JPJC H2 Paper 1 Prelim Exam 2023 worked solution updated

© Jurong Pioneer Junior College [Turn Over NAME CLASS 22S JURONG PIONEER JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION 2023 CHEMISTRY 9729/01 Higher 2 Paper 1 Multiple Choice Questions 19 September 2023 1 hour Candidates answer on the Question paper. 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 exam index number 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 or 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 12 printed pages.

2 © Jurong Pioneer Junior College 9729/01/J2 PRELIM EXAM/2023 Ans Key 1 B 2 C 3 C 4 B 5 D 6 D 7 C 8 A 9 A 10 D 11 D 12 A 13 C 14 A 15 C 16 C 17 D 18 A 19 A 20 D 21 D 22 B 23 C 24 D 25 D 26 B 27 B 28 C 29 A 30 B 1 Element X has eight more protons than element Y. Which statement must be correct? A Atoms of element Y are larger than atoms of element X. B Element X has at least one fully filled shell of electrons. C Element X and element Y are in the same group. D Element X and element Y are in the same period. A: X could be K whereas Y could be Na, K (X) has larger atomic radius than Na (Y). Statement A is not correct all the time. B: Element X must have at least 9 protons, which means element X must have fully filled 1s subshells. C: Y could be K whereas X could be Co, they are in the same period (Period 4) but not in the same group. Statement C is not correct all the time. D: Y could be He whereas X could be Ne, they are in the same group (Group 18) but not in the same period. Statement D is not correct all the time. Ans: B

3 © Jurong Pioneer Junior College 9729/01/J2 PRELIM EXAM/2023 [Turn Over 2 Molecular dimerisation can be described as the process in which two identical molecules combine to give a single product. Examples of dimers are: Al2Cl6 and (CH3CH2CO2H)2. Which statement about the above dimers is incorrect? A Two hydrogen bonds hold the CH3CH2CO2H molecules together in the dimer. B Each aluminium atom is surrounded by four chlorine atoms in the dimer, Al2Cl6. C CH3CH2CO2H exists as dimers when dissolved in water. D Al2Cl6 dimers are non-polar molecules. A: as seen in the diagram, there are a total of 2 hydrogen bond holding the dimer together. B: as seen in the diagram, each alumnium atom is surrounded by 4 chlorine atoms. ✓C: due to the presence of large amount of H 2O, the CH 3CH2CO2H molecule will more likely to form hydrogen bonding with H2O molecules. Thus, they will exist as monomer. D: the dimer is highly symmetrical, all dipole moments cancel each other so that the dimer is non-polar in nature. Ans: C 3 Which statement is correct? A Cl has a relative isotopic mass of 35.5. B Cl2 has a relative molecular mass of 70. C ICl has a relative molecular mass of 162.4. D NaCl has a relative molecular mass of 58.5. A: Cl has a relative (isotopic) atomic mass of 35.5. isotopic mass refer to specific isotopes whereas atomic mass take into account of all different isotopes. B: Cl2 has a relative molecular mass of (70) 71 (35.5 x 2 = 71). ✓C: D: NaCl has a relative (molecular) formula mass of 58.5. NaCl is ionic compound, it does not exist as discreet molecules, thus, it only has relative formula mass. Ans: C

4 © Jurong Pioneer Junior College 9729/01/J2 PRELIM EXAM/2023 4 Use of the Data Booklet is relevant to this question. The most common scuba diving gas tank typically comprises 11.7% of helium gas, 56.2% of nitrogen gas and 32.1% of oxygen gas by volume. What is the mass of oxygen gas inside the diving gas tank of 0.018 m3 and 300 bar at 293K? A 1.14 kg B 2.28 kg C 7.10 kg D 7100 kg 5(300 10 )(0.018 32.1%) 8.31 29332.0 2280 2.28 mpV nRT pV RT Mr m m g kg =  =   =   == Ans: B 5 The following graph shows the behaviour of an ideal gas with fixed mass. Which of following will yield the graph above? X-axis Y-axis condition A p in Pa pV in Pa m3 constant temperature B p in Pa ρ (density) in g m−3 constant temperature C T in K V in m3 constant pressure D T in K 1 p in Pa−1 constant volume A: pV = nRT ⇒ pV = k (since n and T are both constant). Graph is y = k graph. B: pV = nRT ⇒ p = m RTMr V ⇒ p = ρ RT Mr (since T is constant), graph is y = kx graph. C: pV = nRT ⇒ V = nR p T (since n and p are both constant), Graph is y = kx graph. D: pV = nRT ⇒ 1 p T = nR V (since n and V are both constant), graph is yx = k ⇒ y = k x Ans: D Y X

5 © Jurong Pioneer Junior College 9729/01/J2 PRELIM EXAM/2023 [Turn Over 6 Which description of Dalton’s Law is correct? A Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. B The partial pressure of a gas in a mixture is given by the product of its mole fraction and the total pressure. C The partial pressure of a gas in a mixture is given by the product of its percent by mass and the total pressure. D The total pressure of a mixture of gases is equal to the sum of the partial pressures of these gases. A: this is Avogadro’s Law, n ∝ V B: this is application of Dalton’s Law C: mole ratio does not equal to mass ratio, thus, this is the wrong application of Dalton’s Law D: correct description of Dalton’s Law of partial pressure. Ans: D 7 When iodine is oxidised by nitric acid, a white crystalline solid iodine-containing oxide can be isolated from the mixture. 0.001 mole of this oxide reacts with 0.01 0 mole of acidified potassium iodide to give 0.006 mole of iodine, I2. What is the oxidation number of iodine in the oxide? A +1 B +3 C +5 D +7 [R]: iodine oxide + ne−⟶ I2 [O]: 2I− ⟶ I2 + 2e− Since I− : I2 : e− = 2 : 1 : 2, 0.010 mol of I− will produce 0.005 mol of I2 Total amount of I2 produced is 0.006 mol ⇒ amount of I2 produced from [R] = 0.001 mol Since I− : e− = 1 : 1, total amount of electron lost from [O] = 0.010 mol Total amount of electron gain from [R] = 0.010 mol I2 : e− = 0.001 : 0.010 ⇒ 1 : 10 I2 : I = 2 : 1, each iodine will gain 5 electron in [R] to form I2 (with oxidation number 0) So oxidation number of iodine will be +5 in iodine oxide. Ans: C

6 © Jurong Pioneer Junior College 9729/01/J2 PRELIM EXAM/2023 8 Which statements are correct? 1 enthalpy change of combustion of H2 = enthalpy change of formation of H2O 2 enthalpy change of atomisation of H2 = bond energy of H−H 3 enthalpy change of solution of HCl = l− +enthalpy change of hyd aratio nd n H C of A 1 only B 1 and 2 only C 1 and 3 only D 1, 2 and 3 ✓1: it is a correct statement 2: HatH2: ½ H2(g) ⟶ H(g), whereas BE(H-H): H2(g) ⟶ 2H(g) ∴ BE(H-H) = 2 x HatH2 3: Hsol only refer to ionic solid, in which HCl is simple covalent compound. In addition, Hsol involves the breaking of ionic bond, i.e. −L.E also. Hsol is not simply the sum of hydration energy of the respective ions. Ans: A 9 Use of Data Booklet is relevant to this question Carbon can exist naturally in two different crystalline forms, diamond and graphite. carbon (diamond) ⟶ carbon (graphite) Hꝋ = −1.9 kJ mol−1 Sꝋ = +3.4 J K−1 mol−1 Which statements are correct and explain why diamond does not change into graphite at room temperature and pressure? 1 The forwa