CJC H2 CHEM P1 SOL Prelim

1 [Turn over CATHOLIC JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS Higher 2 CHEMISTRY 9647/01 Paper 1 Multiple Choice Wednesday 3 September 2014 1 hour Additional Materials: Multiple Choice Answer Sheet Data Booklet READ THESE INSTRUCTIONS FIRST Write your name and HT group on the Answer Sheet in the spaces provided. Write in soft pencil. Do not use staples, paper clips, highlighters, glue or correction fluid. There are forty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the separate Answer Sheet. Read the instructions on the Answer Sheet very carefully. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this booklet. The use of an approved scientific calculator is expected, where appropriate. This document consists of 18 printed pages and 0 blank page. WORKED SOLUTIONS

2 9647/01/CJC JC2 Preliminary Exam 2014 Section A For each question there are four possible answers, A, B, C and D. Choose the one you consider to be correct and record your choice in soft pencil on the separate Answer Sheet provided. 1 Use of the Data Booklet is relevant to this question. Which one of the following has the same number of the stated partic le as atoms in 127 g of iodine at room conditions? A number of atoms in 79 g of gold B number of anions in 0.5 mol of barium chloride C number of ions in 1 mol of potassium bromide D number of molecules in 10 g of hydrogen fluoride Answer: B 127 g of I2 = 2127 127  mol = 0.5 mol of I2 molecules = 1.0 mol of I atoms = 6.02 × 1023 I atoms A: 79 g of Au = 197 79 mol of Au atoms = 0.401 mol of Au atoms = 2.41 × 1023 Au atoms B: Since there are 2 Cl- anions in 1 BaCl2, there is 1 mol of Cl- in 0.5 mol BaCl2 And thus the no. of anions present = 6.02 × 1023 Cl- C: Since there are 2 ions in 1 KBr, there are 2 mol of ions (K+ and Br-) in 1 mol KBr And thus the no. of ions present = 1.20 × 1024 ions D: 10 g of HF = 191 10  mol of HF molecule = 0.500 mol of HF molecule = 3.01 × 1023 HF molecule

3 9647/01/CJC JC2 Preliminary Exam 2014 [Turn over 2 Use of the Data Booklet is relevant to this question. In polluted air, the white paint pigment in older oi l paintings forms lead(II) sulfide, PbS, that is black in colour. To restore the white colour, an oxidising agent, hydrogen peroxide, H2O2, is used. Given that 0.239 g of PbS requires 25 cm 3 of 0.160 mol dm –3 H2O2 for oxidation, what is the possible identity of the white paint pigment? A S8 B PbS2 C PbSO3 D PbSO4 Answer: D no. of mol of PbS = 1.32207 239.0  = 0.00100 mol no of mol of H2O2 = 1000 25 × 0.160 = 0.00400 mol Thus 4 H2O2 ≡ PbS From data booklet, H2O2 + 2H+ + 2e- → 2H2O Thus, 4 H2O2 ≡ 8 e-  PbS ≡ 8 e- Thus S2- loses 8 e-. S in PbS is oxidised from -2 to +6. Thus a possible identity is PbSO4. 3 Use of the Data Booklet is relevant to this question. Which one of the following specie s has more neutrons than electrons and more electrons than protons? A 37Cl– B 48Ti4+ C 79Br+ D 32S2- Answer: A 37Cl– 48Ti4+ 79Br+ 32S2- Proton: 17 22 35 16 Electron: 18 18 34 18 Neutron: 20 26 44 16

4 9647/01/CJC JC2 Preliminary Exam 2014 4 In recent years, many scientists have been researching the potential of copper complexes as drugs in chemotherapy due to their ability to inhibit cell proliferation and induce apoptosis in cells. An example of such a complex is shown below: Which one of the following best describes the bonds formed with Cu? Cu-N1 Cu-N2 A dative π B π ionic C ionic dative D σ π Answer: C N1 is negatively charged. It can participate in ionic bonding with Cu2+ ion. N2 has a lone pair of electrons . Thus it can only form dative covalent bond with Cu 2+ ion as it is electrically neutral.

