NJC Prelim P1_Solutions

1 NJC H2 Chemistry Prelim Paper 1 Suggested Answers 1 C 6 B 11 B 16 C 21 D 26 D 2 D 7 D 12 D 17 B 22 B 27 B 3 A 8 D 13 A 18 A 23 D 28 A 4 C 9 C 14 C 19 D 24 C 29 C 5 C 10 A 15 C 20 B 25 C 30 C 1 Ans: C SO2 reacts with acidified KMnO4 Total vol of SO2 = 10 cm3 Left over CO2 reacts with NaOH(aq) Vol of CO2 = 2 cm3 Vol of SO2 produced from CS2 = 4 cm3 Vol of SO2 produced from H2S = 10  4 = 6 cm3 Vol of CS2 : Vol of H2S = 2 : 6 mole fraction of H2S in the mixture = 6 8 = 0.750 2 Ans: D Step I: Oxidation state of Cr increases from +3 in Cr2O42− to +6 in CrO42−. Step II: Oxidation state of Cr remains as +6 in both CrO42− and Cr2O72−. 2CrO42− + 2H+  Cr2O72− + H2O Step III: Oxidation state of Cr decreases from +6 in Cr2O72− to +3 in Cr2O3.

2 3 Ans: A 5Fe2+ + MnO4− + 8H+  5Fe3+ + Mn2+ + 4H2O Amount of MnO4− required = 20 1000 × 0.40 = 0.008 mol Amount of Fe2+ = 5 × 0.008 = 0.040 mol At point X, the reaction is only half completed, [Fe2+] = [Fe3+]. Ecell when connected to standard hydrogen electrode = +0.77 V The colour change at end point is from yellow to first permanent pink. Vol KMnO4 added Species present in conical flask Colour of solution 0 Fe2+(aq) Pale green Before eqv pt Fe2+(aq), Fe3+(aq), Mn2+(aq) Yellow-green At eqv pt Fe3+(aq), Mn2+(aq) Yellow Just after eqv pt Fe3+(aq), Mn2+(aq), small amt of MnO4−(aq) Pink In excess Fe3+(aq), Mn2+(aq), large amt of MnO4−(aq) Purple 4 Ans: C Making reference to the I .E. values from Data Booklet, we can conclude that I is potassium. Element C is Al, it is in Group 13. Element F is S and it exists as S8 molecules. The lowest boiling point is Ar gas (element H). Ion of E (P3−, 0.212 nm) is larger than that of J (Ca2+, 0.099nm) Element D and G are Si and Cl respectively. The compound formed is SiCl4. 5 Ans: C The ionic compound can form strong ion-dipole interaction with polar solvent, making it soluble in polar solvents. The long chain hydrocarbon can also form strong temporary dipole-induced dipole interaction with non-polar organic solvents, making it soluble in organic solvents. Around S atom, there are 4 bond pair regions and 0 lone pair, the shape is tetrahedral. All the C atoms are sp3 hybridised with tetrahedral shape and bond angle of 109.5°

3 6 Ans: B pV = nRT, with n and p kept constant. As T increases, Option A: V increases. [ V = nR p ×T] Option B: pV T remains constant. [ pV T = nR] Option C: Density of the gas deceases [ m V = pMr RT ] Option D: pV increases. [pV = nRT] 7 Ans: D Hrxn =  Hfo (products)   Hfo (reactants) −27 = 3×(−394) − [Hfo Fe2O3 + 3×(−110)] Hfo Fe2O3 = −825 kJ mol−1 8 Ans: D The following processes are involved in the Born-Haber cycle for ionic compounds: 1) Enthalpy change of formation of BaF3(s) 2) Enthalpy change of atomisation of Ba(s) 3) Bond energy of F−F 4) 1st + 2nd + 3rd ionisation energies of Ba(g) 5) First el