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ISC Class XII Analysis Of Pupil Performance 2017 : Chemistry

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Analysis of Pupil Performance Research Development and Consultancy Division Council for the Indian School Certificate Examinations New Delhi Year 2017 __________________________________________________________________________________ Published by: Research Development and Consultancy Division (RDCD) Council for the Indian School Certificate Examinations Plot No. 35-36, Sector VI Pushp Vihar, Saket New Delhi-110017 Tel: (011) 29564831/33/37 E-mail: council@cisce.org Copyright, Council for the Indian School Certificate Examinations All rights reserved. The copyright to this publication and any part thereof solely vests in the Council for the Indian School Certificate Examinations. This publication and no part thereof may be reproduced, transmitted, distributed or stored in any manner whatsoever, without the prior written approval of the Council for the Indian School Certificate Examinations. FOREWORD This document of the Analysis of Pupils Performance at the ISC Year 12 and ICSE Year 10 Examination is one of its kind. It has grown and evolved over the years to provide feedback to schools in terms of the strengths and weaknesses of the candidates in handling the examinations. We commend the work of Mrs. Shilpi Gupta (Deputy Head) and the Research Development and Consultancy Division (RDCD) of the Council who have painstakingly prepared this analysis. We are grateful to the examiners who have contributed through their comments on the performance of the candidates under examination as well as for their suggestions to teachers and students for the effective transaction of the syllabus. We hope the schools will find this document useful. We invite comments from schools on its utility and quality. Gerry Arathoon Chief Executive & Secretary November 2017 i PREFACE The Council has been involved in the preparation of the ICSE and ISC Analysis of Pupil Performance documents since the year 1994. Over these years, these documents have facilitated the teaching-learning process by providing subject/ paper wise feedback to teachers regarding performance of students at the ICSE and ISC Examinations. With the aim of ensuring wider accessibility to all stakeholders, from the year 2014, the ICSE and the ISC documents have been made available on the Council s website www.cisce.org. The document includes a detailed qualitative analysis of the performance of students in different subjects which comprises of examiners comments on common errors made by candidates, topics found difficult or confusing, marking scheme for each answer and suggestions for teachers/ candidates. In addition to a detailed qualitative analysis, the Analysis of Pupil Performance documents for the Examination Year 2017 have a new component of a detailed quantitative analysis. For each subject dealt with in the document, both at the ICSE and the ISC levels, a detailed statistical analysis has been done, which has been presented in a simple user-friendly manner. It is hoped that this document will not only enable teachers to understand how their students have performed with respect to other students who appeared for the ICSE/ISC Year 2017 Examinations, how they have performed within the Region or State, their performance as compared to other Regions or States, etc., it will also help develop a better understanding of the assessment/ evaluation process. This will help them in guiding their students more effectively and comprehensively so that students prepare for the ICSE/ ISC Examinations, with a better understanding of what is required from them. The Analysis of Pupil Performance document for ICSE for the Examination Year 2017 covers the following subjects: English (English Language, Literature in English), Hindi, History, Civics and Geography (History & Civics, Geography), Mathematics, Science (Physics, Chemistry, Biology), Commercial Studies, Economics, Computer Applications, Economics Applications, Commercial Applications. Subjects covered in the ISC Analysis of Pupil Performance document for the Year 2017 include English (English Language and Literature in English), Hindi, Elective English, Physics (Theory and Practical), Chemistry (Theory and Practical), Biology (Theory and Practical), Mathematics, Computer Science, History, Political Science, Geography, Sociology, Psychology, Economics, Commerce, Accounts and Business Studies. I would like to acknowledge the contribution of all the ICSE and the ISC examiners who have been an integral part of this exercise, whose valuable inputs have helped put this document together. I would also like to thank the RDCD team of Dr. Manika Sharma, Dr. M.K. Gandhi, Ms. Mansi Guleria and Mrs. Roshni George, who have done a commendable job in preparing this document. The statistical data pertaining to the ICSE and the ISC Year 2017 Examinations has been provided by the IT section of the Council for which I would like to thank Col. R. Sreejeth (Deputy Secretary - IT), Mr. M.R. Felix, Education Officer (IT) ICSE and Mr. Samir Kumar, Education Officer (IT) - ISC. Shilpi Gupta Deputy Head - RDCD November 2017 ii CONTENTS Page No. FOREWORD i PREFACE ii INTRODUCTION 1 QUANTITATIVE ANALYSIS CHEMISTRY 3 QUALITATIVE ANALYSIS Theory (Paper-1) 10 Practical (Paper-2) 31 This document aims to provide a comprehensive picture of the performance of candidates in the subject. It comprises of two sections, which provide Quantitative and Qualitative analysis results in terms of performance of candidates in the subject for the ISC Year 2017 Examination. The details of the Quantitative and the Qualitative analysis are given below. Quantitative Analysis This section provides a detailed statistical analysis of the following: Overall Performance of candidates in the subject (Statistics at a Glance) State wise Performance of Candidates Gender wise comparison of Overall Performance Region wise comparison of Performance Comparison of Region wise performance on the basis of Gender Comparison of performance in different Mark Ranges and comparison on the basis of Gender for the top and bottom ranges Comparison of performance in different Grade categories and comparison on the basis of Gender for the top and bottom grades The data has been presented in the form of means, frequencies and bar graphs. Understanding the tables Each of the comparison tables shows N (Number of candidates), Mean Marks obtained, Standard Errors and t-values with the level of significance. For t-test, mean values compared with their standard errors indicate whether an observed difference is likely to be a true difference or whether it has occurred by chance. The t-test has been applied using a confidence level of 95%, which means that if a difference is marked as statistically significant (with * mark, refer to t-value column of the table), the probability of the difference occurring by chance is less than 5%. In other words, we are 95% confident that the difference between the two values is true. t-test has been used to observe significant differences in the performance of boys and girls, gender wise differences within regions (North, East, South and West), gender wise differences within marks ranges (Top and bottom ranges) and gender wise differences within grades awarded (Grade 1 and Grade 9) at the ISC Year 2017 Examination. The analysed data has been depicted in a simple and user-friendly manner. 1 Given below is an example showing the comparison tables used in this section and the manner in which they should be interpreted. Comparison on the basis of Gender Gender Girls Boys *Significant at 0.05 level N 2,538 1,051 Mean 66.1 60.1 SE 0.29 0.42 The t-value 11.91* table shows comparison between the performances of boys and girls in a particular subject. The t-value of 11.91 is significant at 0.05 level (mentioned below the table) with a mean of girls as 66.1 and that of boys as 60.1. It means that there is significant difference between the performance of boys and girls in the subject. The probability of this difference occurring by chance is less than 5%. The mean value of girls is higher The results have also been depicted pictographically. In this case, the girls performed significantly better than the than that of boys. It can be interpreted that girls are performing significantly better than boys. boys. This is depicted by the girl with a medal. Qualitative Analysis The purpose of the qualitative analysis is to provide insights into how candidates have performed in individual questions set in the question paper. This section is based on inputs provided by examiners from examination centres across the country. It comprises of question wise feedback on the performance of candidates in the form of Comments of Examiners on the common errors made by candidates along with Suggestions for Teachers to rectify/ reduce these errors. The Marking Scheme for each question has also been provided to help teachers understand the criteria used for marking. Topics in the question paper that were generally found to be difficult or confusing by candidates, have also been listed down, along with general suggestions for candidates on how to prepare for the examination/ perform better in the examination. 