RVHS Prelim_H1_Chemistry_P2_Soln

River Valley High School 8873/02/PRELIMS/25 [Turn over 2025 JC 2 Preliminary Examination RIVER VALLEY HIGH SCHOOL JC2 H1 CHEMISTRY 8873 PRELIMINARY EXAMINATION SUGGESTED SOLUTIONS Paper 2 1 (a) One mole of a substance contains exactly 6.02  1023 elementary entities. [1] (b) (i) Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g) [1] (ii) Oxidising agent. The oxidation state of Mn in MnO4− decreases from +7 to +2 in Mn2+ and the oxidation state of Fe in Fe2+ increases from +2 to +3 in Fe3+. [2] (iii) The aliquot changes from yellow to orange. [1] (iv) MnO4 – + 8H+ + 5e → Mn2+ + 4H2O Fe2+ → Fe3+ + e ×5 MnO4 – + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O [1] (v) amount of MnO4− used = 32.00/1000  0.025 = 8.00  10−4 mol [1] (vi) amount of Fe2+ in 25.0 cm3 = 8.00  10−4  5 = 4.00  10−3 mol amount of Fe2+ in 250 cm3 = 4.00  10−3  250/25.0 = 4.00  10−2 mol [1] [1] (vii) Mass of Fe in wire = 4.00  10−2  55.8 = 2.232 g % by mass of Fe in wire = 2.232/3.35  100 % = 66.6 % The % of Fe in the ore is >60%, more ores can be purchased. [3] [Total: 11] 2 (a) (i) [2] (ii) There are 2 lone pairs of electrons and 2 bond pairs of electron s around O in HOCl whereas there are only 1 lone pair of electrons and 3 bond pairs of electrons in around N in NH2Cl. To minimise repulsion, HOCl has a bent shape while NH2Cl has a trigonal pyramidal shape. Since the lone pair – lone pair repulsion is greater in HOCl than the lone pair – bond pair repulsion in NH2Cl, the bond angle in HOCl is smaller at 104.5 than 107 in NH2Cl. [3]

2 River Valley High School 8873/02/PRELIMS/25 2025 JC 2 Preliminary Examination (iii) Yes, HOCl is a polar molecule. Cl is more electronegative than H, O−Cl is less polar than O−H. There is a net dipole moment in the HOCl molecule. [1] (b) CsCl has a giant ionic lattice structure while HOCl has a simple covalent structure. More energy is required to overcome the stronger electrostatic forces of attraction between Cs+ and C l− ions than the weaker hydrogen bonds between HOCl molecules. [2] [Total: 8] 3 (a) Intermediate 1: Intermediate 2: [2] (b) Step 1: HCl, heat or PCl3 or SOCl2 or PCl5 Step 2: ethanolic NH3, heat in a sealed tube Step 3: CH3COOH, DCC [3] (c) [1] [Total: 6]

3 River Valley High School 8873/02/PRELIMS/25 [Turn over 2025 JC 2 Preliminary Examination 4 (a) (i) Maximum temperature = 51.0 C 2 BFL drawn to extrapolate to 3 min [2] (ii) Highest temperature rise = 51.0 − 25.0 = 26.0 C [1] (iii) Heat evolved = 4.18  25.0  26.0 = 2717J = 2.72 kJ (3.s.f) [1] (iv) Amount of Mg in mass of FA 1 added into cup = 1.00 24.3 = 4.115  10−2 mol Amount of CuSO4 in 25.0 cm3 of FA 2 = 1.00 × 25.0 1000 = 2.50  10−2 mol CuSO4 is the l