5 9647/01/CJC JC2 Preliminary Exam 2014 [Turn over 5 What is the most likely bond angle of the sulfur atom in S,S-diphenylsulfilimine? A 90 B 107 C 109 D 120 Answer: B There are 3 bond pairs and 1 lone pair around central atom S. Thus the shape is trigonal pyramidal, 107. 6 Which graph correctly describes the behaviour of fixed masses of the ideal gases I and J, where I has a higher Mr than J? A B C D 1/V P Constant T I J T PV Constant P J I 1/T P Constant V I J Constant T I J PV P

6 9647/01/CJC JC2 Preliminary Exam 2014 Answer: C A B C D 7 Fe3+ and SCN- react in a closed system to give the complex, [Fe(SCN)] 2+, which is blood-red in colour. Fe3+(aq) + SCN-(aq) ⇌ [Fe(SCN)]2+(aq) ΔH < 0 Which one of the following changes will result in the solution turning pale red? A Increase the concentration of SCN-. B Decrease the pressure of the system. C Decrease the temperature of the system. D Add a small amount of dilute NaOH(aq) to the resulting mixture. 1/V P Constant T J I T PV Constant P J I 1/T P Constant V J I Constant T J I PV P

7 9647/01/CJC JC2 Preliminary Exam 2014 [Turn over Answer: D For the solution to turn pale red, there must be a decrease in the concentration of [Fe(SCN)]2+(aq). A Increasing the concentration of SCN - will shift the position of equilibrium to the right, causing more [Fe(SCN)]2+(aq) to be formed. B Decreasing the pressure of the system will have no effect on the position of the equilibrium, as pressure changes only affect gases. C As the forward reaction is exothermic, decreasing the temperature of the system will cause the position of equilibrium to shift to the right. D Adding a small amo unt of NaOH(aq) will cause precipitation of Fe(OH) 3, thus decreasing [Fe3+], resulting in the shift of position of equilibrium to the left. 8 Solid NaC l dissolves in water to give Na +(aq) and C l-(aq) ions under standard conditions of 298 K, 1 atm. NaCl(s) → Na+(aq) + Cl-(aq) Na+(g) + Cl-(g) → NaCl (s) ΔHlatt o = -781 kJ mol-1 Na+(g) → Na+(aq) ΔHhyd o = -390 kJ mol-1 Cl-(g) → Cl-(aq) ΔHhyd o = -381 kJ mol-1 What is the standard enthalpy change of solution, ΔHsol o, for the above reaction? A -10 kJ mol-1 B +10 kJ mol-1 C -20 kJ mol-1 D +20 kJ mol-1 Answer: B ΔHsol o NaCl(s) Na+(aq) + Cl-(aq) ΔHlatt o ΔHhyd o Na+(g) + Cl-(g) ΔHsol o = -ΔHlatt o + ΣΔHhyd o = – (-781) + [(-390) + (-381)] = + 781 – 771 = +10 kJ mol-1

8 9647/01/CJC JC2 Preliminary Exam 2014 9 What is the pH of the resulting solution when 2.50 g of NH4Cl is dissolved in 250 cm3 of 0.100 mol dm-3 NH3(aq)? [Kb of NH3 = 1.8 × 10-5 mol dm-3] A 5.01 B 8.99 C 9.53 D 14.0 Answer: B Mr of NH4Cl = 14 + 4(1) + 35.5 = 53.5 [salt] = 5.53 50.2 × 250 1000 = 0.187 mol dm-3 [base] = [NH3] = 0.100 mol dm-3 pOH = pKb + lg          base salt = –lg(1.8 × 10–5) + lg      100.0 187.0 = 4.74 + 0.27 = 5.01 pH = 14 – pOH = 14 – 5.01 = 8.99 10 Equal volumes of aqueous KI and 0.200 mol dm -3 of Pb(NO 3)2 are mixed together to precipitate PbI2. Given that the K sp value of PbI2 is 8.70 × 10 -9 mol3 dm-9, which one of the following could have been the initial concentration of KI? A 8.70 × 10-8 mol dm-3 B 2.95 × 10-4 mol dm-3 C 5.90 × 10-4 mol dm-3 D 1.50 × 10-2 mol dm-3 Answer: D 2 I-(aq) + Pb2+(aq) → PbI2