2 STATISTICS AT A GLANCE Total Number of Candidates: 37,403 Mean Marks: Highest Marks: 100 65.2 Lowest Marks: 03 3 PERFORMANCE (STATE-WISE & FOREIGN) West Bengal 66.4 Uttarakhand 65.5 Uttar Pradesh 62.3 Tripura 52.5 Tamil Nadu 74.7 Telangana 61.4 Sikkim 60.0 Rajasthan 66.7 Punjab 62.4 Odisha 59.4 Madhya Pradesh 63.5 Manipur 54.1 Meghalaya 72.1 Maharashtra 72.9 Kerala 72.5 Karnataka 71.9 Jharkhand 62.0 Himachal Pradesh 65.6 Haryana 74.4 Gujarat 69.5 Goa 52.4 Delhi 71.9 Chandigarh 63.1 Chattisgarh 54.3 Bihar 66.0 Assam 81.5 Andhra Pradesh 60.9 Foreign 77.0 The States of Assam and Tamil Nadu secured highest mean marks. Mean marks secured by candidates studying in schools abroad were 77. 4 GENDER-WISE COMPARISON BOYS GIRLS Mean Marks: 66.6 Mean Marks: 64.4 Candidates: 14,738 Candidates: 22,665 Number of Number of Comparison on the basis of Gender Gender Girls Boys N Mean SE t-value 14,738 22,665 66.6 64.4 0.13 0.11 12.92* *Significant at 0.05 level 5 REGION-WISE COMPARISON East North Mean Marks: 64.8 Mean Marks: 63.9 Number of Candidates: 11,295 Number of Candidates: 20,795 Highest Marks: 100 Lowest Marks: 06 Highest Marks: 100 Lowest Marks: 03 REGION Mean Marks: 71.2 Mean Marks: 71.1 Number of Candidates: 3,332 Number of Candidates: 1,836 Highest Marks: 100 Lowest Marks: 15 South Mean Marks: 81.0 Number of Candidates: 145 Highest Marks: 100 Lowest Marks: 43 Foreign 6 Highest Marks: 100 Lowest Marks: 23 West Mean Marks obtained by Boys and Girls-Region wise 64.8 63.3 North 66.4 72.6 70.0 63.8 East 73.6 South 81.6 80.5 69.5 West Foreign Comparison on the basis of Gender within Region Region North (N) East (E) South (S) West (W) Foreign (F) Gender Girls Boys Girls Boys Girls Boys Girls Boys Girls Boys N Mean SE 7,968 12,827 4,481 6,814 1,495 1,837 735 1,101 59 86 64.8 63.3 66.4 63.8 72.6 70.0 73.6 69.5 81.6 80.5 0.18 0.15 0.24 0.21 0.39 0.38 0.58 0.50 1.45 1.72 *Significant at 0.05 level The performance of girls was significantly better than that of boys in the northern, eastern, southern and western region. In foreign region no significant difference was observed. 7 t-value 6.06* 8.31* 4.85* 5.42* 0.52 MARK RANGES : COMPARISON GENDER-WISE Comparison on the basis of gender in top and bottom mark ranges Marks Range Top Range (81-100) Bottom Range (0-20) *Significant at 0.05 level Gender Girls Boys Girls Boys N Mean SE 3,428 4,719 19 27 88.2 88.5 16.8 16.6 0.09 0.07 0.96 0.71 Boys Girls t-value -2.54* 0.23 All Candidates Marks Range (81-100) 88.5 81 - 100 88.2 88.3 70.1 61 - 80 70.4 70.2 Marks Range (81-100) 49.7 41 - 60 50.5 50.0 32.4 21 - 40 Marks Range (0 -20) 33.1 32.6 16.6 0 - 20 16.8 16.7 8 GRADES AWARDED : COMPARISON GENDER-WISE Comparison on the basis of gender in Grade 1 and Grade 9 Grades Grade 1 Grade 9 Gender Girls Boys Girls Boys N Mean SE 1,272 1,915 172 383 93.8 93.7 25.7 25.5 2.63 2.14 1.97 1.31 Boys Girls t-value 0.02 0.06 All Candidates 93.7 93.8 93.8 1 In Grade 1 and Grade 9 no significant difference was 84.3 84.3 84.3 2 74.3 74.5 74.3 3 observed between the average performance of girls and boys. 64.5 64.6 64.5 4 57.0 57.0 57.0 5 52.0 52.1 52.1 6 47.1 47.0 47.1 7 42.5 42.5 42.5 8 9 9 25.5 25.7 25.6 THEORY (PAPER-1) Part I (20 marks) Answer all questions Question 1 (a) Fill in the blanks by choosing the appropriate word/words from those given in the brackets: (iodoform, acetaldehyde, positive, greater, acidic, acetone, disaccharide, negative, increases, glucose, decreases, chloroform, polysaccharide, lactose, lesser, basic, cationic hydrolysis, anionic hydrolysis) (i) Calcium acetate on heating gives __________ which gives _________ on heating with iodine and sodium hydroxide solution. (ii) On dilution of a solution, its specific conductance _______ while its equivalent conductance_____________. (iii) Sucrose is a ___________ and yields upon hydrolysis, a mixture of ________ and fructose. (iv) More _______ is the standard reduction potential of a substance, the ________ is its ability to displace hydrogen from acids. (v) An aqueous solution of CH3COONa is_________ due to _________. [5] (b) Complete the following statements by selecting the correct alternative from the choices given: [5] (i) In a face centered cubic lattice, atom (A) occupies the corner positions and atom (B) occupies the face centre positions. If one atom of (B) is missing from one of the face centered points, the formula of the compound is: (1) (2) (3) (4) (ii) A2B5 A2B3 AB2 A2B The half-life period of a first order reaction is 20 minutes. The time required for the concentration of the reactant to change from 0 16 M to 0 02M is: (1) 80 minutes (2) 60 minutes (3) 40 minutes (4) 20 minutes (iii) For a spontaneous reaction will be respectively: 10 (1) (2) (3) (4) -ve and +ve +ve and -ve +ve and +ve -ve and -ve (iv) The conjugate acid of HPO2 4 is: (1) H3PO3 (2) H3PO4 (3) H2P 4 (4) PO43(v) The polymer formed by the condensation of hexamethylenediamine and adipic acid is: (1) Teflon (2) Bakelite (3) Dacron (c) (4) Nylon-66 Answer the following questions: [5] (i) Why the freezing point depression ( ) of 0 4M NaCl solution is nearly twice than that of 0 4M glucose solution? (ii) Identify the order of reaction from each of the following units of rate constant (k): (a) mol L-1 sec-1 (b) mol-1 L sec-1 (iii) Specific conductivity of 0 20 M solution of KCl at 298 K is 0 025 S cm-1. Calculate its molar conductivity. (iv) Name the order of reaction which proceeds with a uniform rate throughout? (v) (d) What are the products formed when phenol and nitrobenzene are treated separately with a mixture of concentrated sulphuric acid and concentrated nitric acid? Match the following: (i) (ii) (iii) (iv) (v) Diazotisation Argentite Thermosetting plastics Electrochemical cell Bidentate ligand [5] (a) (b) (c) (d) (e) 11 Bakelite Nernst equation Aniline Ethylenediamine Froth floatation process Comments of Examiners (a) (i) Many candidates wrote acetaldehyde instead of acetone in the first blank. A few candidates wrote chloroform instead of iodoform in the second blank. (ii) Many candidates interchanged (reversed) the answer instead of decreases and increases they wrote increases and decreases . (iii) Instead of disaccharide many candidates wrote polysaccharide in the first blank. Few wrote lactose in place of glucose for the second blank. (iv) Some candidates wrote positive instead of negative in the first blank. For the second blank, instead of greater a few candidates wrote more which was not given in the list of words. (v) Some candidates wrote acidic in place of basic in the first blank. Instead of anionic hydrolysis some candidates wrote cationic hydrolysis in the second blank. (b) (i) Some candidates wrote A2B3 or AB2 instead of A2B5 . (ii) Most of the candidates were able to calculate the answer correctly i.e. 60 minutes but some candidates wrote 40 minutes or 20 minutes also. (iii) The correct option was -ve and +ve but some candidates gave incorrect options also. (iv) Instead of H2PO4- some candidates wrote other options such as PO43-, H3PO4 etc. which were not correct. (v) Instead of Nylon-66 which was the correct answer, some candidates wrote Dacron or Teflon which was not correct. (c) (i) Many candidates wrote that NaCl is a strong electrolyte and glucose is a non-electrolyte. However, they were not able to explain that colligative property is directly proportional to number of moles in solutions. In a few cases, van t Hoff factor was not given. (ii) Instead of zero order and 2nd order reaction, some candidates wrote 1st order and zero order reaction. (iii) Most of the candidates calculated the value of molar conductivity correctly, but some candidates wrote the answer with incorrect unit. (iv) Instead of zero order reaction some candidates wrote first order and second order reactions. 