EG_Chemistry-12_(22-07-2022) a

12 CBSE 2022-23 CHEMISTRY (As per the Latest Syllabus issued by CBSE on April 21, 2022) Dr. Sulekh Chandra Ph.D, FICCE, MNASc, FICS, FISC Associate Professor Department of Chemistry Zakir Husain Delhi College (University of Delhi), Delhi Full Marks Pvt Ltd (Progressive Educational Publishers) New Delhi-110002

Published by: 9, Daryaganj, New Delhi-110002 Phone: 011- 40556600 (100 Lines) Website: www.fullmarks.org E-mail: [email protected] © Publishers All rights reserved. No part of this publication may be reproduced or transmitted, in any form or by any means, without permission. Any person who does any unauthorised act in relation to this publication may be liable to criminal prosecution and civil claims for damages. Branches: • Chennai • Guwahati Marketing Offices: • Ahmedabad • Bengaluru • Bhopal • Dehradun • Hyderabad • Jaipur • Jalandhar • Kochi • Kolkata • Lucknow • Mumbai • Patna • Ranchi New editioN “This book is meant for educational and learning purposes. The author(s) of the book has/have taken all reasonable care to ensure that the contents of the book do not violate any existing copyright or other intellectual property rights of any person in any manner whatsoever. In the event the author(s) has/have been unable to track any source and if any copyright has been inadvertently infringed, please notify the publisher in writing for corrective action.” Printed at:

Preface Chemistry-12 is based on the latest curriculum guidelines specified by the CBSE. It will certainly prove to be a torch-bearer for those who toil hard to achieve their goal. This All-in-one Question Bank has been developed keeping in mind all the requirement of the students for Board Examinations preparations like learning, practicing, revising and assessing. Salient Features of the Book: ● Each chapter is designed in ‘Topic wise’ manner where each topic is briefly explained with sufficient Examples and Exercise. Exercise which covers all the possible variety of Questions. ● Answers with hints are provided separately after the exercise. ● Importance of Each Topic and Frequently Asked Types of Questions provides an idea to the students on which type they should focus more. ● Assignment is provided at the end of each chapter. ● Previous years’ Board Questions have been covered in every chapter. ● 1 Solved and 2 Unsolved Sample Papers for mock test are given with hints & answers for self assessment. ● Common Errors by the students are provided to make students aware what errors are usually done unknowingly. ● The book has been well prepared to build confidence in students. Suggestions for further improvement of the book, pointing out printing errors/mistakes which might have crept in spite of all efforts, will be thankfully received and incorporated in the next edition. —Publisher (iii)

Syllabus Time : 3 Hours Marks : 70 S.No. Title No. of Periods Marks 1. Solutions 15 7 2. Electrochemistry 18 9 3. Chemical Kinetics 15 7 4. d- and f-Block Elements 18 7 5. Coordination Compounds 18 7 6. Haloalkanes and Haloarenes 15 6 7. Alcohols, Phenols and Ethers 14 6 8. Aldehydes, Ketones and Carboxylic Acids 15 8 9. Amines 14 6 10. Biomolecules 18 7 Total 160 70 Unit II: Solutions 15 Periods Types of solutions, expression of concentration of solutions of solids in liquids, solubility of gases in liquids, solid solutions, Raoult’s law, colligative properties - relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, osmotic pressure, determination of molecular masses using colligative properties, abnormal molecular mass, Van’t Hoff factor. Unit III: Electrochemistry 18 Periods Redox reactions, EMF of a cell, standard electrode potential, Nernst equation and its application to chemical cells, Relation between Gibbs energy change and EMF of a cell, conductance in electrolytic solutions, specific and molar conductivity, variations of conductivity with concentration, Kohlrausch’s Law, electrolysis and law of electrolysis (elementary idea), dry cell-electrolytic cells and Galvanic cells, lead accumulator, fuel cells, corrosion. Unit IV: Chemical Kinetics 15 Periods Rate of a reaction (Average and instantaneous), factors affecting rate of reaction: concentration, temperature, catalyst; order and molecularity of a reaction, rate law and specific rate constant, integrated rate equations and half-life (only for zero and first order reactions), concept of collision theory (elementary idea, no mathematical treatment). Activation energy, Arrhenious equation. Unit VIII: ‘d ’ and ‘f ’ Block Elements 18 Periods General introduction, electronic configuration, occurrence and characteristics of transition metals, general trends in properties of the first row transition metals - metallic character, ionization enthalpy, oxidation states, ionic radii, colour, catalytic property, magnetic properties, interstitial compounds, alloy formation, preparation and properties of K2Cr2O7 and KMnO4. Lanthanoids: Electronic configuration, oxidation states, chemical reactivity and lanthanoid contraction and its consequences. Actinoids: Electronic configuration, oxidation states and comparison with lanthanoids. (iv)

Unit IX: Coordination Compounds 18 Periods Coordination compounds - Introduction, ligands, coordination number, colour, magnetic properties and shapes, IUPAC nomenclature of mononuclear coordination compounds. Bonding, Werner’s theory, VBT, and CFT; structure and stereoisomerism, importance of coordination compounds (in qualitative analysis, extraction of metals and biological system). Unit X: Haloalkanes and Haloarenes 15 Periods Haloalkanes: Nomenclature, nature of C-X bond, physical and chemical properties, optical rotation, mechanism of substitution reactions. Haloarenes: Nature of C-X bond, substitution reactions (Directive influence of halogen in monosubstituted compounds only). Uses and environmental effects of - dichloromethane, trichloromethane, tetrachloromethane, iodoform, freons, DDT. Unit XI: Alcohols, Phenols and Ethers 14 Periods Alcohols: Nomenclature, methods of preparation, physical and chemical properties (of primary alcohols only), identification of primary, secondary and tertiary alcohols, mechanism of dehydration, uses with special reference to methanol and ethanol. Phenols: Nomenclature, methods of preparation, physical and chemical properties, acidic nature of phenol, electrophillic substitution reactions, uses of phenols. Ethers: Nomenclature, methods of preparation, physical and chemical properties, uses. Unit XII: Aldehydes, Ketones and Carboxylic Acids 15 Periods Aldehydes and Ketones: Nomenclature, nature of carbonyl group, methods of preparation, physical and chemical properties, mechanism of nucleophilic addition, reactivity of alpha hydrogen in aldehydes, uses. Carboxylic Acids: Nomenclature, acidic nature, methods of preparation, physical and chemical properties; uses. Unit XIII: Amines 14 Periods Amines: Nomenclature, classification, structure, methods of preparation, physical and chemical properties, uses, identification of primary, secondary and tertiary amines. Diazonium salts: Preparation, chemical reactions and importance in synthetic organic chemistry. Unit XIV: Biomolecules 18 Periods Carbohydrates: Classification (aldoses and ketoses), monosaccahrides (glucose and fructose), D-L configuration oligosaccharides (sucrose, lactose, maltose), polysaccharides (starch, cellulose, glycogen); Importance of carbohydrates. Proteins: Elementary idea of - amino acids, peptide bond, polypeptides, proteins, structure of proteins - primary, secondary, tertiary structure and quaternary structures (qualitative idea only), denaturation of proteins; enzymes. Hormones - Elementary idea excluding structure. Vitamins: Classification and functions. Nucleic Acids: DNA and RNA. (v)

Question Paper Design S.No. Domains Marks % 1. 28 40 2. RemembeRing and UndeRstanding 21 30 Exhibit memory of previously learned material by recalling facts, terms, basic concepts 3. and answers. Demonstrate understanding of facts and ideas by organizing, comparing, 21 30 translating, interpreting, giving descriptions and stating main ideas. applying Solve problems to new situations by applying acquired knowledge, facts, techniques and rules in a different way. analyzing, evalUating and CReating Examine and break information into parts by identifying motives or causes. Make inferences and find evidence to support generalizations. Present and defend opinions by making judgments about information, the validity of ideas or quality of work based on a set of criteria. Compile information together in a different way by combining elements in a new pattern or proposing alternative solutions. (vi)

(vii)

Overview Of a Chapter Chemistry Xii H:\\Full Marks\\Till\\(EG) Chemistry-12\\EG_C-12_Ch-1\\EG_C-12_Ch-1 Reader’s Sign _______________________ Date __________ H:\\Full Marks\\Till\\(EG) Chemistry-12\\EG_C-12_Ch-1\\EG_C-12_Ch-1 \\ 22-Jul-2022     Rajat Aggarwal     Proof-2 \\ 22-Jul-2022     Rajat Aggarwal     Proof-3 Reader’s Sign _______________________ Date __________ 1 Solutions Types of Binary Solution • Molarity: M = WB × 1000 mB solution ]in mLg Type of Solute Solvent Common Volume of solution examples • Molality: m = WB × WA 1000 Topics covered mB ^in gramsh Gas Gas Mixture of 1.1 Solutions: Modes of Expressing Concentration of Solution oxygen and • Mole fraction: xA = nA = WA /mA Gaseous Liquid Gas nitrogen gases nA + nB WA /mA + WB /mB Solutions 1.2 Ideal and Non-ideal Solutions 1.3 Colligative Properties Chloroform for binary solution, xA + xB = 1 Solid Gas mixed with for ternary solution, xA + xB + xC = 1 C hapter map nitrogen gas nA is no. of moles of ‘A’, nB is no. moles of ‘B’ WB = mass of solute, mB = molar mass of solute, Camphor in WA is mass of solvent, mA is molar mass of solvent nitrogen gas If both the components are in gas phase, Gas Liquid Oxygen dissolved in water SOLUTIONS yA = Mole fraction of component in gas phase Ethanol dissolved Partial pressure of that component Liquid Liquid Liquid in water Solutions Solid Liquid = Total vapour pressure Glucose dissolved Types of solutions Concentration of solution Raoult's law Ideal solution in water yA + yB = 1 Non-ideal solution • Parts Per Million (ppm): ppm is the number Molarity, molality, Henry's law Positive deviation Gas Solid Solution of Percent by mass, Negative deviation hydrogen in of grams of a solute dissolved per million grams mole fraction, ppm palladium of the solution or the number of cm3 of solute Solid Liquid Solid Amalgam of mercury with dissolved per million cm3 of the solution. sodium Colligative property Solutions Mass of the component Copper dissolved ppm = Mass of the solution × 106 in gold Solid Solid • Relationship of Molarity (M) and molality Elevation in Depression in • Mass percentage (w/w) : Mass % (m) with mole fraction of solute (xB): boiling point freezing point x × d × 10 M= mB Relative lowering of Osmotic pressure Mass of the component in the solution ⇒ vapour pressure Osmosis Total mass of the solution = × 100 (x = mass percentage of solute, d = density of Reverse osmosis Abnormal molecular mass Isotonic solution • Volume percentage (v/v) : Volume % solution, mB = molar mass of solute) Hypotonic solution Association van’t Hoff factor Dissociation Hypertonic solution = Volume of the component × 100 ⇒ m= xB × 1000 Total volume of the solution (1 − xB ) mA • Mass by volume percentage (w/V): Mass by Mass of solute (xB = mole fraction of solute, Degree of association Degree of dissociation volume % : Volume of solution × 100 mA = molar mass of solvent) ExErcisE 1.1 Multiple Choice Questions (MCQs) (1 Mark) 40 g of it is dissolved in enough water to make a final volume upto 2 L? [M.Wt. of glucose = 180 g mol–1] All concepts are presented in points, which can be1. 2.46 g of NaOH (molar mass = 40) is dissolved in water and the solution is made upto 100 cm3 in a volumetric flask. Calculate the molarity of ✎ (i) 1100 mol L–1 (ii) 0.22 mol L–1 ✎ Topic 1. Solutions: Modes of Expressing Concentration of Solution (iii) 0.44 mol L–1 (iv) 0.11 mol L–1 easily learnt and remember.solution. [Na = 23 u, O = 16 u, H = 1 u]. 4. Calculate the molality of 2.5 g ethanoic acid (i) 0.615 (ii) 0.515 C(CHH3C3COOOOHH=)6i0ng 75 g o f benzene. [M.wt. of • Solution: Solutions are homogeneous mixtures of (or the one that is dissolved in the solvent) is called mol–1] two or more than two components. The component solute. (iii) 0.246 (iv) 1.626 that is present in larger quantity (or the one that (i) 0.515 mol kg–1 (ii) 0.556 mol kg–1 dissolves another component) is known as solvent. • Binary Solution: If a solution is consisting of (iii) 0.665 mol kg–1 (iv) 0.775 mol kg–1 only two components, i.e. a single solute dissolved Each concept is well explain by relevant diagrams,2. A solution is prepared by adding 2 g of glucose The component that is present is smaller quantity in a solvent, then it forms a binary solution. into 18 g of water. Calculate the mass percent tables and illustrations for better understanding.of glucose in the solution. [M.Wt. of glucose = 180 g mol–1]. (i) 9% (ii) 10% 5. iaCsnaddlcicsuaslorabltveoentdhtieentmr1aa2cs2hslgoproeifrdcceea(nrCtbCaogln4e)t,oeiftfbr2ae2cnhgzleoonfrbeide(enC.z6eHn6e) (iii) 1.11% (iv) 2%H:\\Full Marks\\Till\\(EG) Chemistry-12\\EG_C-12_Ch-2\\EG_C-12_Ch-2 (i) 12.27% Reader’s Sig(ni_i_)__1__5__._2__7__%__________ Date __________ \\ 22-Jul-2022     Rajat Aggarwal     Proof-3 11 3. What is the concentration of glucose in mol L–1, if (iii) 22% (iv) 14.4% 12 Chemistry-12 ✎ Each chapter is divided into topics and each C. Fuel cell (3) rechargeable concept is dealt separately. (4) reaction at anode, Zn Æ Zn2+ + 2e– D. Mercury cell ✎ Flowchart representation of the chapter. (ii) A (1) B (4) C (3) D (2) Code: (iv) A (4) B (1) C (2) D (3) (i) A (4) B (3) C (2) D (1) Answers (iii) A (3) B (4) C (1) D (2) 1. (ii) 2. (i) H:\\Full Marks\\Till\\(EG) Chemistry-12\\EG_C-12_Ch-1\\EG_C-12_Ch-1 Quick revision notes • Element with lowest value of reduction potential \\ 22-Jul-2022     Rajat Aggarwal     Proof-3 • An electrochemical cell is a cell which either Reader’s Sign _______________________ Date __________ produces electric current as a result of redox is the best reducing agent whereas the one with reaction or facilitate a chemical reaction (redox higher value of reduction potential is the best reaction) through the introduction of electrical oxidising agent. energy. • If the given electrode when coupled with SHE • Electrolyte is an important component of (Standard hydrogen electrode) acts as anode, ExErcisE 1.2 electrochemical cell. we give negative sign to its standard reduction Multiple Choice Questions (MCQs) (1 Mark) (iii) Powdered sugar in cold water. • In galvanic cell, the chemical energy of a potential, e.g. E0Zn2+/Zn = – 0.76 V. spontaneous redox reaction is converted into • If the given electrode acts as cathode when 1. When 1 mole of benzene is mixed with 1 mole of (iv) Powdered sugar in hot water. electrical energy. toluene. The vapour will contain: coupled with SHE, then we give positive sign to • In electrolytic cell, electrical energy is used to its reduction potential, e.g. E0Cu2+/Cu = + 0.34 V. (Given : vapour of benzene = 12.8 kPa and vapour 7. At equilibrium the rate of dissolution of a solid carry out a non-spontaneous redox reaction. • In the cell representation, anode is on the left pressure of toluene = 3.85 kPa). solute in a volatile liquid solvent is .................... . • The standard electrode potential of any electrode hand side and cathode is on the RHS. (NCERT Exemplar) dipped in a suitable solution is defined with respect to the standard electrode potential of • Pt is an inert electrode used in SHE and as an (i) equal amount of benzene and toluene as it (i) less than the rate of crystallisation hydrogen electrode, taken as zero. electrode in solution containing ions only, e.g. forms an ideal solution Pt(s)/Fe3+/Fe2+. (ii) greater than the rate of crystallisation • The standard potential of cell can be obtained (ii) unequal amount of benzene and toluene as by difference in standard reduction potential of • The conductivity (k) of an electrolytic solution it forms a non ideal solution (iii) equal to the rate of crystallisation cathode and anode, depends upon the concentration of electrolyte, nature of solvent and temperature. (iii) higher percentage of benzene (iv) zero (iv) higher percentage of toluene 8. A beaker contains a solution of substance ‘A’. • E0 = E0 – E0 where SRP stands for • wMhoelnar‘Cc’oins dmuoclatirvitiytyin(Lmmo)lisL–d1eafinnded‘ka’ sin1S00cm0 k–1/C. Precipitation of substance ‘A’ takes place when cell cathode anode (SRP) (SRP) [CBSE S.P. 2020-21] small amount of ‘A’ is added to the solution. The standard reduction potential. • Conductivity decreases but Lm increases with decrease in concentration. 2. What type of liquids form ideal solutions? solution is (NCERT Exemplar) • Cathode is +ve and anode is –ve in Galvanic cell (i) Liquids with different structure but same (i) saturated (ii) supersaturated but in an electrolytic cell anode is +ve, cathode polarity ✎ Provides a complete comprehensive summary• The standard potential of the cells and equilibrium (iii) unsaturated (iv) concentrated is –ve. • oLfmsitnrocrnegaesleescstlroowlyltyews iwthhidleecirnecarseeaisnescovnecreynsttreaetpiolny (ii) Liquids with similar structure but different 9. Maximum amount of a solid solute that can be for weak electrolytes in very dilute solution. polarity dissolved in a specified amount of a given liquid of the chapter along with important definitionsDrG0 = – nE0cellF Dr G0 = – 2.303 RT log Kc and facts.• Concentration dependence of the potential of constant are related to standard Gibbs energy as: • Kohlrausch found that molar conductivity at (iii) Liquids with similar structure and polarity. solvent does not depend upon .................... . infinite dilution for an electrolyte is the sum of (iv) Liquids with different structure and polarity. (NCERT Exemplar) the contribution of the cation as well anions. It 3. What type of intermolecular attractive forces exist (i) Temperature (ii) Nature of solute the electrodes and the cells are given by Nernst is known as law of independent migration of ions among the pair of methanol and acetone forming a solution? equation as: and has many applications. (iii) Pressure (iv) Nature of solvent For a general reaction, aA + bB ne−→ cC + dD • Electrolysis is a process of breaking down of an (i) Dipole-dipole interactions 10. Low concentration of oxygen in the blood and electrolyte into simpler substances by passing tissues of people living at high altitude is due to (ii) Ion-dipole interation Ecell = E0cell – RT ln [C]c [D]d electricity through molten state or in aqueous .................... . (NCERT Exemplar) nF [A]a [B]b solution. (iii) Ion-ion interactions (i) low temperature • The quantitative aspects of electrolysis are (iv) Covalent bonding (ii) low atmospheric pressure • Electrochemical series is a series of elements governed by Faradays laws of electrolysis. 4. 10 mL of a liquid A was mixed with 10 mL of liquid (iii) high atmospheric pressure in which elements are arranged in increasing • The products of electrolysis are decided by B. The volume of resulting solution was found to be 18.9 mL. What do you conclude? or decreasing order of their standard reduction preferential discharge theory as well as reduction (iv) both low temperature and high atmospheric potential. potentials of electrodes. (i) Resulting solution is an ideal solution. pressure 70 Chemistry-12 (ii) Resulting solution shows positive deviation 11. Considering the formation, breaking and strength from Raoult’s law. of hydrogen bond, predict which of the following (iii) Resulting solution shows positive deviation mixtures will show a positive deviation from from ideal behaviour. Raoult’s law? (NCERT Exemplar) (iv) Resulting solution shows negative deviation (i) Methanol and acetone. from Raoult’s law. (ii) Chloroform and acetone. 5. A and B are liquids which on mixing produce a warm solution. Which type of behavior is shown (iii) Nitric acid and water. ✎ Exercise to each topic has been dealt separately by the solution? and important NCERT Textual and NCERT Exemplar (iv) Phenol and aniline. Questions included, segregated into 1 Mark, 2 Marks, 3 Marks and 5 Marks Questions. Also included HOTS (i) Resulting solution is an ideal solution. 12. The value of Henry’s constant KH is .................. . Questions that test the mental ability of the learner. (NCERT Exemplar) (ii) Resulting solution shows positive deviation from Raoult’s law. (i) greater for gases with higher solubility. (iii) Resulting solution shows negative deviation (ii) greater for gases with lower solubility. from Raoult’s law. (iii) constant for all gases. (iv) None of these (iv) not related to the solubility of gases 6. On dissolving sugar in water at room temperature 13. On the basis of information given below mark the solution feels cool to touch. Under which of the correct option. following cases dissolution of sugar will be most Information: rapid? (NCERT Exemplar) (A) In bromoethane and chloroethane mixture (i) Sugar crystals in cold water. intermolecular interactions of A–A and B–B (ii) Sugar crystals in hot water. type are nearly same as A–B type interactions. SolutionS 19 (viii)

