HCI Prelim ANS

HCI (College) / 2009 C2 H1 Chemistry / Prelims  Answers  Paper 1 – MCQ 1 A 6 A 11 B 16 C 21 A 26 D 2 B 7 A 12 B 17 B 22 B 27 C 3 A 8 C 13 C 18 C 23 D 28 D 4 C 9 A 14 B 19 B 24 C 29 B 5 D 10 D 15 A 20 A 25 D 30 D Paper 2A – Structured 1 (a) (i) Relative atomic mass, Ar  average mass of one atom of an element 1 12 x the mass of one atom of 12C (ii) Ar of Cu: (30.8/100)(65) + (69.2/100)(63) = 63.6 (iii) Ion protons neutrons electrons 63Cu2+ 29 34 27 (b) (i) No of moles of excess HCl = 20 x 0.5 x 10-3 No of moles of NH3 neutralized by HCl = (40-20)x0.5x10-3 = 1.0x 10-2 No of moles of NH4 + = mole of NH3 neutralized = 1.0x10-2 Mass of NH 4 + in 2g of salt A = 18( 10-2) = 0.180 g (ii) Mass of SO4 2- ion in 2g of salt A = 2.33 x [( 32.1 +64)/ (137+96.1)] = 0.961 g (iii) No of moles of A: 2.00 / 399.5 = 0.00501 mol No of moles of NH4 + = 0.0100 mol No of moles of SO 4 2- = 0.960 / (32.1 + 16.0 x 4) = 0.0100 mol No of mole of A: NH 4 + : SO4 2- 1: 2 :2 [1] Cu(NH 4)2(SO4)2 . 6H2O 2 (a) XCln  XCln−2 + Cl2 5 3 ) 5 . 35 108 /( 19 . 1 ( ) 5 . 35 108 /( 714 . 0 ( 2       n n OR No. of moles of Cl− for 1st sample = mol 10 8.2935.5 108 1.19 3  (1) No. of moles of Cl− for 2nd sample = mol 10 4.9835.5 108 0.714 3  (2) (1) : (2) = 5 : 3 n = 5 8872/02/S/09 1

HCI (College) / 2009 C2 H1 Chemistry / Prelims  Answers  Ar of X = 124.2 5 ) 10 (8.29 0.294 0.5 3      X is Sb (b) (i) Both exist as giant ionic lattice held by strong electrostatic forces of attraction between oppositely charged ions.       rr q qLE q+, q- and r- are the same. r+ of KCl is larger, hence numerical value of LE is lower. Less energy is required to break the weaker lattice structure. (ii) NaCl exists as giant ionic lattice held by strong electrostatic forces of attraction between oppositely charged ions. A lCl3 is a simple covalent molecule with weak dispersion forces between the molecules. Less energy is required to separate the molecules than the ions.                             (c) (i) The position of equilibrium shifts to the right. By Le Chatelier’s principle, an increase in pressure favours the side of the reaction with less number of moles of gaseous molecules. This results in a decrease in pressure, thus re-establishing the equilibrium. (ii) When temperature is increased, the position of equilibrium shifts to the left. Forward reaction is exothermic as it involves bond formation. By Le Chatelier’s principle, the equilibrium will shift backward to remove the heat. (iii)