12 Suggestions for teachers Insist upon writing organic reactions with conditions. The relationship between specific conductance,equivalent conductance, specific conductivity and molar conductivity should be explained to students. Ask students to learn carbohydrates, types of carbohydrates, chemical properties, etc. Electrochemical series and identification of anode and cathode on the basis of standard electrode potential must be explained thoroughly to students. Salt hydrolysis of all types of salts must be explained with suitable examples. Explain the calculations to find the number of atoms in various types of cubic unit cells. Give practice to find the formula of the compound. More practice should be given in numerical problems based on half-life period. Instruct students to write the answer to any numerical problem with correct unit. Explain the conditions for spontaneous process in terms of G and E to the students. Bronsted-Lowry s concept and conjugate acid base pairs should be explained clearly to the students. Monomers and their polymers should be explained in a tabular form. Explain how the value of van t Hoff factor changes for electrolytes and non-electrolytes by taking different examples. Give adequate practice in class in calculating the units of rate constant (k) for different order reactions. Sufficient practice should be given to students to represent zero, first and second order reactions graphically. Directive influence of various functional groups in aromatic compounds should be clearly explained to students. (v) Most of the candidates failed to explain that OH group is ortho and para directing group. The concept of directive influence was not very clear to many candidates. (d) Most of the candidates attempted this part correctly. A few candidates gave incorrect answers also, such as diazotization was matched with ethylenediamine and bidentate ligand was matched with aniline. MARKING SCHEME Question 1 (a) (i) acetone, iodoform (ii) decreases, increases (iii) disaccharide, glucose (iv) negative, greater (v) basic, anionic hydrolysis (b) (i) (ii) (iii) (iv) (v) (1) or A2B5 (2) or 60 minutes (1) or -ve and +ve (3) or H2PO4 (4) or Nylon-66 (c) (i) NaCl Na+ + Cl0 4M 0 4M 0 4M Glucose 0 4M (is an electrolyte) Glucose (non electrolyte) 0 4M Number of moles increases (doubles) in NaCl solution. (ii) Hence, the depression in freezing point is nearly twice than that of glucose solution. (a) Zero order reaction (b) 2nd order reaction (iii) Molar conductivity = 1000 = or 1000 0 025 0 20 = 125 S cm2 mol-1 (iv) Zero order reaction. (v) OH OH O conc. H2SO4 O2N +3HONO2 conc. NO2 O + HONO2 conc. O NO2 + 3H2O NO2 2, 4 6 trinitro phenol NO2 + H2O conc. H2SO4 O 100oC NO2 m-dinitro benzene 13 (d) Match the following: (i) (ii) (iii) (iv) (v) Diazotisation Argentite Thermosetting plastics Electrochemical cell Bidentate ligand (c) (e) (a) (b) (d) Aniline Froth floatation process Bakelite Nernst equation Ethylenediamine Part II (50 marks) SECTION A Answer any two questions. Question 2 (a) (b) (c) (i) Determine the freezing point of a solution containing 0 625 g of glucose (C6H12O6) dissolved in 102 8 g of water. (Freezing point of water = 273 K, Kf for water = 1 87K kg mol-1, at. wt. C = 12, H = 1, O = 16) [2] (ii) A 0 15 M aqueous solution of KCl exerts an osmotic pressure of 6 8 atm at 310 K. Calculate the degree of dissociation of KCl. (R = 0 0821 Lit. atm K-1 mol-1). [2] (iii) A solution containing 8 44 g of sucrose in 100 g of water has a vapour pressure 4 56 mm of Hg at 273K. If the vapour pressure of pure water is 4 58 mm of Hg at the same temperature, calculate the molecular weight of sucrose. [1] (i) When ammonium chloride and ammonium hydroxide are added to a solution containing both Al3+ and Ca2+ ions, which ion is precipitated first and why? [2] (ii) A solution of potassium chloride has no effect on litmus whereas, a solution of zinc chloride turns the blue litmus red. Give a reason. [2] How many sodium ions and chloride ions are present in a unit cell of sodium chloride crystal? [1] 14 Comments of Examiners (a) (i) The value of Tf was calculated correctly by most of the candidates but some candidates added the value of Tf to 273 instead of subtracting this value from 273. (ii)Many candidates calculated the value of van t Hoff factor (i) correctly but some candidates failed to calculate the value of degree of dissociation ( ) for KCl. Suggestions for teachers - - (iii) Most of the candidates calculated the molecular weight of sucrose correctly but some candidates wrote the incorrect unit. (b)(i) Many candidates wrote that Al3+ ions will be precipitated first. A few candidates wrote that Ca2+ ions will precipitate first. The concept of common ion effect and solubility product was not considered by many candidates. - (ii) Many candidates gave correct explanation for KCl but for ZnCl2 some candidates wrote anionic hydrolysis instead of cationic hydrolysis. (c) Some candidates wrote the coordination number 6:6 instead of 4 Na+ and 4 Cl ions. - - - While teaching, enough practice should be given in solving numerical problems. Ask students to write the formula, do the substitution and give the answer with the correct unit. While teaching abnormal molecular weights, calculation of van t Hoff factor (i) along with degree of dissociation/ association must be clearly explained to the students. More practice should be given to students in solving numerical problems for calculation of molecular weights of nonvolatile substances. Quantitative inorganic analysis should be explained by using solubility product and common ion effect. Explain that the salts of strong acid and strong base such as KCl do not undergo hydrolysis hence the solution is neutral. Whereas salt of strong acid and weak base (ZnCl2) undergo cationic hydrolysis hence solution is acidic Diagram of unit cell should be drawn and explained. The difference between coordination number and number of constituent particles should be explained MARKING SCHEME Question 2 (a) (i) Kf = 1 87 K kg mol-1, = 0 625g , W = 102 8 g, m = 180 1000 1000 1 87 0 625 = 180 102 8 = 0 06316 Freezing point of solution = 273 - 0 06316 = 272.93684 K = (ii) C = 0 15M, = 6 8 , = 310 , = 0 0821 L atm K-1mol-1 KCl K + + Cl (n = 2) 15 = 6 8 = i 0 15 0 0821 310 6 8 i = 0 15 0 0821 310 = 1 7812 1 Degree of dissociation ( ) = 1 = 1 7812 1 =78 12% (iii) (i) = 0.7812 or 0 P = 4 58 mm of Hg, P = 4 56 mm of Hg, w = 8 44g, W = 100 g, M = 18 g mol-1 = 4 58 - 4 56 = 0 02 mm of Hg. m= (b) 2 1 0 = 4 58 8 44 18 0 02 100 = 347 90 1 Al3+ ion will be precipitated first. The Ksp value of Al(OH)3 is less as compared to Ca(OH)2. Therefore, the ionic conc. product of [Al3+] and [OH-] ion will exceed the solubility product. The conc. of OHion is not sufficient (due to C.I.E) to exceed the solubility product of (Ca2+) and [OH] due to high Ksp value. (ii) (c) KCl is a salt of strong acid and strong base hence do not hydrolyse with H+ and OHof water. Hence, the aq solution of KCl is neutral and has no effect on litmus ZnCl2 is salt of strong acid and weak base hence undergo cationic hydrolysis, therefore, the number of H+ ions increases in solution and turns blue litmus red. Number of Na+ = 4 Number of Cl- = 4 Question 3 (a) Lead sulphide has face centered cubic crystal structure. If the edge length of the unit cell of lead sulphide is 495 pm, calculate the density of the crystal. (at. wt. Pb = 207, S = 32) (ii) For the reaction: 2H2 + 2NO 2H2O + N2 , the following rate data was obtained: (i) [NO] mol L-1 [H2] mol L-1 1 0 40 0 40 4 6 10-3 2 0 80 0 40 18 4 10-3 3 0 40 0 80 9 2 10-3 S.No. [1] [3] Rate: mol L-1 sec-1 Calculate the following: (1) The overall order of reaction. (2) The rate law. (3) The value of rate constant (k). (b) (i) The following electrochemical cell is set up at 298 K: 2+ ( )(1 )// 2+ ( )(1 )/ Given EoZn2+/Zn = - 0 761V, EoCu2+/Cu = + 0 339 V (1) Write the cell reaction. (2) Calculate the emf and free energy change at 298 K. 16 [2] (ii) Answer the following: [2] (1) What is the effect of temperature on ionic product of water (Kw)? (c) (2) What happens to the ionic product of water (Kw) if some acid is added to it? Frenkel defect does not change the density of the ionic crystal whereas, Schottky defect lowers the density of ionic crystal. Give a reason. [2] Comments of Examiners (a)(i) Some candidates calculated the density without converting the unit of edge