SAJC Prelim P3 ANS

1 2009 SAJC Prelim Paper 3 MARK SCHEME 1. (a) (i) 5C2O4 2- + 2MnO4 - + 16H+  10CO2 + 2Mn2+ + 8H2O No. of moles of C2O4 2- ions 0.025 x 0.1 = 0.0025 Given C2O4 2- : MnO4 - is 5:2, No. of moles of MnO4 - needed to oxidize C2O4 2- = 0.0025 x 2/5 = 0.001 Thus, Volume of MnO4 - needed = (0.001 / 0.1) x 1000 = 10cm3 (ii) Volume of MnO4 - ions needed to oxidize X2+ ions = 15 – 10 = 5cm3 No. of moles of MnO4 - ions = [(15 – 10)/1000] x 0.1 = 0.0005 No of moles of electrons absorbed by MnO4 - = 0.0005 x 5 = 0.0025 Mole ratio of X2+ : e = 1 : 1 Since 1 mole of X2+ loses 1 mole of electrons, Oxidation state of X in the product = +3 (b) Compounds A and B and C all exist as simple covalent molecules. Compound A is held by hydrogen bonding. Compound B is held by weak induced dipole-induced dipole interactions (id-id)/weak VDW. Less energy is needed to overcome the weaker id-id interactions in B. Although Compound C does not have H-bonding, having more electrons causes stronger id-id (or VDW, but not pd-pd) which compensates for the absence of H- bonding. Hence more energy is required to overcome the intermolecular forces of attraction in C. (c) Since the oxide of Y can react with both NaOH and HCl, it is an amphoteric oxide. Among the possible elements, only the oxide of aluminium is amphoteric in nature: Compound Y is Al. Al 2O3 (s) + 6HCl (aq)  2A lCl3 (aq) + 3H2O (l) Al2O3 (s) + 2NaOH (aq) + 3H2O (l)  2NaAl (OH)4 (aq) In the presence of water, AlCl3 undergoes hydrolysis to form an acidic solution, which reacts with the NaOH. A lCl3 (s) + aq  [A l(H2O)6]3+ (aq) + 3Cl (aq) [Al(H2O)6]3+ (aq) + H2O (l) [Al(H2O)5(OH)]2+ (aq ) + H3O+ (aq)

2 Phosphorus chloride and phosphorus oxide will react with water but silicon oxide is insoluble in water. Compound Z is P. PCl 3(l) +3H2O(l) H3PO3 (aq)+3HCl (aq) or PCl 5 (s)+ 4H2O(l)  H3PO4(aq) + 4HCl (aq) P 4O6 (s) + 6H2O (l)  4H3PO3 (aq) or P 4O10 (s) + 6H2O (l)  4H3PO4 (aq) (d) (i) I2, NaOH(aq), heat No yellow ppt for P, yellow ppt for Q (ii) H2SO4(aq) [HCl(aq) not accepted], heat then add acidified KMnO4 or heat with KMnO4 + H2SO4(aq) R will decolourise purple KMnO4, S will not decolourise KMnO4 (e)

3 2. (a) (i) 25 oC (ii)  Complete the circuit by allowing the ions to flow through  maintain electrical neutrality by supplying ions to neutralize any built up of c solution;  prevent mixing of the two solutions in the half-cells. [Any one] (iii) 3Sn(s) + Cr2O7 2-(aq) + 14 H+(aq)  3Sn2+(aq) + 2Cr3+(aq) + 7H2O(l) E = +1.33 + 0.14 = +1.47 V (iv) Q = It = 0.2 x 9000 = 1800 C Since 96500 x 2 C of electricity consumes 1 mole of Sn, 193000 C of electricity consumes 119 g of Sn 1800 C of electri