RI Prelim P4_Answers

- 1 - 2018 Y6 H2 Chemistry Preliminary Exams Paper 4 (Suggested Solutions) 1(a) Volume of FA 2 needed = 5 30 x 250 = 41.67 cm3 1(b) Dilution of FA 2: Final burette reading / cm3 41.00 Initial burette reading / cm3 0.00 Volume of FA 2 used / cm3 41.00 Titration Results: Titration number 1 2 Final burette reading / cm3 24.80 24.80 Initial burette reading / cm3 0.00 0.00 Volume of FA 1 used / cm3 24.80 24.80 Values used   1(c) Average volume of FA 1 used = 24.80 + 24.80 2 = 24.80 cm3 25.0 cm3 of FA 4 produced an amount of iodine which required 24.80 cm3 of FA 1. 1(d) Amount of thiosulfate ions = 0.0500 × 24.80 1000 = 1.24 × 10−3 mol Mole ratio of Cu2+ : I2 : S2O32− = 2 : 1 : 2 Amount of Cu2+ in 25.0 cm3 of FA 4 = amount of S2O32− = 1.24 × 10−3 mol [Cu2+] in FA 4 = 1.24×10-3 25.0×10-3 = 0.0496 mol dm⁻3 1(e) [Cu2+] in FA 2 = 0.0496 x 250 41 = 0.302 mol dm⁻3 1(f) Amount of NaI in 10 cm3 of 50 g dm−3 sodium iodide = 50 23.0 + 126.9 x 0.010 = 0.00334 mol Amount of NaI required to react with Cu2+ in 25.0 cm3 of FA 4 = 2 x 25.0 x 10−3 x 0.0496 = 0.00248 mol When using 50 g dm −3 sodium iodide, amount of iodide is still in excess. Hence there is no effect on the titre volume as the same amount of I2 will be produced.

- 2 - 2(b)(i) Results: No V FA5/ cm3 VFA6/ cm3 T1/ oC T2/ oC ΔT/ oC 1 10.0 30.0 29.0 32.0 3.0 2 20.0 20.0 29.6 35.2 5.6 3 30.0 10.0 29.6 33.3 3.7 4 35.0 5.0 30.0 31.6 1.6 5 25.0 15.0 29.6 35.4 5.8 6 15.0 25.0 29.0 33.2 4.2 2(b)(ii) 2(b)(iii) maximum temperature change of reaction mixture, ΔTmax = 6.5 0C volume of FA 5 required for complete reaction, Vrxn = 23.5 cm3 2(c)(i) To obtain ΔTmax, VFA6 = 40.0 – 23.5 = 16.5 cm3 Amount of S2O32− = 16.5 1000 x 0.100 = 0.00165 mol Amount of NaClO in 23.5 cm3 of FA 5 = 4 x amount of S2O32− = 0.00660 mol Concentration of NaClO in FA 5 = 0.00660 0.0235 = 0.281 mol dm−3 2(c)(ii) Heat given out = 40.0 x 4.18 x 6.5 = 1086.8 J ΔH1 = − 1086.8 0.00165 J mol−1 = −659 kJ mol−1 2(d) q = mcΔTmax = n × ΔH1  ΔTmax = n  H1 m ൈ c When VFA5 and VFA6 are doubled, both n and m are doubled (ΔH1 and c remain constant). Hence ΔTmax is unaffected. 2(e)(i) Amount of S2O32− = 0.030 x 0.100 = 3.00 x 10−3 mol Amount of ClO− = 4 x amount of S2O32− = 0.0120 mol Mass of NaClO = 0.0120 x (23.0 + 35.5 + 16.0) = 0.894 g 1.0 2.0 3.0 4.0 5.0 6.0 7.0 0.0 5.0 10.0 15.0 20.0 25.0 30.0 35.0 40.0 45.0 ΔT/ oC VFA 5/ cm3 Vrxn ΔTmax

- 3 - 2(e)(ii) Assuming a percentage purity of 80 % NaClO, Mass of FA 7 required = 0.894 0.80 = 1.12 g 2(e)(iii) Using an analytical balance, weigh accurately about 0.400 g of FA 7. Record the mass of the weighing bottle and the mass of FA 7, m1. Using a burette, add 30.00 cm 3 of FA 6 in a clean and dry Styro