length Suggestions for teachers from pm to cm. Some candidates took an - Stress upon that the value of Z incorrect value of Z. The unit of density was changes with type of unit cells. also incorrectly mentioned. - Give more practice in solving (ii) Several candidates calculated the overall numerical problems based on density, order of reaction directly without showing edge length, etc. the calculation. A few candidates were not - More practice must be given on able to calculate the value of rate constant numerical problems based on order of reaction. Emphasize on step by step (k). calculations. (b) (i) Most of the candidates gave the cell reaction Insist that students must practice the and emf of the cell correctly. Some numerical problems based on candidates were not able to calculate the electrochemistry. value of free energy change ( G ) because - Defects in crystals must be explained they didn t know the correct equation. with the help of diagrams. How the Many candidates did not write negative sign density of the crystal changes due to before the value of ( G ). these defects should also be explained. (ii)(1) Many candidates wrote that the ionic product of water (kw) remains the same instead of writing that the value of kw increases with increase in temperature (kw temperature). (2) Instead of writing that the value of kw remains constant on the addition of acid several candidates wrote the value of kw increases. (c) Cation was not specified by many candidates for Frenkel defect. Also for Schottky defect, many candidates wrote atoms instead of cations and anions. MARKING SCHEME Question 3 (a) (i) (ii) Density (d) = 3 or 4 239 6 023 1023 (495 10 10 )3 = 13 087 g/cm3 For reaction 2NO + 2H2 2H2O + N2 Rate = k [NO]p (H2]q (i) 4 6 10-3 = k(0 4)P (0 4)q (ii) 18 4 10-3 = k(0 8)p (0 4 )q (iii) 9 2 10-3 = k(0 4)p(0 8)q Dividing eq (ii) by eq (i) 17 18 4 10 3 4 6 10 3 0 8 = 0 4 4 = (2)p p=2 Dividing eq (iii) by eq(i) 9 2 10 3 4 6 10 3 0 8 = 0 4 2 = (2)q q=1 (i) Over all order of reaction Rate = k [NO]2 [H2]1 Order of reaction = 2 + 1 = 3 (ii) Rate law = rate = k[NO]2[H2]1 (iii) (b) (i) Rate constant (k) = ( )3 = (1) Cell reaction: 4 6 10 3 (0 4)3 k = 0 071875 mol-2L2s-1 Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) (2) E0cell = E0 (cathode) E0(anode) = 0 339 V (- 0 761V) = 1 10 V Go = -nFE0 = -2 96500 1 10 = -212300 J = -212 3 k.J (1) As the temperature is increased, the value of Kw will also increase (Kw temperature) (ii) (2) When acid is added to water, the value of Kw remains constant. (c) Frenkel defect arises due to presence of holes in the cationic lattice site and cation occupies an interstitial position. Hence, density remains the same. Schottky defect arises when same number of cations and anions are missing from their normal site and a pair of holes are formed hence density decreases. Question 4 (a) (i) Name the law or principle to which the following observations conform: (1) (ii) When water is added to a 1 0 M aqueous solution of acetic acid, the number of hydrogen ion (H+) increases. (2) When 9650 coulombs of electricity is passed through a solution of copper sulphate, 3 175 g of copper is deposited on the cathode (at.wt. of Cu = 63 5). (3) When ammonium chloride is added to a solution of ammonium hydroxide, the concentration of hydroxyl ions decreases. What is the difference between the order of a reaction and its molecularity? [3] 18 [2] (b) (i) Explain why high pressure is required in the manufacture of sulphur trioxide by contact process. State the law or principle used. [2] (ii) Calculate the equilibrium constant (Kc) for the formation of NH3 in the following reaction: [1] N2(g) + 3H2(g) 2NH3(g) At equilibrium, the concentration of NH3, H2 and N2 are 1 2 10-2, 3 0 10-2 and 1 5 10-2 M respectively. (c) Explain the following: [2] (i) Hydrolysis of ester (ethyl acetate) begins slowly but becomes fast after some time. (ii) The pH value of acetic acid increases on addition of a few drops of sodium acetate. Comments of Examiners (a)(i) 1. Instead of Ostwald s dilution law many candidates wrote common ion effect . 2. Some candidates wrote only Faraday s Law or Faraday s IInd Law instead of Faraday s 1st Law of electrolysis . 3. Instead of Common ion effect some candidates wrote Le Chatelier s principle . (ii) Some candidates were not able to write one pair of difference correctly. The concept of order of reaction and molecularity was not clear to many candidates. (b)(i) Many candidates mentioned the name of the Law correctly but failed to give the requirement of high pressure in the manufacture of SO3. (ii) Most of the candidates solved the numerical correctly. Some candidates did not substitute the correct values in the formula hence got incorrect value of Kc. The unit of Kc was also not mentioned correctly by many candidates. (c)(i) A number of candidates did not write that acetic acid acts as an auto catalyst. (ii) A few candidates wrote only common ion effect but failed to explain that H+ ion concentration decreases as the dissociation of acetic acid is suppressed. 19 Suggestions for teachers - - - - - Insist upon that students must learn the laws/principles and their applications after understanding thoroughly. Explain Molecularity and order of reaction to students with suitable examples. Le Chatelier s principle and its application in various chemical reactions at equilibrium must be clearly explained to students. Method to calculate equilibrium constants (kc) for different reactions should be given more practice and the answer must be given along with the correct unit. Explain different types of catalysts to students. Explain the relationship of pH value with the concentration of [H+] ion. Common ion effect should be explained clearly. MARKING SCHEME Question 4 (a) (i) (1) Ostwald s dilution law. (2) Faraday s 1st law of electrolysis (3) Common ion effect (ii) (b) (i) Order of reaction Molecularity 1 Order of reaction is the sum of 1 Molecularity is the total exponents in the rate law equation. number of molecules of reactants taking part in a particular step of reaction 2 Order of reaction may have 2 There is always a whole fractional values or it may be zero. number other than zero. 2SO2 + O2 2SO3 Formation of SO3 takes place with decrease in volume, when pressure is increased the equilibrium will shift in that direction where volume is less. Hence the forward reaction is favoured with high pressure. Le Chatelier s principle: When a stress is applied on a system at equilibrium, the system behaves in such a way so as to counteract the stress. (ii) (c) (i) [ 3 ]2 Kc = [ 3 2 ] [ 2 ] = 2 1 2 10 2 2 (1 5 10 2 )(3 0 10 2 )3 Kc = 3 55 10 mol-2 L2 CH3COOC2H5 + H2O CH3COOH + C2H5OH auto catalyst Acetic acid formed during the reaction acts as auto catalyst hence after some time, the rate of reaction increases. (ii) Due to common ion effect, the dissociation of acetic acid decreases hence concentration of H+ ion decreases. The pH value is inversely proportional to H+ ion concentration. Therefore, the pH value of the solution increases. SECTION B Answer any two questions. Question 5 (a) Write the formula of the following compounds: (i) Potassium trioxalatoaluminate(III) (ii) Hexaaquairon(II) sulphate. (b) Name the types of isomerism shown by the following pairs of compounds: (i) [Cu(NH3)4][PtCl4] and [Pt(NH3)4][CuCl4] (ii) [Co(Pn)2Cl2]+ and [Co(tn)2Cl2]+ 20 [2] [1] (c) For the coordination complex ion [Co(NH3)6]3+ (i) Give the IUPAC name of the complex ion. (ii) [2] What is the oxidation number of cobalt in the complex ion? (iii) State the type of hybridisation of the complex ion. (iv) State the magnetic behaviour of the complex ion. Comments of Examiners (a) (i) Many candidates mentioned only K in place of K3 in the formula K3[Al(C2O4)3). In some Suggestions for teachers cases, the orders of central metal atom and - The IUPAC system of nomenclature ligand was reversed. The formula of ligand should be strictly followed while was also not written correctly by many writing the name or formula of any candidates. coordination compound. (ii) Some candidates wrote Fe after H2O in the - Give enough practice to students in formula of the compound e.g. [H2O)6 Fe]SO4 writing the name of coordination instead of [Fe (H2O)6] SO4 which was compounds and their structures. incorrect. - Different types of isomerism in (b)(i) A few candidates wrote coordinate isomerism coordination compounds should be instead of coordination isomerism . Some taught with examples. candidates also