ADSS Prelim P3 Ans

Admiralty Secondary School Marking Scheme 4E Pure Chemistry (Paper 3) PRELIMINARY EXAMINATION 2024 PAPER 3 (overall deduct 1 mark of consistently did not round off to 3 sf) Qn. Answers Mark Remarks Marker’s Report 1(a) Results e x p t volume of P /cm3 volume of Q /cm3 initial temperature of P /°C highest temperature of mixture /°C temperature rise /°C 1 40.00 10 29.5 32.5 3.0 2 35.00 15 29.5 34.0 4.5 3 30.00 20 29.0 36.0 7 4 25.00 25 29.0 37.0 8 5 20.00 30 29.0 35.5 6.5 6 15.00 35 29.0 34.5 5.5 7 10.00 40 29.5 33.0 3.5 1m: volumes must be between 15 & 35 cm3, and must be at least 5 cm3 apart. 5 Well done. Some students miscalculated the temperature rise.

1m: total volume of P and Q must add up to 50 cm3 1m: all volumes of P to nearest 0.05 cm3, all volumes of Q to nearest cm3 or nearest 0.5 cm3) 1m: temp should be recorded to 1 d.p., to the nearest 0.5 °C 1m: correct calculation for temperature rise (b) 1m: correct labels & units: x-axis (vol of Q/cm3); y-axis (temperature rise/°C) 1m: appropriate scale 1m: correct plotted points, within half a small square accuracy 1m: 2 smooth lines that intersect, passing through origin (0,0) 4 Generally okay. Most students were able to plot the points correctly except for a handful. Some students did not draw the line passing through the origin. Only a small number of students drew a curve instead of a line. 0 5 10 15 20 25 30 35 40 2 4 6 8 10 Volume of Q/cm3

(c) 26 cm3 1 Based on graph, at point of intersection of the 2 straight lines Ok (d) 50 – 26 = 24.00 cm3 (to 2 d.p.) 1 Must show working A few students did not show working (e) H2SO4 + 2NaOH  Na2SO4 + 2H2O No. of moles of Q = (26/1000) x 1.20 = 0.0312 mol [M1] Mole ratio of P:Q = 1:2 No. of mole of P in 24 cm3 = 0.0312/2 = 0.0156 mol [M1] Concentration of P = 0.0156 mol / (24/1000) = 0.650 mol/dm3 (to 3 s.f.) [A1] (to 3 s.f.) 1 1 1 allow ecf minus 1m overall if not to 3 s.f. Not very well done. Quite a number of students assumed that the mole ratio of P:Q was 1:1 as they did not write out the chemical equation, wrote the chemical formula for Na2SO4 wrongly or did not balance the equation correctly. (f) To better prevent heat loss as polystyrene is an insulator of heat 1 Reject prevent temperature loss Some students did not explain clearly and just stated that polystyrene reduces heat loss, without explaining why. (g) Maximum temperature will be higher because the same amount of energy was given out but the total volume of mixture was lesser than 50 cm3, hence the mixture will be heated to a higher temperature. OR there are more moles of reacting particles per unit volume, hence there is a higher frequency of collisions and higher frequency of effective collisions, (therefore a higher rate of reaction.) 1 1 1 Max 2 marks Badly done. Many left blank, or said that temperature will be the same as heat given out is the same.

2(a) Mass of empty boiling tube / g 29.92 Mass of boiling tube with C before heating / g 31.95 Mass of boiling tube and its contents after heating / g 31.46 1m correct headings 1m correct units and 2 d.p. 1 1 No table, minus 1m A few students did not draw table. Some wrote weight instead of mass. (b) Change in mass of C = 31.95 – 31.46 = 0.49 g 1 Well done (c) No. of moles of CO2 = 0.49/44 = 0.1114 mol Mass of CuCO3 reacted = 0.1114 x 124 = 1.38 g 1 1 Surprisingly badly done. Many students tried to find the number of moles of CuO even though the question stated that it contains impurities. Some divided by 24. (d) Initial mass of C = 31.95 – 29.92 = 2.03 g Percentage by mass of CuCO3 = (1.38/2.03) x 100 = 70.0 % (3 s.f.) 1 Minus 1m overall if wrong s.f. Allow ecf Most students did not find the initial mass of C and just used 2 g. However, they were not penalised.

(e) Perform strong heating using non-luminous flame / heat for a longer time to ensure that all the copper(II) carbonate is decomposed. 1 Reject repeat experiment Well done 3(a) (i) test observations 1 White ppt formed. Insoluble in excess sodium hydroxide. 2 No ppt formed/no observable change/ no visible change 3 No ppt formed/no observable change/ no visible change/no effervescence 4 Pungent gas evolved. Gas produced turns damp red litmus paper blue. Ammonia gas is produced. 5 White ppt formed 6 Effervescence observed. Gas produced forms white ppt in limewater. CO2 is produced. 1 1 1 1 1 1 1 1 Well done. Some students still wrote ‘white solution formed’ or did not following the correct way of recording observations, resulting in ambiguous answers. (ii) Ca2+ NO3- 1 1 Max 1m if student write both name and formulae Quite a number of students wrote the name instead of chemical formulae (iii) In test 5, a white ppt is formed, which means that the cation (Ca2+) has been removed from the (hard) water Only students who understood how hard water is softened

by precipitation/in the form of a precipitate (as it forms an insoluble carbonate when solution Y was added) 1 were able to explain clearly. Those who were unsure just repeated what test 5 was. (b) Procedure: - Put B into a burette. Pipette 25.0 cm3 of A into a conical flask. - Titrate A using B until one drop of B produces permanent pale pink colour. - Record volume of B used. - Repeat experiment until 2 consistent results are obtained. Data Processing: - Calculate the number of moles of B needed (vol. of B x 0.500 = x mol) - Calculate number of moles of iron(II) ions in 25.0 cm3 of A (5x = y mol) 1 1 1 1 Majority were able to use titration method. Many did not draw the diagram. Some students filled burette with A instead of B. Most were unable to calculate the concentration of iron(II) ions in 50 cm3 of hard water sample as they did not realise that the 50 cm3 of hard water sample was diluted to 100 cm3 Burette filled with B A

- No. of moles of iron(II) ions in 100 cm3 of A = y x 4 = z - No. of moles of iron(II) ions in 50 cm3 of hard water sample = z - Concentration of iron(II) ions in 50 cm3 of hard water sample = z / (50/1000) = ANS Procedure(3m): 1m: what is measured in the experiment 1m: drawing correct titration setup with B in burette and A in conical flask 1m: correct procedure of how to carry out the titration w independent and dependent variables which add to pipette and burette, and taking average of the two best readings Data processing(2m): - correct calculation of no. of moles of B needed & correct calculation of number of moles of iron(II) in 25 cm3 of A and 100 cm3 of A, - (and hence) no. of moles of iron(II) in 50 cm3 of hard water sample, and correct calculation of concentration of iron(II) ions in 50 cm3 of hard water sample 1