VS 2023 O Level Pure Chemistry P3 Answer Scheme

1 Victoria School2023 O Level ChemistrySuggested Answer to Chemistry Paper 3QnSkillAnswerMark1(a)(i)PDO MMO Results table:experimentvolume of A / cm3volume of B / cm3volume of deionized water / cm3reaction time / srate of reaction / s-111.010.00.02934.521.02.08.02733.6631.04.06.09310.841.06.04.05418.551.08.02.03826.31m- volume of B must be 4.0, 6.0 and 8.0 cm31m- total volume of water and B = 10cm31m-volume of water and B and time given to the correct s.f1m- accuracy for first reading 1m- trend between volume of B and time 51(a)(ii)MMOSee table aboveAll values calculated correctlyAll values to three significant figures111(b)(i)ACErate of reaction / s-1 volume of B / cm3

2 1m-Scale chosen is appropriate- no awkward scale(no 3, 8, 1.5, 6, 7, etc) (uniform scale chosen to use more than half of each axis)1m- graph of best fit - equal distribution of points. -To draw curve if it looks like one- (only 1 anomalous point allowed)line must pass through originpoints not on the line must be balanced on either side and any anomalous points ignored) *1m-Axes (include labels with units)1m all Points plotted correctly (to half a small square) 11111(b)(ii)ACEvolume of B / cm3 rate of reaction / s-1 Correct volume of B read from graph with units shown (5.25 cm3 ± 0.10). Working on the graph111(c)(i)ACEMethod 1: conc. of B = 10/11 x 0.1 = 0.0909 mol/dm3OrMethod 2Concentration of B: 0.1 mol/dm3 Volume: 10 cm3No. of moles of B: 0.1 x 10/1000 = 0.001 molVolume of reaction mixture: 10+1= 11cm3Concentration of B = 0.001 = 0.0909 mol/dm3 (11/1000) 1

3 1(c)(ii)ACEPDOConcentration of B = 0.09091 mol/dm3 / 5 = 0.0182 mol/dm3appropriate units in final answers in (c)(i), (ii) (mol/dm3)Rejected answer: concentration of B = 2/11 x 0.1 = 0.0182 as question asked for deduction. So students have to use the answer from ci.111(c)(iii)The rate of reaction increases as the volume of B added increases. If Line graph: graph shows a straight line and has a constant gradientIf Curve: graph shows a curve with an increasing gradient. 1mAs the concentration of B increases, there are more reacting particles per unit volume.Hence there are more effective collisions between reacting particles resulting in a higher rate of reaction. 111(c)(iv)To dilute the sodium thiosulfate solution and ensure that the total volume of reaction mixture remains the same to ensure a fair experiment. 1mVolume of B is proportional to its concentration of B 1m111(d)(i)ACEpercentage uncertainty = 0.2 ÷ 10 x 100 = 2.0%appropriate significant figures in final answers in (d) (i)111(d)(ii)PDOUse a burette to measure volume of B instead of a measuring cylinderOrUse of a light sensor with a data logger to measure reaction time1Total: 27

4 QnSkillAnswerMark2(a)(i)MMOtestobservationsTest 1Remove the stopper from a sample of C in a test tube. Heat the sample in the test tube gently at first then strongly for 1 minute.Green solid turned black on heatingColourless liquid formed on wall of test tube/condensation occuredColourless gas produced white ppt when bubbled into limewaterCO2 was producedTest 2Transfer the other provided 0.25g measure of C to a boiling tube. Carefully add dilute nitric acid to a depth of approximately 5cm. Keep the final solution for use in tests 3-5.Bubbling of colourless, odorless gas Green solid dissolved in acid to form a blue solutionColourless gas produced white ppt when bubbled into limewaterCO2 was produced[marks awarded either in Test 1 or 2]Test 3Add a 1cm depth of the solution from test 2 into a test tube. Add aqueous ammonia slowly with shaking until no further change is seen.Blue ppt was producedThe ppt dissolved in excess aqueous ammonia to form a dark blue solutionTest 4Add a 1cm depth of the solution from test 2 into a test tube. Add a few drops of aqueous barium nitrate.No observable changeTest 5Add a 1cm depth of the solution from test 2 into a test tube. Add a few drops of aqueous silver nitrate.white ppt formed At least 1m per box, Max 6m1111112(a)(ii)ACEchloride / Cl-Test 512(b)ACEMass of conical flask + A before reaction /g175.83Mass of conical flask with solution 176.46

5 after reaction/g1m proper statement/in a table; 1m correct to 2 d.p.12(c)(i)ACECuCO3 (s) + 2HCl (aq)  CuCl2 (aq) + CO2 (g) + H2O (l) 1m for equation, 1m for ss12(c)(ii)Total initial mass of conical flask + A + D = 175.83 + 0.7g = 176.53gMass loss due to gas = Initial mass – final mass = 176.53 – 176.46 = 0.07 g12(c)(iii)No of mol of CO2 = 0.07 / 44 = 0.0015909 mol1 mol CO2 ≡ 1 mol CuCO3Mass of CuCO3 = 0.0015909 x 124 = 0.197 g112(c)(iv)Total mass of D = 0.70Percentage by mass of CuCO3 = 0.197 / 0.70 x 100 = 28.1 %13(a)PDiagram MethodPerform fractional distillation with the reaction mixture.Heat the mixture in the round bottomed flask until it vaporizes and the gases rise up the fractionating column. Propanal with the lower boiling point will condense in the condenser and be collected first at 49 oC (as shown on the thermometer) as a distillate in a beaker/conical flask.Apparatus that has to be labelled in diagram/mentioned in procedure-2mfractionating column, round botted flask, condenser,water in and water out on condenser, thermometer,beaker/conical flaskcollect propanal at 49oC -1mmethod -1m [ include underlined words in procedure] 11 3(b)Add a reactive metal or carbonate1

6 If propanoic acid is present, there will be effervescence of colourless gas[reject esterification]1Total: 13-1m presentation overall if no statements for mole concept questionsSkillMarksPDO8MMO10ACE17P5