wrote linkage and ionisation - Rules of nomenclature must be isomerism which were not correct. explained to students with emphasis (ii) A number of candidates were not able to on correct spelling . identify the ligands in the formula hence, gave - Calculation of oxidation state of the incorrect answers. central metal atom or ion should be explained clearly to students. (c )(i) Many candidates wrote incorrect spelling in the - Explain Valence Bond Theory in IUPAC name of the complex ion. Instead of detail. Give enough practice by taking double aa and double mm , a single a and different examples. a single m was written. Some did not write - How magnetic behavior changes by the word ion . paired and unpaired electrons should (ii) Most of the candidates wrote the correct be discussed. oxidation number i.e. +3 but a few candidates wrote -3 which was not correct. (iii)Many candidates reported sp3d2 hybridisation instead of d2 sp3 hybridisation. (iv) A few candidates wrote the magnetic behaviour of given complex ion as paramagnetic instead of diamagnetic . MARKING SCHEME Question 5 (a) (i) K3[Al(C2O4)3] (b) (ii) (i) [Fe(H2O)6]SO4 Coordination isomerism. (ii) Ligand isomerism. (i) (ii) (iii) (iv) hexaamminecobalt(III) ion. +3 d2sp3 hybridisation diamagnetic (c) 21 Question 6 (a) Give balanced equations for the following reactions: (i) Potassium permanganate is heated with concentrated hydrochloric acid. (ii) Lead sulphide is heated with hydrogen peroxide. [3] (iii) Ozone is treated with potassium iodide solution. (b) Discuss the theory involved in the manufacture of sulphuric acid by contact process. [2] Comments of Examiners (a)(i) Many candidates wrote either incorrect or incomplete chemical equation for potassium permanganate heated with concentrated hydrochloric acid. In some cases, the equations were unbalanced. (ii)A few candidates wrote unbalanced equation for the Lead sulphide heated with hydrogen peroxide. (iii)The chemical equation written for Ozone treated with potassium iodide solution was not balanced and the products written were also incorrect in several cases. In some equations, oxygen was missing. (b) Heat was not mentioned in the equation to obtain SO3. The catalyst was not written. To prepare oleum (H2S2O7), concentrated H2SO4 was not used. Suggestions for teachers - More practice should be given in writing complete and balanced equations. - Oxidising nature of H2 O2 should be emphasized. - How ozone oxidises I- to I2 should be explained. - Emphasis should be given on writing complete and correct equations in the manufacture of sulphuric acid by contact process. MARKING SCHEME Question 6 (a) (b) heat (i) 2KMnO4 + 16HCl 2KCl + 2MnCl2 + 8H2O + 5Cl2 (ii) PbS + 4H2O2 PbSO4 + 4H2O (iii) 2KI + H2O + O3 2KOH + I2 + O2 Theory: S + O2 heat SO2 or Catalytic oxidation of SO2 2SO2 + O2 2 05 4FeS2 + 11O2 2Fe2O3 + 8SO2 2SO3 = SO3 + H2SO4 2S2O7 Conc. oleum H2S2O7 + H2O 2H2SO4 22 Question 7 (a) (i) What are the types of hybridisation of iodine in interhalogen compounds IF3, IF5 and IF7, respectively? Draw the structure of xenon hexafluoride (XeF6) molecule and state the hybridisation of the central atom. (ii) (b) Give the balanced equations for the conversion of argentite (Ag2S) to metallic silver. Comments of Examiners (a)(i) Hybridisation of iodine in interhalogen compounds IF3,IF5 and IF7 respectively was not correctly mentioned by some candidates. The candidates wrote dsp3, d2sp3 and d3sp3 instead of sp3d, sp3d2 and sp3d3. (ii)Many candidates were able to draw six covalent bonds between XeF6, but they did not show the lone pair. Some wrote sp3d2 hybridisation instead of sp3d3 . (b) Many candidates wrote either incorrect or incomplete equations for the conversion of argentite to metallic silver. Candidates did not mention reversible sign for reaction of Ag2S with NaCN. Some candidates wrote incorrect formula of the complex compound. Suggestions for teachers - - - The concept of hybridization should be explained to students by taking various examples in class. Explain the compounds of Xenon with fluorine, their geometry, hybridization and structure in detail. Advise students to write proper balanced equations for extraction of metals. MARKING SCHEME Question 7 (a) IF3 Sp3d IF5 Sp3d2 IF7 Sp3d3 (i) (ii) F F Xe F (b) F F F Sp3d3 hybridisation Ag2S + 4NaCN 2Na[Ag(CN)2] + Na2S 2Na [Ag(CN)2] + Zn Na2[Zn(CN)4] + 2Ag 23 [3] [2] SECTION C Answer any two questions. Question 8 (a) (b) How can the following conversions be brought about: (i) Acetaldehyde to propan-2-ol. [1] (ii) Nitrobenzene to p-aminoazobenzene. [1] (iii) Acetic acid to methylamine. [2] (iv) Aniline to benzene. [1] (i) How will you distinguish between primary, secondary and tertiary amines by Hinsberg s test? [1] Why do alcohols possess higher boiling points as compared to those of corresponding alkanes? Identify the compounds A, B and C: [1] (ii) (c) (i) (ii) Pcl5 H2 Pd/BaSO4 C6 H5 COOH A B H C C H H2 O dilH2 SO4 +Hg SO4 A KCNalc distil H2 [Ni] C B 140o C conc H2 SO4 C Comments of Examiners (a) (i) Most of the candidates wrote the conversion correctly, from acetaldehyde to propan-2-ol by using Grignard s reagent. (ii) Many candidates converted nitrobenzene to aniline but could not correctly complete the conversion upto p-amino azobenzene. (iii) Some candidates did not show heat after the formation of ammonium acetate to give acetamide. (iv) Aniline was directly converted to benzene by some candidates without the formation of benzene diazonium chloride. (b) (i) Solubility of 1 amine in KOH and insolubility of 2 amine was missed by many candidates. For 3 amine, some candidates wrote clear solution , although it does not react. (ii) Many candidates were not able to write the correct reason. (c) (i) Compounds A and B were identified correctly by most of the candidates but compound C i.e. benzoin was not identified correctly by several candidates. 24 Suggestions for teachers - - - - - - More practice should be given in conversion of organic compound s. Every step of conversion must be shown with proper reagent/conditions. Conversion of nitrobenzene to aniline and then C6HsN2Cl is an important aspect of many conversions hence must be practiced in class Insist that candidates give detailed explanation of the reaction by mentioning the solution, precipitate and the effect of addition of KOH How the extent of intermolecular hydrogen bonding affects the boiling point of alcohols should be explained to students. Insist that students must learn all the important name reactions. Stress upon writing the complete balanced with the correct equations conditions/reagents. More practice should be given in identification of organic compounds. [3] (ii) Many candidates identified compounds A and B i.e. CH3CHO and C2H5OH correctly but some were not able to identify compound C. MARKING SCHEME Question 8 (a) (i) CH3 CH3 CH3MgBr + CH3 C = O CH3 C OMgBr H (ii) C6H5NO2 / 6[ ] HOH CH3 C OH + Mg(OH)Br H C6H5NH2 2 + H C6H5N2Cl 0 5 C6H5N2Cl + C6H5NH2 C6H5-N = N - C6H4NH2 + HCl (p-aminoazobenzene) (iii) CH3COOH + 3 CH3COONH4 (b) (iv) C6H5NH2 2 + C6H5N2Cl 5 (i) 1o amine 2o amine 3o amine (ii) (c) (i) 6 5 2 6 5 2 6 5 2 CH3CONH2 3 2, 2 + 2 / CH3NH2 6 6 an insoluble substance ( ) no change(insoluble) clear solution In alcohols, inter molecular hydrogen bonding occurs due to which the boiling point of alcohols are higher than those of corresponding alkanes. Alkanes do not form hydrogen bond. A= C6H5COCl B= C6H5CHO OH O C= C6H5 C C C6H5 H (ii) A - CH3CHO B CH3CH2OH C CH3CH2 O CH2CH3 Question 9 (a) Give balanced equations for the following name reactions: (i) Friedel-Crafts reaction (alkylation) (ii) Williamson s synthesis (iii) Aldol condensation [3] (b) Give chemical test to distinguish: (i) Ethyl alcohol and sec-propyl alcohol (ii) Acetaldehyde and acetic acid [3] 25 (c) (i) Deficiency of which vitamin causes the following diseases: [4] (1) Scurvy (2) Night blindness (ii) Write two differences between globular and fibrous proteins. Comments of Examiners (a) (i) Some candidates wrote Friedel Crafts acylation instead of Friedel-Crafts alkylation. The catalyst use in the reaction was not mentioned in a few cases. (ii)Williamson s synthesis reaction was well attempted by most of the candidates. (iii) Many candidates did not mention the catalyst dilute NaOH. Some wrote the catalyst as, concentrated NaOH; (b) (i)Most of the candidates could give correct observations with Lucas test or with Victor Meyer s test. Some candidates wrote the iodoform test which