CJC 2023 H2 Chem Prelim P1 Worked Solutions (1)

1 9729/01 CJC JC2 Preliminary Examination 2023 [Turn over CANDIDATE NAME CLASS 2T CHEMISTRY 9729/01 Paper 1 Multiple Choice 15 September 2023 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, class and NRIC/FIN number on the Answer Sheet in the spaces provided. 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 16 printed pages. WORKED SOLUTIONS Catholic Junior College JC 2 Preliminary Examinations Higher 2

2 9729/01 CJC JC2 Preliminary Examination 2023 1 What volume of air is required for the complete combustion of 1.0 dm 3 of octane vapour, C8H18 (g), in a car engine? (Air contains 20% oxygen by volume.) A 2.5 dm3 B 12.5 dm3 C 62.5 dm3 D 125.0 dm3 2 The ionisation energies, in kJ mol –1, of three elements are given in the table. 1st ionisation energy 2nd ionisation energy 3rd ionisation energy Ne 2080 3950 6150 Na 494 4560 6940 Mg 736 1450 7740 Which statement(s) about these ionisation energies is/are correct? 1 Ne has the greatest 1st ionisation energy of the three elements because its electrons experience the most interelectronic repulsion. 2 Mg has the lowest 2nd ionisation energy of the three elements because the electron being removed experiences the greatest shielding effect. 3 There is a large increase from the 2nd to 3rd ionisation energy of Na because the electron is being removed from the next inner subshell. A 1 and 2 only B 2 only C 1 and 3 only D 3 only Topic: Mole Concept Combustion of octane: C8H18 + 𝟐𝟐𝟐𝟐 𝟐𝟐 O2  8CO2 + 9H2O Volume of O2 needed = 1 x 𝟐𝟐𝟐𝟐 𝟐𝟐 = 12.5 dm3 Volume of air needed = 𝟏𝟏𝟏𝟏𝟏𝟏 𝟐𝟐𝟏𝟏 × 𝟏𝟏𝟐𝟐. 𝟐𝟐 = 62.5 dm3 Answer: C Topic: Atomic Structure Electronic configuration of Ne: 1s22s22p6; Na: 1s22s22p62s1; Mg: 1s22s22p62s2 Statement 1 – Wrong. Ionization energy (IE) depends on nuclear charge, distance of electron from nucleus, and shielding effect. Out of the three elements, Ne has the smallest nuclear charge, but the electron being removed is from the 2nd principal quantum shell compared to Na and Mg (3rd principal quantum shell), and experiences less shielding effect, hence greatest attraction from nucleus, and greatest amount of energy required to remove it. Statement 2 – Correct. Out of the three elements, Mg has the greatest nuclear charge but this is outweighed by greater shielding effect (2 inner shells vs 1 inner shell for Ne + and Na+), hence weakest attraction from nucleus on the valence electron being removed, and lowest 2nd IE. Statement 3 – Wrong. 2nd and 3rd electron removed are both from the 2p subshell (electrons are removed from 2p before 2s subshell as 2p is higher in energy). 8th electron to be removed

3 9729/01 CJC JC2 Preliminary Examination 2023 [Turn over 3 Equimolar amounts of the liquids hexane, CH3(CH2)4CH3, and triethylamine, (CH3CH2)3N, are mixed together at 20 °C. The original intermolecular forces are disrupted and stronger intermolecular forces between CH3(CH2)4CH3 and (CH3CH2)3N are formed simultaneously. The boiling point of hexane is 69 °C. The boiling point of triethylamine is 90 °C. Which row is correct? initial temperature of mixture boiling point of mixture A above 20 °C below 69 °C B below 20 °C below 69 °C C below 20 °C above 90 °C D above 20 °C above 90 °C 4 In the gaseous state, phosphorus(V) chloride exists as a molecule with the formula PC l5. However, when it is a solid, it is ionic with the formula PCl4+PCl6–. Which one of the following statements is correct? A The bond angle in PCl4+ is smaller than that in PCl6–. B There is a dative covalent bond in PCl6–. C The P atom in PCl4+ has an expanded octet. D There is a net dipole moment in PCl5. Topic: Chemical Bonding (modified from 2020/I/5) Since the question stem states that stronger intermolecular forces (IMF) are formed after mixing, energy released from forming the new IMF greater than the energy absorbed to break original IMF (exothermic process overall), hence the initial temperature of mixture should rise to above 20 °C, before equilibrating back to room temperature. Since the resulting IMF are s tronger, the energy required to separate the molecules during boiling is also larger than that in pure hexane and triethylamine respectively. Thus, the boiling point of the new mixture is above 90 °C. Answer: D Topic: Chemical Bonding Option A: Wrong. The bond angle in PCl4+ is 109.5o while the bond angle in PCl6– is 90o. Option B: Correct. There is a dative bond from Cl– to P. Option C: Wrong. There are only 8 electrons surrounding P in PCl4+. Option D: Wrong. The dipole moments cancel out in PCl5. Cl Cl Cl Cl ClP 90o 120o Cl P Cl Cl Cl 109.5o 90o P Cl Cl Cl Cl ClCl trigonal bipyramidal tetrahedral octahedral Answer: B

4 9729/01 CJC JC2 Preliminary Examination 2023 5 Which pairs of compounds contain a polar and a non-polar molecule? 1 CO2, H2O 2 SO2, NO2 3 CH2Cl2 , SiCl4 A 1, 2 and 3 B 1 and 3 only C 2 and 3 only D 1 only Topic: Chemical Bonding Option 1: Yes, CO2 is non-polar while H2O is polar CO O O H H Option 2: No, both SO2 and NO2 have bent shapes and are polar. S O O N O O Option 3: Yes, CH2Cl2 is polar while SiCl4 is non-polar C H H Cl Cl Si Cl Cl Cl Cl Answer: B (1 and 3 only)

5 9729/01 CJC JC2 Preliminary Examination 2023 [Turn over 6 A gaseous dimer, X2, dissociates into its gaseous monomer, X, at 400 K and 1 atm pressure. Dissociation is complete at 450 K. X2 (g) → 2X (g) Which of the following graphs shows the variation of volume with temperature when one mole of X2 is heated from 350 K to 500 K at a constant pressure of 1 atm? Assume that the gases behave ideally. (R = 0.082 mol−1 dm3 atm K−1) A B C D Topic: Gaseous State Concept: Ideal Gas equation pV = nRT Answer: D PV=nRT V = (nR/P) T Pressure is kept constant ∴ V initially varies linearly with T with constant gradient (nR/P). Then at 400 K, dissociation occurs, and n increases as well, hence gradient increases, until 450 K. The V at 450 K should be 900R, as V=(2R/1)(450)= 900R. 350 400 450 500 T/ K Vol/ dm3 400R 350 400 450 500 T/ K Vol/ dm3 800R 350 400 450 500 T/ K Vol/ dm3 800R 350 400 450 500 T/ K Vol/ dm3 800R 800R 400R 400R 400R

6 9729/01 CJC JC2 Preliminary Examination 2023 7 Use of the Data Booklet is relevant to this question. At 400 K, the following species behave as ideal gases: H2, CH4, NO2, Ar. 5.00 g of each of these gases is put separately into a gas syringe kept at 400 K. The volume of each gas is adjusted to be the same, and pressure is measured. Which gas will have the lowest pressure? A H2 B CH4 C NO2 D Ar 8 The graph below shows the variation in the melting point for 8 consecutive elements in the Periodic Table, all with atomic number ≤ 2