NYJC 2020 Prelim 9729 P4 Answers

1 NYJC J2 Prelim Exam 2020 Paper 4 Suggested Mark Scheme 1(c) Results 1 2 initial burette reading / cm3 0.00 10.25 final burette reading / cm3 10.25 20.50 volume of FA 1 added / cm3 10.25 10.25 [2]  correct headers, units and results recorded in a single table  results recorded to 2dp and no 50.00 in the table and results recorded under the correct header  correct evaluation of volume of FA 1 added  3 pts – 2m, 2pts – 1m (d) volume of FA 1 = 10.25 10.25 2  = 10.25 cm3 volume of FA 1 = 10.25 …………………….. cm3 [1] Student obtains appropriate “average”, to 2 d.p., from any experiments with uncorrected end-point titre values within 0.10 cm3. Do not award this mark if the titres used are not identified either in the table (by, for example, a tick) or in a calculation. (e) (i) Volume of FA 1 = 10.25 × 50.0 10.0 (or × 5) = 51.25 cm3 volume of FA 1 = 51.25……………….. cm3 [1] (ii) Equilibrium amount of C2H5CO2H(aq) = amount of NaOH that reacts with the acid remaining in 50 cm3 of the aqueous layer = 51.25 0.0501000  = 0.002562  0.00256 mol [1] C2H5CO2H(aq) Ý C2H5CO2H(org) initial amount / mol 50 0.200 0.010001000 (1) 0 change in amount / mol  0.007437 (3) + 0.007437 (4) equilibrium amount / mol 0.002562 (2) 0.007437  0.00744 [1] Equilibrium amount of C2H5CO2H(aq) = 0.00256……………………. mol [1] Equilibrium amount of C2H5CO2H(org) = 0.00744……………………. mol [1]

2 (iii) 0.007437401000 0.002562501000 cK  = 3.628  3.63 Kc = 3.63……………………. [2] finding conc. – 1m, substituting the values into the Kc expression correctly – 1m (f) Increases [1] Solubility of propanoic acid in 2 -methylpropane decreases [1] [because energy released when propanoic acid forms instantaneous dipole-induced dipole forces of attraction with 2- methylpropane is not really enough to overcome the hydrogen bond between the propanoic acid molecules]. So more FA 1 is required to neutralise the prop anoic acid that remains in the aqueous phase. 2(a) Results [7] 1 mark: headers and units 1 mark: All temperature recorded to 1 decimal place and all vol umes recorded to whole number or 1 decimal place 1 mark: volumes of water added is correct 1 mark: correctly calculate average initial temperature and T for all expts to 1 decimal place. 1 mark: Selected four additional volumes of FA 3 and the selected volumes must be sequential and well-spaced i.e. 5 cm3 or more from any other volume. 1 mark: Student selects one FA 3 volume less than 35 cm3, except 0 cm3 and 30 cm3 1 mark: Student selects three FA 3 volumes more than 35 cm3 and less than or equal 60 cm3 Experiment 1 2 3 4 5 6 Volume of FA 4 (or H2SO4) / cm3 25 25 25 25 25 25 Volume of water (or H2O) / cm3 65 45 55 35 25 15 Volume of FA 3 (or NaOH) / cm3 10 30 20 40 50 60 Initial temperature of FA 4 and water / C 30.0 30.0 30.0 30.0 30.0 30.0 Initial temperature of FA 3 (or NaOH) / C 30.0 30.0 30.0 30.0 30.0 30.0 Average initial temperature / C 30.0 30.0 30.0 30.0 30.0 30.0 Maximum temperature obtained / C 32.0 37.0 34.4 38.0 38.0 38.0 T / C 2.0 7.0 4.4 8.0 8.0 8.0

3 (b) [2] 1 mark Axes correct way round + correct labels + units + scale. Note: Sensible scales must be used and must allow for the lines to be extrapolated to cross each other. Awkward scales (e.g. 3:10) are not allowed. The plotted points must occupy at least half of the graph grid in both x and y directions. 1 mark Plotting – all points correctly plotted within ±½ small square. 36.5

