2023 ACJC H2 Chem Prelim Paper 1 Worked Solutions

1 © ACJC 2023 9729/01/Prelim/2023 [Turn over Anglo-Chinese Junior College JC2 Preliminary Examination Higher 2 CHEMISTRY Paper 1 Multiple Choice Additional Materials: Multiple Choice Answer Sheet Data Booklet 9729/01 13 September 2023 1 hour READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. Write your name and index number on the Answer Sheet in the spaces provided unless this has been done for you. There are thirty questions in this section. 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 xx printed pages and xx blank page.

2 © ACJC 2023 9729/01/Prelim/2023 [Turn over 1 Which types of chemical bonds listed below are present in solid ammonium nitrate? 1 ionic bonds 2 dative bonds 3 hydrogen bonds A 1 only B 1 and 2 only C 1 and 3 only D 1, 2 and 3 Ans: B Identify structure to be giant ionic, with covalent bonds within the polyatomic cation, NH4+, and polyatomic anion, NO3–. There is a dative covalent bond between N and H in NH4+, and N and O in NO3–. AOA 2 1.00 mol of gaseous molecules A take up a volume of 50 dm 3 at a pressure of 2 bar and a temperature of 50 oC. Which statements explain the above observation? 1 Gaseous molecules of A are in constant rapid random motion. 2 Gaseous molecules of A have significant molecular volume. 3 Gaseous molecules of A experience strong intermolecular forces of attraction. A 1 only B 1 and 2 only C 2 and 3 only D 1, 2 and 3 Ans: B For ideal gas, 𝑉𝑖𝑑𝑒𝑎𝑙 = 𝑛𝑅𝑇 𝑝 = 1.00×8.31×(273+50) 200 000 = 13.4 dm3 ∴ Vobserved > Videal All gases are in constant rapid random motion and exerts pressure as a result. Since there is a positive deviation (Vobserved > Videal), it is possibly due to significant molecular volume. Strong IMF would lead to a negative deviation. AOB

3 © ACJC 2023 9729/01/Prelim/2023 [Turn over 3 Two properties relating to silicon and sulfur which are non–metallic elements and their atoms are as follows. • property 1 – the oxide forms a strong acid in water • property 2 – no paired 3p electrons Which properties do silicon and sulfur have? silicon sulfur A 1 and 2 1 only B 1 and 2 1 and 2 C 2 only 1 only D 2 only 1 and 2 Ans: C Property 1 SiO2 does not dissolve in water because of strong covalent bonds between the Si and O atoms in the giant molecular structure. SO3 readily soluble to form an acidic solution. SO3(l) + H2O(l) → H2SO4(aq) Property 2 Si – 1s22s22p63s23p2 No paired 3p electrons S – 1s22s22p63s23p4 1 paired 3p electrons AOA

4 © ACJC 2023 9729/01/Prelim/2023 [Turn over 4 The Group 2 metals have higher melting points than the Group 1 metals. Which factors could contribute towards the higher melting points? 1 There are smaller interatomic distances in the metallic lattices of the Group 2 metals. 2 Two valence electrons from each Group 2 metal atom are available for bonding in the metallic lattice. 3 Group 2 metals have the higher first ionisation energy. A 1, 2 and 3 B 1 and 2 only C 2 and 3 only D 1 only Ans: B Option 1 is correct – As the interatomic distances is smaller in Group 2 metals, the charge density (since radius is smaller) is higher, hence, stronger metallic bond and higher mp. Option 2 is correct – As Group 2 metals contribute 2 valence electrons to the sea of delocalised electrons, it has stronger metallic bond, and hence, higher mp. Option 3 is incorrect – Higher first IE relates to metals in gaseous state. Hence, does not explain the mp. AOA 5 Which statement about the trend in the property of the halogens down the group is correct? A The electronegativity increases. B The reactivity as oxidising agents increases. C The enthalpy change of reaction with hydrogen becomes more endothermic. D The volatility increases. Ans: C The electronegativity and volatility decrease down the group. As can be observed from the EӨ values, the oxidising strength of the halogens decreases down the group. Halogens react with hydrogen in the gaseous phase to give hydrogen halides. As reactivity/oxidising power of the halogens decreases down the group, the vigour of the reaction also decreases down the group. Hence, the enthalpy change of reaction with hydrogen becomes more endothermic. AOA Recall: • Energy is needed to overcome the metallic bonds during melting of metals • Metallic bond strength is dependent on: (1) charge density, (2) number of valence e- contributed to delocalised e-