H:\\Full Marks\\Till\\(EG) Chemistry-12\\EG_C-12_Ch-1\\EG_C-12_Ch-1 Reader’s Sign _______________________ Date __________ H:\\Full Marks\\Till\\(EG) Chemistry-12\\EG_C-12_Ch-1\\EG_C-12_Ch-1 Reader’s Sign _______________________ Date __________ \\ 22-Jul-2022     Rajat Aggarwal     Proof-3 \\ 22-Jul-2022     Rajat Aggarwal     Proof-3 Important Formulae Revision ChaRt 1. Mass percentage of a component (w/w) = Mass of the component in solution × 100 Total mass of the solution 2. Volume percentage of component (v/v) = Volume of the component × 100 Total volume of solution 3. Mole fraction of a component (x) = Number of moles of the component Total number of moles of all the components Henry’s law: The partial 4. Parts per million = Total Number of parts of component of solution × 106 pressure of gas dissolved in a Ideal solutions follow There are 9 types of solution: number of the parts of all the components Raoult’s law liquid is directly proportional to (i) Solid in solid, (ii) Solid in liquid, (iii) Liquid in 5. Molarity = Number of moles of solute = WB × 1000 liquid, (iv) Solid in gas, (v) Gas in gas, (vi) Liquid Volume of solution (in litres) mB × V ]in mLg its mole fraction. in solid, (vii) Liquid in gas, (viii) Gas in solid, (ix) Gas in liquid. 6. Molality = Number of moles of solute = WB × 1000 Raoult’s Law: The vapour Solution iS a HomogeneouS Molarity is the no. of moles of solute per Mass of solvent (in kilograms) mB × WA ^in gramsh pressure of each component in mixture of two or more litre of solution. a solution is directly proportional 7. Normality = Number of gram equivalent of solute to its mole fraction. SubStanceS, e.g. Salt Solution Molality is the no. of moles of solute per Volume of solution (in litres) Kg of solvent. 8. Molarity (M) = xB × density of solution # 10 Mole fraction is the ratio of no. of moles mB of a component to the total no. of moles of all the components. 9. Molality (m) = xB × 1000 Colligative property depends Non-ideal solution do not ^1–xBhmA upon the number of particles follow Raoult’s law of solute. 10. Henry’s Law, P = KHx 11. According to Raoult’s law–‘For a solution of volatile liquids the partial vapour pressure of each component Positive deviation Negative deviation in the solution in directly proportional to its mole fraction’, i.e. DH = + ve, DV = + ve DH = –ve, DV = –ve Minimum boiling azeotropes Maximum boiling azeotropes P1 = P10 x1; P2 = P20 x2 12. Using Dalton’s law of partial pressures the total pressure of solution is calculated. + Ptotal = P0 x1 P0 x2 1 2 13. Relative lowering of vapour pressure P10 – P1 = x2 P10 14. ∆T = Tb – T 0 ; ∆Tb = Kb × 1000 × w2 b M2 × w1 15. ∆T = T 0 – Tf ; ∆Tf = Kf × 1000 × w2 Depression in freezing point is directly Relative lowering of vapour pressure f M2 × w1 ✎ Important formulae, Important reactions in the16. Osmotic Pressure, p = CRT = is equal to the mole fraction of solute. n proportional to molality V RT DTf = i × Kf × w2 # 1000 P 0 –P1 =i# w2 # 1000 M2 # w1 1 M 2 # w1 chapterM2 = are provided at one place for recapitulation. P0 w2 RT 1 πV 17. Osmotic pressure is the extra pressure 18. van’t HH\\:2\\oF2u-fJlulfMl-2fa0ar2k2sc\\ Tti lol\\(r EG, ) iC h=eRmajiasttArAyg-N1bg2a\\rnoEwGrao_l mCr-1m 2a_ Cla hm-l1 \\mEPoGr_oloCoa-fl1-ar32_rCmhm-1aassss = Observed colligative property that needs to be applied on the solution Elevation in boiling point is directly Calculated colligative propertyReader’s Sign _______________________ Date __________ side to stop osmosis. Colligative ProPerty proportional to molality. Kf × w2 # 1000 = Total number of moles of particles after association/dissociation p=i× n2 RT DTb = i × M2 # w1 Total number of moles of particles before association/dissociation V 19. Inclusion of van’t Hoff factor modified the equations for colligative properties as: Observed colligative property Association, i < 1 Calculated colligative property Dissociation, i > 1 P10 – P1 = i. n2 Common Errors van’t Hoff factor = P10 n1 (i) StuDdTebnt=s usiu×alKlyEbrg×Mre1to20cr×0os0wnf×1uswe2d between w/w%, CorrECtions Normal molecular mass (i) Students have to understand clearly the i= Observed molecular mass (ii) wM/avDn%yTafsn=tuddvei/nv×%tsK.mf i×Msi1n20t×0er0wp×1rewt 2the value of ‘i’ for difference between these concepts. SolutionS 45 aDsisroeccipatti=ioonn/ido×ifsnsoo2sVcRmiaTtoisoni sfofrr a solute. (ii) i > 1 (dissociation) and i < 1 (association). om hyp ✎ Have the complete essence of the chapter; Quite (iii) o t o n i c to (iii) In both cases, solvent will flow from its higher effective for a quick revision before exams. hypertonic solution not clearly understood by concentration to its lower concentration many students. solution. (iv) Concept of osmosis and reverse osmosis is not (iv) Conditions need to be ScolelaurtliyonleSarn4t3 for clear to many students. understanding the occurrence of these phenomena. (v) Students do not convert mg into g while solving (v) Mass should be taken in grams. numericals. (vi) Students change DTf or DTb in °C to K (vi) DTf or DTb remains the same in °C or K. H:\\Full Marks\\Till\\(EG) Chemistry-12\\EG_C-12_Ch-1\\EG_C-12_Ch-1 Reader’s Sign _______________________ Date __________ (vii) Students do not convert unit of volume while (vii) Volume should be taken in litres if ‘p’ is in bar. \\ 22-Jul-2022     Rajat Aggarwal     Proof-3 solving numericals. If p is in Pa, volume should be taken in m3. (viii) Students do not consider van’t Hoff factor ‘i’ (viii) ‘i’ must be used in case of electrolytes and in case of electrolyte and solutes undergoing solutes undergoing association/dissociation. association/dissociation. Assignment Time: 45 Minutes M.M.: 25 Multiple Choice Questions (MCQs) (1 Mark) ✎ Common errors and corrections have been tagged 1. Which of the following aqueous solutions should have the highest boiling point? to clear the confusion with cautions answers for (i) 1.0 M NaOH (ii) 1.0 M Na2SO4 (iii) 1.0 M NH4NO3(iv) 1.0 M KNO3 2. We have three aqueous solutions of NaCl labelled as ‘A’, ‘B’ and ‘C’ with concentrations 0.1M, 0.01M and 0.001M, respectively. The value of van’t Hoff factor for these solutions will be in the order .................... . productive learning. (i) iA < iB < iC (ii) iA > iB > iC (iii) iA = iB = iC (iv) iA < iB > iC H:\\Full Marks\\Till\\(EG) Chemistry-12\\EG_C-12_Ch-1\\EG_C-12_Ch-1 Assertion Reason Type Questions (1 Mark) \\ 22-Jul-2022     Rajat Aggarwal     Proof-3 Reader’s Sign _______________________ Date __________ In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (i) Assertion and reason both are correct statements and reason is correct explanation for Assertion. (ii) Assertion and reason both are correct statements but reason is not correct explanation for Assertion. Chapter trend—Based on past Years’ CBSe exams (iii) Assertion is correct statement but reason is wrong statement. (iv) Assertion is wrong statement but reason is correct statement. ➣ It has been observed from this chapter that the weightage of topics ‘Ideal and Non-ideal Solutions’ and ‘Colligative 3. Assertion: Volume percentage of a solute is volume of solute divided by volume of solvent, multiplied by 100. P0 –P1 n2 Reason: Solvent is the substance present in excess in a solution Properties’ (especially, numericals based on colligative properties i.e. 1 = n1 + n2 ; DTb = Kbm; DTf = Kf m; p = CRT) are maximum. Hence, these are most important topics. P0 4. Assertion: On rise in temperature, the solubility of gas in a liquid decreases. 1 Reason: Dissolution of gas in a liquid is an endothermic reaction. ➣ From this chapter , generally 3 marks questions were asked from the topic ‘Colligative Properties’. 5. Assertion: When blood cells are placed in pure water, they swell up due to osmosis. Reason: Reverse osmosis is the process used in desalination of sea water. ➣ Most of the 1, 2 and 3 marks questions from this chapter belong to the topic ‘Colligative Properties’. Very Short Answer Type Questions (1 Mark) 44 Chemistry-12 Questions For practice 6. State Henry’s law. 7. Define azeotropes. Very Short Answer Type Questions (1 Mark) Short Answer Type Questions-I (2 Marks) 1. Give an example of maximum boiling azeotrope. 8. Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. 2. What type of intermolecular interaction exist between methanol and acetone? [Delhi 2014] 142 g mol–1) was dissolved in 50 g Cyclohexane, KCl, CH3OH, CH3CN. 3. oCfawlcautleart,eatshseumboiinlignNg pa2oSinOt4ouf nsodleurtgiooenswcohmenpl2etgeoifoNnias2aStiOon4 .(m(koblaforrmHa2sOs = 0.52 K kg mol–1) 9. (i) On mixing liquid X and Y, the volume of resulting solution decreases. What type of deviation from [AI 2016] Raoult’s law is shown by the resulting solution? What change in temperature would you observe? Short Answer Type Questions-I (2 Marks) (ii) How can the direction of osmosis be reversed? 4. Vapour pressure of water as 20°C is 17.5 mm Hg. Calculate the vapour pressure of water at 20°C when 15 g of glucose (Molar mass 180 g mol–1) is dissolved in 150g of water. [CBSE 2015] Short Answer Type Questions-II (3 Marks) 5. Calculate the boiling point of 1 molar solution of KCl (molar mass 74.5 g mol–1). The density of solution is 10. A solution is prepared by dissolving 5 g of non-volatile solute in 95 g of water. It has vapour pressure 23.375 1.04 g cm–3 and kb for H2O is 0.52 K kg mol–1. mm Hg at 250°C. Calculate the molar mass of solute (pA0 = 23.75 mm Hg). 11. Determine the osmotic pressure of a solution prepared by dissolving 2.5 × 10–2 g of K2SO4 in 2 L solution 6. Calculate osmotic pressure of 0.1 M K4[Fe(CN)6] solution if it is 50% ionised at 27°C. at 25°, assuming it is completely ionised. Short Answer Type Questions-II (3 Marks) {R = 0.0821 L atm K–1 mL–1, K2SO4 molar mass = 174 g mol L–1} 12. A 10% solution of urea is isotonic with 20% solution of an unknown solute ‘X’ at the same temperature. 7. A 10% solution (by mass) of sucrose in water has freezing point of 269.15 K. Calculate the freezing point of 10% solution of glucose in water, if freezing point of pure water is 273.15 K. [Delhi 2017] Calculate molar mass of X. [Given: Molar mass of glucose = 180 g mol–1, Molar mass of sucrose = 342 g mol–1] Long Answer Type Questions (5 Marks) 8. Calculate the van’t Hoff factor in each of the following: 13. Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water? (i) 50% ionised K4[Fe(CN)6] (ii) 60% ionised CaCl2 (i) phenol (ii) toluene (iii) formic acid (iv) ethylene glycol (v) chloroform (iii) Benzoic acid which is 50% dimerised in benzene. 9. A solution of glucose (molar mass = 180 g mol–1) in water has a boiling point of 100.20°C. Calculate the Answers freezing point of the same solution. Molal constants for water, kf and kb are 1.86 K kg mol–1 and 0.512 K kg mol–1, respectively. [Delhi 2017] 2. (iii) iA = iB = iC 1. (ii) 1.0 M Na2SO4 5. (ii) 3. (iv) 10. fSroemezeinegthpyoliennteogf lwycaotle,rH-gOlyCcoHl2s–oluCtHio2nOiHs,–i1s5a°dCd. eWdhtaotyiosutrhCe abro’silirnagdipaotionrt along with 5 kg of water. If the 4. (iii) 10. 120 g mol–1 8. 60 g mol–1 of the solution? 9. 5.27 × 10–3 atm mol–1, and kf = 1.86 K kg mol–1 for H2O] [kb = 0.52 K kg [AI 2014] Answers SolutionS 47 1. CHCl3 and Acetone 2. H-bonding and dipole-dipole interaction 3. 373.44 K 6. p =7.47 bar 4. 17.325 mm Hg 5. 374.077 K ✎ Chapterwise Assignment appended for self evaluation. 9. –0.73 °C 7. 265.55 K 8. (i) i = 3, (ii) i = 2.2, (iii) i = 0.75 10. 377.19 K ✎ Questions from important topics has been given separately for better preparation. 46 Chemistry-12 (ix)

Contents 1. Solutions ....................................................................................................................... 11 2. Electrochemistry .......................................................................................................... 48 3. Chemical Kinetics ........................................................................................................ 75 4. The d- and f-Block Elements ..................................................................................... 114 5. Coordination Compounds .......................................................................................... 144 6. Haloalkanes and Haloarenes..................................................................................... 175 7. Alcohols, Phenols and Ethers .................................................................................... 216 8. Aldehydes, Ketones and Carboxylic Acids ................................................................ 263 9. Amines ........................................................................................................................ 304 10. Biomolecules ............................................................................................................... 339 • Sample Paper-1 (Solved)............................................................................................. 366 • Sample Paper-2 (Unsolved) ......................................................................................... 376 • Sample Paper-3 (Unsolved) ......................................................................................... 380 (x)

1 Solutions Topics covered 1.1 Solutions: Modes of Expressing Concentration of Solution 1.2 Ideal and Non-ideal Solutions 1.3 Colligative Properties C hapter map SOLUTIONS Types of solutions Concentration of solution Raoult's law Ideal solution Non-ideal solution Molarity, molality, Henry's law Positive deviation Percent by mass, Negative deviation mole fraction, ppm Colligative property Elevation in Depression in boiling point freezing point Relative lowering of Osmotic pressure vapour pressure Osmosis Abnormal molecular mass Reverse osmosis Isotonic solution Association van’t Hoff factor Dissociation Hypotonic solution Hypertonic solution Degree of association Degree of dissociation Topic 1. Solutions: Modes of Expressing Concentration of Solution • Solution: Solutions are homogeneous mixtures of (or the one that is dissolved in the solvent) is called two or more than two components. The component solute. that is present in larger quantity (or the one that dissolves another component) is known as solvent. • Binary Solution: If a solution is consisting of only two components, i.e. a single solute dissolved The component that is present is smaller quantity in a solvent, then it forms a binary solution. 11

Types of Binary Solution • Molarity: M = WB × 1000 ]in mLg mB solution Type of Solute Solvent Common Volume of solution examples • Molality: m = WB × WA 1000 mB ^in gramsh Gas Gas Mixture of oxygen and • Mole fraction: xA = nA = WA /mA Gaseous Liquid Gas nitrogen gases nA + nB WA /mA + WB /mB Solutions Chloroform for binary solution, xA + xB = 1 Solid Gas mixed with for ternary solution, xA + xB + xC = 1 nitrogen gas nA is no. of moles of ‘A’, nB is no. moles of ‘B’ WB = mass of solute, mB = molar mass of solute, Camphor in WA is mass of solvent, mA is molar mass of solvent nitrogen gas If both the components are in gas phase, Gas Liquid Oxygen dissolved in water Liquid Liquid Liquid yA = Mole fraction of component in gas phase Solutions Solid Liquid Ethanol dissolved Partial pressure of that component in water = Total vapour pressure Glucose dissolved in water yA + yB = 1 • Parts Per Million (ppm): ppm is the number Gas Solid Solution of hydrogen in of grams of a solute dissolved per million grams Solid Liquid Solid palladium of the solution or the number of cm3 of solute dissolved per million cm3 of the solution. Amalgam of Solutions mercury with Mass of the component sodium ppm = Mass of the solution × 106 Solid Solid Copper dissolved • Relationship of Molarity (M) and molality in gold (m) with mole fraction of solute (xB): • Mass percentage (w/w) : Mass % x × d × 10 ⇒ M= mB Mass of the component in the solution = Total mass of the solution × 100 (x = mass percentage of solute, d = density of • Volume percentage (v/v) : Volume % solution, mB = molar mass of solute) = Volume of the component × 100 ⇒ m= xB × 1000 Total volume of the solution (1 − xB ) mA • Mass by volume percentage (w/V): Mass by Mass of solute (xB = mole fraction of solute, volume % : Volume of solution × 100 mA = molar mass of solvent) ExErcisE 1.1 Multiple Choice Questions (MCQs) (1 Mark) 40 g of it is dissolved in enough water to make a final volume upto 2 L? [M.Wt. of glucose = 180 g mol–1] 1. 2.46 g of NaOH (molar mass = 40) is dissolved in water and the solution is made upto 100 cm3 (i) 1100 mol L–1 (ii) 0.22 mol L–1 in a volumetric flask. Calculate the molarity of solution. [Na = 23 u, O = 16 u, H = 1 u]. (iii) 0.44 mol L–1 (iv) 0.11 mol L–1 4. Calculate the molality of 2.5 g ethanoic acid (i) 0.615 (ii) 0.515 C(CHH3C3COOOOHH=)6i0ng 75 g o f benzene. [M.wt. of mol–1] (iii) 0.246 (iv) 1.626 (i) 0.515 mol kg–1 (ii) 0.556 mol kg–1 2. A solution is prepared by adding 2 g of glucose into 18 g of water. Calculate the mass percent (iii) 0.665 mol kg–1 (iv) 0.775 mol kg–1 of glucose in the solution. [M.Wt. of glucose = 180 g mol–1]. 5. iaCsnaddlcicsuaslorabltveoentdhtieentmr1aa2cs2hslgoproeifrdcceea(nrCtbCaogln4e)t,oeiftfbr2ae2cnhgzleoonfrbeide(enC.z6eHn6e) (i) 9% (ii) 10% (iii) 1.11% (iv) 2% (i) 12.27% (ii) 15.27% 3. What is the concentration of glucose in mol L–1, if (iii) 22% (iv) 14.4% 12 Chemistry-12

