SAJC_2008 H2 Prelim Paper3 Solutions

Suggested solutions for SAJC 2008 Prelim Paper 3 (H2 Chemistry) 1(a)(i) CrO4 3- + 8H+ + 2e Î Cr3+ + 4H2O (ii) 2CrO4 3- + 2H+ Î Cr2O7 2- + H2O + 2e (iii) 3CrO4 3- + 10H+ Î Cr3+ + Cr2O7 2- + 5H2O (b) No. of mole of S2O3 2- = 0.30 x 32.00/1000 = 0.0096 mol No. of mole of I2 = ½ x 0.0096 = 0.0048 mol Since Cr2O7 2- ≡ 3I2 From equation (iii), 3CrO4 3- ≡ Cr2O7 2- Thus, Cr2O7 2- ≡ 3 I2 ≡ 3CrO4 3- No. of mole of CrO4 3- = No. of mole of K3CrO4 = No. of mole of I2 = 0.0048 mol Mass of K3CrO4 used = 0.0048 x 233.3 = 1.12 g 1(c) Mass = density x volume = 7.3 x 0.5 x 1.0 x 10-5 x 106 = 36.5 g No. of mole of Cr = 52 5 . 36 = 0.702 mol Cr2O7 2- (or Cr+6) + 6e Î Cr(s) 6 F ≡ 1 mole Cr 6 x 0.702 mol x 96500 C ≡ 40 x time Time = 10161.45 sec = 2.82 hr 1(d) H 2O is preferentially reduced rather than Al3+ ions. 1(e) Cr3+(aq) ion is coloured due to the splitting of the d-orbitals. The electronic repulsion between the water ligands and Cr 3+ ion causes the degenerate 3d orbitals to undergo splitting, resulting in 2 groups of non-degenerate orbitals, with a small energy gap between them. By absorption of light energy, electrons can be promoted from a lower energy d- orbital to a higher energy d-orbital. This energy is related to the wavelength of the light absorbed, and the light not absorbed is thus seen as the colour of the complex. 1fi) CH 3CHC lCHC lCH 3 CH 3CHCNCH(CH 3)CN CH3CHCOOHCH(CH 3)COOH KCN alcoholic, heat C CH3C H3C O O O O HCl(aq), heat conc H 2SO4, heat CH 2(OH) 2 (ii) Geometric isomerism 1

H3C C=C CH3 H H H3C C=C H CH3H (iii) CH3 NCH3CH2 CH3 CH3 Cl- CH3 NCH3CH2 CH3 CH3 Cl- CH3 NCH3CH2 CH3 + CH3Cl 2(a)(i) 2 assumptions for Ideal gas are : - molecules in gaseous state do not exert any force or negligible forces of attraction; - volume of molecules is negligibly small compared with that of the container. ii) Set II because carbon dioxide will deviate most from ideal gas behaviour at low temperature and high pressure due to presence of attractive forces and significant molecular volume. iii) Attractive forces operate among molecules at relatively short distances. At high pressure, density of gas increases, the molecules are much closer to one another and hence, gases condense into liquid. 2(b)(i) Let the solubility of Ca(H2PO4)2 = s mol dm-3 Ca(H2PO4)2(s) Ca 2+(aq) + 2 H2PO4 -(aq) s s 2s KSP = 1.0 x 10-5 = (s) x (2s)2 = 4s3 s = 3√ [(1.0 x 10-5 ) /4] = 1.36 x 10-2 mol dm-3 No. of moles of Ca(H2PO4)2 that can be dissolved in 100 cm3 of water = 1.36 x 10-2 x 100/1000 = 1.36 x 10-3 mol 2(b)(ii) I. P. = [Ca2+] [H2PO4 -]2 = [ 30/90 x 3.0 x 10-3] x [ 60/90 x 3.0 x 10-2]2 = (1 x 10-3) x (2 x 10-2)2 =