) – (p) 133. (a) – (p, t), (b) – (s), (c) – (q), (d) – (p, q, r, s, t) 134. (a) – (q, s); (b) – (r, t); (c) – (p); (d) – (q, s) Integer Type 135. (3) 142. (3) 136. (5) 143. (1) 137. (2) 144. (2) 138. (1) 139. (0) 140. (4) 141. (2) Previous Years' Questions 145. (c) 146. (d) 155. (a) 156. (a) Objective_PCM_Chapter 03.indd 27 147. (c) 148. (d) 157. (??) 158. (d) 149. (d) 150. (a) 159. (d) 160. (b) 151. (b) 152. (a) 153. (b) 154. (d) 13/05/16 1:05 PM 3.28 Chapter HINTS AND SOLUTIONS Single Option Correct Type _ 730 380 = 365 mL 760 NH3 and HCl react to form solid NH4Cl By definition, pressure is force per unit area Ideal gas cannot be liquefied K.E is directly proportional to √T If T is same, K.E is also same Avogadro’s law helps us to arrive at this conclusion As the condition of constant pressure, or constant volume is not mentioned so the gas may be monoatomic or diatomic 11 At high temperature and low pressure, real gas act as an ideal gas _ 0.082 298 nRT 0.6 14 P = = = 4.89 atm V _ 16 Total mol of gases = = 32 _ 0.082 273 nRT _ P = = V = 25.18 atm = _ _ W RT 0.0821 300 760 18 V = = MP 28 750 = 8.0225 L = 1.97 atm ≈ atm 20 As molecular weight is same, so rms √T, that is, for 3, temperature must be 16 times _ W 22 PV = RT M _ _ WRT 7.5 0.082.62 273 M = = = 29.98 PV 5.6 that is, NO (14 + 16 = 30) _ PV 24 Z = RT ZRT So, V = P 0.5 0.082 273 = = 0.1119 L 100 _ WRT 27 M = PV _ 0.082 546 = = 16 5.6035 28 KE = nRT = 3 400 = 4800 K nCH × 720 30 P(CH4) = Total moles = 720 = 180 mm 31 V2 = P1V1 P2 Objective_PCM_Chapter 03.indd 28 33 P1V1 P2 V2 = T1 T2 V2 _ P 20 2P = 600 300 V2 = 20 L 35 Vav √(1/M) V (SO ) V (gas) On solving = √[M(gas)/M(SO2)] M (gas) = 16 that is, CH4 0.082 540 nRT _ 36 P = = V 44.8 37 r1 : r2 = √(16:1) or : rates are in the inverse ratio of square root of their densities 38 More the dipole moment more is the critical temperature 39 P = nRT/V _ 0.082 546 P = 44.8 = atm 96 56 =2 n O2 = =3 28 32 PN = × 10 = 4atm P = ´ 10 = atm N PO2 = ´ 10 = atm 40 n N2 = 43 As ammonia will have a higher diffusion rate than NH3 due to its lower molecular weight 44 As temperature is constant, pressure decreases during expansion while kinetic energy remains same 45 Vrms a T In order to make Vrms doubled T is increased by four times i.e T = (50+273) K = 1292 K = 1019°C 13/05/16 1:05 PM Gaseous State 3.29 48 As liquification of gas ‘a’ that is, NH3 CH4 N2 O2 49 As P = dRT i.e P a d a M M PA d A M B 2 = ´ = ´ = PB d B M A 1 51 As d pressure 1/temperature 53 Follow Avogadro’s law 57 6.4 g of SO2 at NTP (that is, at 0°C temperature and atm pressure) occupies a volume of 2.24 L 64 g of SO2 that is, mole SO2 at NTP occupies a volume of 22.41 L Now mole of an ideal gas occupies 22.4 L at NTP condition Therefore SO2 acts as an ideal gas 59 PV = P ′V ′ T T′ Here V = 2V, T = 2T PV = P ′ × 2V or P = P T 2T that is, pressure remains same 60 Moles of H2 = 3/2 = 1.5 65 Given number of moles of an ideal gas (n) = 2; temperature of the gas (T) = 546 volume of gas (V) = 44.8 L we know from the ideal gas equation the pressure (P) nRT = V 0.0821 546 = = atm 44.8 (where R = gas constant