6. 4L of 0.02 M aqueous solution of NaCl was diluted Reason: Copper dissolved in gold is a solid-solid by adding one litre of water. The molarity of the solution. resultant solution is .................... . 15. Assertion: Volume percentage of a solute is volume of solute divided by volume of solvent, (NCERT Exemplar) multiplied by 100. Reason: Solvent is the substance present in (i) 0.004 (ii) 0.008 excess in a solution (iii) 0.012 (iv) 0.016 7. The molarity of a solution which contains 32.0 g Very Short Answer Type Questions (1 Mark) of methyl alcohol in 200 mL of aqueous solution is [At mass of C = 12 u, H = 1 u, O = 16 u] 16. Which of the two molality or molarity is better to express the concentration of solutions? Why? (i) 0.5 M (ii) 5 M 17. Under what conditions molarity = molality? (iii) 50 M (iv) 20 M Justify your answer. 8. Calculate the moles of methanol in 5 litre of Short Answer Type Questions-I (2 Marks) its 2 m solution. If density of the solution is 0.981 kg L–1. [Molar mass of methanol = 32 g mol–1] (i) 9.22 (ii) 4.66 18. gMmolLar–i1t.yWohf aHt 2wSiOll4bies 0.8 and its density is 1.06 the concentration in terms (iii) 18.4 (iv) 1.64 of molality and mole fraction. [Atomic masses of 9. A 6.9 m solution of KOH in water contains 30% H = 1 u, S = 32 u, O = 16 u] by mass of KOH. The density of KOH solution is [Atomic masses of K = 39 u, O = 16 u, H = 1 u] 19. Find out the molarity and molality of a 15% (i) 2.48 g mL–1 (ii) 1.28 g mL–1 solution of H2SO4 (density = 1.02 g cm–3). 20. (ii) 5.6 g mL–1 (iv) 3.1 g mL–1 Concentr acotendsisHteNdOo3f, a laboratory reagent generally 69% by mass of HwNhOic3h. 10. If the mole fraction of water in the solution Calculate the volume of this solution of sulphuric acid is 0.85. Then the molality of contains c1m9.–93)g[AoftoHmNicOm3. a(DsseenssoitfyHof=c1onuc,.NHN= O143 sulphuric acid is [M.wt. of H2SO4 = 98 g mol–1]. is 1.41 g (i) 8.5 (ii) 9.0 u, O = 16 u] (iii) 9.8 (iv) 1.5 21. Calculate the masses of cane sugar and water Assertion Reason Type Questions (1 Mark) required to prepare 250 g of 25% cane sugar In the following questions a statement of assertion solution. [M. Wt. of C12H22O11, Cane sugar = 342 followed by a statement of reason is given. Choose g mol–1] the correct answer out of the following choices. (i) Assertion and reason both are correct statements 22. Which one molarity or molality is dependent on and reason is correct explanation for Assertion. temperature and why? (ii) Assertion and reason both are correct statements but reason is not correct explanation for Assertion. 23. Calculate the amount of benzoic acid required for (iii) Assertion is correct statement but reason is wrong preparing 250 mL of 0.15 M solution in methanol. statement. [M.Wt. of benzoic acid = 122 g mol–1] [NCERT] (iv) Assertion is wrong statement but reason is correct statement. Short Answer Type Questions-II (3 Marks) 11. Assertion: Chemists prefer to refer the concentration of solutions in terms of molality. 24. Define molality and molarity. Reason: Molality does not change with temperature. [NCERT Exemplar] 12. Assertion: For a concentrated solution molarity can be equal to molality. 25. Define mole fraction. A solution contains 36.0 g Reason: Molarity can be equal to molality when wthaetmeroalenfdra4c6tigoentohfyelaaclhcochoomlp(Con2Hen5tOiHn t),hdeestoelrumtiionne. the density of the solution is 1 g cm–3 so that [M.Wt. of H2O = 18 g mol–1, C2H5OH = 46 g mol–1] 1000 mL of solution is equal to 1000 g or 1 kg of 26. A solution contains 25% water, 25% ethanol solvent. and 50% acetic acid by mass. Calculate the mole 13. Assertion: Molarity of a solution in liquid state fraction of each compound. [6M0.gWmt.ool–f1H] 2O = 18, changes with temperature. C2H5OH = 46, CH3COOH = Reason: The volume of a solution changes with 27. Calculate the molarity of the following solutions: change in temperature. (NCERT Exemplar) 14. Assertion: Binary solution has two solutes (i) 4 g NaOH dissolved in 200 mL of the aqueous dissolved in a solvent. solution. (ii) 5.3 gofaanqhuyedoruosussoNluat2iCoOn.3 is dissolved in 100 mL [ Atomic Masses of Na = 23, C = 12,O = 16 u] 28. A solution was prepared by mixing equal volume of 30% by mass of H2SO4 (density 1.218 g cm–3 solutions 13

aCnadlc7u0la%tebythmeamssoloaflHity2SoOf 4th(disensosliutyti1o.n6.10 g cm–3). (i) express this in percent by mass. 29. 214.2 g of sugar (molar mass 342 u) syrup contains (ii) determine molality of chloroform in the 34.2 g of sugar. Calculate (i) molality of the solution water sample. [Atomic masses of C = 12 u, H = 1 u, Cl = 35.5 u] [NCERT] and (ii) mole fraction of sugar in the syrup. Long Answer Type Questions (5 Marks) 30. A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. 33. What is the molality of sguclpmh–u3 rciconatcaiidni(nHg2S5O0%4) solution of density 1.20 Calculate the mass percentage of the resulting H2SO4 by mass. [H = 1 u, S = 32 u, O = 16 u] solution. [NCERT] 34. A solution of glucose in water is labelled as 10% 31. An antifreeze solution is prepared from 222.6 g w/w, what would be the molality and mole fraction Cofaelcthuylaletneethgleycmolo, l(aCli2tHy6Oof2)tahneds2o0lu0tigonof. water. If the of each component in the solution? If the density of the solution is 1.2 g mL–1, then what shall be density of the solution is 1.072 g mL–1, then what the molarity of the solution? [C = 12 u, H = 1 u, shall be the molarity of the solution? [C = 12 u, H O = 16 u] [NCERT] = 1 u, O = 16 u] [NCERT] 35. How many millilitres of 0.1 M HCl are required to 32. A sample of drinking water was found to be react completely with 1 g mixture of Na2CO3 and severely contaminated with chloroform, l(eCvHelClo3f) NaHCO3 containing equimolar amounts of both? supposed to be a ca rcinogen. The [Na = 23 u, C = 12 u, O = 16 u, H = 1 u] contamination was 15 ppm (by mass). Answers 1.1 1. (i) 2. (ii) 3. (iv) Density = 1.02 g cm–3 4. (ii) 5. (ii) 6. (iv) Molar mass of solute is 98 g mol–1 7. (ii) 8. (i) 9. (ii) So the molarity of the solution 10. (iii) 11. (i) 12. (iv) M= WB × 1000 mB # V 13. (i) 14. (iv) 15. (iv) 15 × 1.02 × 1000 98 # 100 16. Molality does not change with temperature = = 1.56 whereas molarity changes with temperature If WB = 15 g then WB = 85 g, then because of the volume changes with temperature. Molality can be calculated as: Therefore molality is better and is usually referred by chemists for expressing the concentration of m= WB × 1000 mB × WA solutions. 17. Molality can be equal to molarity, if the solution = 15 × 1000 = 1.8 is very dilute and its density is 1gcm–3 so that 98 × 85 20. 69% by mass of HHNNOO33 means 100 g of solution 1000mL of solution = 1000g or 1 kg of solvent. contains 69 g of 18. Mass of H2SO4 in 1 litre (WB) ⇒ 69 g of HNO3 + 31 g of water contains 69 g of = 0.8 × 98 = 78.4 g HNO3 Mass of 1 litre of solution 69 g 31 g 1.41 g cm–3 1 g cm–3 = 1000 × 1.06 = 1060 g ⇒ of HNO3 + Mass of solvent (WA) = 1060 – 78.4 = 981.6 of water contains 69 g of HNO3 contains According to formula 48.94 mL of HNO3 + 31 mL of water 69 g of HNO3 m= WB × 1000 = 78.4 × 1000 = 0.815 mB × WA 98 × 981.6 79.94 mL of solution contains 69 g of HNO3 Mole fraction of solute 79.94 # 19.9 69 (XA) = 981.6 = 54.53 So, mL 18 0.8 of solution contains 19.9 g of HNO3 54.53 + 0.8 = 0.014 fi Thus, volume of solution = 23.05 mL 19. % w = 15 21. Mass % of cane sugar = 25% W WB = 15 g, WA = 100 – 15 = 85 g Mass % = Mass of solute × 100 Mass of solution Mass of solution = 100 g 25 = Mass of cane sugar ×100 100 g 250 Volume of solution = 1.02 g cm−3 fi Mass of cane sugar 14 Chemistry-12

= 25 × 250 = 62.5 g Molecular mass of C2H5OH = 46 100 Mass of CH3COOH = 50 g \\ Mass of water will be Molecular mass of CH3COOH = 60 \\ Number of moles of : = 250 – 62.5 = 187.5 g 22. Molarity is dependent on temperature. Molarity Water = 25 = 1.38 of a solution is the number of moles of solute 18 dissolved per litre of the solution. The volume of solution changes with the change in temperature. C2H5OH = 25 = 0.543 46 Hence molarity changes with temperature. However, molality is independent of the changes CH3COOH = 50 = 0.833 in temperature, as it is the number of moles of 60 Hence, mole fraction of Water the solute dissolved per kg of solvent and mass 1.38 of solvent does not vary with temperature. = 1.38 + 0.543 + 0.833 = 0.501, 23. M= WB × 1000 Mole fraction of C2H5OH mB × V 0.543 Here, M = 0.15, mB = 122 and V = 250 mL = 1.38 + 0.543 + 0.833 = 0.197 \\ 0.15 = WB × 1000 Mole fraction of CH3COOH 122 × 250 ⇒ WB = 4.575 g 0.833 24. Molality is defined as the number of moles of a = 1.38 + 0.543 + 0.833 = 0.302 solute dissolved per kg of solvent. 27. (i) Molecular mass of NaOH = 23 + 16 + 1 = 40g mol–1 Molarity is defined as the number of moles of solute dissolved per litre of the solution. Number of gram moles present in 4g of NaOH 25. Mole fraction of a component in a solution is the ratio of its number of moles to the total number = 4 = 0.1 40 of moles of all the components present in the Volume of solution in litres solution. = 200 = 0.2 For example, when 2 moles of alcohol and 3 moles 1000 of water are added, then mole fraction of alcohol is 2/5 and water is 3/5. Molarity = Number of moles of solution Volume of solution (in litres) We know, mole fraction, nA = 0.1 = 0.5 M xA = nA + nB 0.2 (ii) Molecular mass of anhydrous Na2CO3 wcohnesrteituneAnatnsdAnaBnadrBe the number of the moles of = 2 × 23 + 12 + 16 × 3 = 106g respectively. Number of gram moles in 5.3 g of Na2CO3 Molar mass of water = 18 g mol–1 5.3 = 106 = 0.05 Molar mass of alcohol = 46 g mol–1 No. of moles of water Volume of solution (in litres) = 36 g = 2.0 mol = 100 = 0.1 18 g mol–1 1000 No. of moles of C2H5OH M= 0.05 = 0.5 M 46 g 0.1 = 46 g mol–1 = 1.0 mol 28. Suppose we have taken 100 mL solution of each Total no. of moles = 2.0 + 1.0 = 3.0 H2SO4. 2 Mass of 30% H2SO4 \\ Mole fraction of water = 3 = 0.67, and = 100 × 1.218 = 121.8 g Mole fraction of C2H5OH = 1 = 0.33 Mass of H2SO4 in 121.8 g of 30% H2SO4 solution 3 30 26. Firstly, we calculate the number moles of each = 100 × 121.8 = 36.54 component. Mass of water Let the total mass of the solution be 100 g = 121.8 – 36.54 = 85.26 g Mass of H2O = 25 g, Mass of 70% H2SO4 Molecular mass of water = 18 = 100 × 1.610 = 161.0 g Mass of C2H5OH = 25 g Mass of H2SO4 in 161 g of 70% H2SO4 solution solutions 15

= 70 × 161 = 112.7 g Volume of solution = Mass 100 Density Mass of water = 161 – 112.7 = 48.30 g = 422.6 = 394.21 mL Total mass of H2SO4 on mixing 1.072 = 36.54 + 112.7 = 149.24 g Molarity of the solution can be calculated by using Number of moles of H2SO4 the following relation: = 149.24 M= WB × 1000 98 mB × V Mass of water = 222.6 × 1000 = 9.1 M 62 × 394.21 = 85.26 + 48.30 = 133.56 g 32. (i) Let mass of water sample = 106 g = 133.56 kg Then, mass of chloroform = 15 g 1000 Mass percent of chloroform m= 149.24 × 1000 = 11.4 m \\ 98 133.56 = Mass of chloroform × 100 Total mass of solution 29. (i) Mass of sugar = 34.2 g 15 Number of moles of sugar = 106 × 100 = 34.2 = 0.1 = 1.5 × 10–3 % 342 Mass of water (ii) Molality, m = WB × 1000 ...(1) = 214.2 – 34.2 = 180 g where, MB WA 180 Number of moles of water = 18 = 10 WB = mass of solute = 15 g mB = molar mass of solute = 119.5 g Molality = 0.1 × 1000 = 0.555 m WA = mass of solvent = 106 g 180 Substituting these value in eqn. (1), we get (ii) Total number of moles = 10.0 + 0.1 = 10.1 Mole fraction of sugar = 0.1 = 0.0099 m= 15 × 100 = 1.25 × 10–5 10.1 119.5 × 106 30. Mass of solute in 300 g of solution 33. Mass of 1 cm3 H2SO4 solution = 1.20 g = 25 × 300 = 75 g Mass of 1 litre H2SO4 solution 100 = 1.20 × 1000 = 1200 g \\ Mass of solvent = 300 – 75 = 225 g Mass of H2SO4 in 100 g solution of H2SO4 = 50 g Similarly, mass of solute in 400 g of solution = 40 × 400 = 160 g Mass of H2SO4 in 1200 g solution of H2SO4 100 50 and Mass of solvent = 100 × 1200 = 600 g = 400 – 160 = 240 g \\ Mass of water in solution \\ Total mass of the solute = 1200 – 600 = 600 g = 75 + 160 = 235 g Molar mass of H2SO4 Total mass of solution = 98 g mol–1 = 300 g + 400 g = 700 g No. of moles of H2SO4 \\ Mass percentage of solute = Mass in grams = 600 g Molar mass 98 g mol–1 = Mass of solute × 100 \\ Molality Mass of solution = 235 × 100 = 33.6% = No. of moles of H2 SO4 × 1000 700 Mass of water (in grams) 31. Molality ‘m’ = WB × 1000 = 600 × 1 × 1000 = 10.20 m mB × WA 98 600 34. Molality of the solution can be calculated Here, WB = 222.6 g; mB = 62; WA = 200 g \\ Molality, as: m= WB × 1000 ...(1) mB × WA m= 222.6 × 1000 = 17.95 m 62 × 200 WB = 10 g (mass of solute); Total mass of solution WA = 90 g (mass of solvent ) mB = 180 (Molar mass of solute) = 222.6 + 200 = 422.6 g 16 Chemistry-12

m= 10 × 1000 = 0.612 x = 0.56 180 × 90 We know, Thus, number of moles of Na2CO3 xB × 1000 = 5.283 × 10–3 (1 – xB) mA m= ...(2) approximately number of moles of NaHCO3 = 5.283 × 10–3 where, xB = mole fraction of solute xB × 1000 During the process of neutralisation, following \\ 0.62 = (1 – xB) × 18 reactions will take place. ⇒ xB = 0.011 Na2CO3 + 2HCl Æ 2NaCl + H2O + CO2≠ xA = 0.989 NaHCO3 + HCl Æ NaCl + H2O + CO2≠ Number of moles of HCl required Molarity of solution can be calculated by using the following relation, = 2 × Number of moles of Na2CO3 + Number of moles of NaHCO3 M= x × d × 10 mB = 2 × 5.283 × 10–3 + 5.238 × 10–3 (x is mass % of glucose) = 0.0158 = 10 × 1.2 × 10 = 0.67 Molarity may be given as, 180 35. Let there be x g Na2CO3 and (1 – x) g of NaHCO3 nB × 1000 in the mixture. M= V Molar mass of Na2CO3 = 106 g/mol (nB = Number of moles of solute) nB × 1000 Molar mass of NaHCO3 = 84 g/mol \\ V= M Number of moles of Na2CO3 = Number of moles of NaHCO3 0.0158 × 1000 x (1 – x) = 0.1 106 84 \\ = = 158 mL of 0.1 HCl is required. Topic 2. Ideal and Non-ideal solutions Solubility of a gas in liquid: Gatenrmdeapvteiercraettuvhereers,Kah.HeKnvcHaelvtuahele,useloowilnuecbrrieliwatysilelosfbwaeigtthahseininscoraleualbisqieluitiindy (i) Increases with increase of pressure. decreases with increase of temperature. (ii) Decreases with increase of temperature. Raoult’s Law: Raoult’s law in its general form can Henry’s law: At constant temperature, the solubility be stated as: of a gas in a liquid is directly proportional to the pressure of the gas over the solution. For any solution, the partial vapour pressure of each volatile component in the solution is directly Or proportional to its mole fraction in the solution. The partial pressure of the gas in vapour phase (P) Component 1, P1 ∝ x1 ⇒ P1 = P10 x1 is directly proportional to the mole fraction of the gas Component 2, P2 ∝ x2 ⇒ P2 = P20 x2 (x) in the solution. cwfroahmcetprieoonnPe10onfattsnhdienPcio20tmsaprpeountreheneftovsraompf o,tuharendpsorxelu1stsaiuonrnde. of volatile x2 = mole According to Dalton’s law for a solution of two volatile liquids, Psoln = P10 x1 + P20 x2 When the non-volatile solute is dissolved in pure solvent, the vapour pressure of the pure solvent tdheecrneaosne-vs,olia.et.ilPesosluotliuonte< pPa0rsotlivcelnet.s This occurs because are present on the surface of the solution, so the rate of evaporation of molecules of the volatile solvent from the surface of the solution decreases. P = KHx KH = Henry’s law constant. Liquid-liquid volatile solutions are classified The value of KH depends upon the nature of gas. into two types on the basis of Raoult’s law : Ideal Solutions and Non-Ideal Solutions: solutions 17

Ideal Solution: The solution which obeys Raoult’s Non-ideal solutions showing negative deviations: law at all concentration and at all temperature, i.e. • In this case the intermolecular attractive forces between the solute and solvent particles are P1 = P10 x1 and P2 = P20 x2 stronger than those between the solvent – solvent and the solute – solute particles, i.e. = P1 + P2 III P0 P Total 2 Vapour pressure of solution II Vapour pressure P1 P2 0 P1 Vapour pressure P1 I P2 x1 = 1 Mole fraction x1 = 0 x2 = 0 x2 = 1 • If the intermolecular attractive forces between the Mole fraction xx12 solute – solvent particles are nearly equal to those xx21 = 0 xx21 = 1 = 1 = 0 between the solvent – solvent particles and solute – solute particles then, they form ideal solution. • P1 > P10 x1 • P2 > P20 x2 • Enthalpy of mixing, Dmixing H = 0, Volume change • Dmixing H = –ve on mixing, Dmixing V = 0. • Dmixing V = –ve • Dissolution is exothermic, so heating decreases • Examples: n-hexane and n-heptane. Bromoethane and chloroethane. Benzene and toluene. Non-Ideal Solution: The solution which does not obey the solubility. the Roult’s law over the entire range of concentration is • Examples: Chloroform and acetone, Nitric acid non-ideal solution. These type of solutions show either and water, Phenol and aniline. positive or negative deviation from ideal behavior. • Azeotropes are the binary solutions (liquid Non-ideal solutions showing positive deviations: mixtures) having the same composition in liquid Vapour pressure and vapour phase and it is not possible to separate of solution the components of an azeotrope by fractional distillation. Vapour pressure Types of azeotropes: • Minimum boiling azeotrope: ã The non-ideal solutions showing positive P1 P2 deviations form minimum boiling azeotrope at a specific composition. ã Example: 95% ethanol and 5% water (by Mole fraction volume). xx12 xx21 = 0 xx12 = 1 ã Ethanol with b.p. 351.3 K, Water with b.p. = 1 = 0 373 K forms azeotrope with 351.1 K as the • In this case, intermolecular attractive forces boiling point. between the solute and solvent particles are ã P1 > P10 x1 and P2 > P20 x2 • Maximum boiling azeotrope: weaker than those between the solvent – solvent and the solute – solute particles, i.e. ã The non-ideal solutions showing negative • P1 > P10 x1 deviations form maximum boiling azeotrope • P2 > P20 x2 • Dmixing H = +ve at a specific composition. • Dmixing V = +ve • Dissolution is endothermic, so heating increases ã Example: 68% nitric acid and 32% water (by mass). ã Nitric acid with b.p. 359 K, Water with b.p. the solubility. 373 K forms azeotrope with 393.5 K as the • Ethanol and acetone, Ethanol and water, CS2 and boiling point. acetone are a few examples. ã P1 > P10 x1 and P2 > P20 x2 18 Chemistry-12