4 (c) z, the volume of FA 3 = 36.5 cm3………………… maximum temperature change, Tmax = 8 oC................. 1 mark Graph lines must be best-fit lines, drawn so as to best reflect the distribution of points before and after the equivalence point. The line before and after the equivalence point should be extrapolated until it crosses. Markers must not alter or adjust a poorly drawn student’s line. 1 mark for the followings:  Correctly read V(FA 3) at Tmax and Tmax from graph 2 marks on Accuracy Accuracy based on V(FA 3) at Tmax and on value of Tmax If a student has incorrectly read either/both of the values of V(FA 3) and Tmax, the teacher will read the correct values from the student’s graph and use these values in awarding accuracy marks. However, if the student’s line is not best -fit, teacher will not re -draw the best-fit line and re-read V(FA 3) and Tmax. Comparison of student’s V( FA 3) at Tmax with teacher’s FA 3 at Tmax: if within 2 cm3, award 1 mark, otherwise 0 mark Comparison of student’s Tmax with teacher’s Tmax: if within or equal to 1 oC, award 1 mark, otherwise 0 mark (d) (i) H2SO4 + 2NaOH  Na2SO4 + 2H2O n(NaOH) = 36.5 1.501000 = 0.05475 mol n(H2SO4) = 0.05475 / 2 = 0.02737 mol [H2SO4] = 0.02737 251000 = 1.095  1.10 mol dm3 concentration of H2SO4 in the FA 4 solution = 1.10 mol dm3…….. [1] (ii) H1 =  3100 4.18 10 8 0.02737    =  122 kJ mol1 H1 =  122 kJ mol1 …….. [2] 1 mark Calculate [H2SO4] correctly 1 mark Calculate H1 correctly, including giving the correct sign 1 mark Shows appropriate significant figures and units

5 (e) (i) Before the equivalence point, NaOH is the limiting reactant. The amount of water produced is equal to the amount of NaOH. As the total volume of the reaction mixture is kept constant (100 cm3), the volume of NaOH (FA 3) is proportional to T. As volume of NaOH increases, T increases proportionally. [1] H =  mc∆T n(H2O) =  Vc∆T n(H2O) , where V is the total volume of the solution. (Assuming 1 cm3 of solution  1 g.) When NaOH is the limiting reagent, H =  Vc∆T n(H2O) =  Vc∆T n(NaOH) =  Vc∆T [NaOH] × V(NaOH) 1000  V(NaOH) =  1000Vc∆T [NaOH] × ∆H Since V, c, [NaOH] and H are constants, V(NaOH) is proportional to T. (ii) After the equivalence point, H2SO4 is the limiting reactant . The amount of water produced is constant as volume of H 2SO4(FA 4) is constant at 25 cm 3. As the total volume of the reaction mixture is kept constant (100 cm 3), no matter how much volume of NaOH increases, T is constant/will not change. [1] When H2SO4 is the limiting reagent, H =  Vc∆T n(H2O) =  Vc∆T 2n(H2SO4) =  Vc∆T 2 × [H2SO4] × V(H2SO4) 1000  V(H2SO4) =  1000Vc∆T 2 × [H2SO4] × ∆H Since V(H2SO4) is constant, T will be constant.

6 3(a) 1. Weigh the mass of a clean, dry and empty 250 cm 3 conical flask on an electronic balance. 2. Using a clean and dry 250 cm3 conical flask/beaker, weigh the mass of the 100 cm 3 of water on an electronic balance. Or Use a 50 cm3 burette to measure 100.00 cm3 of distilled water and transfer the water into a clean and dry 250 cm3 conical flask/beaker. 3. Place the beaker in the tub containing the ice cubes and controlled the temperature in the tub using the thermometer, ensuring that the temperature is roughly maintained at 10 oC. 4. (Using a spatula, add a large spatula of) Add (solid) Ca(OH)2 into the flask/beaker and stir with a glass rod. OR Using a dry, empty weighing bottle, weight accurately 5.00g of Ca(OH) 2. Add (solid) Ca(OH)2 into the flask/beaker and stir with a glass rod. Continue to add more solid and stir, until some Ca(OH) 2 remains undissolved to ensure a saturated solution is obtained. Allow the mixture to stand in the ice -bath for 30 minutes to establish equilibrium. 5. Weigh the mass of the mixture obtained from step 4 OR Reweigh the weighing bottle with the residual Ca(OH)2 to find out the mass of Ca(OH)2 added to the beaker. 6. Weigh the mass of a piece of dry filter paper. Filter the mixture using the dry filter paper and a dry filter funnel into another dry 250 cm3 conical flask/beaker. 7. Dry the filt er paper and residue under an IR lamp. (After 10 minutes, allow the filter paper to cool down in a desiccator.) 8. Using an electronic balance, weigh the