Copy of 2024 ANDSS 4E Chem Prelim P3 ANS

2024 ANDSS 6092 Prelim Paper 3 - Mark Scheme 1 (a) Read all the instructions carefully before starting the experiments in Question 1. You are provided with salt solution W in a boiling tube. You will carry out tests on W to deduce its identity. You should test and identify any gases evolved. Record all your observations in the table. The volumes given below are approximate and should be estimated rather than measured, unless instructed otherwise. test observations Test 1 Put about 1 cm depth of solution W in a clean test-tube. Add 1 cm depth of aqueous sodium hydroxide. Gently warm the mixture. - no ppt / no visible change / no apparent change - pungent gas turns damp red litmus blue [1] - ammonia / NH3 is produced [1] Test 2 Put about 1 cm depth of solution W in a clean test-tube. Add an equal depth of dilute nitric acid and then add a few drops of aqueous barium nitrate. - white ppt [1] Test 3 Put about 1 cm depth of solution W in a clean test-tube. Add an equal depth of dilute nitric acid and then add a few drops of aqueous silver nitrate. - no ppt / no white ppt / no visible change / no apparent change / solution remains colourless [1] [4] (b) Deduce the identity of salt W. ammonium [1] sulfate [1] OR (NH4)2SO4 [2] [2]

2 (a) (i) shift titration 1 titration 2 average volume used for accuracy mark [SS results] 01 29.80 29.70 29.75 02 30.00 29.80 29.90 03 29.80 29.80 29.80 04 29.80 29.80 29.80 [5] Marking Points format Record initial burette readings, final burette readings and volume added with correct headings and units in a titration table. [1] decimal places All burette readings for all accurate titres in titration table are recorded to nearest 0.05 cm3. [1] accuracy Supervisor’s result 1. For the average titre (of consistent readings) within 0.20 cm3 of supervisor’s average value scores 2 marks. 2. For the average titre (of consistent readings) within 0.30 cm3 of supervisor’s average value scores 1 mark. [2] concordance At least two titre values are within 0.20 cm3. [1] (ii) 1m awarded only when working for average volume is shown, correct to 2 decimal places (reject 3 sf) 1 (b) No. of moles of KOH = 0.100 x 25.0 1000 = 0.00250 mol / 2.5 x 10-3 mol 1 (c) ratio of KOH : H2C4H4O5 = 2 : 1 no. of moles of malic acid = 0.00250 ÷ 2 = 0.00125 mol 1 (d) concentration of P = 0.00125 ÷ [(ave. vol. of P) ÷ 1000)] Note: allow ecf from (c) 1

2 (e) average concentration of malic acid in apple juice = 4.50 ÷ [2(1) + 4(12) + 4(1) + 5(16)] = 4.50 ÷ 134 either step shown as working [1] = 0.03358208955 (at least 5 sf)  0.0336 mol/dm3 [1] 2 (f) Method 1: average volume of apple juice = [(d) x 200 1000)] ÷ (e) (at least 5 sf in working) [1] (final answer to 3sf) dm3 [1] 2 Meth