ExErcisE 1.2 Multiple Choice Questions (MCQs) (1 Mark) (iii) Powdered sugar in cold water. 1. When 1 mole of benzene is mixed with 1 mole of (iv) Powdered sugar in hot water. toluene. The vapour will contain: 7. At equilibrium the rate of dissolution of a solid (Given : vapour of benzene = 12.8 kPa and vapour solute in a volatile liquid solvent is .................... . pressure of toluene = 3.85 kPa). (NCERT Exemplar) (i) equal amount of benzene and toluene as it (i) less than the rate of crystallisation forms an ideal solution (ii) greater than the rate of crystallisation (ii) unequal amount of benzene and toluene as it forms a non ideal solution (iii) equal to the rate of crystallisation (iii) higher percentage of benzene (iv) zero (iv) higher percentage of toluene 8. A beaker contains a solution of substance ‘A’. [CBSE S.P. 2020-21] Precipitation of substance ‘A’ takes place when 2. What type of liquids form ideal solutions? small amount of ‘A’ is added to the solution. The (i) Liquids with different structure but same polarity solution is (NCERT Exemplar) (i) saturated (ii) supersaturated (ii) Liquids with similar structure but different (iii) unsaturated (iv) concentrated polarity 9. Maximum amount of a solid solute that can be (iii) Liquids with similar structure and polarity. dissolved in a specified amount of a given liquid (iv) Liquids with different structure and polarity. solvent does not depend upon .................... . 3. What type of intermolecular attractive forces exist (NCERT Exemplar) among the pair of methanol and acetone forming a solution? (i) Temperature (ii) Nature of solute (iii) Pressure (iv) Nature of solvent (i) Dipole-dipole interactions 10. Low concentration of oxygen in the blood and tissues of people living at high altitude is due to (ii) Ion-dipole interation (iii) Ion-ion interactions .................... . (NCERT Exemplar) (iv) Covalent bonding (i) low temperature 4. 10 mL of a liquid A was mixed with 10 mL of liquid (ii) low atmospheric pressure B. The volume of resulting solution was found to be 18.9 mL. What do you conclude? (iii) high atmospheric pressure (iv) both low temperature and high atmospheric (i) Resulting solution is an ideal solution. pressure (ii) Resulting solution shows positive deviation 11. Considering the formation, breaking and strength from Raoult’s law. of hydrogen bond, predict which of the following (iii) Resulting solution shows positive deviation mixtures will show a positive deviation from from ideal behaviour. Raoult’s law? (NCERT Exemplar) (iv) Resulting solution shows negative deviation (i) Methanol and acetone. from Raoult’s law. (ii) Chloroform and acetone. 5. A and B are liquids which on mixing produce a warm solution. Which type of behavior is shown (iii) Nitric acid and water. by the solution? (iv) Phenol and aniline. (i) Resulting solution is an ideal solution. 12. The value of Henry’s constant KH is .................. . (ii) Resulting solution shows positive deviation (NCERT Exemplar) from Raoult’s law. (i) greater for gases with higher solubility. (iii) Resulting solution shows negative deviation (ii) greater for gases with lower solubility. from Raoult’s law. (iii) constant for all gases. (iv) None of these (iv) not related to the solubility of gases 6. On dissolving sugar in water at room temperature 13. On the basis of information given below mark the correct option. solution feels cool to touch. Under which of the following cases dissolution of sugar will be most Information: rapid? (NCERT Exemplar) (A) In bromoethane and chloroethane mixture (i) Sugar crystals in cold water. intermolecular interactions of A–A and B–B (ii) Sugar crystals in hot water. type are nearly same as A–B type interactions. solutions 19

(B) In ethanol and acetone mixture A–A or B–B (i) At specific composition methanol-acetone type intermolecular interactions are stronger mixture will form minimum boiling azeotrope than A–B type interactions. and will show positive deviation from Raoult’s law. (C) In chloroform and acetone mixture A–A or B–B type intermolecular interactions are (ii) At specific composition methanol-acetone weaker than A–B type interactions. mixture forms maximum boiling azeotrope (NCERT Exemplar) and will show positive deviation from Raoult’s law. (i) Solution (B) and (C) will follow Raoult’s law. (ii) Solution (A) will follow Raoult’s law. (iii) At specific composition methanol-acetone (iii) Solution (B) will show negative deviation mixture will form minimum boiling azeotrope and will show negative deviation from from Raoult’s law. Raoult’s law. (iv) Solution (C) will show positive deviation from (iv) At specific composition methanol-acetone Raoult’s law. mixture will form maximum boiling azeotrope 14. Two beakers of capacity 500 mL were taken. One and will show negative deviation from Raoult’s law. of these beakers, labelled as “A”, was filled with 400 mL water whereas the beaker labelled “B” was 17. aKrHev4a0l.u3e9,fo1r.6A7r,(1g.)8,3C×O120(–g5)a, HndC0H.4O13(gr)easnpdecCtiHve4(lgy). filled with 400 mL of 2 M solution of NaCl. At the Arrange these gases in the order of their same temperature both the beakers were placed increasing solubility. (NCERT Exemplar) in closed containers of same material and same capacity as shown in figure. (i) HCHO < CH4 < CO2 < Ar (ii) HCHO < CO2 < CH4 < Ar AB (iii) Ar < CO2 < CH4 < HCHO (iv) Ar < CH4 < CO2 < HCHO water NaCl solution Assertion Reason Type Questions (1 Mark) At a given temperature, which of the following In the following questions a statement of assertion statement is correct about the vapour pressure followed by a statement of reason is given. Choose of pure water and that of NaCl solution. the correct answer out of the following choices. (NCERT Exemplar) (i) Assertion and reason both are correct statements and reason is correct explanation (i) vapour pressure in container (A) is more than for Assertion. that in container (B). (ii) vapour pressure in container (A) is less than (ii) Assertion and reason both are correct that in container (B). statements but reason is not correct explanation for Assertion. (iii) vapour pressure is equal in both the containers. (iv) vapour pressure in container (B) is twice the (iii) Assertion is correct statement but reason is vapour pressure in container (A). wrong statement. 15. If two liquids A and B form minimum boiling (iv) Assertion is wrong statement but reason is azeotrope at some specific composition then correct statement. .................... . (NCERT Exemplar) 18. Assertion: Aquatic species are more comfortable in cold water rather than in warm water. (i) A–B interactions are stronger than those between A–A or B–B. Ratetahseosna:mDeiffteermenptergaatsuerseh. av[eCdBiSffEereSn.Pt .K2H0v2a0l-u2e1s] 19. Assertion: Nitric acid and water form maximum (ii) vapour pressure of solution increases because more number of molecules of liquids A and B boiling azeotrope. can escape from the solution. (iii) vapour pressure of solution decreases because Reason: Azeotropes are binary mixtures having less number of molecules of only one of the the same composition in liquid and vapour phase. liquids escape from the solution. [CBSE S.P. 2020-21] (iv) A–B interactions are weaker than those 20. Assertion: Two liquids, say X and Y boil at between A–A or B–B. 380 K and 400 K respectively, then liquid Y is more volatile. 16. On the basis of information given below mark the correct option. Reason: Liquid Y has higher boiling point thus it evaporates less readily. Information: On adding acetone to methanol some of the hydrogen bonds between methanol 21. Assertion: On rise in temperature, the solubility of gas in a liquid decreases. molecules break. (NCERT Exemplar) 20 Chemistry-12

Reason: Dissolution of gas in a liquid is an 36. The cutting of onions at low temperature is endothermic reaction. more comfortable than cutting of onions at room temperature. Why is it so? Very Short Answer Type Questions (1 Mark) 37. The vapour pressure of two liquids A and B are 22. A binary solution contains two components A and 120 mm and 180 mm Hg at a given temperature. B having the same mole fraction. Will their mole If 2 moles of A and 3 moles of B are mixed to form fraction in vapour phase be the same or different? an ideal solution, calculate the vapour pressure 23. How does pressure cooker reduces cooking time? of the solution at the same temperature. 24. When two liquids are mixed, the resulting solution 38. The vapour pressure of pure liquids M and N at is found to be cooler. What do you conclude about 20°C are 22 mm Hg and 75 mm Hg respectively. the deviation from ideal behaviour. A solution is prepared by mixing equal number of moles of M and N. Assuming the solution to be 25. Acetone and chloroform are mixed together with ideal, calculate vapour pressure of the solution. reduction in volume, what type of deviation will be observed on the basis of Raoult’s law? 39. The vapour pressure of water at 298 K is 19.8 mm Hg. 0.1 mol of glucose is dissolved in 178.2 g 26. What role does molecular interactions play in a of water. Calculate the vapour pressure of water solution of alcohol and water? [NCERT] for this resultant solution. 27. What is the significance of Henry’s law constant, kH? 40. The vapour pressure of water at room temperature is 23.8 mm Hg. What will be the vapour pressure of an aqueous solution of sucrose with mole Short Answer Type Questions-I (2 Marks) fraction equal to 0.12? 28. Give an example of a miscible liquid pair showing 41. What happens to the vapour pofreKsIs?uErexpiflaHing.I2 is positive deviations from the Raoult’s Law. Give added to an aqueous solution reason. 42. If a teaspoon full of salt is added to water, what will happen to the vapour pressure? Explain. 29. Henry’s law constant for CO2 dissolving in water is 1.67 × 108 Pa at 298 K. Calculate the quantity of 43. What type of liquids form azeotropic mixtures and CO2 in 500 mL of soda water when packed under why? 128.5gamtmolC–1O] 2 pressure at 298 K. [M.wt. of H2O = [NCERT] 44. The partial pressure of ethane over a solution containing 6.56 × 10–3 g of ethane is 1 bar. If the solution contains 5 × 10–2 g of ethane, then what 30. (i) Why an increase in temperature is observed shall be the partial pressure of the gas. on mixing chloroform and acetone? [NCERT] (ii) Why does an aqueous sodium chloride solution freezes at a lower temperature than Short Answer Type Questions-II (3 Marks) water? 45. tCOohfa2aklictsHuOblfauo2trbeebOxtl2ehedriesttss3ho4raluo.6ubp8gialhkirttbywiaaoarlft.epOrr2eaistns2ug9r3Le K–o1..f Assuming 0.98 bar. 31. It has been found that one litre sample of sea The value water contains 5.8 × 10–3 g of dissolved O2. Find out the concentration of dissolved O2 in sea water in ppm. (Density of sea water = 1.03 g cm–3) [M.wt. 46. Two liquids A and B on mixing form an of H2O = 18 g mol–1] ideal solution. The vapour pressure of a solution 32. What concentration of nitrogen should be present containing 3 moles of A and 1 mole of B is 550 mm Hg. However, when 4 moles of A and in a glass of water at room temperature? Assume 1 mole of B are mixed, the vapour pressure of the temperature as 25°C, a total pressure of 1 solution thus formed is 560 mm Hg. What will atmosphere and mole fraction of nitrogen in air be the vapour pressure of pure A and B at this is 0.78, (kH for nitrogen = 8.42 × 10–7 M/mm Hg.) temperature? [AI 2009] 47. Vapour pressure of pure water is 40 mm Hg. If a non-volatile solute is added to it then the vapour 33. In an experiment 27 g of acetylene (C2H2) is pressure falls by 4 mm Hg. Calculate the molality dissolved in 1 L of acetone at 1.0 atm pressure. If the partial pressure of acetylene (C2H2) is of the solution. increased upto 12 atm, what is its solubility in 48. A 298 K the vapour pressure of water is 23.75 acetone (CH3COCH3)? mm Hg. Calculate the vapour pressure at the 34. State Henry’s law and list the conditions necessary same temperature for 5% aqueous solution of urea (NH2CONH2). for the validity of Henry’s law. 49. iCnamlcualkaitnegth2e.5mkagsosfo0f .u2r5eam(oNlaHl2aCqOuNeoHu2s) required solution. 35. What type of liquid pairs show (i) Positive deviation, (ii) Negative deviation from Raoult’s [NCERT] law? solutions 21

50. Calculate (a) molality, (b) molarity, (c) mole total pressure of the solution. [Given: PKC].2H5OH = 51 mm Hg, PCH3OH = 90 mm Hg at 300 fraction of KI, if the density of 20% (mass/mass) aqueous KI solution is 1.202 g mL–1. [NCERT] 55. If N2 gas is bubbled through water at 293 K, 1 litre 51. The vapour pressure of water is 12.3 kPa at 300 K. of how many millimoles of N2 gas would dissolve in Calculate the vapour pressure of 1 molal solution water? Assume that N2 exerts a partial pressure of a non-volatile solute. of 0.987 bar. Given that Henry’s Law constant for 52. Calculate the mass of a non-volatile solute (molar N2 at 293 K is 76.48 kbar. [NCERT] mass = 40 g mol–1) which should be dissolved in 56. Vapour pressure of CHCl3 and CH2Cl2 at 298 K 114 g of octane to reduce its vapour pressure to are 200 mm Hg and 415 mm Hg respectively. 80%. [NCERT] Calculate (i) Vapour pressure of solution prepared 53. (i) Explain why on addition of 1 mol of glucose by mixing 25.5 g of CHCl3 and 40 g of CH2Cl2 at to 1 litre of water, the boiling point of water 298 K, (ii) mole fraction of each component in increases. vapour phase. [NCERT] (ii) Henry’s law constant cfaolrcuClaOt2e in water is 57. What is meant by positive and negative deviations 1.67 × 108 Pa at 298 K, the number from Raoult’s law and how is the sign of Dmix H of moles of CuOnd2eirn2.55030×m10L5 of soda water related to positive and negative deviations from when placed Pa at the same Raoult’s law. [NCERT] temperature. [Delhi 2017(C)] 58. Heptane and octane form an ideal solution. At Long Answer Type Questions (5 Marks) 373 K, the vapour pressure of two liquid components 54. A nearly ideal solution is prepared by mixing are 105.2 kPa and 46.8 kPa, respectively. What 32 g methanol and 23 g ethanol at 300 K. Calculate the partial pressure of its constituents and will be the vapour pressure of a mixture of 26 g of heptane and 35 g of octane? [NCERT] Answers 1.2 1. (iii) 2. (iii) 3. (i) closer to each other and hence reduction in volume 4. (iv) 5. (iii) 6. (iv) ntaekgeastivpeladceev,iai.tei.onDVfrsoomlutitohn e<id0eaolrbtehhearveiowuril.l be 7. (iii) 8. (ii) 9. (iii) 10. (ii) 11. (i) 12. (ii) 26. Alcohols dissolve in water due to the formation of 13. (ii) 14. (i) 15. (i) intermolecular hydrogen bond. But on increasing 16. (i) 17. (iii) 18. (ii) the length of carbon chain of alcohol the attractive 19. (ii) 20. (iv) 21. (iii) force becomes weak thus, the mixture shows 22. The relation between Raoult’s law and Dalton’s positive deviations from ideal behaviour. law: 27. Higher tihnethvaelluiqeuoifdk. H, lower will be solubility of the gas P 0 × xA = yA × P ...(a) A 28. An example of a miscible liquid pair showing positive deviations from the Raoult’s law is a PB0 × xB = yB × P ...(b) wcohmeproenexnAtsaAndanxdBBairnethtehseolmutoiolen, fractions of mixture of ethyl alcohol and cyclohexane. the mole fraction of components A aynA danBd iynB are the H C2H5 — vapour phase. Dividing equation a by equation b, — — — we get: C2H5—O-----H—O-----H—O-----H—O----- P 0 × xA = yA C2H5 C2H5 A yB When cyclohexane is added to ethyl alcohol, the PB0 × xB molecules of cyclohexane try to occupy the spaces So mole fraction in the vapour phase will not be the same because it depends upon the vapour in between ethyl alcohol molecules. Consequently pressure of the pure components also. some of the hydrogen bond in between alcohol 23. In a pressure cooker the boiling point of solution in molecules breaks up and attractive forces water increases. So the heat content is increased. Higher is heat content of water, lesser will be the between the molecules get weakened. The escaping tendency of the molecules of ethanol cooking time. and cyclohexane from the solution increases, 24. Since heat is absorbed, therefore DH has +ve due to which the vapour pressure of the solution value. So there will be positive deviation from the ideal behaviour. increases. 25. Due to the formation of intermolecular hydrogen 29. KH = 1.67 × 108 Pa bonding, the molecules of two constituents come P = 2.5 atm = 2.5 × 1.01325 × 105 Pa 22 Chemistry-12

P = KHx2 2.533 × 105 Pa m2 = ?, p2 = 12 atm 1.67 × 108 Pa By putting the value in formula x2 = P = KH 27 1 = m2 = 27 × 12 = 324 g L–1 m2 = 12 = 1.517 × 10–3 Moles of water 34. Henry’s law: The mass of a gas dissolved in a solvent is directly proportional to its partial = 500 = 27.78 18 pressure. x2 = n2 = n2 The essential conditions are: n1 + n2 27.78 + n2 (i) Pressure should not be too high. 1.517 × 10–3 (27.78 + n2) = n2 (ii) Temperature should not be too low. ⇒ n2 = 0.0419 + 0.0015 n2 (iii) The gas should not associate or dissociate. 0.9985 n2 = 0.0419 35. Molecules for which A – B liquid pairs interactions n2 = 0.0420 moles are: \\ Amount of CO2 dissolved (i) Weaker than A – A and B – B interactions = 0.0420 × 44 = 1.85 g show positive deviation. 30. (i) When chloroform and acetone are mixed, the (ii) Stronger than A – A and B – B interactions new interactions between chloroform and show negative deviation. acetone molecules will be stronger than in between the pure component molecules, i.e. 36. At low temperature the vapour pressure is low, DH is negative, so heat is evolved and hence so the onions will not produce too much vapours. the escaping tendency of the molecules from At room temperature vapour pressure is high, the mixture decreases. so more vapours are produced, bringing tears on cutting onions. (ii) On adding a non-volatile solute like NaCl to water, the vapour pressure of solution is 37. Given: P 0 = 120 mm Hg, lowered. The vapour pressure of this solution A becomes equal to that of pure solid solvent at lower temperature than the freezing point of PB0 = 180 mm Hg, nA = 2 mol, nB = 3 mol pure solid solvent. This is the freezing point of the solution, so freezing point of NaCl xA = nA = 2 , xB= 3 solution is lower than that of pure solvent. nA + nB 5 5 P = PA + PB 2 3 5 5 PA0 xA + PB0 xB = 120 × + 180 × 31. Mass of sea water = 48 + 108 = 156 mm Hg = Volume × Density 38. As you know equal number of moles of both liquids = 1000 × 1.03 = 1030 g M and N are mixed, Concentration in ppm 1 2 = MMaassssoof fsosolulutitoen0×1 6 xM = xN = = 5.8 × 10–3 × 106 = 5.63 ppm \\ P = PM + PN = PM0 xM + PN0 xN 1030 32. Partial pressure of nitrogen in air can be = 22 × 1 + 75 × 1 2 2 = 48.5 mm Hg calculated as: 39. Given: nB = 0.1 PN2 = 0xN.728××P760 178.2 = = 592.8 mm Hg 18 nA = = 9.9 Molarity of N2 in a glass of water = 8K.H42××P1N02 –7 xA = nA = 9.9 = 0.99 = nA + nB 9.9 + 0.1 M/mm Hg × 592.8 We know = 4.99 × 10–4 M PA = PA0xA = 19.8 × 0.99 33. According to Henry’s Law = 19.6 mm Hg m1 p1 m2 = p2 40. Given: xB = 0.12 xA = 1 – xB = 1 – 0.12 = 0.88 where m1 is the solubility of the gas at pressure hp1eiagnhdt mat2 As per Raoult’s law is the solubility at pressure, p2 over the PA = P0xA equilibrium. = 23.8 × 0.88 = 20.94 mm Hg. Given: m1 = 27 g L–1, p1 = 1 atm, solutions 23