equal to 0.0821 litre atm/K/atm) 67 For A P1 = 2P, V1 = 2V, T2 = 2T For B P2 = P, V2 = V, T2 = T According to ideal gas equation, P1V1/n1RT1 = P2V2/n2RT2 2P 2V/n1R 2T = P V/n2RT 2/n1 = 1/n2 n1/n2 = : 68 3H2 N2 2NH3 30 L 30 L H2 is limiting reagent Expected volume of NH3 =20 L Actual volume of NH3 = 10 L H2 consumed = 15 L ; H2 left = 15 L N2 consumed = L ; N2 left = 25 L NH3 formed = 10 L Moles of O2 = 4/32 = 1/8 kinetic energy n (moles of gas) 70 V1/V2 = √(T1/T2) K.E (H ) 1.5 = = 12 : K.E (O ) 1.8 1/9 = 100/T2 w dRT RT 62 M = = V P P _ 1.964 10–3 82.1 273 that is, M = = 44, that is, the gas is CO2 63 According to Boyle’s law, PA VA = 0.4PA VB 100 PA = 0.4 PA VB VB = 250 cc Before opening the stopcock, volume of gas in bulb B must be (250 – 100) = 150 mL 64 V1T2 = √ (M2 / M1) V2 T2 500 × T2 = √ 64/4 1000 × 30 T2 = 1000 × 30 × = 240 500 Objective_PCM_Chapter 03.indd 29 104/3 104 = √(100/T2) T2 = 100 = 900 71 Given P (pentane) = 440 m P (hexane) = 120 mm From Dalton’s law of partial pressure P (total) = P (pentane) + P (hexane) = (440 + 120) mm = 560 mm Mole fraction, X (pentane) = 440/560 = 0.786 73 r u u = √(3RT/M) or r1/r2 = √(T1M2 / T2M1 ) r (N2)/r (SO2) = √(T1 64/323 28) = 1.625 or (1 625 )2 323 28 _ T = = 373 K 64 76 Using Graham’s law, To2 is 20 seconds 77 μ1/μ2 = √(T1M2/M1T2) = √(50 32/800 2) = 13/05/16 1:05 PM 3.30 Chapter 79 Vrms a VH T = so VO M TH TO × MO MH _ 149 a 92 TB = = = 453.7 K Rb 0.0812.3 0.04 = on solving VO2 = 454 K 94 For gas A, the volume increases from one litre to four = 0.95 × 103 m/s 80 V1/V2 = √(T1/T2)t 0.3/V2 = √(300/1200) V2 = 0.6 m/sec 81 At a given temperature, rate of effusion P and 1/√M Rate of effusion of hydrogen gas _ 2000 = 10 torr min–1 3 √(4 / 2) 1000 litre This would reduce the pressure but the temperature increases from 300 K to 600 K 600 pA = = atm 300 = 20√2 torr 96 In 15 L of H2 gas at STP, the number of molecules –1 82 rCH /rx = √(Mx/MCH ) = √(Mx/16) Mx = 64 84 Applying Dalton’s law _ W/2 PH = Ptotal W/2 + W/16 On solving, Similarly, 600 pB = 3 = atm 400 So, total pressure is (1 + 2.25) atm = 3.25 atm _ 6.023 3.1023 = 15 = 4.033 1023 22.4 In L of N2 gas at STP 6.023 1023 3 5 _ = = 1.344 1023 22.4 In 0.5 g of H2 gas 6.023 31023 0.5 = = 1.505 1023 PH = 8/9 Ptotal 85 Average kinetic energy depends only on temperature and does not depend upon the nature of the gas 87 P (drop in pressure ) = 760 – 570 190 = 190 mm = atm 760 In 10 g of O2 gas PV = nRT P (drop in pressure) V n = RT 109 32.24 = = 0.0249 760 0.8213 273 urms = √(3RT/M) Average K.E is E = RT or 2E = 3RT urms = √(2E/M) PV/nRT PV = nRT + nPb PV/nRT = + Pb/RT PV/nRT = 0.025 moles 88 It is the most probable velocity given by √2RT/M 89 Mass of the filled balloon = 50 + 685.2 = 735.2 kg Pay load = Mass of displaced air – Mass of balloon = 5108 – 735.2 = 4372.8 kg 90 For n mol of a real gas, van der Waals equation is _ n2a P + 2 (V – nb) = nRT V n2a When 2 ≠ V n2 a P+ 2 V = nRT V _ n2a PV + 2 = nRT V n2a PV = nRT 2 . V 91 This is because rate of diffusion of H2 is maximum (r 1/√M) ( ) Objective_PCM_Chapter 03.indd 30 _ 6.023 1023 = = 1.882 1023 32 Hence, maximum