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] Plotting – all points within ± ½ small square. Check all points and put ticks if correct. [1] Draws best fit straight line through plotted points. [1] From 2(c) Correctly extrapolates ΔT for M g of potassium iodide, ΔTM, from plotted graph at the point M = mass FA 6 dissolved 2 Or correctly calculates using ΔT per gram of potassium iodide by finding gradient Working on the graph, e.g. dotted lines or otherwise, must be shown. [1] (b) Record mass of empty weighing bottle, mass of weighing bottle + FA 6 , mass of weighing bottle + residual solid, mass of FA 6 used. and Table has correct headers and units. [1] Follows instructions – weighs 5.95 g to 6.04 g of FA 6 Final mass of FA 6 dissolved may be out of range, but initially weighed mass must be within range. [1] Records initial water temperature and final temperature of solution, and Table has correct headers and units. [1] All temperature recorded and calculated ΔTx value are to 0.5 oC, and all mass recorded calculated mass of FA 6 used are to 2 d.p. [1] Correctly calculates ΔTx. [1] (c) Method 1: Correctly calculates change in temperature due to lithium chloride, ΔTx + ΔTM = ΔTLiCl [1] Correctly calculates the value of change in temperature of water per gram of lithium chloride =

2018 HCI C2 H2 Chemistry Preliminary Exam / Paper 4 ΔTLiCl where M = mass of FA 6 dissolved ÷ 2 M [1] Method 2: Calculates the value of change in temperature of water per gram of FA 6, ΔT x mass of FA 6 dissolved, m Since FA 6 contains equal masses of lithium chloride and potassium iodide, ΔTx/m gives the total temperature change for 0.5 g of lithium chloride and 0.5 g of potassium iodide. Therefore, to obtain the total temperature change for 1 g of each compound, ΔT x/m needs to be multiplied by two. The temperature change per gram of lithium chloride = 2(ΔT x/m) − (−ΔT per gram of potassium iodide) [1 mark for addition; 1 mark for correct computation] 3 (a) Magnesium [1] (b) OR Correct energy cycle [1] From correct energy cycle, ΔHr = ΔH2 – ΔH1 [1] (c) M1: [Justification of quantities] Either choose a suitable mass of metal powder or suitable volume of HC l, and show calculations to obtain the minimum quantity of the other reagent needed, and state chosen quantities of reagents. [1] M2: [Data to be collected] CORRECT temperature taking procedure. Describes taking temperature readings over time from before addition of metal until at least 6 extra readings after maximum temperature is reached. [1] Mg(s) + Cu(s) + 2HCl (aq) CuCl2(aq) + H2(g) + Mg(s) MgCl2(aq) + Cu(s) + H2(g) ΔH2 ΔH1 ΔHr MgCl2(aq) + Cu(s) + 2HCl (aq) CuCl2(aq) + H2(g) + MgCl2(aq) 2HCl(aq) + CuCl2(aq) + Mg(s) ΔH1 ΔH2 ΔHr

M3: [Sketch of graph] Sketch of cooling curve and show correctly how to extrapolate min/max or ΔT temperatures from sketch. Axes must be temperature against time. [1] M4 & M5: [Details in procedure] – dot-tick marking, up to 2m, no half marks • weighing balance for weighing metal powder • suitable apparatus (burette/pipette/measuring cylinder) for measuring volume of HCl(aq) • thermometer for measuring temperature • measures taken to ensure accuracy of expt / minimise hazards, e.g. stirring/ use lid/ double styrofoam cup / beaker for support [2] M6: [Processing of results] Correct sample calculation for obtaining ΔH2. Answer may differ depending on whether metal or acid chosen as limiting reactant, and students’ chosen metal. Q = 1.0 × V HCl × 4.18 × (Tmax – Tmin) = a J [if metal is limiting] ΔH2 = − a ÷ 1000 kJ mol–1 mM ÷ Ar [if acid is limiting] ΔH2 = − a × 2 ÷ 1000 kJ mol–1 VHCl ÷ 1000 × 1.0 [1]

2018 HCI C2 H2 Chemistry Preliminary Exam / Paper 4 Sample procedure: Amount of Mg in 1 g of magnesium = 1 ÷ 24.3 = 0.0412 mol Volume of HCl required for complete reaction = (0.0412 × 2) × 1.0 × 1000 = 82.3 cm3 Volume of HCl to be used = 100 cm3 (excess) Using a weighing balance and weighing bottle, weigh accurately about 1.0 g of Mg metal. Using a measuring cylinder, transfer 100 cm 3 of HCl(aq) into a Styrofoam cup. Place the Styrofoam cup into another Styrofoam cup and place both into a 250 cm 3 beaker for stability. Cover with a lid and insert a thermometer through the lid. Record the temperature of the solution, and start the stopwatch. Take the temperature every 30 seconds until 2.5 min. At 3 min, add the magnesium from the weighing bottle to the HCl(aq) in the Styrofoam cup. From 3.5 min, take temperature readings every 30 seconds until there are at least 6 readings after the temperature starts decreasing linearly. Reweigh the weighing bottle to obtain the mass of magnesium used, m M. Plot a graph of temperature /oC against time /min. (graph sketch) Q = 1.0 × 100 × 4.18 × (T max – Tmin) = a J ΔH2 = − a ÷ 1000 kJ mol–1 mM/24.3 (d) H2(g) / metal powder is flammabl