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 p