2024 Prelims NJC H2 Chem P1 (Ans)

NJC/H2 Chem Preliminary Examination/01/2024 [Turn over NATIONAL JUNIOR COLLEGE SH2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME SUBJECT CLASS REGISTRATION NUMBER CHEMISTRY Paper 1 Multiple Choice Additional Materials: Optical Answer Sheet Data Booklet 9729/01 19 September 2024 1 hour READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, highlighters, glue or correction fluid. Write your name, subject class and registration 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 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. Instructions on how to fill in the Optical Mark Sheet Shade the index number in a 5 digit format on the optical mark sheet: 2nd digit and the last 4 digits of the Registration Number. Example: Student Examples of Registration No. Shade: 2305648 35648 This document consists of 15 printed pages and 1 blank page.

2 NJC/H2 Chem Preliminary Examination/01/2024 Suggestion solution for P1 (MCQ) 1 C 7 A 13 C 19 A 25 D 2 A 8 C 14 A 20 C 26 A 3 B 9 B 15 A 21 B 27 C 4 D 10 D 16 C 22 C 28 B 5 D 11 C 17 D 23 B 29 A 6 B 12 B 18 B 24 B 30 D

3 NJC/H2 Chem Preliminary Examination/01/2024 [Turn over 1 Use of the Data Booklet is relevant to this question. A sample of 35.6 g of hydrated sodium carbonate contains 25.84% sodium ions by mass. When this sample is heated, anhydrous sodium carbonate and water vapour are formed. What is the mass lost? A 7.2 g B 10.6 g C 14.4 g D 21.2 g Ans: C Na2CO3.xH2O → Na2CO3 + xH2O 46 106 + 18(𝑥) × 100% = 25.84 x = 4 Mass of water given off = 35.6 178 × 18 × 4 = 14.4g 2 Use of the Data Booklet is relevant to this question. Sodium and fluorine are both reactive elements. Which statements are correct? 1 One Na atom has two more protons than one F− ion. 2 One Na atom has two more neutrons than one F atom. 3 One Na+ ion has the same number of electrons as one F− ion. Ans: A F → F- Na → Na+ p 9 9 11 11 n 10 10 12 12 e 9 10 11 10 A 1, 2 and 3 B 1 and 2 only C 2 and 3 only D 1 only 3 When iodine is oxidized by nitric acid, a white crystalline solid oxide can be isolated from the mixture. 0.001 mole of this oxide reacts with 0.01 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 +5 C +6 D +10 Ans: B

4 NJC/H2 Chem Preliminary Examination/01/2024 0.001 mol of iodine oxide reacts with 0.01 mol of I − to gives 0.006 mol of I 2. Balancing number of I atoms on both sides, 0.006 mol of I 2 contains 0.012 mol of I atoms. Hence, 0.001 mol of iodine oxide contains 0.002 mol of iodine atoms. 0.002 mol Ix+ reacts with 0.010 mol I− to produce 0.006 mol I2. [O] 2I− ⎯→ I2 + 2e− 0.010 mol I− gives 0.010 mol of e− [R] Ix+ + ne− ⎯→ I2 0.002 mol of Ix+ gains 0.010 mol of e− 1 mol of Ix+ gains 5 mol of e− to give I2 During reduction, oxidation state of Ix+ decreases by 5 units from +5 to 0. 5 Hydrogen peroxide solution decomposes. The equation for this reaction is shown. 2H2O2 (aq) 2H2O (l) + O2(g) A 300 cm3 sample of hydrogen peroxide solution is warmed. After 150 minutes, 10.00 dm 3 of oxygen gas, measured at r.t.p., is collected. Under these conditions, the reaction has a constant half-life of 50 minutes. 4 Which pair of compounds meets the criteria below? • The first compound has a larger bond angle than the second compound. • The second compound is more polar than the first compound. A CO2, BCl3 B IF3, H2O C HCN, SO3 D CO2, NCl3 Ans: D molecules bp lp shape angle Polar? CO2 2 0 Linear 180o Non polar BCl3 3 0 Trigonal planar 120 o Non polar IF3 3 2 T-shape ~90o Polar H2O 2 2 Bent 105o Polar HCN 2 0 Linear 180o Polar SO3 3 0 Trigonal Planar 120o Non polar NCl3 3 1 Trigonal pyramidal 107.5o Polar

