2020 RI Prelim P4 Ans (students)

1 © Raffles Institution 2020 9729/04/S/20 2020 RI H2 Chemistry Prelim Paper 4 – Suggested Solutions 1(a)(i) Table 1.1 final burette reading / cm3 25.40 35.40 Note: (also applies to Table 1.2)  Record burette readings to 2 d.p.  Get consistent results (i.e. at least two titre values which are within 0.10 cm3 of each other). initial burette reading / cm3 0.00 10.00 volume of FA 2 used / cm3 25.40 25.40 Comments:  Generally well done. 1(a)(ii) V1 = 25.40 + 25.40 2 = 25.40 cm3 Note:  Average volume, V1, should be calculated using consistent titres (within 0.10 cm3 of each other) obtained in Table 1.1.  Working must be shown.  Average volume to 2 d.p. Comments:  Generally well done. 1(b)(i) Amount of KMnO4 used = 25.40 1000  0.0200 = 0.000508 mol From equation 2, mole ratio of MnO4− : Fe2+ = 1 : 5 Amount of Fe2+ = 0.000508  5 = 0.00254 mol [Fe2+] = 0.00254 10  1000 = 0.254 mol dm−3 Note: For all questions involving calculations,  statements and working should be clearly shown.  all final answers to 3 s.f. Comments:  Generally well done. 1(b)(ii) Concentration of hydrated halotrichite in FA 1 = 22.26 100  1000 = 222.6 g dm−3 Since each formula unit of the hydrated halotrichite contains only one Fe2+, from (b)(i), concentration of hydrated halotrichite in FA 1 = concentration of Fe2+ = 0.254 mol dm−3 Relative formula mass of hydrated halotrichite = 222.6 0.254 = 876 Comments:  Some students did not read the question carefully – FA 1 is prepared by dissolving 22.26 g of hydrated halotrichite in only 100 cm3 of water.  A few students incorrectly gave units for the Mr of the hydrated halotrichite. Note that o Relative formula mass (Mr) and relative atomic mass (Ar) have no units o Units for molar mass is g mol−1

2 © Raffles Institution 2020 9729/04/S/20 1(c) Table 1.2 Volume of FA 3 added / cm3 0.00 6.00 12.00 18.00 final burette reading for FA 2 / cm3 19.30 33.40 42.50 initial burette reading for FA 2 / cm3 0.00 20.00 35.00 volume of FA 2 used / cm3 25.40 19.30 13.40 7.50 Comments:  Generally well done.  Some students did not follow the question and failed to select volumes of FA 3 which differ by at least 6.00 cm3. 1(d) Comments:  Generally well done.  The scale chosen for x-axis should allow room for extrapolation to obtain V2.  Avoid odd scales, e.g. one large square for 4 cm 3 or 6 cm 3. Such odd scales will be penalised as it makes the plotting and/or reading of coordinates difficult. 1(e)(i) From graph, V2 = 25.50 cm3 Comments:  Generally well done.  To obtain V2 (volume of FA 3 required to react completely with 10.0 cm 3 of FA 1 with no FA 2 added), the extrapolation of graph to obtain the x-intercept must be clearly shown. Note:  The x -axis and y -axis should be correctly labelled with the correct units.  Appropriate scale must be used (plots should span at least half the grid in both x - and y-direction) and interval markings clearly shown.  Data points should be plotted correctly.  A best-fit straight line should consider all the plotted points.  Anomalous points, if any, should be circled.  Coordinates can be read to half the smallest division.