5 © ACJC 2023 9729/01/Prelim/2023 [Turn over 6 Group 1 and Group 2 ionic hydrides react with water. H‒ + H2O → OH‒ + H2 In an experiment, 1 .01 g of a sample of an ionic hydride is dissolved in excess H2O. The resulting solution required 24.0 cm3 of a 2.0 mol dm–3 HCl solution for complete neutralisation. What is the formula of the hydride? A LiH B NaH C MgH2 D CaH2 Ans: D Amt of HCl required = 0.048 mol Amount of OH‒ = 0.048 mol = Amount of H‒ If Group 1 ionic hydride, amount of MH = 0.048 mol and Mr = 21.0 If Group 2 ionic hydride, amount of MH2 = 0.024 mol and Mr = 42.1 Mr of CaH2 = 40.1 + 2.0 = 42.1 AOB 7 Two identical iron(II) sulfate solutions were separately titrated with acidified K 2Cr2O7 and acidified KMnO4 solutions of equal concentrations. Which statement describes the required volumes of K 2Cr2O7 and KMnO4 solutions needed to completely oxidise the iron(II)? A The volume of KMnO4 solution is 0.83 times that of the volume of K2Cr2O7. B The volume of KMnO4 solution is 1.20 times that of the volume of K2Cr2O7. C The volume of KMnO4 solution is 1.82 times that of the volume of K2Cr2O7. D The volume of KMnO4 solution is 2.22 times that of the volume of K2Cr2O7. Ans: B 6 moles of electrons per mole of dichromate. 5 moles of electrons per mole of manganate(VII). Let no of moles of Fe2+ be x No of moles of dichromate = x/6, no of moles of manganate(VII) = x/5 Given the fact that the dichromate and manganate(VII) solutions are of the same concentrations → vol of manganate is 6/5 that of vol of dichromate. AOB

6 © ACJC 2023 9729/01/Prelim/2023 [Turn over 8 Which reaction represents a standard enthalpy change at 298 K? A 1 2F2(g) ⟶ F(g) B C(g) + O2(g) ⟶ CO2(g) C C(s) + 4F(g) ⟶ CF4(g) D C3H8(g) + 5O2(g) ⟶ 3CO2(g) + 4H2O(g) Ans: A A ✓: Standard enthalpy change of atomisation of F2 B : At 298 K, C should be a solid C : Fluorine exists as F2 under std state D : At 298 K, H2O should be a liquid AOA 9 Ethane undergoes combustion as shown. C2H6(g) + 7 2O2(g) 2CO2(g) + 3H2O(l) ∆Hco = –1561 kJ mol–1 Some relevant data are given below. ∆Hco of C(s) = –394 kJ mol–1 ∆Hfo of C2H6(g) = –85 kJ mol–1 ∆Hfo of H2O(g) = –243 kJ mol–1 What is the standard enthalpy change of vaporisation of H2O(l)? A +43 kJ mol–1 B –43 kJ mol–1 C +129 kJ mol–1 D –129 kJ mol–1 Ans: A By Hess Law, –1561–85 = 2(–394)+3(–243) –3 ×∆HvapƟ of H2O(l) ∆HvapƟ of H2O(l) =+43 kJ mol-1 AOB

7 © ACJC 2023 9729/01/Prelim/2023 [Turn over 10 The product [X][Y] of the concentrations of X and Y is plotted against time, t, for the following second-order reaction. X + Y ⟶ Z Which graph would be obtained? A B C D Ans: B X + Y → Z The question states the reacti