41. Vapour pressure decreases: KI and HgI2 react 550 = 3 P 0 + 1 PB0 together to form a complex 4 A 4 2KI + HgI2 K2 HgI4 2200 = 3P 0 + 1PB0 ...(a) A Total vapour pressure of the solution (4 moles of Therefore vapour pressure of HgI2 will decrease. A + 1 mole of B) 42. Vapour pressure decreases. It is due to the fact that addition of non-volatile solute molecules 4 1 decreases the surface area of the solvent molecules 560 = 5 P 0 + 5 PB0 A which is vapourising. 2800 = 4P 0 + PB0 ...(b) A Solving (a) and (b) 43. Two types of liquid mixtures form azeotropic 0 mixtures which distil at a particular temperature P A = 600 mm Hg such as pure liquids. = Vapour pressure of pure A (i) Minimum boiling azeotropes: Mixtures PB0 = 400 mm Hg which have mole fraction corresponding to = Vapour pressure of pure B the maximum vapour pressure or minimum boiling point or we can say that those 47. P0 =40 mm Hg, P = 36 mm Hg According to Raoult’s law solutions which show positive deviations from the Raoult’s law. P = P0xA, 36 = 40 × xA ⇒ xA = 0.9 xB ×1000 0.1 × 1000 (ii) Maximum boiling azeotropes: Mixtures m= xA × mA = 0.9 × 18 which have mole fraction corresponding to the minimum vapour pressure or maximum = 6.17m boiling point, i.e. mixtures which show 48. In the solution, 5 g of urea is present and 95 g of water is there. negative deviations from the Raoult’s law. Number of moles of urea will be 44. If solubility of a gas at 1 bar pressure is known, then it can be calculated at other pressure, using = 5 = 0.083 the following relation: 60 Number of moles of water S1 P1 S2 = P2 , = 98 = 5.445 18 wanhderPe2,Sr1esapnedctSiv2ealyre solubilities at pressure P1 Mole fraction of solvent 6.56 × 10–3 1 xA = 5.445 = 0.985 5 × 10–2 P2 (5.445 + 0.083) = Vapour pressure of solution = xAP0 P2 = 7.62 bar = 0.985 × 23.75 45. According to Henry’s Law = 23.39 mm Hg PO2 = kPHOx2O2 49. Let WB mass of solute is required. xO2 = kH \\ Mass of solvent (WA) = (2500 – WB) g We know, WB × 1000 = 0.98 = 2.8 × 10–5 Molality, m = mB × WA 34.68 × 103 nO2 Here, mB = 60 (Molecular mass of urea) xO2 = nO2 + n(water) 0.25 = WB × 1000 60 × (2500 – WB) nO2 nO2 = n(water) = 2.8 × 10–5 = 55.55 WB = 36.95 g 50. 20% (mass/mass) KI solution means 20 g KI (1 L water has 55.55 moles H2O) dissolved in 80 g of water ConsunmOp2t=io1n.5o6f × 10–3 mol L–1 O2 (in g L–1) in water Total weight of the solution = 100 g = 1.56 × 10–3 × 32 = 0.05 g L–1 Volume of the solution = 100 g 1.202 g mL−1 46. Suppose the vapour pressure of A be P 0 and the vapour pressure of B be PB0 . A WB × 1000 20 × 1.202 × 1000 MB # V 166 # 100 Total vapour pressure of the solution (3 moles of M= = A + 1 mole of B) = 1.45 mol L–1 = xA P 0 + xB PB0 Molality (m) = WB × 1000 = 20 × 1000 A MB # WA 166 # 80 xreAspaencdtivxeBlya.re the mole fractions of A and B = 1.50 mol kg–1 24 Chemistry-12

We know, 20 = 60 mm Hg xB = nB = 166 PB = PB0 xB = 51 × 1 nA + nB 80 20 3 18 166 = 17 mm Hg + As we know xB = 0.0263 xB × 1000 P = PA + PB 51. m= (1 – xB) × mA = 60 + 17 = 77 mm Hg 1= xB × 1000 55. As per Henry’s Law (1 – xB) × 18 PN2 = k0.H98×7xbNa2 r, xB = 0.0176 PN2 = xA = 1 – xB = 0.9824 PA = PA0 xA = 12.3 × 0.9824 kH = 76.48 k bar PN2 xN2 = kH , 0.987 76.48 × 103 = 12.083 kPa = 1.29 × 10–5 52. According to Raoult’s law; PA = PA0 xA xN2 = Mole fraction of N2 in aqueous ...(1) solution If PA0 = 100, then PA = 80 = nN2 From equation (1), xA = 0.80 nN2 + nH2O \\ nA \\ nA + nB = 0.80 where nH2O = 55.5 (for 1 L of water) 1.29 × 10–5 = nN2 ª nN2 114 55.5 114 nN2 + 55.5 ⇒ = 0.80 WB + 114 nN2 = 0.000717 mol or 40 114 = 0.717 mmol 1 = 0.80c1 + WB m 56. P0A (CHCl3) = 200 mm Hg 40 P0B (CH2Cl2) = 415 mm Hg WB = 10g nA(CHCl3) = WA = 25.5 = 0.213 53. (i) Vapour pressure of the solvent decreases mA 119.5 when non-volatile solute (glucose) is added nB(CH2Cl2) = WB = 40 = 0.47 to the solvent. Hence the boiling point mB 85 increases. nA xA = nA + nB (ii) 10p5 CPOa2 = k1.H6x7C×O1208 Pa × xCO2 = 2.53 × = 0.213 = 0.31 0.213 + 0.47 xCO2 = 2.53 × 105 1.67 × 108 xB = 1 – 0.31 = 0.69 Vapour pressure of the solution may be calculated = 1.51 × 10–3 as nCO2 = 1.51 × 10–3 × 500 P = PA0 xA + PB0 xB 18 = 200 × 0.31 + 415 × 0.69 = 75.5 × 10–2 = 62 + 286.35 18 = 4.2 × 10–2 mol = 348.35 nCO2 = 0.042 moles 54. Suppose CH3OH is liquid A and C2H5OH is liquid B. Mole fraction of each component in vapour phase may be calculated as: nA = WA = 32 =1 mA 32 PA = yA× P 62 = yA × 348.35 nB = WB = 23 = 0.5 yA = 0.18 mB 46 yB = 1 – 0.18 = 0.82 57. In a non-ideal solution with positive deviations, xA = nA = 1 = 2 the partial pressure of each component in the nA + nB 1 + 0.5 3 solution is greater than the vapour pressure as expected according to the Raoult’s law, xB = 1 – 2 = 1 3 3 PA = PA0 xA = 90 × 2 i.e., PA > PA0 xA ; PB > PB0 xB ; P > PA + PB 3 solutions 25

In a non-ideal solution with negative deviations, nA = 26 = 0.26 the partial pressure of each component in the 100 solution is smaller than as expected according to the Raoult’s law, nB = 35 = 0.307 114 PA < PA0 xA ; PB < PB0 xB ; P < PA + PB (i) DmixH > 0, i.e. endothermic solution formation, xA = nA = 0.26 = 0.46 nA + nB 0.26 + 0.307 shows positive deviation. xB = 1 – xA = 0.54 (ii) Dmix H < 0, i.e. exothermic solution formation, Vapour pressure of the solution may be calculated shows negative deviation. as, 58. Let liquid A be heptane (C7H16), liquid B be octane and their molar masses are 100 (C7H16) and 114 P = PA0 xA + PB0 xB (C8C18), respectively = 105.2 × 0.46 + 46.8 × 0.54 = 48.392 + 25.272 = 73.664 kPa Topic 3. Colligative Properties The properties of dilute solutions containing non- CE volatile solute which depends only upon the number of particles of the non-volatile solute present in the solution are termed as colligative properties. Vapour pressure B Solution The main colligative properties are: Vapour PressureFrozen Liquid solventsolvent (i) Relative lowering of vapour pressure = p10 – p1 = n2 D Freezing point p10 n2 + n1 A DTf depression (ii) Elevation in boiling point: DTb = Kb.m Tf T0 BD M = Molar mass of the solvent in g mol–1 Tf0 = Freezing point of the solvent 1.0 atm (1.013 bar) X DHfus = Molar enthalpy of fusion of the solvent Y (in kg mol–1) A SolvSenoltution iKt fisdeinpednepdesnodnelny toonf the nature of the solvent but the nature of the solute. Boiling point DTb elevation (iv) Osmotic Pressure: It may be defined as the extra pressure that must be applied to the C T0b Tb solution compartment to prevent the flow of solvent molecules into the solution from the Temperature (K) solvent side, when the two are separated by a SPM (semipermeable membrane). Kb = Molal elevation constant (K kg mol–1) Kb can be calcu lated from thermodynamic relationship: Kb = MRTb2 Principle of measuring osmotic pressure. The pressure in excess of atmospheric pressure that M = Molar m(1a0s0s0o.f∆tHhevasp)olvent (in g mol–1) must be applied to the solution side to prevent it from rising in the tube is the osmotic pressure. Tb0 = Boiling point of the solvent of This will also be equal to the hydrostatic pressure DHvap = Molar enthalpy of vapourisation of the liquid column of height, h, So the solvent in kg mol–1. p p=hrg Kit bisdeinpdeenpdesnodnelnytoonf the nature of the solvent but h the nature of the solute. (iii) Depression in freezing point: DTf = Kf .m Kf = Molal depression constant (K kg mol–1) or Cryoscopic constant. Solvent Kreflactainonbshe ipca: lculated from the thermodynamic Solution Kf = MRT 2 Semipermeable f membrane (1000.∆H fus) 26 Chemistry-12

p = CRT i= Observed value of colligative property Calculated value of colligative property C = Molarity R = Gas constant or i= Normal molar mass Abnormal molar mass = 0.082 litre atm K–1 mol–1 Modified Formulae: For substances undergoing MB = WB .R.T π.V dissociation or association in the solution: Reverse osmosis occurs when a pressure larger p10 – p1 = n2 than the osmotic pressure is applied towards the p10 ie n1 + n2 o solution side. DTb = iKbm Reverse Osmosis: If a pressure greater than DTf = iKfm the osmotic pressure is applied to the solution n V side, then pure solvent (or water) flows out of the p= i RT solution through the semi-permeable membrane. In this way the direction of osmosis is reversed Degree of Dissociation: If one mole of a substance A dissociates to form n molecules or and hence the process is called reverse osmosis. ions and a is the degree of dissociation, then Thus, we can say that reverse osmosis is just opposite to the process of osmosis. Reverse A Æ np osmosis is used in the desalination process to get (1 mole) pure water from sea water. At equilibrium, 1 – a na Piston After dissociation, number of moles = 1 – a + na Pressure Fresh Salt = 1 + (n – 1)a water water so i= 1 + (n – 1) α 1 Water outlet SPM or a= i–1 n–1 Isotonic solutions: Two solutions are said to be isotonic when they exert the same osmotic Degree of Association: If among n molecules, pressure, because they have the same molar ‘A’ associates to form one molecule and a is the concentration. degree of association, then It is used in making injections as all intravenous nA Æ An α injections must be isotonic with the body fluids. At equilibrium, 1 – a n Isosmotic solutions: When two isotonic After association the total number of molecules = solutions are separated by a semi-permeable α 1–a+ n membrane, then no osmosis will occur as they a have same osmotic pressure. These solutions are 1 – a + n called isosmotic solutions. ⇒ i= 1 Hypotonic solutions: A solution having lower or a= (1 – i) n n 1 osmotic pressure than the other solution is said – to be hypotonic with respect to the other solution. Hypertonic solutions: A solution having higher Values of R in different units: osmotic pressure than the other solution is said to (i) R = 8.314 J K–1 mol–1, (when V = m3, P = Pascal or N/m2) be hypertonic with respect to the other solution. (ii) R = 0.082 L atm K–1 mol–1, (when V = litre or Abnormal Molecular Mass: When the molecular ml, P = atm) mass of a substance obtained by using colligative (iii) R = 0.083 L bar K–1 mol–1, (when V = litre, P = properties does not tally with the theoretical bar) value, then the molecular mass is known as Interconversion of units of volume And 1 litre = 1000 mL = 1000 cm3, 1 m3 = 1000 litres abnormal molecular mass. Interconversion of units of Pressure: 1 atm = 101325 Pascals = 1.01325 × 105 Pascals This happens when the substance in the solution = 760 torr = 760 mm of Hg = 1.01325 bar undergoes association or dissociation. 1 bar = 105 pascals = 0.987 atm Van’t Hoff Factor (i): van’t Hoff factor is the ratio of the experimental value of the colligative property to the calculated value of the colligative property. solutions 27

ExErcisE 1.3 Multiple Choice Questions (MCQs) (1 Mark) 9. For carrying reverse osmosis for desalination of water the material used for making semipermeable 1. In isotonic solutions .................... . (i) solute and solvent both are same. membrane is (ii) osmotic pressure is same. (i) potassium nitrate (iii) solute and solvent are same (iv) solute is always same solvent may be different. (ii) parchment membrane (iii) cellulose acetate 2. Colligative properties are observed when (iv) cell membrane .................... . 10. Colligative properties depend on .................... . (i) a non volatile solid is dissolved in a volatile (NCERT Exemplar) liquid. (i) the nature of the solute particles dissolved (ii) a volatile liquid is dissolved in another in solution. volatile liquid. (ii) the number of solute particles in solution. (iii) a gas is dissolved in non volatile liquid. (iii) the physical properties of the solute particles dissolved in solution. (iv) a volatile liquid is dissolved in another volatile liquid. (iv) the nature of solvent particles. 3. Osmotic pressure of a solution is 0.0821 atm at a 11. The unit of ebulioscopic constant is ................... . temperature of 300 K. The concentration in moles/ (NCERT Exemplar) litre will be (i) K kg mol–1 or K (molality)–1 (i) 0.33 (ii) 0.666 (ii) mol kg K–1 or K–1 (molality) (iii) 0.3 × 10–2 (iv) 3 (iii) kg mol–1 K–1 or K–1(molality)–1 4. Which relationship is not correct? (iv) K mol kg–1 or K (molality) K b.1000.W2 (i) ∆T b = M2 .W1 12. An unripe mango placed in a concentrated salt solution to prepare pickle, shrivels because (ii) M2 K f .1000.W1 .................... . (NCERT Exemplar) W2 .∆Tf = (i) it gains water due to osmosis. n2 p0 − ps W2 M1 (ii) it loses water due to reverse osmosis. V p0 M2 W1 (iii) π = (iv) = × (iii) it gains water due to reverse osmosis. (iv) it loses water due to osmosis. 5. A 5% solution of cane-sugar (molecular weight 13. At a given temperature, osmotic pressure of a = 342 g mol–1) is isotonic with 1% solution of concentrated solution of a substance ................. . substance A. The molecular weight of X is (NCERT Exemplar) [CBSE 2022] (i) is higher than that of a dilute solution. (i) 342 (ii) 171.2 (ii) is lower than that of a dilute solution. (iii) 68.4 (iv) 136.8 (iii) is same as that of a dilute solution. 6. A plant cell shrinks when it is kept in a (iv) cannot be compared with osmotic pressure of (i) hypotonic solution dilute solution. (ii) hypertonic solution 14. Consider the figure and mark the correct option. (iii) isotonic solution (NCERT Exemplar) (iv) pure water (i) water will move from side (A) to side (B) if 7. What weight of glycerol should be added to a pressure lower than osmotic pressure is 600 g of water in order to lower its freezing point applied on piston (B). by 10°C ? (ii) water will move from side (B) to side (A) if (i) 496 g (ii) 297 g a pressure greater than osmotic pressure is (iii) 310 g (iv) 426 g applied on piston (B). 8. The osmotic pressure of a solution can be increased (iii) water will move from side (B) to side (A) if a by pressure equal to osmotic pressure is applied (i) increasing the volume on piston (B). (ii) increasing the number of solute molecules (iv) water will move from side (A) to side (B) if (iii) decreasing the temperature pressure equal to osmotic pressure is applied (iv) removing semipermeable membrane on piston (A). 28 Chemistry-12

15. If 0.1 m solution has boiling point of 100.052°C (iv) Assertion is wrong statement but reason is then calculate molal elevation constant. correct statement. (i) 0.52 K kg mol–1 21. Assertion: Elevation in boiling point is a colligative property. (ii) 5.2 K kg mol–1 (iii) 273.05 K kg mol–1 Reason: Elevation in boiling point is directly proportional to molarity. (iv) 2730.5 K kg mol–1 22. Assertion: When blood cells are placed in pure 16. A solution containing 12.5 g of a non-electrolytic water, they swell up due to osmosis. substance in 175 g water gave boiling point elevation of 0.70 K. The molar mass of the substance is (Kb for water = 0.52 K kg mol–1) Reason: Reverse osmosis is the process used in desalination of sea water. (i) 17.5 g mol–1 (ii) 53.06 g mol–1 23. Assertion: It is advised to add ethylene glycol (iii) 530.6 g mol–1 (iv) 175 g mol–1 to water in a car radiator while driving in hill 17. The radiator of a vehicle which holds ten kg of station. water is to be used at a place where the lowest temperature is –10°C (263.15 K). The amount of Reason: It acts as an antifreeze to lower the ethylene glycol (molar mass 62.1 g mol–1) which freezing point of water. Thus, prevents the water to freeze in the car radiator. should be added to the water is (Kf for water = 1.86 K kg mol–1). 24. Assertion: Sodium chloride and calcium chloride are used to clear snow from roads. (i) 33.38 g (ii) 333.8 g Reason: The role of sodium chloride and calcium (iii) 3338.7 g (iv) 3.3 g chloride is to lower the freezing point of water, 18. At 25°C the vapour pressure of pure water is hence ice melts. 23.76 mm Hg and that of an aqueous solution of urea is 22.98 mm Hg. The molality of the solution is 25. Assertion: When methyl alcohol is added to water, boiling point of water decreases. (i) 1.88 (ii) 2.88 Reason : When a volatile solute is added to (iii) 18.8 (iv) 0.18 a volatile solvent elevation in boiling point is 19. The mole fraction of benzene in a solution observed. containing 30% by mass in carbon tetrachloride is 26. Assertion : When NaCl is added to water a (i) 0.54 (ii) 0.38 depression in freezing point is observed. (iii) 0.40 (iv) 0.46 Reason: The lowering of vapour pressure of a solution causes depression in freezing point. 20. Hus2eSdifsora toxic gas with rotten egg like smell, is qualitative analysis. If the solubility of cHo2nSstinanwtaitser at STP is 0.195 m, then Henry’s law (NCERT Exemplar) (i) 350 bar (ii) 282 bar 27. Assertion: When a solution is separated from the pure solvent by a semipermeable membrane, the solvent molecules pass through it from pure (iii) 195 bar (iv) 557 bar solvent side to the solution side. Assertion Reason Type Questions (1 Mark) Reason : Diffusion of solvent occurs from a region of high concentration solution to a region of low In the following questions a statement of assertion concentration solution. (NCERT Exemplar) followed by a statement of reason is given. Choose the correct answer out of the following choices. Very Short Answer Type Questions (1 Mark) SPM 28. What is the expected van’t Hoff factor ‘i’ value for K3[Fe(CN)6]? (i) Assertion and reason both are correct statements and reason is correct explanation 29. The molar mass of biomolecules is determined for Assertion. by osmotic pressure and not by other colligative properties. Why? (ii) Assertion and reason both are correct statements but reason is not correct 30. State the following: explanation for Assertion. (i) Raoult’s law in its general form in reference to solutions. (iii) Assertion is correct statement but reason is wrong statement. (ii) van’t Hoff factor [Delhi and A.I. 2010, 2013] 31. Define colligative property. List two colligative properties. 32. Outer hard shell of two eggs are removed. One of the egg is placed in pure water and the other is placed in saturated solution of NaCl. What will you observe and why? solutions 29

