2020 ACJC Prelim Paper 1 worked solutions

1 © ACJC2020 9729/01/Prelim/2020 ANGLO-CHINESE JUNIOR COLLEGE DEPARTMENT OF CHEMISTRY Preliminary Examination CHEMISTRY 9729/01 Higher 2 Paper 1 Multiple Choice 16 September 2020 1 hour Additional Materials: Multiple Choice Answer Sheet Data Booklet READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. Write your name, index number and tutorial class on the Answer Sheet in the spaces provided unless this has been done for you. There are thirty 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 15 printed pages and 1 blank page. 9729/01/Prelim/2020 ANGLO-CHINESE JUNIOR COLLEGE © ACJC 2020 Department of Chemistry [Turn over

2 © ACJC2020 9729/01/Prelim/2020 1 B 11 B 21 D 2 D 12 B 22 D 3 A 13 D 23 C 4 B 14 C 24 B 5 C 15 C 25 C 6 A 16 B 26 A 7 C 17 C 27 D 8 D 18 B 28 A 9 A 19 A 29 B 10 D 20 D 30 C 1 In July 2020, some households in Singapore reported a Pandan smell in their tap water. This was caused by an organic solvent tetrahydrofuran, THF , which has the following structure. The laboratory found that the concentration of THF in the Pandan smelling water was less than 10 ppb (10 parts per billion) by mass, which was significantly lower than the levels of THF that would cause health concerns. What is the concentration of 10 ppb of THF in mol dm3? Given:  1 billion = 109  density of water = 1.0 g cm3 A 1.39 x 1010 B 1.39 x 107 C 1.56 x 107 D 6.40 x 104 Answer: B Mr of THF = (12 x 4) + (1 x 8) + 16 = 72.0 In 109 g of water, there is 10g of THF (10ppb) 1 dm3 = 1000 cm3 of water Since density is 1.0 gcm3 1000 cm3 of water is 1000 g Hence, in 1000 g of water, mass of THF = 9 10 100010  = 1.00 x 105 g

3 © ACJC2020 9729/01/Prelim/2020 Amount of THF = 1.00 x 105 72.0 = 1.39 x 107 mol dm3 2 Use of the Data Booklet is relevant to this question. The approximate percentage composition of the atmosphere s on four planets are given below. The density of a gas may be defined as the mass of 1 dm3 of the gas. Which mixture of gases has the greatest density? planet gases / % by volume A Jupiter H2 88.8, He 10.2, Ar 1.0 B Neptune H2 80.0, He 19.0, CH4 1.0 C Saturn H2 92.3, He 7.2, CH4 0.5 D Uranus H2 82.5, He 15.2, CH4 2.3 Answer: D Density of mixture in option A = 88.8 × 2.0 + 10.2 × 4.0 + 1.0 × 39.9 100 = 2.583 Density of mixture in option B = 80.0 × 2.0 + 19.0 × 4.0 + 1.0 × 16.0 100 = 2.52 Density of mixture in option C = 92.3 × 2.0 + 7.2 × 4.0 + 0.5 × 16.0 100 = 2.214 Density of mixture in option D = 82.5 × 2.0 + 15.2 × 4.0 + 2.3 × 16.0 100 = 2.626 3 Use of the Data Booklet is relevant to this question. An element E can exist in a few oxidation states. 0.01 mol of E2+ is completely reacted with 0.004 mol of acidified KMnO4. What is the final oxidation state of E? A +4 B +3 C 0 D −1 Answer: A MnO4− + 8H+ + 5e− ⇌ Mn2+ + 4H2O 0.004 mol MnO4− gains 0.004 × 5 = 0.020 mol of e−. 0.01 mol E2+ loses 0.020 mol of e−.  1 mol E2+ loses 2 mol of e−. O.S. of E increases from +2 to +4. 4 The successive ionisation energies (IE) of two elements, M and N, are given below. IE / kJ mol–1 1st 2nd 3rd 4th 5th 6th 7th 8th

4 © ACJC2020 9729/01/Prelim/2020 M 550 1065 4138 5500 6910 8760 10230 11800 N 1140 2103 3470 4560 5760 8550 9940 18600 What is the likely formula of the compound that is formed between M and N? A MN B MN2 C M2N D M3N2 Answer: B M shows the largest increases from the 2nd to 3rd IE, so it has 2 valence electrons and it is from group 2. N shows the largest increases from the 7th to 8th IE, so it has 7 valence electrons and it is from group 17. M2+ and N– ions form MN2. 5 Which of the following statements best explains why the first ioni sation energy of argon is higher than that of sulfur? A Sulfur is more electronegative than argon. B Argon has a complete octet, while sulfur does not. C The nuclear charge in argon is greater than that in sulfur. D The electron removed from sulfur is unpaired, while that removed from argon is paired. Answer: C Statement A is wrong because electronegativity describes the attraction of the nucleus and a bonding pair of electrons between two covalently bonded atoms, and not the energy needed for removing a valence electron from a gaseous atom. Statement B is wrong the complete octet does not describe the strength of attraction between the nucleus and its valence electrons. Statement C is correct because stronger nuclear charge and approximately constant shielding effect in Ar leads to stronger effective nuclear charge. Thus Ar has greater nuclear attraction and more energy is needed to remove an electron from the valence shell, thus 1st IE is greater in Ar than S. Statement D is wrong because the first electron removed from sulfur and argon are both from an electron pair.

5 © ACJC2020 9729/01/Prelim/2020 6 Two glass bulbs X and Y are connected by a closed valve. X contains neon at 20 C at a pressure of 1 x 105 Pa. Y is initially empty, and has three times the volume of X. In an experiment, the valve is opened and the temperature of the whole apparatus is raised to 100 C. What is the final pressure in the system? A 3.18 × 104 Pa B 4.24 × 104 Pa C 1.25 × 105 Pa D 5.09 × 105 Pa Answer: A Since amount of Ne remains constant, pV= nRT simplifies to 1 1 2 2 12 5 1 2 1 4 2 1 10 4 20 273 100 273 3.18 10 pV p V TT V p V p       7 Which of the following species is not planar? A XeF4 B ICl3 C PCl4+ D C2Cl4 Answer: C A: Square planar B: T-shaped Y X

6 © ACJC2020 9729/01/Prelim/2020 C: Tetrahedral D: Planar (Trigonal planar about each C atom 8 Which diagram best represents the structure of sodium fluoride? A C B D Answer: D A and B are wrong because they are made up of atoms of Na and F. A and C also show Na & F (or Na+ & F–) particles in pairs, instead of an alternating lattice structure. The giant ionic lattice does not consist of ionic pairs or atomic pairs with intermolecular forces. D is the only diagram that correctly depict all three features coorectly - Na+ & F– ions (not Na & F atoms), - Na+ & F– ions that are larger than Na+ ions, and - Alternating arrangement of Na+ & F– ions in a giant ionic lattice structure Na F Na F Na F Na F Na F Na F Na F F– Na+ F– Na+ F– Na+ Na F Na F F– Na+ F– Na+ F– Na+ Na F Na F Na F Na F Na F Na F Na F Na F F– Na+ F– Na+ F– Na+ F– Na+ Na F F– Na+ F– Na+ F– Na+ F– Na+ F– Na+ F– N