cannot be used to distinguish ethyl alcohol and sec-propyl alcohol. (ii)Tollens reagent test and Fehling s solution test were mentioned in many answers but the observations in several cases were not correct. (c)(i) (1)This part was answered correctly by most of the candidates. (2)Some candidates wrote vitamin D/ vitamin B instead of Vitamin A . (ii)Many candidates could not differentiate between globular protein and fibrous protein correctly. They wrote that globular proteins are globule like and fibrous proteins are fibre like. A few candidates gave examples only. Suggestions for teachers - - - - - - Advise students to read the question carefully and then answer as per the requirement. Insist that the students learn named organic reactions with the conditions and should write balanced equations with proper conditions/reagents. Insist that students use proper reagents and write correct observations to distinguish between the organic compounds. Instruct students to write a positive test for one compound, its observation and reagent, along-with a negative test for the other compound. Sources and deficiency related diseases caused by different vitamins should be discussed in detail. Differences should be given in a tabular form in terms of action, pH and solubility . Merely examples are not considered as differences. MARKING SCHEME Question 9 (a) Friedel Crafts reaction (alkylation) (i) O + CH3Cl (ii) AlCl3 anly) CH3 O + HCl Williamson s synthesis CH3CH2Br + NaOCH3 CH3CH2 O CH3 + NaBr 26 (iii) (b) O CH3 C + HCH2CHO H dil.NaOH OH CH3 C CH2CHO H aldol Chemical test: (i) Ethyl alcohol and sec-propyl alcohol Victor Meyer s Test Ethyl alcohol gives red colour solution Sec-propyl alcohol gives blue colour solution Lucas Test Ethyl alcohol does not react with Lucas reagent at room temperature and does not produce turbidity. Sec-propyl alcohol reacts with Lucas reagent and produce turbidity within 5 minutes. (or any other suitable test) (ii) Acetaldehyde and acetic acid: Silver mirror test: Acetaldehyde gives silver mirror test with Tollen s reagent. Acetic acid does not give this test. Fehling s solution test: Acetaldehyde gives red precipitate with Fehling s solution. Acetic Acid does not give this test. NaHCO3 Test: Acetic acid gives brisk effervescence of CO2, Acetaldehyde does give this test. FeCl3 Test: On adding aq. FeCl3 to acetic acid, blood red colouration is produced. Acetaldehyde does not give this test. (or any other suitable test) (c) (i) (ii) (1) Vitamin C (2) Vitamin A Globular Protein Fibrous protein 1 These proteins are soluble in water These proteins are insoluble in water 2 They are very sensitive to changes in pH, temperature They are stable to moderate changes in pH and temperature 3 Possess folded spheroids structure Possess thread like structures. Question 10 (a) An aliphatic unsaturated hydrocarbon (A) when treated with HgSO4/H2SO4 yields a compound (B) having molecular formula C3H6O. (B) on oxidation with concentrated HNO3 gives two compounds (C) and (D). Compound (C) when treated with PCl5 gives compound (E). (E) when reacts with ethanol gives a sweet-smelling liquid (F). Compound (F) is also formed when (C) reacts with ethanol in the presence of concentrated H2SO4. (i) Identify the compound A, B, C, D, E and F. (ii) Give the chemical equation for the reaction of (C) with chlorine in the presence of red phosphorous and name the reaction. 27 [4] (b) (c) Answer the following: [3] (i) What is the common name of the polymer obtained by the polymerization of caprolactum? Is it an addition polymer or a condensation polymer? (ii) Name the two organic compounds which have the same molecular formula C2H6O. Will they react with PCl5? If they react, what are the products formed? Give balanced equations for the following reactions: (i) Methyl magnesium bromide with ethyl alcohol. (ii) Acetic anhydride with phosphorous pentachloride. [3] (iii) Acetaldehyde with hydroxylamine. Comments of Examiners (a)(i) Identification of compound A , an aliphatic Suggestions for teachers unsaturated hydrocarbon was not done correctly by some candidates. Many candidates failed to - More practice should be given to identify compound D also. students to solve such question in (ii) HVZ reaction was identified correctly by most which the identification of of the candidates. Some candidates wrote the compounds is based on different reaction with higher homologue of CH3COOH. chemical reactions. (b)(i) Many candidates identified the polymer - Teach polymers, their monomers as correctly as Nylon 6, but the type of polymer well as the type of polymerization in was mentioned as addition instead of the chemical detail. Explain condensation polymer . A few candidates wrote Nylon 66 . properties of alcohols and ether. (ii) Some candidates mentioned that dimethyl ether - Insist that students learn the reactions will not react with PCl5. involving Grignard s reagent. (c)(i) Most of the candidates wrote C2H6 instead of - Stress upon writing complete and CH4. balanced equations. (ii) Chemical equations written by many candidates were not balanced. In some cases, the formula of acetic anhydride was incorrect. (iii) Some candidates failed to write the correct formula of hydroxylamine. Several candidates did not mention that water is formed in the reaction. MARKING SCHEME Question 10 (a) (i) A = CH3C CH B = CH3COCH3 C = CH3COOH D = HCOOH E = CH3COCl F = CH3COOC2H5 28 (ii) CH3COOH + Cl2 HVZ reaction. (b) (i) Red P ClCH2COOH + HCl Nylon 6 Condensation polymer. (ii) CH3CH2OH + PCl5 CH3 CH2 Cl + POCl3 + HCl Ethyl alcohol CH3 O CH3 + PCl5 2CH3Cl + POCl3 Dimethyl ether (c) Balanced equations for the asked reactions: (i) (ii) Br OC2H5 CH3MgBr + C2H5OH CH4 + Mg CH3CO.O.OCCH3 + PCl5 2CH3COCl + POCl3 (iii) CH3CHO + H2NOH CH3 CH = NOH + H2O Note: For questions having more than one correct answer/solution, alternate correct answers/solutions, apart from those given in the marking scheme, have also been accepted. 29 Topics found difficult by candidates Concepts in which candidates got confused Suggestions for candidates Relative molecular mass and mole: Numerical problems of freezing point, van t Hoff factor, degree of dissociation. Chemical kinetics: Calculation of order of reaction, unit of different order reaction, value of rate constant (k), difference between order of reaction and molecularity. Solid state: Calculation of density of unit cells, number of ions in the unit cell of NaCl. Ionic equilibria: Solubility product, salt hydrolysis, ionic product of water. Chemical equilibrium: Le Chatelier s principle Electrochemistry: Electrolytic conductance, calculation of emf and free energy change. Coordination compounds: Nomenclature, isomerism, hybridisation etc. Balanced chemical equations of inorganic compounds. Conversion of organic compound, named organic reactions, chemical test to distinguish between organic compounds. Biomolecules, deficiency of vitamins. Depression in freezing point and the freezing point of solution. Abnormal molecular weights and degree of dissociation. Common ion effect and solubility product. Order of reaction and unit of rate constant (k) for different order reactions. Schottky defect, Frenkel defect and density of crystals. Coordination isomerism and Ligand isomerism. Specific conductance and equivalent conductance. Gibbs free energy and emf of cell in terms of spontaneity. Named organic reactions Polymers and monomers, addition and condensation polymers. Globular and fibrous proteins. Read the question properly and then answer accordingly Avoid selective study Practice numerical problems regularly, solve the numerical stepwise with correct formula and write the answer with correct unit. Practice writing balanced chemical equations with necessary conditions. Nomenclature should always be as per IUPAC norms. Learn proper tests to distinguish different organic compounds. Be neat in your work. Write the correct question numbers. Learn to write the key words in your answer. Time your paper, keep time for rechecking to avoid careless mistakes. Be very regular in your studies. 