33. aCnadlccualrabteonthteetmraacshslopreirdcee(nCtCagl4e),oiffb2e2ngzeonf bee(nCz6eHn6e) 45. Calculate the freezing point depression for 0.0711 is dissolved in 122 g of carbon tetrachloride. molal aqueous solution DofoNesat2hSiOs 4s.oIltuitsiocnomacptluetaelllyy [NCERT] ionised in the solution. freeze at –0.32°C? What is the value of van’t Hoff 34. Out of 0.1 molal solution of glucose and sodium factor for it at its freezing point? ([KFforfoerigwna2t0e0r9=] chloride, which one will have higher boiling point? 1.86 K kg mol–1) [DOE] 46. 45g wofaettehr.yCleanlecugllaytceol(i()Cfr2Hee6zOin2)gipsominixt eddepwrietshsi6o0n0, g of Short Answer Type Questions-I (2 Marks) (ii) freezing point of solution. 35. (i) Define the following terms: 47. 1g of a non-electrolyte solute dissolved in 50 g of (a) Ideal solution benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of (b) Osmotic pressure benzene is 5.12 K kg mol–1. Find the molar mass (ii) Calculate the boiling point elevation for a of the solute. [NCERT] solution prepared by adding 10 g of CaCl2 to 2mio0no0li–sg1e,odfmwinoalawterartm, earas.sss[uKmobfinfCograCCHal22CO=l2=i1s10c1.o5mg12pmlKeotlek–l1gy] 48. Calculate the mole fraction of benzene in a solution [Delhi 2017(C)] containing 30% by mass in carbon tetrachloride. [NCERT] 49. Define the following terms: [Delhi 2017] 36. How does sprinkling of salt help in clearing (i) Colligative properties snow covered roads in hilly areas? Explain the phenomenon involved in the process. (ii) Molality (m) 50. Account for the following: [DOE, NCERT Exemplar] (i) Aquatic species are more comfortable in cold water than in warm water. 37. If 0.36 g of glucose (C6H12O6) is added to 100 g of water then what will be the boiling point of the (ii) A person suffering from high blood pressure solution (kb for water = 0.52 kg mol–1). is advised to take less amount of common salt. [DOE] 38. A solution of glycerol (sCo3mHe8Og3l)ycienrowl ainter45w0ags prepared by dissolving 51. (i) On mixing liquid X and liquid Y, the volume of the resulting solution increases. What of water. This solution has a boiling point at type deviation from Raoult’s law is shown 100.42°C. What mass of glycerol was dissolved by the resulting solution? What change in to make this solution[?A(Ik, Db feolrhiw2a0t1er2,=D0e.l5h1i22K01k0g] temperature would you observe after mixing mol–1) liquids X and Y? [AI 2015] 39. The normal freezing point of nitrobenzene (C6H5NO2) is 278.82 K. A 0.25 molal solution of (ii) How can the direction of osmosis be reversed? certain solute in nitrobenzene causes a freezing Write one use of reverse osmosis. point depression of 2 degree. Calculate the value of Kf for C6H5NO2. 52. Define the following terms: 40. What care must be taken to give intravenous (i) Molarity (ii) Molal elevation constant injection to the patients? 53. Explain why on addition of 1 mole of NaCl to one litre of water, boiling point of water increases, while addition of 1 mole of methyl alcohol to water 41. If 0.1 mole of sugar or 0.1 mole of glucose is dissolved decreases its boiling point. in one litre of water, the depression in freezing point will be the same or different. Explain. [HOTS][NCERT Exemplar] 42. Decinormal solution of NaCl developed an osmotic 54. If the vapour pressure of pure water at 298 K is pressure of 4.6 atmosphere at 300 K. Calculate 23.8 mm Hg, then calculate the lowering in vapour the degree of dissociation (R = 0.0821 litre atm pressure caused by addition of 50 g of sucrose K–1 mol–1) (molecular mass = 342 g mol–1) in 500 g water. 43. The freezing point depression of 0.1 molal solution 55. The vapour pressure of pure benzene at a certain temperature is 0.85 bar. When a non-volatile, non- electrolytic solid weighing 0.5 g is added to 39 g of C5H.132COKOkHginmboeln–1z.enWehisa0t .2co5n6cKlu. Ksifofnorcbaennzyeonue of benzene (molar mass 78 g mol–1) then vapour is pressure of solution becomes 0.845 bar. What is draw about the molecular state of CH3COOH in the molar mass of solid substance? [NCERT] benzene? 44. The osmotic pressure of a 0.0103 molar solution Short Answer Type Questions-II (3 Marks) of an electrolyte is found in the 0.70 atm at 300 K. Calculate the van’t Hoff factor. (R = 0.082 56. Define the terms: Osmosis and Osmotic pressure. L atm mol–1 K–1). Is osmotic pressure of a solution a colligative property? Explain. 30 Chemistry-12

57. A solution containing 15 g of urea (molar mass = 70. At 300 K, 36 g glucose present per litre of solution 60 g mol–1) per litre of solution in water has the has an osmotic pressure of 4.98 bar. If osmotic same osmotic pressure (isotonic) as the solution of pressure of the solution is 1.52 bar at the same glucose (molar mass = 180 g mol–1) in 1 L water. temperature, what would be its concentration? Calculate the mass of glucose present in one litre of its solution. 71. Calculate the osmotic pressure of a solution containing 17.1 g of cane sugar (molecular mass 342 58. 15 g of an unknown molecular substance was g/mol) in 500 g of water at 300 K, (R = 0.0821 litre dissolved in 450 g of water. The resulting solution atm deg–1). (Density of solution of 1.034 g cm–3). freezes at – 0.34°C. What is the molar mass of the substance? (Kf for water = 1.86 K kg mol–1) 72. 3A%4%sosloultuitoinonooffasnucuronsken(oCw2Hn2oOr1g1a) nisicisocotomnpicouwnitdh. [AI 2010, Delhi 2012] Calculate the molar mass of the unknown substance. 59. A solution containing 8g of a substance in 100g of diethylether boils at 36.86°C whereas pure ether 73. A solution containing 10.2 g of glycerin per litre of boils at 35.60°C. Determine the molecular mass a solution is found to be isotonic with 2% solution of solute. (For ether Kb = 2.02 K kg mol–1) of glucose (molar mass 180). Calculate the molar [AI, Foreign 2008] mass of glycerin. 60. A solution prepared from 1.25 g of oil of 74. Under what conditions van’t Hoff factor is (i) equal wintergreen (methylsalicylate) in 99 g of benzene to 1, (ii) greater than 1, (iii) less than 1? has a boiling point of 80.31°C. Determine the molar mass of this compound (Boiling point of 75. Equimolal wsoaltuert.ioFnresezoifngNpaoCinl taonf Nd aMCgl sColl2utaiorne prepared in pure benzene is 80.1°C and[Dkeblhfoi,rFboernezigenne2021.503] K kg mol–1). is found to be –2°C. What is the freezing point of 61. An aqueous solution of glucose boils at 100.02°C. MgCl2 solution? What is the number of glucose molecules in the 76. KI and sucrose solutions with 0.1 M concentration have osmotic pressure of 0.465 atm and 0.245 atm solution containing 100g water [kb for H2O = 0.5 Km–1 NA = 6.023 × 1023]. respectively. Find out the van’t Hoff factor and 62. Addition of 0.643 g of a compound to 50 mL of the degree of dissociation of KI. benzene (density 0.879 g cm–3) lowers the freezing 77. What mass of NaCl must be dissolved in 65 g of water to lower the freezing point of water by mpooiln–t1,fcraolmcul5a.5te1°thCetmo o5l.a0r3m°Ca.ssIfoKf tfhies 5.12 K kg compound. 7.50°C? The freezing point depression constant f(aKcft)ofrorfowr aNtaeCr lisis11.8.867K(Mmo–la1.rAmsasussmoef 63. The freezing point of pure nitrobenzene is 278.8 K. van’t Hoff When 2.5 g of an unknown substance is dissolved NaCl = 23 in 100 g of nitrobenzene, the freezing point of + 23.5 = 58.5 g) [AI 2011,10; Foreign 2010] solution is found to be 276.8 K. If the freezing 78. 2.0 g gofofbebnenzoziecnae,ciwdh(iCch6Hs5hCoOwOs Ha )deisprdeissssioolnveind in 25 point depression constant for nitrobenzene is 8 K kg mol–1, what is the molar mass of the freezing point equal to 1.62 K. Molal depression constant for benzene 4.9 K kg mol–1. What is the unknown substance? percentage association of the acid if it forms a 64. Calculate the temperature at which a solution wcoanttearinwiinllgfr5e4ezgeo. f(Kgflufocroswea(tCer6H=112O.866) in 250 g of dimer in the solution? [NCERT] K kg mol–1) 79. An aqueous solution containing 1.248 g of BwaaCtelr2 (molar mass= 208.34 g mol–1) in 100g of 65. What mass of ethylene glycerol (molar mass 62.1 g mol–1) must be added to 5.50 kg of water to lower boils at 100.0832°C. Calculate the degree of dissociation of BaCl2. (Kf = 0.52 K kg mol–1). the freezing point of water from 0°C t[oD–el1h0i°C20. 1(K0]f 80. 0.5 g KCl was dissolved in 100 g of water and the for water = 1.86 K kg mol–1). solution originally at 293 K, freeze at –0.24°C. 66. At 298 K, 100 cm3 of the solution containing 3.002 g of an unknown solute exhibits an osmotic pressure Calculate the percentage ionisation [oDf eslahlti.2(0K1f0=] 1.86 kg mol–1) of 2.55 atm. What should be the molecular mass of the solute? 81. Calculate the mole fraction of |CH2OH in aqueous 67. At 27°C the osmotic pressure of 7 g protein per CH2OH 100 mL solution is 25 mm Hg. Calculate the molecular mass of protein. solution containing 20% (C2H6O2) by mass. [NCERT] 68. A 5% solution of cane sugar is isotonic with 0.877% solution of urea. Calculate the molecular mass of 82. 18 g of Agltuwcohsaet(Cte6mHp12eOra6)tuisrediwssiolllvwedatienr 1 kg of water. boil at urea. [Molecular mass of cane sugar is 342 g mol–1] 1.013 bar? Kb for water = 0.52 K kg mol–1. [NCERT] 69. Find out the osmotic pressure of M/20 solution of urea at 27°C, (R = 0.0821 litre atm K–1 mol–1). solutions 31

83. The boiling point of benzene is 353.23 K. When 61 g of benzoic acid per 500 g benzene when 1.80 g of a non-volatile solute was dissolved in 90g the vapour pressure of pure benzene at that of benzene, the boiling point raised to 354.11 K. temperature of experiment is 66.6 torr? What will Calculate the molar mass of solute, Kb for benzene be the vapour pressure in normal state? is 2.53 K kg mol–1. [NCERT] [AI 2010] 84. 200 cm3 of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic 92. An aqueous solution of 2% non-volatile solute pressure of such a solution at 300 K is found to exerts a pressure of 1.004 bar at the normal be 2.57 × 10–3 bar. Calculate the molar mass of boiling point of the solvent. What is the molar protein. [NCERT] mass of the solute? [NCERT] 85. 2 g of benzoic acid dissolved in 25 g of benzene 93. Calculate the osmotic pressure in pascals shows a depression in freezing point equal to exerted by a solution prepared by dissolving 1.62 K. Molal dmeporl –e1s.siWonhcaotnsitsantth(eKfp) eorf cbeenntzaeng ee 1.00 g polymer of molar mass 185,000 in 450 mL is 4.9 K kg association of acid if it forms dimer in solution? of water at 37°C. [NCERT] [NCERT] 94. Calculate the boiling point of 1 molar aqueous solution of KCl (density = 1.04 g mL–1). 3(K5.b5=a0m.5u2) 86. Calculate the molarity of each of the following K kg mol–1, atomic mass of K = 39, Cl = solutions: 95. (i) What is hypertonic solution? (i) 30 g of Co(NO3)2.6H2O in 4.3 L of solution. (ii) 30 mL of 0.5 M H2SO4 diluted to 500 mL. (ii) Why should intravenous injection be isotonic [NCERT] with body fluid? 87. 15H00e80nPrmyaL’satslao2wd9a8cKwon.asCtteaarlncwtuhfloaertneCptOhaec2kqienudawunanttidteyerroisf2C.15.O6a27timn× (iii) What will happen if cell a human is placed in a hypotonic solution? CO2 pressure at 298 K. [NCERT] Long Answer Type Questions (5 Marks) 88. Boiling point of water at 750 mm Hg is 99.63°C. 96. Phenol associates in benzene to a certain extent to form a dimer. A solution containing 2 × 10–2 How much sucrose is to be added to 500 g of water such that it boils at 100°C. [NCERT] kg of phenol in 1 kg of benzene has its freezing 89. Calculate the mass of ascorbic acid (Vitamin C, point lowered by 0.69 K. Calculate the degree of lCo6wHe8rOit6)s to be dissolved in 75 g of acetic acid to association of phenol. (Kf for benzene = 5.12 K kg melting point by 1.5°C. mol–1). (Kf of acetic acid = 3.9 K kg mol–1) [NCERT] 97. 0.6 mL of acetic acid (CH3COOH) having density 1.06 g mL–1, is dissolved in 1 litre of water. The 90. A solution containing 30 g of non-volatile solute in exactly 90 g of water has a vapour pressure depression in freezing point observed for this of 21.85 mm Hg at 298 K. Further 18 g of water strength of acid was 0.0205°C. Calculate the van’t is then added to the solution, the new vapour pressure becomes 22.15 mm Hg at 298 K. Hoff factor and the dissociation constant of acid. Calculate: [NCERT] (i) molecular mass of solute, 98. The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find (ii) vapour pressure of water at 298 K. out the composition of the liquid mixture if total 91. Benzoic acid is completely dimerised in benzene. What will the pressure of the solution containing vapour pressure is 600 mm Hg. Also find the composition of vapour phase. [NCERT] Answers 1.3 1. (ii) 2. (i) 3. (iii) temperature and biomolecules are generally 4. (ii) 5. (iii) 6. (ii) unstable at higher temperatures. 7. (ii) 8. (ii) 9. (iii) 10. (ii) 11. (i) 12. (iv) 30. (i) For a solution of volatile solute the partial 13. (i) 14. (ii) 15. (i) vapour pressure of each component in the 16. (ii) 17. (iii) 18. (i) solution is directly proportional to its mole 19. (iv) 20. (ii) 21. (i) fraction. 22. (ii) 23. (i) 24. (i) 25. (iii) 26. (i) 27. (ii) (ii) van’t Hoff factor is defined as the ratio of 28. i = 4 observed molecular mass to the theoretical molecular mass of the solute in a solution. 29. Osmotic pressure is measured around room 31. Properties that depend upon the number of particles of solute and not on the nature of solute, 32 Chemistry-12

e.g. Elevation in boiling point, depression in 39. DT = Kf × m freezing point etc. Given: DT =2K, m = 0.25, 32. When the egg is placed in water, it swells due to 2 = Kf × 0.25 osmosis, by the entry of pure water into the egg 2 (endo-osmosis). On the other hand, when egg is fi Kf = 0.25 = 8 K kg mol–1 placed in saturated solution of NaCl, it shrinks 40. Concentration of the solution to be inject- ed due to exo-osmosis process, i.e. loss of water out should be similar to that of blood, i.e. the solution of the egg. should be isotonic with blood. In case of non- 33. Mass percentage of benzene isotonic solutions, either exosmosis, i.e. shrinking of blood cells will take place or they swell to burst = Mass of benzene0×10 = 22 × 100 in case if the concentration of the solution being Total mass 144 = 15.27% injected is very less. 34. Sodium chloride solution will have higher boiling 41. Since both glucose and sugar are non- electrolytes point due to greater number of particles. and they do not associate or dissociate in the solution, therefore the molarity will be the same 35. (i) (a) The solution which obeys Raoult’s law and so the depression in freezing point will also over the entire range of concentration and be the same. temperature. 42. Given: C = 0.1 M, p = 4.6 atm, T = 300 K (b) It is the extra pressure to be applied to R = 0.0821 litre atm K–1 mol–1 solution to stop osmosis. (ii) CaCl2 → Ca2+ + 2Cl– By putting the above values in the following i= 3 equation: DTb = iKb × WB × 1000 i= π = 4.6 MB WA CRT 0.1 × 0.0821 × 300 = 1.87 = 3 × 0.512 K kg mol–1 10 g For dissociation 111 g mol–1 1000 × × 200 g a= i–1 for NaCl, n = 2 n–1 3 × 0.512 × 50 = 111 \\ 1.87 – 1 = 0.87 2–1 76.80 % dissociation is 87%. = 111 43. van’t Hoff factor will be DTb = 0.69 K 36. When salt is spread over snow covered roads, its DT = i × Kf × m freezing point is depressed which helps in clearing DT = 0.256 K, i = ?, of roads. Kf = 5.12 K kg mol–1, m = 0.1 ∆T 0.256 37. As we know WB × 1000 i= Kf × m = 5.12 × 0.1 = 0.5 mB × WA DT = Kb × i < 1, it indicates that the CH3COOH is present in associated state, i.e. as Given: WB = 0.36, mB = 180 (for C6H12O6) WA = 100 g, Kb = 0.52 K m–1 By putting these values in the formula DT = 0.52 × 0.36 × 1000 44. p = iCRT 180 × 100 = 0.0104 K 0.70 = i × 0.0103 × 0.082 × 300 Thus elevated boiling point i = 2.76 = 373.0104 K 45. Assuming complete ionisation of Na2SO4 DT = 3 × Kf × m = 3 × 1.86 × 0.0711 = 0.396 38. DT = kb × WB × 1000 mB × WA Assuming partial ionisation, WB = ∆T × mB × WA DT = i × Kf × m Kb × 1000 0.32 = i × 1.86 × 0.0711 DT = 100.42 – 100 = 0.42 ⇒ i = 2.42 WB × 1000 mB × WA Given, mB = 92 g, WA = 450 g, 46. (i) DT = Kf × Kb = 0.512 K kg mol–1 WB = 0.42 × 92 × 450 45 × 1000 \\ 0.512 × 1000 = 33.96 g = 1.86 × 62 × 600 = 2.25 solutions 33

Depression in freezing point = 2.25°C Given: nB = 50 = 0.146, 342 (ii) T0 – T = 2.25, 273.15–T = 2.25 nA = 500 = 27.78 18 T = 270.9 K and P0 = 23.8 mm Hg \\ Freezing point of the solution DP = nB × P0 = 270.9 K. nA + nB 47. DT = Kf × WB × 1000 = 0.146 × 23.8 mB × WA 0.146 + 27.78 0.40 = 5.12 × 1 × 1000 = 0.124 mm Hg mB × 50 55. ∆P = WB × mA mB = 5.12 × 1000 = 256 g mol–1 P0 mB × WA 50 × 0.40 0.85 – 0.845 0.5 × 78 48. Let A Æ Benzene : B Æ Carbon tetrachloride 0.85 = mB × 39 WA = 30 g, WB = 70 g ⇒ mB = 0.5 × 78 × 0.85 = 170g MA = 78 MB = 154 39 × 0.005 WA 30 56. Osmosis: It is the process of movement of solvent nA = MA = 78 = 0.38 molecules from the solvent to the solution when nB = 70 = 0.45 separated by a semi permeable membrane. 154 nA 0.38 Osmotic Pressure: It is the extra pressure that xA = nA + nB = 0.38 + 0.45 = 0.46 needs to be applied on the solution in order to stop \\ Mole fraction of benzene = 0.46. the entry of solvent molecules from solvent side 49. (i) Colligative properties: Those properties into the solution when the two are separated by which depends upon the number of particles a semipermeable membrance. and not on nature of particles. Yes, osmotic pressure is a colligative property as (ii) Molality: It is defined as number of moles it depends on the concentration of solute and not of solute dissolved per kg of solvent. on the nature of solute. 50. (i) Cold water contains more dissolved oxygen 57. For urea solution: than warm water as solubility of gas in liquids decreases with increase in temperature. W2 = 15 g, M2 = 60 g, V = 1 L W2 RT 15 × R × T p= M2 V = 60 × 1 (ii) Na+ gets mixed up in blood and increases For glucose solution: the osmotic pressure of body fluid which increases the blood pressure. W2 = ?, M2 = 180 g mol–1, V = 1 L W2 RT W2 × R × T 51. (i) Positive deviation from Raoult’s law is shown p= M2 V = 180 × 1 by solution. The tempera- ture will decrease. (ii) When we apply pressure on solution side As per question p (glucose) = p (urea) more than osmotic pressure, the direction of osmosis can be reversed. \\ 15 × R × T = W2 × R × T 60 × 1 180 × 1 It is used for desalination of sea water and removal of dissolved salts in water. ⇒ W2 = 45 g 58. DTf = Tf° – Tf 52. (i) Molarity: It is defined as number of moles of solute dissolved per litre of solution. = 0°C – (–0.34°C) = 0.34 K (ii) Mtooelleavl aetlieovnaintibooni lcinognpstoainnttw (Khebn):sIotliustieoqnuianl W1 = 450 g = 0.45 kg, one molal. W2 = 15 g, Kf = 1.86 K kg mol–1 Kf × W2 # 1000 53. NaCl is a non-volatile solute. When NaCl is DTf = M2 × W1 added to water, the vapour pressure of solution decreases, hence boiling point increases. ⇒ 1.86 # 1 #51000 = 0.34 M2 # 450 Water and methanol form minimum boiling azeotropes due to decrease in forces of attraction, \\ M2 = 182.35 g mol–1 hence boiling point decreases. 59. Elevation in boiling point 54. According to Raoult’s law = 36.86 – 35.60 = 1.26 P0 – P1 P0 = ∆P = nB WB = 8 g, WA = 100 g, Kb = 2.02 K kg mol–1 P0 nA + nB As we know 34 Chemistry-12

