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 2 September 2015 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 30 printed pages and 0 blank page. WORKED SOLUTIONS

2 9647/01/CJC JC2 Preliminary Exam 2015 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 When 20 cm 3 of a gaseous hydrocarbon was completely burnt in 130 cm 3 of oxygen, the volume of ga s remaining after the reaction wa s 100 cm 3. This volume was decreased to 40 cm 3 when the resulting mixture wa s passed through aqueous sodium hydroxide. All measurements were made at room temperature and pressure. What is the formula of this hydrocarbon? A C2H2 B C3H6 C C3H8 D C4H10 Answer: B Since temperature and pressure are constant, mol ratio of gas ∝ vol ratio CxHy (g) + )( 4 yx O2 (g) → xCO2 (g) + 2 y H2O (l) Initial vol/ cm3: 20 130 0 0 Final vol/ cm3: 0 40 (100-40) Vol reacted or formed/ cm3: 20 (130-40) 60 = 90 1 vol 4.5 vol 3 vol  By inspection, x = 3. 2 9) 4 y(3   y = 6  molecular formula of the hydrocarbon is C3H6 2 Consider the following half-equations: C2O4 2–  2CO2 + 2e– Fe2+  Fe3+ + e– MnO4 – + 8H+ + 5e–  Mn2+ + 4H2O What volume of 0.01 mol dm –3 potassium manganate( VII) is required to oxidise completely 25.0 cm3 of an acidified solution of 0.01 mol dm–3 FeC2O4? A 10 cm3 B 15 cm3 C 25 cm3 D 42 cm3 Answer: B Both Fe2+ and C2O4 2– ions can be oxidised by MnO4 – ions which is reduced. Hence, there are two oxidation half-equations to consider: Fe2+  Fe3+ + e– --- 1 and C2O4 2–  2CO2 + 2e– --- 2 So combine 1+2 first by summing up both half-equations to get: Fe2+ + C2O4 2–  Fe3+ + 2CO2 + 3e– --- 3

3 9647/01/CJC JC2 Preliminary Exam 2015 [Turn over Combine this equation with the reduction half-equation of MnO4 – MnO4 – + 8H+ + 5e–  Mn2+ + 4H2O To get: 3MnO4 – ≡ 5 FeC2O4 Amount of FeC2O4 = 25/1000 x 0.01 = 2.50 x 10–4 mol Amount of KMnO4 reacted = (2.50 x 10-4 / 5) x 3 = 1.50 x 10–3 mol Vol. of KMnO4 required = (1.50 x 10–3) / 0.01 = 1.50 x 10–2 dm3 = 15 cm3 3 X and Y are elements with atomic numbers between 6 and 15. Their first seven ionisation energies in kJ mol–1 are shown below. X 580 1800 2700 11600 14800 18400 23300 Y 1310 3400 5300 7500 11300 13300 71300 Which of the following best describes the compound formed between X and Y? A basic B acidic C neutral D amphoteric Answer: D Calculate the difference between successive I.E. For X, there is a sharp increase from the 3 rd to 4 th I.E. This indicates that the 4 th electron is removed from an inner electron shell, closer and more tightly bound by the positively-charged nucleus. Thus, there are 3 valence electrons; X belongs to Group III, and is Al. Using the same concept, ther e is a sharp increase from the 6th to 7 th I.E. for Y. Y belongs to Group VI, and is O. The compound formed from X and Y has the formula A l2O3 which is amphoteric in nature. Alternatively, you can match the I.E. given to those given in the data booklet to determine the identity of X and Y.

4 9647/01/CJC JC2 Preliminary Exam 2015 4 Which pair of compounds fits the following descriptions? (i) The first compound has a larger bond angle about the central atom than the second compound and, (ii) The second compound is more polar than the first compound. first compound second compound A B C D BCl3 CF4 HCN PH3 SO2 XeF4 BeCl2 NH3 Answer: A If necessary, draw the dot-and-cross diagrams of the compounds first. Option A: SO2 has a bent shape and a smaller bond angle than 120˚ due to the presence of the lone pair of electrons. (i) is satisfied. (ii) is also satisfied as BCl 3 is non -polar while SO2 has a net dipole moment and is polar. Option B: (i) is satisfied but (ii) is not as both are symmetrical and have no net dipole moment and are hence non-polar. Option C: (i) is not satisfied. (ii) is not satisfied as HCN is polar while BeCl2 is not. Option D: The more electronegative the central atom, the greater the repulsion between the bond pairs. N is more electronegative than P and pulls the bonded electron pairs closer to itself. Hence, the bonded electron pairs get closer to each other and repel much more, resulting in a larger bond angle for NH 3 (even though NH 3 and PH3 are both trigonal pyramidal in shape).

5 9647/01/CJC JC2 Preliminary Exam 2015 [Turn over 5 Which of the following statements regarding covalent compounds is true? A Tetrachloromethane is a volatile liquid because the C –Cl bond can be broken easily. B CHCl3 has a higher boiling point than CHF 3 because it has more electrons than CHF3. C Hydrogen chloride is soluble in water because it can form permanent dipole - permanent dipole forces of attraction with water molecules. D Each hydrogen bond formed between H 2O molecules is stronger than that formed between HF molecules. Answer: B Option A: For a simple covalent molecule like CCl4, covalent bonds are not broken when a liquid boils/vaporizes. CC l4 is non-polar and the weak intermolecular van der Waals’ forces of attraction are overcome instead. Option B: boiling point ∝ strength of van der Waals’ forces ∝ no. of electrons Option C: HC l when dissolved in water undergoes complete dissociation to form H + and Cl- ions which form ion-dipole attractions with water molecules. Option D: strength of hydrogen bond ∝ difference in electronegativity of the 2 atoms involved in the hydrogen bonding. O–H bond has a smaller electronegativity difference than H –F bond as O is less electronegative than F. Note: H 2O has a higher boiling point than HF as there are, on average, more hydrogen bonds between H2O molecules than ther e are between HF molecules (H 2O has more extensive intermolecular hydrogen bonding than HF). 6 Which of the following diagrams does not describe the behaviour of a fixed mass of an ideal gas? ( = density of the gas, T = temperature is measured in K) A B C D 1/p T  0 p pV  0 0 V constant T constant T constant T constant V 0 p

6 9647/01/CJC JC2 Preliminary Exam 2015 Answer: C For a given mass of gas, the no. of mo les of gas, n, will be constant since Mr unchanged. pV = nRT Option A: V = nRT ( 1 P) V = constant x ( 1 P) , at constant T for a given mass of gas. Hence, V  ( 1 P) Option B: pV = nRT pV = constant, at constant T for given mass of gas, regardless of changes in P, V or  Option C: pV = nRT = m Mr RT ρ = m V = Mr RT (P) = constant x (P), at constant T for a given mass of gas