2025 Prelim P4 Mark Scheme (for exchange) H2 Chem HCI

1 HCI C2 H2 Chemistry Preliminary Examination/Paper 4 Mark Scheme Hwa Chong Institution 2025 C2 H2 Chemistry Prelim Exam Paper 4 Mark Scheme Qn Mark Scheme 1(a) test observations To 1 cm depth of FA 1, add 1 cm depth of dilute H2SO4. No effervescence To 1 cm depth of FA 1, add aqueous barium nitrate (dropwise). White ppt To 1 cm depth of FA 1, add 1 (or 2) drops of aqueous silver nitrate. followed by aqueous NH3, until there is no further change Pale cream/ cream/ cream-yellow/ pale yellow ppt ppt partially soluble/ insoluble in NH3(aq) [1] for each test and correct observation M1 M2 M3 M4 anion: SO42– [1] and Br– [1] M5 M6 1(b)(i) test observations Transfer all of the solid sample of FA 4 into a boiling tube. Add all of FA 3 to this boiling tube. Gently warm the boiling tube. Turn off the Bunsen flame. White solid dissolves to give a colourless solution. [½] (–½ overall if “effervescence/ gas/ gas bubbles” is written here) Handle the hot boiling tube with care. Add FA 2 to the mixture in three separate portions using a spatula, shaking after each separate addition. Add all the remaining FA 2 to the mixture. Using the glass filter funnel , filter the mixture into a test tube and leave the filtrate to stand. The filtrate contains compound X. Retain this filtrate for use in (b)(ii). Effervescence [½] Gas gives white ppt in limewater [½] Gas is CO2 [½] Colourless/ pale pink/ reddish-pink/ red/ purple-red filtrate [½] Black residue [½] M7 M8 M9 1(b)(ii) test FA 5 filtrate from (b)(i) To 1cm depth of FA 5 in a test tube, add aqueous ammonia dropwise, until in excess. Repeat using filtrate from (b)(i). off-white ppt [½] insoluble in excess [½] turns brown on standing/ on contact with air [1] light brown / brown ppt [½] insoluble in excess [½] turns darker brown on standing/ on contact with air M10 M11 M12 1(b)(iii) Ion present is: Mn2+ [½] Explanation: off-white ppt formed with aqueous NH 3, ppt insoluble in excess aqueous NH3 and turns brown on standing/ on contact with air [½] M13

2 HCI C2 H2 Chemistry Preliminary Examination/Paper 4 Mark Scheme 1(b)(iv) Identity of ion: Mn3+ [½] (Incomplete) reduction of MnO2 (ZO2) (FA 2) to Mn3+ / some of the MnO 2 were reduced (in (b)(i)) to Mn3+ only. Mn3+ formed the brown ppt when aqueous NH3 was added. [½] M14 1(ci) test Observations 1 Add 2 cm depth of aqueous silver nitrate to a clean boiling tube. Add 1 cm depth of aqueous sodium hydroxide slowly to the same tube. brown/black ppt formed Add aqueous ammonia slowly, with shaking, until the precipitate just dissolves. You may use a clean glass rod to help dissolve the precipitate. brown ppt dissolved to form a colourless solution Add 2 cm depth of FA 6 to this mixture and shake the boiling tube. Place the boiling tube in the test-tube rack and leave it for 3 minutes. silver mirror formed 2 Add about 1 cm depth of FA 6 in a test-tube. To this test -tube add 8 drops of sodium hydroxide solution followed by iodine solution, dropwise until no further change is seen. yellow ppt formed 4 correct observations: 3m 3 correct observations: 2m 1 or 2 correct observations: 1m M15 M16 M17 1(c)(ii) The reagent undergoes reduction [½] from [Ag(NH3)2]+ to Ag since oxidation state of Ag decreases from +1 to 0. [½] M18 1(c)(iii) M19

