NJC Prelim P3_Solutions

NJC 2018 SH2 H2 Chemistry Prelim Paper 3 Solutions: Question 1 (a)(i) Suggested solution Both octan-1-ol and iodine have simple covalent structure Strength of Td-Id in iodine due to its large elec tron cloud size is comparable to the stronger hydrogen bonding between octan-1-ol molecules. (ii) Suggested solution Graphite has a giant molecular structure with strong covalent bonds between carbon atoms to be overcome during melting. Fullerene has a simple covalent structure with weaker td-id interactions between fullerene molecules to be overcome during melting. Hence, more energy is required to melt graphite and a higher temperature is required. (b) Suggested solution Step 1: reaction 1 ;(-COCHO) Step 2: K2Cr2O7/KMnO4 with H2SO4(aq) and heat; (-COCOOH) Step 3: NaBH4 in ethanol or H2 with Nickel/Pt ; (compound B) c(i) Suggested solution PCO2 = ଴.଴ଷହ ଵ଴଴ × 101.3kPa = 0.03546kPa KH = [஼ைమ(௔௤)] ௉಴ೀమ [CO2(aq)]= 1.17 x 10-5 mol dm−3 c(ii) Suggested solution Kc = [ுమ஼ைమ(௔௤)] [஼ைమ] [H2CO3] = 1.52x 10-8 mol dm−3 (iii) Suggested solution Ka1 is larger than Ka2. H2CO3 will dissociate accordingly to eqm 1 first and [H +] from eqm 1 suppresses the dissocation of HCO3− in eqm 2. OR HCO 3− from eqm 1 is low as Ka1 is small. Hence [H+] from HCO3− is negligible. [H+] is largely from eqm 2. c(iv) Suggested solution Ka1 = [ு஼ைయష][ுశ] [ுమ஼ைయ] [H+]=1.915 x 10-6 moldm−3 pH = 5.71 d Suggested solution

Question 2 (a) Suggested solution: At anode: Co has ௢௫ ௢ (Co/Co2+) = +0.28V more positive that ௢௫ ௢ (Cu/Cu2+) = −0.34V, hence Co will be oxidised together with Cu and dissolve as Co 2+, becoming part of the electrolyte. Ag on the other hand has more negative ௢௫ ௢ (Ag/Ag+) of −0.80V, hence will not be oxidised. It will fall below anode as sludge. At cathode: Co2+ has a more negative ௥௘ௗ ௢ (Co2+/Co) = −0.28 V than ௥௘ௗ ௢ (Cu2+/Cu) = 0.34 V, hence Co2+ is not reduced at the cathode, remain dissolved as electrolyte. (b) (i) Suggested solution • mass of Ag = 0.07 g • No of moles of Co oxidised at anode = no of moles of Co(C 4H7N2O2)2 = ଴.ହହ ହ଼.ଽାଶ(ସ×ଵଶା଻ାଶ × ଵସାଶ × ଵ଺) = 0.0019038 mol Mass of Co in alloy (anode) = 0.0019038 x 58.7 = 0.112 g Mass of Cu = 1.25 – 0.07 – 0.112 = 1.068 g b(ii) % purity = 1.068 /1.25 x 100 = 85.4% b(iii) Suggested solution: Any one of the suggestion below is acceptable: (i) Mass of Ag and Co(C 4H7N2O2)2 should be heated to consistent mass to ensure all water is driven off, so that mass measured is that of dry mass of Ag and Co(C4H7N2O2)2. (ii) As mass of Ag (sludge ) is rather small, use electronic balance of higher precision in order to reduce % uncertainty in mass measurement. (iii) Repeat the experiment using fresh sa mples so that an average of the mass measurements can be taken to reduce random error. b(iv) Suggested solution:

Total no of moles of e used for e