2025 ASRJC Prelim H1Chem P1 Soln

ASRJC JC2 Prelim 2025 8873/H1 [Turn over ANDERSON SERANGOON JUNIOR COLLEGE 2025 JC 2 PRELIMINARY EXAMINATION NAME:______________________________ ( ) CLASS: 25 / ___ CHEMISTRY 8873/01 Paper 1 Multiple Choice 2 September 2025 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 and class 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 __ printed pages. Multiple Choice Answer Sheet Write your name, class and NRIC / FIN number, including the reference letter. Shade the NRIC / FIN number. Exam Title: JC2 Preliminary Exam Exam Details: H1 Chemistry / Paper 1 Date: 02/09/2025

2 ASRJC JC2 Prelim 2025 8873/H1 1 Use of the Data Booklet is relevant to this question. L is the Avogadro constant. Which statement is correct? A 71.0 g of chlorine gas contains 2 L molecules. B 1.00 mol of sodium sulfate contains 3 L ions. C 6.9 g of lithium ion, Li+, contains 3 L electrons. D 48 dm3 of hydrogen gas, measured at r.t.p, contains 2 L atoms. Answer: B A n(Cl2) = 71.0/71.0 = 1 mol No. of molecules = 1 x L B 1 mol of Na2SO4 contains 2 mol of Na+ and 1 mol of SO42-. No. of ions = 3 x L C n(Li+) = 6.9/6.9 = 1 mol 1 Li+ contains 3 − 1 = 2 e. No. of electrons = 2 x L D Molar volume of a gas at r.t.p is 24 dm3. n(H2) = 48/24 = 2.0 mol No. of H atoms = 2 x 2 x L = 4 L 2 Use of the Data Booklet is relevant to this question. Iodine-131 is radioactive and is used in thyroid cancer treatment. Which statements about this element are correct? 1 There are p electrons present in 5 quantum shells. 2 There are 78 neutrons, 53 protons and 54 electrons in the negative ion of 131 53I . 3 All electrons in the negative ion of 131 53I are paired. A 1, 2 and 3 B 2 and 3 only C 1 and 3 only D 2 only Answer: B reason Statement 1 is incorrect There are p electrons present in 4 quantum shells (2p, 3p, 4p, 5p). Statement 2 is correct For a singly negatively charged ion, number of electrons is = 53 + 1 = 54 No. of neutrons = 131 – 53 = 78 Statement 3 is correct All electrons in negative ion of 131 53I are paired.

3 ASRJC JC2 Prelim 2025 8873/H1 [Turn Over 3 The first seven ionisation energies of an element Y are shown in the sketch graph. What could be the identity of element Y? A Bromine B Carbon C Phosphorus D Silicon Answer: D Option D is correct. The biggest energy gap is between the fourth and fifth electron. This suggested there are 4 valence electrons in the outermost shell. Although both carbon and silicon has 4 valence electrons, carbon which has only two 1s inner electrons cannot be Y. This is observed from the data that Y has three electrons from next inner quantum shell. Hence, D is silicon.

4 ASRJC JC2 Prelim 2025 8873/H1 4 A beam of particles contains He4 2+ and 𝑒−0 ions. All particles approach an electric field at the same speed. Which diagram indicates the deflection of these particles as they pass through the electric field? A B C D Answer: D Negatively charged electrons deflect towards positively charged plate while positively charged protons deflect towards negatively charged plate. Angle of deflection, θ, is proportional to the charge of the particle, but inversely proportional to its mass. 𝑞 𝑚 for e−0 = ∞ 𝑞 𝑚 for He4 2+= 2 4 = 0.5 Thus, the angle of deflection for e−0 > He4 2+. He4 2+ e−0 e−0 e−0 e−0 He4 2+ He4 2+ He4 2+

5 ASRJC JC2 Prelim 2025 8873/H1 [Turn Over 5 Use of the Data Booklet is relevant to this question. Under acidic conditions, potassium manganate(VII) reacts with potassium iodide. The iodine that was liberated can be determined by titrating with sodium thiosulfate solution. The relevant equations are shown but not balanced. Reaction I: 2 MnO4–(aq) +16 H+(aq) + a I–(aq) … Mn2+(aq) + … H2O(l) + … I2(aq) Reaction II: b I2(aq) + 2 S2O32−(aq) … I−(aq) + …S4O62−(aq) What are the values of a and b in the correctly balanced equations? a b A 10 1 B 10 2 C 5 1 D 5 2 Answer: A There are various ways to solved a and b. The easiest is to use Data Booklet to identify the number of electrons involved in half- equation. Since no. of electron gain = no. of electron lose, one can determine the mole ratio between the reactants. In reaction I: Mn in MnO4– is reduced from +7 to +2 (ie takes in 5 e). I- lose 1 e to oxidise to element I2 from -1 to 0. (ie lose 1 mol e). Hence, reducing 2 moles of MnO4– require 10 moles of iodide. a = 10. In reaction II: Observing 2 mol of S2O32−(aq) form 1 mol of S4O62−(aq), one can use Data Booklet to know 2 mol of e are lost and taken by 1 mole of iodine. Hence, b = 1.

6 ASRJC JC2 Prelim 2025 8873/H1 6 Use of the Data Booklet is relevant to this question. The complete combustion of 0.0050 mol of hydrocarbon W requires 0.540 dm3 of oxygen measured at r.t.p. 0.27 g of water is produced. What is the molecular formula of W? A C2H6 B C3H6 C C3H8 D C4H8 Answer: B n(O2)=0.540/24 = 0.0225 mol n(H2O)= 0.27/18 = 0.015 mol CxHy + (x+y/4)O2 → xCO2 + y/2 H2O Mole ratio 1 4.5 x 3 No. of mole of O2 reacting to one mole of CxHy, y= 0.0225/0.005 = 4.5 No. of mole of H in one mole of CxHy, y/2 = 0.015/0.005 => y = 6 x + y/4 = 4.5 => x + 6/4 = 4.5 => x = 3 Hence, CxHy is C3H6. 7 Which equation shows a reaction in which co-ordinate (dative covalent) bonds are formed? A NO2 + NO2 N2O4 B 2NO + 2H2 N2 + 2H2O C H2O + H+ H3O+ D Cl2 + CH4 CH3Cl + HCl Answer: C A Each N in NO2 molecule has an unpaired electron. They come together to form N─N covalent bond. O2N O2N NO2NO2+ B This is a redox reaction which involved bonds breaking and bonds forming. C Lone pair on O in water donated to empty orbital in H+, resulting in dative formation.

7 ASRJC JC2 Prelim 2025 8873/H1 [Turn Over H2O H2O H+ H+ + D This is a substitution reaction of alkane involving bonds breaking and bonds forming. 8 Sulfur dioxide is a very important compound in winemaking. Sulfur dioxide can be represented as O S O . Which statements about sulfur dioxide is true about its polarity and its bond angle. A Sulfur dioxide is polar and has a bond angle of 90o exactly. B Sulfur dioxide is non-polar and has a bond angle of 104o approximately. C Sulfur dioxide is polar and has a bond angle of 118o approximately. D Sulfur dioxide is non-polar and has a bond angle of 180o exactly Answer: C SO2 molecule