molecules are in 15 litre H2 at STP 97 Rms of one mole of monoatomic gas is 98 For positive deviation 99 For ideal gases, w PV = nRT = RT M (w) _ _ RT _ RT P = = d M V M M = RT (d/P) Given d = 2.00 + 0.020 P2 (for a real gas) d/P = 2.00 + 0.020 P P t ∫ d / P = 200 which is d/P for an ideal gas So, M = RT = 25 = 50 g mol–1 13/05/16 1:05 PM Gaseous State 3.31 100 1L = 1000 mL = 1000 cm3 mass = density volume = (0.0006 g cm–3) (1000 cm3) = 0.6 g 18 g of water = 18 cm3 0.6 g of water = 0.6 cm3 Hence, actual volume occupied by molecules = 0.6 cm3 101 Z = PV/nRT Given Z < PV/nRT < or PV < nRT (1 atm) V< mol (0.0821 L atm K–1 mol–1) (273 K) Vm < 0.0821 273 L Vm < 22.4 L 102 At the same temperature, oxygen and hydrogen molecules will have the same average energy; weight of H2 molecules is 1/16 of O2 molecules 103 μH2/μN2 = √[(TH /MH ) (MN /TN )] 2 2 √7 = √(TH 28/TN 2) = TH 14/TN 2 2 TN = 2TH 2 So, TN TH 2 104 At constant temperature V = K/P (Boyle’s law) So, P = K/V As such, a graph plotted between P and 1/V is linear at a constant temperature 105 Vc = 3b, assuming the gas to obey van der Waals equation _ 0.072 b (the covolume) = = 0.024 L mol–1 3 24 cm b = per molecule, (NA = 1023) 1023 b = 10–23 cm3 per molecule = πr3 –23 πr = 10 _ r3 = 3 10–23 4π r = (3/4π 10–23 )1/3 cm 106 Applying Dalton’s law _ W/2 PH = Ptotal W/2 + W/30 On solving, PH = 15/16 Ptotal 108 Pressure exerted by oxygen will proportional to mole fraction of O2 W/32 Mole fraction of O2 = = 1/3 W/16 + W/32 109 According to ideal gas equation, PV = nRT P n 1 = n (V and T are constant) P2 P w _ w 1 or = w as n = P2 M Hence, 10/8 = 15/w2 or w2 = 12 kg Gas leaked in one day = 15 – 12 = kg Gas leaked in days = 3 = 15 kg _ 33 110 Moles of O2 = = 32 _ 32 Moles of SO2 = = 0.5 64 Total moles = + 0.5 = 1.5 Let total pressure = P Partial pressure of O2 = XO2 P (where XO2 = mole fraction of oxygen) _ _ P = 3 P = = P 1.5 3/2 112 C3H8 5O2 mL mL mL In this reaction, mL of reactants give mL of products So, 10 mL of propane + 50 mL of oxygen would yield 30 mL of CO2 Unreacted O2 = (70 – 50) mL = 20 mL So, volume after explosion = 30 + 20 = 50 mL After exposure to alkali only O2 remains = 20 mL 113 Most probable speed = √[2RT/M] So √[2RT1/M1] : √[2RT2/M2] = 0.715 : T1/M1 M2/T2 = (0.715)2 T2 M1 M2/T1 = (0.954)2 So (M2/M1)2 = (0.715)2 (0.954)2 M1/M2 that is, MA/MB = 0.715 0.954 = 1.466 370 114 Initial moles of O2 = = 11.6 32 11.6 0.082 298 nRT _ V = = P 30 √(3RT/M) : √(8RT/πM) √3 : √(8/π) = 1.086 : Objective_PCM_Chapter 03.indd 31 = 9.43 L As volume is same and P is atm so at 348 K moles _ PV of O2 (n) = RT 107 urms : uav 3CO2 4H2O (l) _ 3 9.43 = 0.33 0.082 3 348 = So, weight of O2 left = 0.33 32 = 10.6 g Hence O2 escaped = 370 – 10.6 = 358.4 g ≈ 359 g 13/05/16 1:05 PM 3.32 Chapter More than One Option Correct Type 116 At higher altitudes the pressure is low and the boiling point is also low 119 Al low pressures, b is negligible, so van der Waal’s equation becomes 117 Let the no of moles of H2, O2 and Ne be a, b and c respectively P + a + b + c = 2.0 Pressure exerted by moles of gaseous mixture = 50 atm ( Z= 1st spark