30 PRACTICAL (PAPER-2) Attempt all questions Question 1 [7] You are provided with two solutions as follows: C-10 is a solution containing 3 2 gms of potassium manganate (VII) (KMnO4) per litre. C-11 is a solution prepared by dissolving 6 85gms of impure sample of oxalic acid crystals (H2C2O4 2H2O) per litre. PROCEDURE: Rinse and fill the burette with potassium manganate(VII) solution C-10 (KMnO4). Pipette out 20 ml or 25 ml of the oxalic acid solution C-11 (H2C2O4.2H2O) in a clean conical flask. To this, add 20 ml of dilute sulphuric acid (H2SO4) C-12, specially provided for this purpose. Warm the contents of the flask to 60oC 70oC. The heating should be continued till the first bubble appears at the bottom of the flask. Remove the conical flask from fire and titrate this solution by running solution C-10 from the burette. Shake the solution constantly till a permanent pale pink colour is obtained. Ensure that the pink colour obtained does not disappear on shaking the contents of the conical flask. Repeat the above procedure to get at least two concordant readings. Tabulate your readings. State: (a) The capacity of the pipette used. (b) The titre value you intend to use in your calculations. Show the titre value to the Visiting Examiner. The equations for the above reactions are as follows: 2KMnO4 + 3H2SO4 + 5H2C2O4 K2SO4 + 2MnSO4+ 8H2O + 10CO2 2 + 2+ 2MnO 4 + 5C2O4 + 16H 2Mn + 10CO2 + 8H2O Relative atomic masses: K = 39 Mn = 55 C = 12 O = 16 H=1 Calculate the following: (i) The molarity of potassium manganate (VII) solution C-10. 31 (ii) The molarity of oxalic acid solution C-11. (iii) The strength of oxalic acid solution in gms per litre. (iv) The percentage purity of the sample of oxalic acid solution. Note: Molarity must be calculated upto at least 4 decimal places. Comments of Examiners (i) A number of candidates did not seem to be aware of the significance of tabulating the readings. Some common errors observed were as follows: Initial and final readings were not written. Many candidates just gave one titre value; they did not have an understanding of concordant readings . Some used average value with a difference between two readings of more than 0.2. Some candidates gave burette readings up to 2 decimal places, which was absurd. Some candidates did not read the question paper carefully and used incorrect solutions in the burette and pipette. Overwriting in the titre value was also observed. Candidates used incorrect formulae and incorrect atomic weights for calculating molecular weight of KMnO4. (ii) Many candidates used incorrect formulae to calculate molarity of hydrated oxalic acid i.e. grams per litre/ molecular weight instead of M1 V1/M2 V2 = n1/n2 = 2/5. Some candidates rounded off the value of molarity in questions (i) and (ii) and expressed only up to two places of decimal instead of four. (iii) Some candidates reported molarity of C-10 and C-11 correctly upto 4 places of decimal but when they substituted the value of C-11 for the calculation of part (iii), they rounded off the value. (iv) Some candidates used incorrect formula for calculating percentage purity of oxalic acid i.e. impure/pure into 100 instead of pure/ impure into 100. 32 Suggestions for teachers Instruct students to use the reading time judiciously and read the question paper carefully. Insist that students tabulate the titre value correctly. Teach them the tabular form and explain the significance of each column. Insist on one trial run and two concordant readings. Instruct students that the average should not be taken and overwriting in the readings should be strictly avoided. Give sufficient practice in calculating molarity (not normality), percentage purity, and water of crystallization for all oxidation -reduction titrations prescribed in the syllabus. Conduct experiments in the laboratory throughout the year under your guidance. Ask students to strictly follow the instructions given in the question paper (pertaining to express the calculated molarities up-to desired number of decimal places). Also, advise the students to round off the value of water of crystallization to the nearest whole number. Instruct students to refer to the formula of the substances and relative atomic masses as given in the question paper. Follow the chemical equation given in the question paper and apply that for the number of moles. MARKING SCHEME Question 1 Calculations: Let the titre value be 24 3 mL (i) Molarity of the solution C 10 (KMnO4) Molarity = (ii) = 3 2 158 = 0 0202 M Molarity of the solution, C - 11 (Oxalic Acid H2C2O4.2H2O) 1 1 = 2 2 1 1 2 2 = 1 M1 Molarity of C-10 2 V1 Volume of C-10 n1 Number of moles of C-10 M2 Molarity of C-11 V2 Volume of C-11 n2 Number of moles of C-11 2 5 Let the titre value be 24 3 mL 0202 24 3 2 25 2 =5 M2 = 0 0491M (iii) Strength in grams per litre of Oxalic acid = molarity molecular weight = 0 0491 126 = 6 1866 g / L (iv) % purity of Oxalic acid = ( 1 ) =90 32% ( 1 ) 33 = 6 1866 6 85 100 Question 2 [4] You are provided with two organic compounds, C-13 and C-14. Perform the experiments given below on each of the two compounds. Record the changes taking place at every step of the experiment. Note the smell of the substance formed, if significant, the colour of the solution obtained, the colour of the precipitate produced and any other observations you may have. State the identity of each compound on the basis of the experiments and observational changes. PROCEDURE: (a) Substance C - 13 (i) Take 2 ml of saturated solution of sodium (NaHSO3) in a test tube. To this, add a few drops of C-13. Shake well and gently warm the contents. (ii) Take 1 ml of C-13 in a test tube and add 1 ml of freshly prepared sodium nitroprusside solution, followed by 4 to 5 drops of dilute sodium hydroxide solution. (iii) Take 2 ml of C-13 in a test tube. To this, add a few crystals of iodine followed by a few drops of sodium hydroxide solution till the colour disappears. Warm the contents gently and cool. (b) Substance C-14 Dissolve 2gm of the substance in 10 ml of hot water and perform the following tests: (i) Take 2 ml of C-14 in a test-tube and add a few drops of sodium bicarbonate solution. (ii) Take 2 ml of C-14 in a test-tube and add a few drops of concentrated sulphuric acid (H2SO4), followed by 5 ml of ethanol and heat the contents on water-bath. (iii) Take 2 ml of C-14 in a test-tube and add neutral ferric chloride (FeCl3) solution. 34 Comments of Examiners (a) C-13 Most candidates did not seem to have practical knowledge of reporting observations of organic compounds. They did not understand the significance of adding a particular reagent. (i) Some candidates reported evolution of a suffocating gas instead of white precipitate obtained. (ii) Many candidates reported red precipitate instead of red /violet solution . It seems that they did not know the difference between solution and precipitate. (iii) A number of candidates wrote brown precipitate or yellow solution or yellow solution disappeared on adding sodium hydroxide instead of yellow precipitate of iodoform . (iv) C-13 was reported as ketone instead of acetone by several candidates. (b) C 14 (i) White precipitate was reported by many candidates instead of effervescence . (ii) Some candidates reported formation of ethyl benzoate instead of fruity smell of ester. (iii)Many candidates reported brown red precipitate instead of buff coloured precipitate . Suggestions for teachers In the laboratory, for identification of organic compounds, observations of iodoform test, ester test etc., explain on the basis of the theory. Advise students to write their experimental observation and inference in tabular form so that observations are sequentially reported instead of reporting the final observation only. Explain the difference between precipitate and solution. Familiarise the students with the concept of colours by demonstrating the experiments in the laboratory. Tell the student to identify the specific organic compound as mentioned in the syllabus and not the family to which they belong. MARKING SCHEME Question 2 (Identification of organic compounds) (a) Substance C-13 (i) White crystalline precipitate is obtained (residue/ solid/mass/ deposit). (ii) A violet / red / wine red colouration is produced (solution/colour). (iii) Yellow precipitate of iodoform having characteristic smell is obtained (residue/solid deposit/mass). (iv) C-13 is identified as acetone. (based on any two correct tests.) (b) Substance C-14: Effervescences / gas evolved/ bubbles evolved. Pleasant/ fruity/ sweet/ smell of Ester was obtained/ peppermint. A buff coloured precipitate is formed. solid/ mass/ residue/ deposit Deduction - C-14 is identified as Benzoic acid. (based on any two correct tests.) 