DT = Kb × WB × 1000 64. As we know WB × 1000 mB WA mB × WA DT = kf × Kb WB × 1000 mB = ∆T × WA Kf = 1.86 K kg mol–1, WB = 54 g, WA = 250 g, = 2.02 × 8 × 1000 mB = 180 g mol–1 1.26 100 DT = 1.86 × = 128.25 g mol–1 54 × 1000 = 2.232 180 × 250 T0 – T = 2.232 60. Here: DT = 80.31 – 80.1 = 0.21 273.15 –T = 2.232 Kb = 2.53 K kg mol–1 T = 270.918 K WB = 1.25 g (Solute) So, the solution of glucose in water will freeze at WA = 99 g (Solvent) 270.918 K. DT = Kb × WB × 1000 mB × WA 65. As we know WB × 1000 By putting these values in the above formula: DT = Kf × mB × WA 2.53 × 1.25 × 1000 = mB Given: DT = 10, Kf = 1.86 Kkg mol–1, 99 × 0.21 WB = ?, mB = 62.1, WA = 5.5 × 103g mB = 152.1 g mol–1 WB × 1000 By putting these values in formula 61. DT = Kb × mB × WA WB × 1000 DT = 100.02 – 100 = 0.02 10 = 1.86 × 62.1 × 5.5 × 103 mB = 180, WA = 100 g, Kb = 0.5 K m–1 WB = 1836.3 g By putting the values in above formula 66. T = 298 K, V = 100 cm3 = 0.100 litre, 0.02 = 0.5 × WB × 1000 WB = 3.002 g, p = 2.55 atm, MB = ?, 180 × 100 R = 0.0821 litre atm K–1 mol–1 WB = 0.72 g Number of glucose molecules in the solution We know, = Number of moles × Avogadro’s number pV = < WB FRT MB = WB × NA = 0.72 × 6.023 × 1023 \\ 2.55 × 0.1 = 3.002 g × 0.0821 × 298 mB 180 < MB F = 2.409 × 1021 fi MB = 30.02 g × 0.0821 # 298 2.55 # 0.1 62. As we know = 288.03 g mol–1 mB = Kf × WB × 1000 67. As we know that ∆T WA pV = nRT Kf = 5.12 K kg mol–1, DT = 5.51 – 5.03 = 0.48 K pV = WB × RT mB WB = 0.643 g, WA = V × d 25 × 100 = 7 × 0.082 × 300 760 1000 mB = 50 × 0.0879 = 43.95 g mB = 52348.8 amu mB = 5.12 × 0.643 ×1000 68. Suppose the molecular mass of urea is m2, then 0.48 43.95 Molar conc. of sugar = 156.06 g mol–1 = W1 = 5 0.1 , and m1 × V1 342 × 63. For calculating molecular mass Molar concentration of urea mB = Kf × WB × 1000 = W2 = 0.877 ∆T WA m2 × V2 m2 × 0.1 Kf = 8 K kg mol–1, WB = 2.5 g, WA = 100 g, As both the solutions are isotonic DT = 278.8 – 276.8 = 2K \\ W1 = W2 By putting the values in the above formula m1 × V1 m2 × V2 mB = 8 × 2.5 × 1000 = 100 g mol–1 \\ 5 = 0.877 2 100 342 × 0.1 m2 × 0.1 \\ Molar mass of unknown substance, m2 = 0.8772× 34 = 59.987 g mol–1 mB = 100g mol–1 ⇒ 5 solutions 35

69. According to the equation: (iii) When the solute undergoes association in the solution. p = CRT  C= 1 = 0.05, R = 0.0821, 75. DT (NaCl) = i × kf × m = 2 × kf × m1 ...(1) 20 DT (MgCl2) = i × Kf × m = 3 × Kf × m2 ...(2) T = 273 + 27 = 300 K On dividing eqn. (2) with eqn. (1) p = 0.05 × 0.0821 × 300 TT (NaCl) = 2 × m1 TT (MgCl2) 3 × m2 = 1.2315 atm 70. As we know WB or ° 2C = 2 [Q m1 = m2] mB ∆T (MgCl2) 3 pV = × RT or DT (MgCl2) = 3 Case I: 4.98 × 1 = 36 ×R×T ...(1) So, solution of MgCl2 will freeze at –3°C. 180 ...(2) 76. For KI, p1 = iCRT ⇒ 0.465 = i × 0.1 × R × T Case II: 1.52 × 1 = WB ×R×T ...(1) 180 On dividing eqn. 1 by 2, we get For sucrose, p2 = CRT ⇒ 0.245 = 0.1 × R × T ...(2) 4.98 = 36 1.52 WB On dividing eqn. (1) by eqn. (2), we get ⇒ WB = 10.987 g i= 0.465 = 1.898 \\ Molar conc. = 0.061 g/L 0.245 For dissociation of KI = K+ + I– 71. Volume of solution i–1 1.898 – 1 Mass 500 + 17.1 a= n–1 = 2–1 = 0.898 = Density = 1.034 77. As we know = 500 mL WB ×1000 DT = i × Kf × mB × WA Osmotic pressure can be calculated as: pV = nRT Given: DT =7.5, i = 1.87, ⇒ pV = ( WB 2 RT Kf = 1.86 K m–1, mB WA = 65 g, mB = 58.5 g mol–1, \\ p × 0.5 = 17.1 × 0.082 × 300 WB = ? WB ×1000 342 7.5 = 1.87 × 1.86 × 58.5 × 65 p = 2.46 atm \\ 72. For isotonic solutions WB = 7.5 × 58.5 × 65 = 8.2 g 1.87 × 1.86 × 1000 p1 = p2 C1RT = C2RT C1 = C2 78. As we know WB × 1000 ( WB × 1000 2 = ( WB × 1000 2 DT = i × Kf × mB × WA mB ×V sucrose mB ×V unknown 1.62 = i × 4.9 × 2 × 1000 4 × 1000 = 3 × 1000 122 × 25 342 × 100 mB × 100 fi i = 0.504 mB = s32o×lu43t4ions= 256.5g mol–1 For association 1 – 0.504 isotonic 1–i 73. For a= 1 = 1 = 0.992 1– n 1 – 2 p1 = p2, i.e. C1RT = C2RT, C1 = C2 WB × 1000 WB × 1000 (n = 2 for dimerisation) mB ×V mB ×V ( 2 = ( 2 \\ % of association = 0.992 × 100 = 99.2%. glycerin glu cos e 79. DT = i × Kb × WB × 1000 mB × WA ( 10.2 × 1000 2 = ' 2× 1000 1 mB × 1000 glycerin 180 × 100 glu DT = 0.0832, Kb = 0.52 K kg mol–1, cos e WB = 1.248 g, mB = 208.34 g mol–1, mB = 10.2 × 180 = 91.8 g mol–1 WA = 100 g 2 × 10 0.0832 = i × 0.52 × 1.248 × 1000 74. (i) When the solute does not dissociate or 208.34 × 100 associate in the solution. (ii) When the solute is an electrolyte and i = 2.67 undergoes dissociation in solution. As we know 36 Chemistry-12

a= i–1 = 2.67 – 1 = 0.835 After putting these values in the above formula: n–1 3–1 2 × 1000 (n = 3 for BaCl2) 1.62 = i × 4.9 × 122 × 25 80. DT = i × Kf × WB × 1000 i = 0.504 mB × WA a= 1–i = 1 – 0.504 = 0.992 1 1 0.24 = i × 1.86 × 0.5 × 1000 1– n 1 – 2 74.5 × 100 i = 1.922 Percentage of association As we know = 0.992 × 100 = 99.2%. a= i–1 = 1.922 – 1 = 0.922 86. Given: WB = 30 g, MB = 291, n–1 2–1 V = 4.3 L = 4300 mL % ionisation = 0.922 × 100 = 92.2%. (i) We know, 81. Number of moles of ethyl glycol WB × 1000 20 Molarity, M = MB × V = 62 = 0.322 30 × 1000 Mass of water = 80 g = 291 × 4300 = 0.024 M Number of moles of water (ii) Molarity of solution after dilution may be calculated as: (nA) = 80 = 4.44 18 M1V1 (Concentrated) = M2V2 (diluted) Mole fraction of |CH2OH 0.5 × 30 = M2 × 500 0.5 × 30 CH2OH M2 = 500 = 0.03 M xB = nB 87. According to Henry’s Law: nA + nB P = KH × x ...(1) 0.32 P = 2.5 atm = 2.5 × 101325 Pa, = 4.44 + 0.322 = 0.067 KH = 1.67 × 108 Pa 82. For elevation in boiling point Putting these values in eqn. (1), we get DT = Kb × WB × 1000 2.5 × 101325 = 1.67 × 108 × x mB × WA x = 1.517 × 10–3 18 × 1000 nCO2 nCO2 = 0.52 × 180 × 1000 = 0.052 xCO2 = nH2O + nCO2 ª nH2 O ...(2) \\Boiling point of solution will be 500 18 T – T0 = 0.052 ⇒ nH2 O = = 27.77 ⇒ T = T0 + 0.052 \\ From eqn. (2), we get = 373.15 + 0.052 = 373.202 K nCO2 27.77 83. DT = Kb × WB × 1000 1.517 × 10–3 = mB × WA ⇒ nCO2 = 0.042 mol 354.11 – 353.23 = 2.53 × 1.80 × 1000 88. DT = 100 – 99.63 = 0.37K mB × 90 Kb = 0.52 K kg mol–1, mB = 2.53 × 1.80 × 1000 mB = 342 g mol–1 0.88 × 90 WA = 500 g = 57.5 g mol–1 Amount of sucrose required may be calculated by using the following relation: 84. pV = WB × RT mB mB = WB × RT DT = Kb × WB × 1000 ...(1) V π mB × WA ...(1) = 1.26 × 0.083 × 300 Putting these values in eqn. (1), we get 0.2 2.57 × 10–3 0.52 × WB × 1000 342 × 500 [ 200 cm3 = 0.2 L] 0.37 = = 61038.9 g mol–1 \\ WB = 121.67 g 85. DT = i × Kf × WB × 1000 89. We know, WB × 1000 mB × WA mB × WA DT = Kf × DT = 1.62 K, Kf = 4.9 K kg mol–1, WB = 2 g, mB = 122 g mol–1, WA = 25 g Given: DT = 1.5K, Kf = 3.9 K kg mol–1 solutions 37

WA = 75 g, = 3.057 × 10–4 × 101325 Pa mB = Molar mass of ascorbic acid p = 30.975 Pa = 176 g mol–1 94. Volume of solution = 1000 mL Putting these values in eqn. (1), we get Mass of solution = 1000 × 1.04 = 1040 g 1.5 = 3.9 × WB × 1000 The mass of solute present in 1 molar solution is 176 × 75 \\ WB = 5.077 g WB = 74.5; WA = 1040 – 74.5 = 965.5 g 90. Let the vapour pressure of water at 298 K be P0 KCl being a strong electrolyte is dissociated and molecular mass of solution be m. completely, so, i = 2 P0 – P1 = WB × mA \\ DT = i × Kb × WB × 1000 P0 mB × WA mB × WA Set I: P0 – 21.85 = 30 × 18 = 2 × 0.52 × 74.5 × 1000 P0 90 × mB 74.5 × 965.5 = 1.077 K Set II: P0 – 22.15 = 30 × 18 \\ DT = T – T0 = 1.077 P0 108 × mB or T – 373 = 1.077, i.e., T = 374.077 K 95. (i) It has concentration more than the body On dividing set I by set II fluids. (P0 – 21.85) × P0 = 108 = 6 P0 P0 – 22.15 90 5 (ii) Intravenous should always be isotonic with body fluids. Otherwise, cell will shrink and P0 = 23.65 mm Hg burst in hypertonic and hypotonic solutions respectively. If we substitute the value of P0 in set I, then 23.65 – 21.85 = 30 × 18 (iii) The cell will swell and may burst in hypotonic 23.65 90 × mB solution. mB = 78.95 g 96. We know, 91. Mol. wt of Benzoic acid = 122g mol–1 Mol. wt of Benzene = 78g mol–1 i × Kf × WB × 1000 DT = mB × WA ...(1) \\ nA = 500 = 6.41, nB = 61 = 0.5 Given: DT = 0.69 K, Kf = 5.12 K Kg mol–1, 78 122 WA = 1 kg, WB = 2 × 10–2 kg, mB = 94 g/mol Putting these values in eqn. (1), we get \\ xB = nB = 0.5 = 0.0724 nA + nB 6.41 + 0.5 ∆P = i × xB 0.69 K = i × 5.12 × 2 × 10–2 × 1000 P0 94 × 1 \\ 0.69 × 94 = i × 102.4 For complete dimerisation, 0.69 × 94 1 i= 102.4 = 0.633 i= 2 \\ By putting the value in the formula, we get For association, a = 1–i 1 ∆P = 1 × 0.0724 1– n 66.6 2 DP = 2.410 Here, n = 2 P0 – P = 2.410 (Q phenol associates in benzene to form dimer) 66.6 – P = 2.410 or P = 69.01 torr \\ a= 1 – 0.633 = 0.734 1 92. ∆P = xB = nB 1 – 2 P0 nA + nB Now percentage of association For very dilute solution (mnAA >>> nB) \\ ∆P nB WB WA = 0.734 × 100 = 73.4%. nA mB P0 = = × 97. Mass of CH3COOH is 0.6 × 1.06 = 0.636 g, i.e. 1.013 – 1.004 = 2 × 18 WB = 0.636 g, Also mB = 60 g, 1.013 mB × 98 WB × 1000 Now, DT = i × Kf × mB × WA mB = 41.37 g mol–1 93. Applying the eqn., pV = nRT, we get 0.0205 = i × 1.86 × 0.636 × 1000 60 × 1000 450 1 p× 1000 = 185000 × 0.0821 × 310 i = 1.04 p = 3.057 × 10–4 atm Now, Degree of ionisation = a = i–1 n–1 38 Chemistry-12

= 1.04 – 1 = 0.04 (n = 2) 98. We know, 2–1 Acetic acid undergoes ionisation as P = PA + PB = P 0 xA + PB0 xB A CH3 COOH CH3 COO – + H+ 600 = 450 × xA + 700(1 – xA) Ca xA = 0.4 At equilibrium C—Ca Ca \\ xB = 0.6 Dissociation constant, Ka [CH3 COO–] [H+] PA = PA0 xA = 450 × 0.4 = 180 mm = 6CH3 COOH@ PB = PB0 xB = 700 × 0.6 = 420 mm Ca2 Ca × Ca = 1–a P = PA + PB = 180 + 420 = 600 mm C – Ca Mole fraction of the component in vapour phase As, C= 0.636 = 0.0106 mol L–1 may be calculated by using Dalton’s law 60 PA = yA × P and a = 0.04 yA = PA = 180 = 0.3 Ka = 0.0106 × (0.04)2 = 1.76 × 10–5 P 600 \\ 1 – 0.04 \\ yB = 1 – yA = 1 – 0.3 = 0.7 C ase based questions (Reference: Nada Abdulrazzaq, Baseem Al- 1. Case Study Sabbagh, Julia M. Rees, William B. Zimmerman, Separation of Azeotropic Mixtures Using Air An azeotropic mixture is a liquid mixture Microbubbles Generated by Fluidic Oscillation, comprising two or more components whose Separations: Materials, Devices and Processes, proportion cannot be altered by conventional April 2016 Vol. 62, No. 4, 1192-1199) atmospheric distillation. The feasibility of separating the azeotropic mixture of ethanol-water The following questions are multiple choice using microbubble-mediated batch distillation is questions. Choose the most appropriate answer: presented. The effects of the depth of the liquid mixture in the bubble tank and of the inlet (i) Why it is not possible to separate the air microbubble temperature on the process components of azeotrope by fractional efficiency were investigated. The enrichment of ethanol in the vapour phase was higher than that distillation? achieved at equilibrium conditions for all liquid ethanol mole fractions considered, including the (a) They have same composition in solid and azeotrope. On decreasing the depth of the liquid liquid phase. mixture and increasing the temperature of the air microbubbles, the separation efficiency of ethanol (b) They have same composition in liquid and was improved. Ethanol with purity of about vapour phase. 98.2 vol % was obtained using the lowest liquid level (3 mm) in conjunction with the highest air (c) They have same composition in solid and microbubble temperature (90°C). Separation was vapour phase. achieved with a small rise in the temperature of the liquid mixture (4°C) at a depth of 3 mm and (d) They have very high boiling point. evaporation time of 90 min making this system suitable for treating thermally sensitive mixtures. (ii) Ethanol with purity of about 98.2% by volume Microbubble distillation is a new technology is obtained from ethanol-water mixture using that is still being developed. However, on scale up it promises to contribute effectively to the microbubble mediated batch distillation. The purification of numerous chemicals. The findings of this work demonstrate that this technique has approximate percent volume of ethanol obtained major advantages over traditional methods and has great potential to be adapted and applied into by fractional distillation is any energy industry where distillation is involved. (a) 90% (b) 75% (c) 95% (d) 85% OR What type of mixture is the mixture of ethanol and water? (a) Maximum boiling azeotrope (b) Minimum boiling azeotrope (c) Negative deviation from Raoult’s law (d) None of these solutions 39

(iii) In microbubble-mediated batch distillation, itself on both sides of the membrane, the solvent enrichment of ethanol in vapour phase is done flows initially, however, to dilute the solution, by and this flow will continue untill sufficient excess (a) decreasing the depth of the liquid mixture hydrostatic pressure is generated on the solution in the bubble tank to 3mm. side to block the net flow of solvent. This excess (b) increasing the temperature of air pressure is the osmotic pressure, at thermodynamic microbubbles to 90°C. equilibrium also, the chemical potential of solvent (c) small rise in temperature of the liquid will be the same on both sides of the membrane. mixture to 4°C. (d) all of these (Source: Alfred Rudin, Phillip Choi, The Elements of Polymer Science and Engineering (Third (iv) Microbubble method is suitable for treating Edition), 2013, Pages 584) thermally sensitive mixtures means (a) this method does not involve heating the In these questions, a statement of assertion mixture followed by a statement of reason is given. (b) this method does not involve any Choose the correct answer out of the such reagent which is sensitive to the following choices. components of mixture (a) Assertion and reason both are correct (c) this method involves variation in pressure to separate the components of mixture statements and reason is correct explanation (d) t h i s m e t h o d i n v o l v e s s u c h a l o w for assertion. temperature so that mixtures which are unstable at high temperature can also be (b) Assertion and reason both are correct separated statements but reason is not correct explanation for assertion. 2. Case Study Colligative properties reflect the chemical (c) Assertion is correct statement but reason is potential of solvent in solution. Alternatively, wrong statement. a colligative property is a measure of the depression of the activity of the solvent in (d) Assertion is wrong statement but reason is solution, compared to the pure state. Colligative correct statement. properties include vapour pressure lowering, boiling point elevation, freezing point depression (i) Assertion: When a solution is separated and membrane osmometry. from the pure solvent by a semipermeable membrane, the solvent molecules pass A solution is separated from semipermeable through it from pure solvent side to the membrane, which allows solvent molecule to pass solution side. but blocks solute. The solute cannot distribute Reason: Diffusion of solvent occurs from a region of high concentration solution to a region of low concentration solution. (ii) Assertion: Osmotic pressure is a colligative property. Reason: Osmotic pressure is directly proportional to molarity. (iii) Assertion: The solute cannot distribute itself on both the sides of membrane. Reason: The semipermeable membrane is impermeable to the solute. (iv) Assertion: Elevation in boiling point is a colligative property. Reason: Elevation in boiling point is directly proportional to molarity. OR Assertion : When NaCl is added to water a depression in freezing point is observed. Reason: The lowering of vapour pressure of a solution causes depression in freezing point. Answers 1. (i) (b) (ii) (c) OR (b) (iii) (d) (iv) (d) 2. (i) (c) (ii) (a) (iii) (a) (iv) (c) OR (a) 40 Chemistry-12

