HCI Prelim P4_Mark_Scheme

2018 HCI C2 H2 Chemistry Preliminary Exam / Paper 4 Paper 4 1 (a) (b) (i) Tabulates initial and final burette readings and volume added in each of the tables. Tables have correct headers and units. [1] Mark is lost if any final and initial burette readings are inverted. All the burette readings are recorded to the nearest 0.05 cm3. [1] Has at least two uncorrected titres (refers to student’s computed titres) for end-point within 0.10 cm3. [1] (b) (ii) Student correctly obtains appropriate ‘average’ from any experiments with uncorrected end-point titre values within 0.10 cm3. [1] Mark is lost if the titres used are not identified either in the table (by, for example, a tick) or in a calculation or identified wrongly or no titre within 0.10 cm3. Mark is lost if there are arithmetic errors in the titration results table. Mark is lost if the candidate made an arithmetic error in the calculation of the mean titre. Calculate the Supervisor’s and the student’s scaled mean titre values using the following expression: Scaled mean titre = 49.25 × mean titre volume of FA 1 diluted Calculate the difference between the student’s scaled mean titre value and the Supervisor’s scaled mean titre value. Give 3 marks if this difference is ≤ 0.2 Give 2 marks if this difference is > 0.2 but ≤ 0.3 Give 1 mark if this difference is > 0.3 but ≤ 0.5 Give 0 marks if this difference is > 0.5 [3] (c) (i) nO 2 produced by 1cm3 of H2O2 = 0.75 ൊ 24000 nH2O2 in 1 cm3 = 0.75 ൊ 24000 × 2 = 6.25 x 10−5 mol [1] [H2O2] = ans ÷ 1/1000 = 0.0625 mol dm−3 [1] (ii) nH2O2 in 25.0 cm3 = 0.0625 x 25/1000 = 1.56 x 10−3 mol = nI2 nS2O32− = 1.56 x 10−3 × 2 = 3.125 × 10−3 mol [S2O32−] = 3.125 × 10−3 ÷ 31.50/1000 = 0.0992 mol dm−3 [1] (iii) 1. nS 2O32− = 0.0992(or ecf) × (titre in (b) ÷ 1000) 2. nKIO 3 = ANS ÷ 6 3. [KIO 3] in FA 5 = ANS ÷ 25/1000 Award 1 mark for the correct use of the mole ratio in step 2 and 1 mark for steps 1 and 3 if both correct (allow ecf for step 3 if step 2 is wrong). [2] HWA CHONG INSTITUTION 2018 C2 H2 CHEMISTRY PRELIMINARY EXAM SUGGESTED SOLUTIONS

(iv) [KIO3] in FA 1 = ANS(c(iii)) x (250 ÷ volume of FA1 used) [1] (d) The transition metal would cause the hydrogen peroxide to decompose leading to less hydrogen peroxide reacting with the iodide. [1] Less iodine would be produced, so volume of sodium thiosulfate used will be lower than expected. Hence, it would give the false impression that the concentration of thiosulfate is higher than what it actually is. So the concentration of potassium iodate calculated will be higher than expected. [1] 2 (a) (c) Axes correct way round + correct labels + units + scale (must be chosen so that plotted points occupy at least half the graph grid in both x and y directions). Sensible scales must be chosen. Awkward scales (e.g. 3:10) are not allowed. [1] Plottin