3 HCI C2 H2 Chemistry Preliminary Examination/Paper 4 Mark Scheme Qn Mark Scheme 2(a) actual time t /min initial burette reading /cm3 final burette reading /cm3 volume of FA 11 added /cm3 min s 1 4 1 4.0 0.60 17.60 17.00 2 8 1 8.0 17.60 32.50 14.90 3 12 30 12.5 3.95 16.80 12.85 4 16 1 16.0 16.80 28.00 11.20 5 20 1 20.0 28.00 37.70 9.70 1. [1] Headers and units (Reject “sec” / “minutes” / “seconds” as the units) This mark is lost if t (the decimal time) is recorded in s. The question stated that t is to be recorded in min. M20 2. [1] Records all burette readings and volume of FA 11 to 0.05 cm3 AND actual times to min and s (whole numbers) This mark is lost if final burette reading < initial burette reading / inverted. This mark is lost if final burette reading > 50.00 cm3. M21 3. [1] Correctly calculates t /min to 1 d.p. This mark is lost if t (the decimal time) is calculated in s. This mark is lost if the actual time in min and s is not recorded. This is because the marker is then unable to check whether the student has correctly calculated t M22 4. [1] 5 sets of titration results AND first aliquot taken at 3.5–4.5 min AND student chooses well -spaced values of chosen times (minimum 3 min intervals) AND longest time is at most 20.4 min This mark is lost if any titres remain the same or rise as t increases. This mark is lost if there is any error in calculating the volume of FA 11. M23

4 HCI C2 H2 Chemistry Preliminary Examination/Paper 4 Mark Scheme 2(b)(i) Volume of Na2S2O3 /cm3 t /min 1. [1] Axes correct way round AND correct axes labels including units M24 2. [1] Scale is chosen such that the plotted points occupy at least half the graph grid in the x direction AND in the y direction. M25 3. [1] Plots all points correctly (give ±½ small square allowance). M26 4. [1] Draws best-fit straight line. Judge best-fit by scatter of points about the candidate’s line, there must be a fair scatter of points on either side of the line AND allow only 1 point to be >1 small square diagonally away from the best -fit straight line. This mark is lost if the graph line is not straight. M27 2(b)(ii) The order with respect to [I2] is zero as graph line is straight, so no change in rate (or rate remains constant) as [I2] decreases. This mark is lost if the graph line is not straight. M28 2(b)(iii) Clear indication of correct coordinates from graph (give ±½ small square allowance) AND gradient correctly calculated. e.g. gradient = 18.8 −9.8 0−19 = –0.474 cm3 min–1 This mark is lost if the graph line is not straight. M29

5 HCI C2 H2 Chemistry Preliminary Examination/Paper 4 Mark Scheme 2(b)(iv) 1. [1] Line correctly extrapolated to meet y-axis. AND Vmax correctly read (give ±½ small square allowance) This mark is lost if the x-axis does not start at t = 0 min. This mark is lost if the y-intercept (Vmax) is not within the grid. Reject use of “y = mx + c” method to calculate Vmax because the question stated that the Vmax is to be read at the y-intercept. M30 2. [1] Accuracy Mark: 1m if student’s Vmax is 17.80 – 18.80 cm3 (If the student’s x-axis does not start at t = 0 min or student draws a curve for the graph, excel was used to determine the student’s V max value, using the student’s data table. Accuracy Mark was then awarded accordingly.) M31 3. [½] n(S2O32–) = Vmax ÷ 1000 × 0.0100 [½] n(I2) in 10 cm3 aliquot = n(S2O32–)  ½ e.g. n(S2O32–) = 18.8 1000 × 0.0100 = 1.88  10–4 mol n(I2) in 10 cm3 aliquot = 1.88  10–4  ½ = 9.40  10–5 mol M32 4. [½] n(I2) in 100 cm3 reaction mixture (or in 50 cm3 FA 8) = n(I2) in 10 cm3 aliquot × 100/10 [½] [I2] in FA 8 = n(I2) in 50 cm3 FA 8 ÷ (50/1000) e.g. n(I2) in 100 cm3 = 9.40  10–5  10 = 9.40  10–4 mol [I2] in FA 8 = 9.40 × 10−4 50/1000 = 0.0188 mol dm–3 M33 2(b)(v) Correct calculation of tmax = Vmax / | gradient | e.g. tmax = 18.8 0.474 = 39.7 min M34 2(c) rate = k[CH3COCH3][H+] or rate = k[propanone][H+] M35 2(d) If the concentration of iodine is very much lower than that of propanone (and that of acid), very little propanone (and acid) is/are removed from the reaction mixture. So the concentration(s) of propanone (and acid) remain effectively constant. M36 2(e) The excess NaHCO3 neutralises / removes the acid (H+ or H2SO4) catalyst. So reaction 1 stops / drops