5 NJC/H2 Chem Preliminary Examination/01/2024 [Turn over What is the initial concentration of the hydrogen peroxide solution? A 0.79 mol dm−3 B 1.6 mol dm−3 C 2.8 mol dm−3 D 3.2 mol dm−3 Ans: D For first order reaction, half-life is constant Time [Reactant] [Product] 0 100% 0% 1st t1/2 50% 50% 2nd t1/2 25% 75% 3rd t1/2 12.5% 87.5% After 3 half−lifes (50 mins × 3), 10.00 dm3 of O2 = 87.5% of O2 produced. Maximum volume of O2 = 10.00 × 100 87.5 = 11.43 dm3 Maximum amount of O2 at r.t.p. = 11.43 24 = 0.4762 mol Initial amount of H2O2 = 0.4762 × 2 = 0.9524 mol Initial conc of H2O2 = 0.9524 0.3 = 3.17 mol dm−3 6 In order to determine the enthalpy of neutralisation of a strong acid and a strong alkali, 25.0 cm 3 of 2.00 mol dm −3 sodium hydroxide is added to 25.0 cm 3 of 2.00 mol dm −3 hydrochloric acid. The increase in temperature is 12°C. In a second experiment, the same method is used, but 50.0 cm 3 of 2.00 mol dm −3 sodium hydroxide is added to 50.0 cm3 of 2.00 mol dm−3 hydrochloric acid. What is the increase in temperature in the second experiment? A 6°C B 12°C C 24°C D 48°C Ans: B H+ + OH- → H2O 0.05 0.05 0.05 q = mc∆T = (25+25) x 4.18 x 12 = 2508 J H+ + OH- → H2O 0.1 0.1 0.1

6 NJC/H2 Chem Preliminary Examination/01/2024 q = mc∆T 2508 x 2 = (50+250) x 4.18 x ∆T ∆T = 12°C 7 X and Y react together to form Z in a reversible reaction. The equilibrium yield of Z at different conditions are shown in the following table. Conditions Equilibrium yield of Z High Temperature Decreased High Pressure Increased Which equation could represent this reaction? A X(g) + Y(g) ⇌ Z(g) ∆H = –100 kJ mol-1 B X(g) + Y(g) ⇌ Z(g) ∆H = +100 kJ mol-1 C X(s) + Y(g) ⇌ 2Z(g) ∆H = –100 kJ mol-1 D X(s) + Y(g) ⇌ 2Z(g) ∆H = +100 kJ mol-1 Ans: A Higher temperature, equilibrium shifts left which favour endothermic reaction. Therefore, the forward reaction is an exothermic reaction. Lower pressure, equilibrium shifts left since L.H.S has greater number of moles of gases. 8 PCl5 decomposes as shown. PCl5(g) PCl3(g) + Cl2(g) 1.0 mole of PCl5(g), 1.0 mole of PCl3(g) and 1.0 mole of Cl2(g) are placed in a container of volume 2 dm3 at 250 °C and allowed to reach equilibrium. At this temperature, the equilibrium mixture contains 1.8 moles of PCl3. What is the value of Kc at 250 oC? A 0.12 B 1.8 C 8.1 D 16.2 Ans: C PCl5(g) PCl3(g) + Cl2(g) I (mols) 1 1 1

7 NJC/H2 Chem Preliminary Examination/01/2024 [Turn over C (mols) -0.8 +0.8 +0.8 E (mols) 0.2 1.8 1.8 Kc = (𝟎.𝟗)(𝟎.𝟗) 𝟎.𝟏 = 8.1 9 Ammonium carbonate is a crystalline solid. On gentle warming a reaction occurs, forming ammonia as one product. How are the carbonate ions behaving during this reaction? A Brønsted-Lowry acid B Brønsted-Lowry base C oxidising agent D r