35 Question 3 [4] Analyse qualitatively the substance C-15 which contains two anions and two cations. Identify these ions. (a) While testing for anions you must mention: (i) How the solution/soda extract was prepared. (ii) How the gases were identified. (iii) The confirmatory test for each anion. Show the results as required to the Visiting Examiner. (b) While testing for cations you must mention: (i) How the original solution for group analysis was prepared. (ii) The formal group analysis with pertinent group reagents. (iii) The confirmatory test for each cation. Show the results as required to the Visiting Examiner. Note: Use of qualitative analysis booklet/table is not allowed. 36 Comments of Examiners (a) Wet tests for anions were performed by many candidates using either the aqueous solution or soda extract, instead of neutralized soda extract. Some other errors observed were as follows: (i) Sodium carbonate extract was incorrectly prepared with sodium bicarbonate instead of sodium carbonate. (ii) The confirmatory test for chloride was performed with salt mixture, concentrated sulphuric acid, manganese dioxide and heat instead of neutral sodium carbonate extract. (iii)The confirmatory test for chloride was left incomplete since the solubility of the precipitate in excess of ammonium hydroxide was not done. (iv) A few candidates reported Chromyl chloride test for chloride which is not acceptable. (v) Test for carbonate ion was done with original solution instead of salt mixture. (vi) Many candidates reported heat for carbonate ion which is not required. (vii) Some candidates added lime water to the test solution instead of passing the evolving gas through lime water. (b) Many candidates prepared original solution for cation detection incorrectly. The salt mixture, or if residue of sodium carbonate extract was used, should have been dissolved in hot dilute HCI/ dilute HNO3. Other errors observed were as follows: (i) Absence of groups I, II and III was not reported. (ii) H2S was not boiled off before group III reagents were added. (iii) Most of the candidates did not add concentrated nitric acid in group III, boil and cool before adding group reagents. (iv) The order of preparing the buffer medium in group III was incorrect. 37 Suggestions for teachers Teach students to make the sodium carbonate extract correctly; and tell them the reason for preparing it. Insist that the wet tests for the anion should be performed with neutralized sodium carbonate extract, even if the salt mixture is soluble in water. The correct reagents must be used for test for anions and the test must be completed. Instruct students to never add solution to gas, pass gas into solution. Teach students the steps for preparing the original solution. Concepts of formal group analysis like common ion, buffer and solubility product must be taught thoroughly to the students before doing salt analysis. For mixture analysis guide the students on how to record formal group analysis correctly and meaningfully with pertinent group reagents. Ask students to use reagents and tests that are acceptable. Explain to the students the importance of adding concentrated nitric acid and boiling to convert ferrous to ferric. Removal of H2S before group III and V must be taught clearly. For systematic analysis of given mixture, clearly explain the importance of reporting of both the groups which are present and which are absent. Advise students to follow correct procedure for doing confirmatory test for each cation (It must be done with the group precipitate obtained for that cation and not with original solution). (v) The dirty white precipitate obtained in group IV was supposed to be used for confirmatory test of Zinc by dissolving in minimum quantity of dilute HCl and boiling, instead, candidates used the original solution. (vi) In group zero several candidates used original solution instead of salt mixture and omitted heat in the test. (vii) Many candidates performed the confirmatory test for ammonium ion incorrectly by adding Nessler s reagent to the test solution instead of passing the gas evolved into Nessler s reagent. MARKING SCHEME Question 3 Substance C-15 Mixture C-15 contains ammonium chloride and zinc carbonate in the ratio (1:1) by mass. Carbonate ion Chloride ion Presence of Group Zero Confirmation of ammonium ion Presence of group IV Confirmation of zinc ion Preparation of sodium carbonate extract Preparation of original solution Details of tests: Test for carbonate: Sample/ C-15 /Salt/ mixture + dil. Sulphuric acid brisk effervescence. The gas when passed through lime water turns it milky/ turbid CO32- ion present Preparation of sodium carbonate extract. 1 part salt + 4 parts sodium carbonate + distilled water boil/warm/heat, cool and filter Test for chloride: To the Na2CO3 extract acidified with dil HNO3 / acetic acid/ dil H2SO4 Add silver nitrate solution. Curdy white precipitate is obtained which is soluble in excess of ammonium hydroxide, chloride ion is confirmed. Preparation of original solution in dilute HCl / dilute HNO3. Group Zero: C15/ Sample/ Salt /mixture when heated with sodium hydroxide, gives pungent smelling gas. Zero group present. Confirmatory test for NH4+ On placing a glass rod dipped in concentrated HCl near the mouth of the test tube gives dense white fumes. OR 38 Gas evolved in zero group is passed through Nessler s reagent which turns brown. *Group IV: (This is accepted only if absence of group I, II and III is correctly shown). Passing H2S gas through the third group solution gives dirty white precipitate. Group IV is present. Confirmatory test for Zn2+: Filter the above precipitate and dissolve it in dilute/ concentrated HCl with boiling. To the clear solution add NaOH solution drop by drop white precipitate appears which dissolves in excess of NaOH. OR add potassium ferrocyanide solution bluish white precipitate is obtained. (Zn2+ ion is confirmed). Original solution + dilute HCl *To No precipitate No precipitate Group I absent Group II absent *Boil No precipitate Group III absent Dirty white precipitate Group IV present the above solution of Group I pass H2S gas OR If OS is made in HNO3/ OS in dil HCl/ OS in HCl pass H2S gas off H2S gas from the above solution (add concentrated HNO3 and boil). Cool and add NH4Cl solid and NH4OH solution OR OS + NH4Cl solid and NH4OH solution only if 1 and 2 are correctly mentioned *Pass H2S gas through the above solution. OR OS + NH4Cl solid and NH4OH solution + Pass H2S gas only if 1,2 and 3 are correctly done. 39 Topics found difficult by candidates Concepts in which candidates got confused Concepts of molarity based on (grams/litre)/ molecular weight for pure substances and morality based on titre value. Difference between precipitate and solution. Principles of formal group analysis. Certain colours like buff precipitate. Solubility of mixture or preparation of original solution for group separation. Use of neutralized sodium carbonate extract for confirmatory test of anion. Identification of gases. Systematic analysis from group zero to IV and reporting absence and presence of groups. 40 Suggestions for candidates Learn thoroughly the basics of volumetric calculation before performing titrations. Titre value must be concordant and not average. Understand the chemistry involved in the precipitation of respective groups, and the importance of formal analysis. Learn formal group separation thoroughly with principles like buffer, common ion and solubility products. Do not forget the use of concentrated nitric acid in group III. Also understand why it is being used. Plan before writing formal group analysis. Listen to the teacher s instructions carefully, read the experiment thoroughly and then perform it. Develop a habit of observation and note them down correctly and to the point. Remember to tabulate your readings neatly, keeping in mind concordant readings not average and avoid overwriting in the tabular column. Do not round off molarity values, report to minimum four decimal places (check scope of syllabus). Follow the molecular formula given in the question paper, whether it is hydrated or anhydrous. Do not heat in the test of carbonate ion. Use of nitric acid and heating before adding reagents of group III and performing the confirmatory test for cation with group precipitate is a must. Remember to pass gas through a reagent and never add reagent to gas. 41

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Additional Info : ISC Class XII Analysis Of Pupil Performance 2017 : Chemistry
Tags : ISC Board, Class 11th, Class 12th, NDA/NA Entrance Examination  

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