ACJC H2 Chem 2021 Prelim Paper 1 Solutions - Final

2 ANGLO-CHINESE JUNIOR COLLEGE DEPARTMENT OF CHEMISTRY Preliminary Examination CHEMISTRY 9729/01 Higher 2 Paper 1 Multiple Choice 14 September 2021 1 hour Additional Materials: Multiple Choice Answer Sheet Data Booklet READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. Write your name, index number and tutorial class on the Answer Sheet in the spaces provided unless t his has been done for you. There are thirty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the separate Answer Sheet. Read the instructions on the Answer Sheet very carefully. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this booklet. The use of an approved scientific calculator is expected, where appropriate. This document consists of 15 printed pages and 1 blank page. 9729/01/Prelim/21 ANGLO-CHINESE JUNIOR COLLEGE © ACJC 2021 Department of Chemistry [Turn over

2 © ACJC 2021 9729/01/Prelim/2021 [Turn over 1 In an organic synthesis, a 62% yield of product is achieved. Which conversions are consistent with the given information? 1 74 g of butan-2-ol (Mr = 74.0)  44.64 g of butanone (Mr = 72.0) 2 72 g of butanal (Mr = 72.0)  45.88 g of butan-1-ol (Mr = 74.0) 3 56 g of but-2-ene (Mr = 56.0)  37.20 g of ethanoic acid (Mr = 60.0) A 1, 2 and 3 B* 1 and 2 only C 2 and 3 only D 1 only Answer B 62% yield ⇒ actual yield theoretical yield ×100 = 62% Option 2 CH3CH2CH2CHO  CH3CH2CH2CH2OH 72 72 = 1 mol 45.88 74 = 0.62 mol % yield = 0.62 1 ×100 = 62 % correct Option 1 CH3CH(OH)CH2CH3  CH3COCH2CH3 74 74 = 1 mol 44.64 72 = 0.62 mol % yield = 0.62 1 ×100 = 62 % correct Option 3 CH3CH=CHCH3  2CH3COOH 56 56 = 1 mol 37.2 60 = 0.62 mol % yield = 0.62 2 ×100 = 31 % Incorrect

3 © ACJC 2021 9729/01/Prelim/2021 [Turn over 2 Use of the Data Booklet is relevant to this question. Sodium percarbonate, x(Na2CO3).y(H2O2) is a chemical that is used in cleaning powders. When dissolved in water, it releases hydrogen peroxide and sodium carbonate, both of which are useful cleaning agents. A sample of 10.0 cm 3 of 0.100 mol dm ‒3 sodium percarbonate requires 20.0 cm 3 of 0.100 mol dm‒3 of acidified KI before the end point of titration is reached. Another identical sample releases 48.0 cm 3 of carbon dioxide at room conditions on reaction with excess acid. What are the values of x and y respectively? x y A* 2 1 B 3 2 C 2 3 D 3 1 Answer A Amount of sodium percarbonate = 10/1000 x 0.10 = 0.001 mol As 1 mole of sodium percarbonate will produce y mole of H2O2 and x mole of Na2CO3 Amount of H2O2 = 0.001y mol Amount of Na2CO3 = 0.001x mol H2O2 + 2H+ + 2I-  I2 + 2H2O Amount of KI reacted = 20/1000 x 0.100 = 0.002 mol Since KI: H2O2 is 2:1, Amount of H2O2 = 0.001 mol, thus y = 1 Na2CO3 + 2H+  H2O + CO2 + 2Na+ Amount of carbon dioxide released = 48.0/24000 = 0.002 mol Since Na2CO3: CO2 is 1:1 Amount of Na2CO3 = 0.002 mol, thus x =2