Decrease in pressure after first electric spark = 50 – 12.5 = 375 atm Decrease in number of moles = 1.5 Also H (g) + O (g) •• PV RT ) × V = RT of Z > 1, PV > RT Molecular diameter of O2 is greater than that of H2 Hence, mean free path of O2 is lesser than that of H2 At constant temperature, kinetic energies of one mole of all gases are the same •• H2 and O2 react in the ratio : So, 0.5 mole of O2 should have reacted with mole of H2 V i.e., at a given pressure , Vreal > Videal H2O (l) a 121 V ∝ T at constant pressure or V = kT dV =k dT P 2nd spark Therefore Change in pressure = 25 – 12.5 = 12.5 atm T ∝ V at constant pressure or T = kV No of moles of O2 added = 12.5 × = 0.5 mole 50 Now H2 will completely react No of moles of H2 left after 1st spark = a – 1 H (g) + O (g) H2O (l) (a − 1) (a − 1) Change in pressure = 25 – 10 = 15 atm Change in no of moles = (a − 1) + 15 × 50 = 0.6 a −1 Therefore Also V T dT = k dV P = K or (V)= K T Differentiating w.r.t V at constant pressure 1 − V dV = T T P Hence − V = T T2 P 8a = 0.7 a = 1.4 (no of moles of H2) 123 Critical temperature, TC = 1.4 X H2 = = 0.7 2.0 Boyle’s temperature, TB = No of moles of Ne (b) = – (1.4 + 0.5) = 0.1 Inversion temperature, Ti = xNe = 0.1 = 0.5 2.0 0.5 X O2 = = 0.25 2.0 Kinetic energy = at zero kelvin temperature At constant temperature, change in pressure of the gas does not affect kinetic energy of the gas PV − pnb = RT Objective_PCM_Chapter 03.indd 32 a Rb 2a Rb Unit of gas constant, R = 0.082 lit Atm K-1 mol-1 = 8.31 joules mol-1 K-1 = 1.987 cal mol-1 K-1 124 P(– nb) = nRT PV – Pnb = nRT 118 Kinetic energy ∝ absolute temperature 27Rb n = 8.31 kPa dm3 K-1 mol-1 n 13/05/16 1:05 PM Gaseous State 3.33 MP − pb = RT W d p = RT M MP − pb = RT d P RT + pb = d M d p d d M = p RT + pb p Pb + M RT M M M bP + RT RT = = = −1 = ( 1+ PB M , M RT ) M bP 1 − RT RT MbP − RT ( RT )2 y = c + mx Passage Based Questions 126 Ideal gas equation is PV = nRT For mole of gas 129 MPV PVm = RT Vm = Z= RT P PV = PVm nRT RT Vm < 22.4 lit when Z is less than unity a Rb 2a ; Ti = Tb : Ti : Tc = : : Rb and Tc = 128 PV = nRT = W M RT W PM = RT dRT V P = dRT : M √2 8a 27Rb 130 u = or P α dT = u= = 1.429 1000 g/cc = 0.001429 g/cc × 76 13.6 × 981 0.001429 3042 1.429 103 cm/s = 46.1 × 103 cm/s 131 RMS velocity ∝ √T Initial u1 = × 104 cms-1 Final u2 = 10 × 104 cms-1 P1 d T1 = P2 d T2 But × × 2 or P1 : P2 = : Objective_PCM_Chapter 03.indd 33 : √3 3P When we compare two gases = 3RT : M πM π u 10 × 10 =2 = u2 × 10 M RMS 8RT : : P = atmosphere d = 76 × 13.6 × 981 dynes sq cm-1 d = 1.429 g /L 27 = 27 : 54 : Average velocity (ii) Vm = Z RT = 22.4 lit for ideal gas at STP and Z = 127 Since Tb = 2RT (i) : u2 u2 = T2 T1 T2 T2 ; =4 T1 T2 T2 = × T1 i.e., the temperature would increase by times 2= 13/05/16 1:05 PM 3.34 Chapter Match the Column Type 133 According to Boyle’s law, PV = constant r= log P + log V = constant K×P M PV = nRT log P = – log V + constant The rate of diffusion of the gas is directly proportional to its pressure and inversely proportional to the square root of its molecular mass P = W × RT = dRT M V PM d= RT M Integer