HCI 2012 ANS

1 C1 HCI Chemistry Block Test ANSWER Paper 1: MCQ 1 B 11 B 2 D 12 A 3 A 13 C 4 B 14 A 5 B 15 C 6 D 16 D 7 B 17 C 8 D 18 A 9 C 19 D 10 D 20 A Paper 2: 1 (a) Mg(NO2)2•xH2O  Mg(NO2)2 + xH2O  mass of water = 2.00 ‒ 1.24 = 0.76 g n(H2O) = 0.76 / 18.0 = 0.04222 mol n(Mg(NO2)2) = 1.24 / 116.3 = 0.01066 mol x = 0.04222 / 0.01066 = 3.96 = 4 (whole no.) (b) The number of water of crystallization is higher than in the given information. The likely reason for the discrepancy is that on heating, the magnesium nitrite decomposed. Therefore the mass loss is higher than the actual value and this results in a higher value of x. (c) (i) 5NO2 – + 2MnO4 – + 6H+  5NO3 – + 2Mn2+ + 3H2O (ii) Using a fixed mass of magnesium nitrite, dissolve in deionised water in a volumetric/standard/graduated flask (iii) The solution in the flask turns from pink to colourless at the endpoint. (iv) Repeat the titration until two titre values within ± 0.10 cm 3 of each other are obtained. 2 (a) (i) 2  and 2  (ii) or or – +

2 (b) (i) CN– + H+  HCN (HCN) = (NaCN) = 14.012.023.0 1.0  = 49.0 1.0 = 0.02041 mol Mass of HCN formed = 0.02041 x 27.0 = 0.551 g (ii) [HCN] = 1000 75 10000.551   mg dm–3 = 7.35 x 10–3 ppm The concentration of HCN has not exceeded the exposure limit. (c) (i) ∆S is positive since the number of moles of gaseous products is more than that of reactants (or there is an increase in the number of gaseous molecules). Hence, there are more ways in which the particles and their energies can be distributed in a larger volume. (ii) (CH4) left = 0.5 – 0.25 = 0.25 mol (HCN) formed = 0.25 mol (H2) formed = 3 x 0.25 = 0.75 mol Total no. of moles of gases = 1.25 mol p = 3-105)(5 0)1(2738.311.25   0 = 387000 Pa = 387 kPa (d) 1. The ions (N3– to Al3+) are isoelectronic. 2. Nuclear charge increases (while shielding effect remains constant). 3. Effective nuclear charge increases Or The electrons are more strongly attracted to the nucleus and hence size of ions decreases. 3 (a) (i) trigonal planar (ii) sp2 (iii) (iv) N O F Ne Na Mg Al Ionic radius

3 (b) (i) Bond energy refers to the average energy required to break 1 mole of a covalent bond in the gaseous atoms. (ii) ΔHr = Σ B.E. (bonds broken) - Σ B.E. (bonds formed) = [(C=C) x 1 + (H-H) x 1] – { [(C-H) x 2] + C-C} = [(610) x 1 + (436) x 1] – { [(410) x 2] +350 } = 1046 – 820 – 350 = – 124 kJ mol–1 (c) (i) Vitamin C can form hydrogen bonding with water molecules in urine. (ii) Dehydroascorbic acid is l ess soluble due to less extensive hydrogen bonding ( or f