Analogy type questions Ideal solution : A : : Non-ideal solution : B 1. Complete the following analogy: Minimum boiling azeotrope: A :: Maximum boiling (i) A : Dmix H < 0, B : Dmix H > 0 azeotrope : B (ii) A : PTotal > 0, B : PTotal < 0 (iii) A : Follows Raoult’s law, B : Do not follow (i) A : Dmix H < 0, B : Dmix H > 0 (ii) A : Positive deviation from Raoult’s law, B : Raoult’s law Negative deviation from Raoult’s law (iv) A : Dmix V < 0 or Dmix V > 0, B : Dmix V = 0 (iii) A : HNO3 and H2O, B : CH3CH2OH and H2O 1. (ii) Answers (iv) A : Dmix V = 0, B : Dmix V > 0 2. Complete the following analogy: 2. (iii) Multiple type questions 1. Match the items given in Column I with the type of solutions given in Column II. Column I Column II A. Soda water (1) A solution of gas in solid B. Sugar solution (2) A solution of gas in gas C. German silver (3) A solution of solid in liquid D. Air (4) A solution of solid in solid E. Hydrogen gas in palladium (5) A solution of gas in liquid (6) A solution of liquid in solid Code: E (2) (i) A (5) B (3) C (4) D (2) E (1) (ii) A (4) B (3) C (5) D (1) E (2) (iii) A (2) B (3) C (4) D (5) E (1) (iv) A (5) B (4) C (3) D (1) 2. Match the laws given in Column I with expresions given in Column II. Column I Column II A. Raoult’s law (1) DTf = Kfm B. Henry’s law (2) p = CRT C. Elevation of boiling point (3) p = x1p10 + x2p20 (4) DTb = Kbm D. Depression in freezing point (5) p = KH.x E. Osmotic pressure D (2) E (4) (ii) A (5) B (3) C (2) D (1) E (4) D (3) E (4) (iv) A (3) B (5) C (4) D (1) E (2) Code: (i) A (3) B (5) C (1) (iii) A (1) B (2) C (5) Answers 1. (i) 2. (iv) Quick revision notes solution, e.g. salt in salt solution. • Solution: It is a homogeneous mixture of two of • Solvent: It is present in larger amount in more substance, e.g. salt solution. • Solute: It is present in smaller amount in solution, e.g. water in salt solution. SolutionS 41

• Mass percentage: It is the amount of solute • Freezing point: The temperature at which dissolved per 100 g of solution. vapour pressure of solid solvent becomes equal to that of the liquid solvent. • Volume percentage: It is amount of solute dissolved per 100 mL of solution. • Boiling point: The temperature at which vapour pressure of the liquid becomes equal to the • Parts per million (ppm): It is amount of atmospheric pressure. substance (in grams) per 106g of solution. • Vapour pressure: The pressure exerted by a • Molarity: It is number of moles of solute vapour in thermodynamic equilibrium with its dissolved per litre of solution. condensed phase at a given temperature is called its vapour pressure. The vapour pressure of pure • Solutions are classified as solids, liquids and solvent is more than the solution. gaseous solutions. • Lowering of vapour pressure: When a non- volatile solute is added to the solvent to form • Molality: It is number of solute dissolved per kg a solution, vapour pressure of the solution of solvent. decreases. It is because escaping tendency of the solvent molecules into vapour form decreases. • The concentration of solution is expressed in terms • Osmosis: The process of passage of solvent of molarity, molality, mole fraction, in percentages molecules into the solution through a semipermeable membrane is called osmosis. (mass and volume percentage) and ppm. • Osmotic pressure: It is the extra pressure that • Mole fraction is the ratio of the number of moles needs to be applied on the solution side so as to stop the process of osmosis. of component to the total number of moles of all • Reverse osmosis: It is the process in which the components of the solution. solvent molecules from the solution passes through SPM (semipermeable membrane) by • Henry’s law: It states that solubility of a gas applying pressure more than the osmotic pressure in a liquid is directly proportional to the partial on the solution side. pressure of the gas. • Isotonic solution: Those solutions which have the same osmotic pressure. • Raoult’s law: The vapour pressure of each component of a solution is directly proportional to • Hypertonic solution: Those solutions whose concentration is more than the body fluids and the mole fraction of each component if both solute has higher osmotic pressure and lower vapour pressure. and solvent are volatile. • Hypotonic solution: Those solutions whose • Ptotal = P0AxA + P0BxB concentration is less than the body fluids and • Raoult’s law for non-volatile solutes: The has lower osmotic pressure but higher vapour pressure. relative lowering of vapour pressure is equal to the • Colligative properties are used to determine the mole fraction of the solute if only solvent is volatile. molecular weight of solute. • Ideal solutions follow Raoult’s law. • Solute particles which associate in the solution exhibits molecular mass more than the actual • Non-ideal solutions deviates from Raoult’s law. molecular mass. • The deviation can be positive as well as negative. • Solute particles which dissociate into ions in aqueous solution exhibits molecular mass less • Azeotropes arises due to very large deviations than the actual molar mass. from Raoult’s law. • The degree to which a solute associate or dissociate can be expressed in terms of van’t Hoff • Positive deviations leads to minimum boiling factor. Azeotropes, e.g. methanol and water, ethanol and • van’t Hoff factor: It is defined as the ratio of normal molecular mass to the experimentally water, cyclohexane and ethanol. determined molecular mass. • Negative deviations leads to the formation of maximum boiling azeotropes, e.g. CHCl3 + Acetone, H2O + HCl, H2O + HNO3. • Colligative properties of a solution depends upon the number of solute particles and is independent of the chemical identity of the solute. • Relative lowering of vapour pressure, elevation in boiling point, depression in freezing point, osmotic pressure are the colligative properties. • eElbevualtiioosncoinpbico iclionngsptoainntt i(nK1b)m: Iotlaisl equal to the solution. • Cryoscopic cfroenesztinagnpt o(iKntf)i:nI1t is equal to the depression in molal solution. 42 Chemistry-12

Important Formulae 1. Mass percentage of a component (w/w) = Mass of the component in solution × 100 Total mass of the solution 2. Volume percentage of component (v/v) = Volume of the component × 100 Total volume of solution 3. Mole fraction of a component (x) = Number of moles of the component Total number of moles of all the components 4. Parts per million = Total Number of parts of component of solution × 106 number of the parts of all the components 5. Molarity = Number of moles of solute = WB × 1000 Volume of solution (in litres) mB × V ]in mLg 6. Molality = Number of moles of solute = WB × 1000 Mass of solvent (in kilograms) mB × WA ^in gramsh 7. Normality = Number of gram equivalent of solute Volume of solution (in litres) 8. Molarity (M) = xB × density of solution # 10 mB 9. Molality (m) = 0xB × 1 00 ^1–xBhmA 10. Henry’s Law, P = KHx 11. According to Raoult’s law–‘For a solution of volatile liquids the partial vapour pressure of each component in the solution in directly proportional to its mole fraction’, i.e. P1 = P10 x1; P2 = P20 x2 12. Using Dalton’s law of partial pressures the total pressure of solution is calculated. Ptotal = P10 x1 + P20 x2 P10 – P1 13. Relative lowering of vapour pressure P10 = x2 14. ∆T = Tb – T 0 ; ∆Tb = Kb × 1000 × w2 b M2 × w1 15. ∆T = T 0 – Tf ; ∆Tf = Kf × 1000 × w2 f M2 × w1 16. Osmotic Pressure, p = CRT = n RT V 17. M2 = w2 RT πV Observed colligative property 18. van’t Hoff factor, i = Normal molar mass = Calculated colligative property Abnormal molar mass = Total number of moles of particles after association/dissociation Total number of moles of particles before association/dissociation 19. Inclusion of van’t Hoff factor modified the equations for colligative properties as: P10 – P1 = i. n2 P10 n1 DTb = i × Kb × 1000 × w2 M2 × w1 DTf = i × Kf × 1000 × w2 M2 × w1 p= i × n2 RT V solutions 43

Common errorS errorS CorreCtIonS (i) Students usually get confused between w/w%, (i) Students have to understand clearly the w/v% and v/v%. difference between these concepts. (ii) Many students misinterpret the value of ‘i’ for (ii) i > 1 (dissociation) and i < 1 (association). association/dissociation for a solute. (iii) In both cases, solvent will flow from its higher (iii) Direction of osmosis from hypotonic to concentration to its lower concentration solution. hypertonic solution not clearly understood by many students. (iv) Conditions need to be clearly learnt for (iv) Concept of osmosis and reverse osmosis is not understanding the occurrence of these clear to many students. phenomena. (v) Students do not convert mg into g while solving (v) Mass should be taken in grams. numericals. (vi) DTf or DTb remains the same in °C or K. (vi) Students change DTf or DTb in °C to K (vii) Volume should be taken in litres if ‘p’ is in bar. (vii) Students do not convert unit of volume while If p is in Pa, volume should be taken in m3. solving numericals. (viii) ‘i’ must be used in case of electrolytes and (viii) Students do not consider van’t Hoff factor ‘i’ solutes undergoing association/dissociation. in case of electrolyte and solutes undergoing association/dissociation. 44 Chemistry-12

revISIon Chart Henry’s law: The partial Ideal solutions follow There are 9 types of solution: pressure of gas dissolved in a Raoult’s law (i) Solid in solid, (ii) Solid in liquid, (iii) Liquid in liquid is directly proportional to liquid, (iv) Solid in gas, (v) Gas in gas, (vi) Liquid in solid, (vii) Liquid in gas, (viii) Gas in solid, its mole fraction. (ix) Gas in liquid. Raoult’s Law: The vapour Solution iS a HomogeneouS Molarity is the no. of moles of solute per pressure of each component in mixture of two or more litre of solution. a solution is directly proportional to its mole fraction. SubStanceS, e.g. Salt Solution Molality is the no. of moles of solute per Kg of solvent. Colligative property depends Non-ideal solution do not upon the number of particles follow Raoult’s law Mole fraction is the ratio of no. of moles of solute. of a component to the total no. of moles of all the components. Positive deviation Negative deviation DH = + ve, DV = + ve DH = –ve, DV = –ve Minimum boiling azeotropes Maximum boiling azeotropes Depression in freezing point is directly Relative lowering of vapour pressure proportional to molality is equal to the mole fraction of solute. DTf = i × Kf × w2 # 1000 P10 –P1 = i# w2 # 1000 M2 # w1 P M 2 # w1 0 1 Osmotic pressure is the extra pressure Elevation in boiling point is directly that needs to be applied on the solution Colligative ProPerty proportional to molality. side to stop osmosis. DTb = i × Kf × w2 # 1000 n2 RT M2 # w1 p=i× V van’t Hoff factor = Observed colligative property Association, i < 1 Calculated colligative property Dissociation, i > 1 i= Normal molecular mass Observed molecular mass solutions 45

Chapter trend—Based on past Years’ CBSe exams ➣ It has been observed from this chapter that the weightage of topics ‘Ideal and Non-ideal Solutions’ and ‘Colligative P 0 –P1 n2 Properties’ (especially, numericals based on colligative properties i.e. 1 = n1 + n2 ; DTb = Kbm; DTf = Kf m; p = CRT) are maximum. Hence, these are most important topics. P10 ➣ From this chapter , generally 3 marks questions were asked from the topic ‘Colligative Properties’. ➣ Most of the 1, 2 and 3 marks questions from this chapter belong to the topic ‘Colligative Properties’. Questions For practice Very Short Answer Type Questions (1 Mark) 1. Give an example of maximum boiling azeotrope. 2. What type of intermolecular interaction exist between methanol and acetone? [Delhi 2014] 142 g mol–1) was dissolved in 50 g 3. oCfawlcautleart,eatshseumboiinlignNg pa2oSinOt4ouf nsodleurtgiooenswcohmenpl2etgeoifoNnias2aStiOon4 .(m(koblaforrmHa2sOs = 0.52 K kg mol–1) [AI 2016] Short Answer Type Questions-I (2 Marks) 4. Vapour pressure of water as 20°C is 17.5 mm Hg. Calculate the vapour pressure of water at 20°C when 15 g of glucose (Molar mass 180 g mol–1) is dissolved in 150g of water. [CBSE 2015] 5. Calculate the boiling point of 1 molar solution of KCl (molar mass 74.5 g mol–1). The density of solution is 1.04 g cm–3 and kb for H2O is 0.52 K kg mol–1. 6. Calculate osmotic pressure of 0.1 M K4[Fe(CN)6] solution if it is 50% ionised at 27°C. Short Answer Type Questions-II (3 Marks) 7. A 10% solution (by mass) of sucrose in water has freezing point of 269.15 K. Calculate the freezing point of 10% solution of glucose in water, if freezing point of pure water is 273.15 K. [Delhi 2017] [Given: Molar mass of glucose = 180 g mol–1, Molar mass of sucrose = 342 g mol–1] 8. Calculate the van’t Hoff factor in each of the following: (i) 50% ionised K4[Fe(CN)6] (ii) 60% ionised CaCl2 (iii) Benzoic acid which is 50% dimerised in benzene. 9. A solution of glucose (molar mass = 180 g mol–1) in water has a boiling point of 100.20°C. Calculate the freezing point of the same solution. Molal constants for water, kf and kb are 1.86 K kg mol–1 and 0.512 K kg mol–1, respectively. [Delhi 2017] 10. fSroemezeinegthpyoliennteogf lwycaotle,rH-gOlyCcoHl2s–oluCtHio2nOiHs,–i1s5a°dCd. eWdhtaotyiosutrhCe abro’silirnagdipaotionrt along with 5 kg of water. If the mol–1, and kf = 1.86 K kg mol–1 for H2O] of the solution? [kb = 0.52 K kg [AI 2014] Answers 1. CHCl3 and Acetone 2. H-bonding and dipole-dipole interaction 3. 373.44 K 6. p =7.47 bar 4. 17.325 mm Hg 5. 374.077 K 9. –0.73 °C 7. 265.55 K 8. (i) i = 3, (ii) i = 2.2, (iii) i = 0.75 10. 377.19 K 46 Chemistry-12

aSSIgnment Time: 45 Minutes M.M.: 25 Multiple Choice Questions (MCQs) (1 Mark) 1. Which of the following aqueous solutions should have the highest boiling point? (i) 1.0 M NaOH (ii) 1.0 M Na2SO4 (iii) 1.0 M NH4NO3(iv) 1.0 M KNO3 2. We have three aqueous solutions of NaCl labelled as ‘A’, ‘B’ and ‘C’ with concentrations 0.1M, 0.01M and 0.001M, respectively. The value of van’t Hoff factor for these solutions will be in the order .................... . (i) iA < iB < iC (ii) iA > iB > iC (iii) iA = iB = iC (iv) iA < iB > iC Assertion Reason Type Questions (1 Mark) In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (i) Assertion and reason both are correct statements and reason is correct explanation for Assertion. (ii) Assertion and reason both are correct statements but reason is not correct explanation for Assertion. (iii) Assertion is correct statement but reason is wrong statement. (iv) Assertion is wrong statement but reason is correct statement. 3. Assertion: Volume percentage of a solute is volume of solute divided by volume of solvent, multiplied by 100. Reason: Solvent is the substance present in excess in a solution 4. Assertion: On rise in temperature, the solubility of gas in a liquid decreases. Reason: Dissolution of gas in a liquid is an endothermic reaction. 5. Assertion: When blood cells are placed in pure water, they swell up due to osmosis. Reason: Reverse osmosis is the process used in desalination of sea water. Very Short Answer Type Questions (1 Mark) 6. State Henry’s law. 7. Define azeotropes. Short Answer Type Questions-I (2 Marks) 8. Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH3OH, CH3CN. 9. (i) On mixing liquid X and Y, the volume of resulting solution decreases. What type of deviation from Raoult’s law is shown by the resulting solution? What change in temperature would you observe? (ii) How can the direction of osmosis be reversed? Short Answer Type Questions-II (3 Marks) 10. A solution is prepared by dissolving 5 g of non-volatile solute in 95 g of water. It has vapour pressure 23.375 mm Hg at 250°C. Calculate the molar mass of solute (pA0 = 23.75 mm Hg). 11. Determine the osmotic pressure of a solution prepared by dissolving 2.5 × 10–2 g of K2SO4 in 2 L solution at 25°, assuming it is completely ionised. {R = 0.0821 L atm K–1 mL–1, K2SO4 molar mass = 174 g mol L–1} 12. A 10% solution of urea is isotonic with 20% solution of an unknown solute ‘X’ at the same temperature. Calculate molar mass of X. Long Answer Type Questions (5 Marks) 13. Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water? (i) phenol (ii) toluene (iii) formic acid (iv) ethylene glycol (v) chloroform 1. (ii) 1.0 M Na2SO4 Answers 3. (iv) 4. (iii) 8. 60 g mol–1 9. 5.27 × 10–3 atm 2. (iii) iA = iB = iC 5. (ii) 10. 120 g mol–1 solutions 47

2 Electrochemistry Topics covered 2.1 Electrochemical cells and Electrode Potentials 2.2 Conductivity and its Variation for Electrolytic Solutions 2.3 Quantitative Aspects of Electrolysis 2.4 Batteries, Fuel cells, Electrochemical Aspects of Corrosion and Hydrogen Economy C hapter map Topic 1. Electrochemical Cells and Electrode Potentials Electrochemical cells Electrochemical cells involve conversion of electric energy to chemical energy or vice-versa. They are of two types viz. Galvanic cells or Voltaic cells and Electrolytic cell. 48

Differences between galvanic cell and electrolytic cell S.No. Galvanic Cell or Voltaic Cell Electrolytic Cell (i) A device in which electrical energy is produced A device in which electrical energy is used to (ii) from chemical reaction. bring about a chemical reaction. E.g. Daniel cell, dry cell, lead storage battery. E.g. Electrolysis of molten NaCl, Electrolysis of dil. aq. H2SO4 solution using Pt electrodes. Electrolyte Electron flow Substances which allow passage of current through – Current + their molten state or in aqueous solutions by Salt bridge movement of ions are called electrolytes. Zinc Copper Electrolytes are of two types: Solution containing Solution containing • Strong electrolytes: They dissociate into ions salt of Zinc salt of Copper completely. • Weak electrolytes: They dissociate into ions only partially. Daniell Cell: A galvanic cell made up of zinc and copper metals is known as Daniell cell. It is represented according to the convention. Cathode made up of copper electrode dipped in an aqueous solution of Cu2+ salt is represented on the RHS and Anode made up of zinc electrode dipped in aqueous solution of Zn2+ salt on the LHS. Functions of salt bridge: • It completes the circuit and allows the flow of current. • It maintains the electrical neutrality on both the sides. Salt-bridge generally contains solution of a strong K+ Cl– electrolyte stuhcehsaams Ke.NO3, KCl etc. KCl is preferred because the transport numbers of both and ions are almost Eext < 1.1 V Eext = 1.1 V Eext > 1.1 e– e– Anode Current Cathode I=0 Cathode Anode –ve +ve Current Zn Salt Cu Zn Cu Cu –ve bridge +ve Zn ZnSO4 CuSO4 ZnSO4 CuSO4 Functioning of Daniell cell when external voltage Eext opposing the cell potential is applied. Transport number or transference number: The • It is expressed in volts. It is an intensive property, current flowing through an electrolytic solution is i.e., independent of the amount of species in the carried by the ions. The fraction of the current carried reaction. by an ion is called its transport number or transference number. • Electrode potential in general, is the the potential difference between electrode (metal) and the Electrode Potential: electrolyte (metal ion solution). • When an electrode is in contact with the solution Oxidation potential: The tendency to lose electrons in the above case in known as oxidation potential. of its ions in a half-cell, it has a tendency to lose Reduction potential: or gain electrons, which is known as electrode potential. • The tendency to gain electrons in the above case is known as reduction potential. According to eleCtroChemistry 49