Type 135 Weight of H2 in each balloon = 50 × g Volume of each balloon filled with H2 = = nRT 50 × 0.0821 × 300 P 0.05 = 24.6 × 103 L Thus volume of displaced air will be = 24.6 × 103L Weight of displaced air = 24.6 × 103 × 1.25 = 30.75 × 103 g Let n balloons are used by amn, then for flying in air Weight of total displaced air = wt of amn + wt of balloons + wt of H2 30.75 ×103 × n = 72.15 × 103 + 20 × 103 + 50 × 137 Number of moles of C2H6 = Number of moles of CO = W2 = 60 30 28 141 Z= 28 Z= For O2, P = O V2 3R (50) × 10 = 0.8 × 0.5 0.7 × 2.0 −5 143 m/s = 0.40 bar = 1.4 bar = 0.88 8RT πM M Ar M He 40 = = 3.16 ro rH 32 = m 2 = 32 4m m = 1g 144 N2 + 3H2 Initial mole Final mole 0 = 1.8 bar, 1.8 × 100bar 140 P1V = n1RT and P2V = n2RT P n W / M W1 = = = P2 n W2 / M W2 Objective_PCM_Chapter 03.indd 34 Vm = Ptotal = PH + P0 = 0.4 + 1.4 Vm perfect Vm perfect V He = VAr =1 139 P1V1 = P2V2 Vm VHe = VAr (approx) 138 RMS velocity of H2, VH = For H2, P = H RT = Given, Vm = Vm perfect – 0.12 Vm perfect = 0.88 Vm perfect P1 V1 PVm 142 Vav = =2 2 = 16 10 Hence gas lost, 20 – 16 = kg p C H = x C H × Pt = × = atm 160 = 88 × 10-2 n=3 x C2 H6 = 10 20 = W2 Ratio of initial and final volumes = 2NH3 initial mole final mote =2 = 13/05/16 1:05 PM Gaseous State 3.35 Previous Years' Questions RT 148 As average K.E = N0 that is, (K.E) T 154 As van der Wall’s constant ‘a’ is a measurer of intermolecular force of attraction while ‘b’ is a measure of size of the gaseous molecule Hence (K.E)313 / (K.E)293 = 313/293 155 P + 149 Van der Waals equation for one mol of a gas is given as [P + a/V2] (V – b) = RT where b is volume correction It arises due to finite size of gaseous molecules Mass 150 Density = Volume Due to ideal behaviour at a given temperature and pressure volume and molar mass are same so they have same vapour density 151 Most probable velocity increases and fraction of molecule possessing most probable velocity decreases 152 Work done is reversible process is maximum Thus T2 PV = RT + Pb PV Pb = 1+ RT RT Z = 1+ Pb RT – 156 C* : C : C = : 1.128 : 1.225 Cmps = 2RT M 8RT πM 3RT = M CAS = Let the total pressure be P C rms – C* : C : C = : Objective_PCM_Chapter 03.indd 35 a V2 P(V – b) = RT; _ w/32 Mole fraction of oxygen = w/32 + w/16 1/32 3/32 = = = 1/3 1/32 + 1/16 3/32 The pressure exerted by oxygen (partial pressure) = XO Ptotal = P 1/3 a (V − b) = RT for mole V2 : π = : 1.128 : 1.225 13/05/16 1:05 PM This page is intentionally left blank Objective_PCM_Chapter 03.indd 36 13/05/16 1:05 PM ... Dronacharya), Rajneesh Shukla (Allen, Kota), Anupam Srivastav, Rajeev Jain (M V N), Sandeep Singhal, Chandan Kumar (Mentor, Patna), P.S Rana (Vidya Mandir, Faridabad) I am indebted to my father,... 5/10/2017 11:19:37 AM A Complete Resource Book in chemistry for JEE Main 2019 A. K Singhal U.K Singhal Dedicated to my grandparents, parents and teachers Copyright © 2018 Pearson India Education Services... I am thankful to my colleagues and friends, Deepak Bhatia, Vikas Kaushik, A. R Khan, Vipul Agarwal, Ankit Arora (ASO Motion, Kota), Manoj Singhal, Yogesh Sharma, (Director, AVI), Vijay Arora,