3 © Raffles Institution 2020 9729/04/S/20 1(e)(ii) From (b)(i), Amount of Fe2+ = 0.000508  5 = 0.00254 mol From equation 2, mole ratio of H2O2 : Fe2+ = 1 : 2 Amount of H2O2 = 0.00254 2 = 0.00127 mol [H2O2] = 0.00127 25.50  1000 = 0.0498 mol dm−3 Comments:  Generally well done.  The calculation should make use of the V2 obtained in (e)(i), instead of using a specific point on the graph, e.g. at 6 cm3 of FA 3 added. 1(f) With a higher concentration of KMnO4 in FA 2, a smaller volume of FA 2 will contain the same amount of KMnO4 required to react with the Fe2+ present in 10.0 cm3 of FA 1 if no FA 3 is added (i.e. smaller V1). Since the volume and concentration of Fe2+ in FA 1 is the same, the volume of FA 3 required to react completely with the Fe2+ if no FA 2 was added remains unchanged (i.e. V2 is unchanged). Comments:  Most students showed understanding that with a higher [KMnO4], volume of FA 2 required to completely react with the (remaining) Fe2+ would be lower than that in (d). However, a number of students did not realise that V2 (x-intercept) will remain unchanged as this is the volume of FA 3 required to react with Fe 2+ with no FA 2 added . Since the amount of Fe 2+ (volume and concentration of FA 1) and concentration of FA 3 remains unchanged, V2 will be the same for both (d) and (f). (d) (f) volume of FA 2 used / cm3 volume of FA 3 added / cm3 0

4 © Raffles Institution 2020 9729/04/S/20 2(a) volume of FA 4 / cm3 10.0 15.0 20.0 25.0 30.0 35.0 volume of FA 5 / cm3 40.0 35.0 30.0 25.0 20.0 15.0 T2 / °C 34.1 36.0 38.1 37.3 35.4 33.5 T 1 / °C 29.0 29.0 29.0 29.0 29.0 29.0 T / °C +5.1 +7.0 +9.1 +8.3 +6.4 +4.5 Comments:  Generally well done.  Some students did not read, understand or follow the instructions stated in the question. For this question, students are required to: o construct an appropriate table with proper headings and units. o record all measurements of volume , temperature and temperature change. Hence, volume of FA 5 needs to be included. o use an appropriate volume of FA 5 so that the total volume of ea ch mixture is 50.0 cm3. o record all data to the appropriate number of decimal places, typically to half the smallest division or according to the question. It is stated in the question that temperature readings are to be recorded to 0.1 oC. Volumes of reagents should also be recorded to 1 d.p. o calculate all ∆T correctly. 2(b) Plot a graph of T vs volume of FA 5. Tmax = +9.65 C Volume of FA 5 = 28.5 cm3 Volume of FA 4 = 50 – 28.5 = 21.5 cm3 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0 10.0 15.0 20.0 25.0 30.0 35.0 40.0 T / C 9.65 28.5 volume of FA 5 added / cm3 Note:  The x-axis and y-axis should be correctly labelled with the correct units.  Appropriate scale must be used (plots should span at least half the grid in both x- and y- direction) and interval markings clearly shown.  Data points should be plotted correctly.  2 best-fit, intersecting, straight lines drawn using data points.  Anomalous points, if any, should be circled.  Coordinates can be read to half the smallest division.

5 © Raffles Institution 2020 9729/04/S/20 Comments:  Students are required to plot the points and draw the best-fit lines on the grid provided.  The scale chosen for y-axis should allow o room for extrapolation for the two best-fit lines to intersect o the plots to span at least half the grid in the y-direction  Students are required to know that a best -fit line is one with equal distribution of data points above and below the line. Poor experimental technique would affect the experimental results and hence the quality of the best-fit lines.  Some students tried to manipulate their data points to obtain a straight line. As a result, their plotted points do not match their experimental data in Q2(a).  The determination of ∆T and volumes of FA 4 and FA 5 were generally well done. 2(c) Amount of H2SO4 in FA 4 = 0.800 (21.5/1000) = 0.0172 mol Amount of M(OH)x in FA 5 = 1.20 (28.5/1000) = 0.0342 mol Comments:  Generally well done. 2(d) Ratio of M(OH)x to H2SO4 = 1.988:1 ≈ 2:1  x = 1 Comments:  Some students forgot that H2SO4 is a dibasic acid and did not include that in the determination of x.  Note that x must be a whole number. 2(e) [M(OH)x] = 16.8 x (1000/250) = 67.2 g dm–3 Mr of M(OH)x = 67.2 / 1.2 = 56.0 Ar of M = 56.0 – 16.0 – 1.0 = 39.0 Comments:  Generally well done . 2(f) q = mcT = 50(4.18)(+9.65) = +2016.8 J H