4 © ACJC 2021 9729/01/Prelim/2021 [Turn over 3 An ion X2+ contains 24 protons. What is the electronic configuration of X3+? A* 1s2 2s2 2p6 3s2 3p6 3d3 B 1s2 2s2 2p6 3s2 3p6 3d4 C 1s2 2s2 2p6 3s2 3p6 3d1 4s2 D 1s2 2s2 2p6 3s2 3p6 3d5 4s1 Answer A X contains 24 electrons. Electronic configuration of X is 1s2 2s2 2p6 3s2 3p6 3d4 4s2 X2+ contains 22 electrons. Electronic configuration is 1s2 2s2 2p6 3s2 3p6 3d4 Loss of 1 electron from X2+ leads to the formation of X3+. Hence, electronic configuration is 1s2 2s2 2p6 3s2 3p6 3d3 4 A sample of the element calcium was vaporised, ionised and passed through an electric field. It was observed that a beam of 40Ca2+ particles gave an angle of deflection of +4o. Assuming an identical set of experimental conditions, by what angle would a beam of 19F¯ particles be deflected? A +2o B ‒ 2o C +4o D* ‒ 4o Solution Angle of Deflection ∝ 𝐶ℎ𝑎𝑟𝑔𝑒 𝑀𝑎𝑠𝑠 For 40Ca2+, +4⁰ ∝ + 2 40 and therefore for 19F‾, ‒ 1 19 ∝ ‒4o Source +4o 40Ca2+ + ‒

5 © ACJC 2021 9729/01/Prelim/2021 [Turn over 5 Use of Data Booklet is relevant to this question. X, Y and Z are isotopes of three different elements. An isotope of Argon has a mass number of 41 and the same number of neutrons in the nucleus as each of the isotopes X, Y and Z. The species Ar, X+, Y2+ and Z3+ all contain the same number of electrons. What are the mass numbers of X, Y and Z? X Y Z A 42 41 40 B 40 41 42 C 44 43 42 D* 42 43 44 Solution 𝐴𝑟18 41 𝐾+ 19 42 𝐶𝑎2+ 20 43 𝑆𝑐3+ 21 44 No. of electrons 18 18 18 18 No. of neutrons 23 23 23 23 6 Equimolar amounts of the liquids, water and ethanol, are mixed together at 20 oC. The original intermolecular forces in the pure liquids are disrupted an d weaker intermolecular forces between water and ethanol are made simultaneously. The boiling point of water is 100 oC. The boiling point of ethanol is 78 oC. Which row is correct? initial temperature of mixture boiling point of mixture A above 20 oC above 78 oC B above 20 oC below 78 oC C below 20 oC above 100 oC D* below 20 oC below 100 oC

6 © ACJC 2021 9729/01/Prelim/2021 [Turn over Solution The reaction is endothermic because weaker intermolecular forces are made. Therefore the initial temperature of mixture is below 20 oC. Molecules will vaporise more easily since weaker intermolecular forces are made. Therefore the boiling point of mixture is below 100 oC. 7 The pKa and solubility values of maleic acid and fumaric acid are as follows: maleic acid fumaric acid pKa1 1.94 3.03 pKa2 6.22 4.54 solubility / g per 100g water 78.8 0.49 Which statements are correct? 1 The pKa1 value of maleic acid is less than the pKa1 value of fumaric acid because the conjugate base of maleic acid can form intramolecular hydrogen bond, while that of fumaric acid cannot form such intramolecular hydrogen bond. 2 The pKa1 value of maleic acid is less than the pKa1 value of fumaric acid because maleic acid is much more soluble in water than fumaric acid. 3 The pKa2 value of maleic acid is greater than the pKa2 value of fumaric acid because intramolecular hydrogen bond prevents the conjugate base of maleic acid from deprotonating the second hydrogen atom. A 1, 2 and 3 B 1 and 2 only C* 1 and 3 only D 2 only Solution Option 1 is correct: Conjugate base of maleic acid can form intramolecular hydrogen bond. Therefore it is stable, and more readily formed than that of fumaric acid. Maleic acid is more acidic than fumaric acid. Hence pKa1 is lower. Option 3 is correct: Conjugate base of maleic acid can form intramolecular hydrogen bond. It is stable, and not likely to lose the second hydrogen atom as H+. Maleic acid is less acidic than fumaric acid. Hence pKa2 is higher. Option 2 is wrong: Acid Strength does not depend on solubility.

7 © ACJC 2021 9729/01/Prelim/2021 [Turn over 8 Which diagram does not describe the behaviour of a fixed mass of an ideal gas? ( = density of the gas, T = temperature measured in K) A C* B D Answer C For a gi