2022 CJC H2 CHEM Prelim P1 QP w Ans FINAL COPY

1 9729/01 CJC JC2 Preliminary Examination 2022 CANDIDATE NAME CLASS 2T CHEMISTRY 9729/01 Paper 1 Multiple Choice 15 September 2022 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 31 printed pages. Catholic Junior College JC 2 Preliminary Examinations Higher 2 WORKED SOLUTIONS

2 9729/01 CJC JC2 Preliminary Examination 2022 + – R S T 1 Which statement about 27 g of Al is always correct? A It contains the same number of atoms as 1 12 g of 12C. B It contains the same number of atoms as 24 dm3 of krypton gas at room temperature and pressure. C It contains the same number of hydrogen ions as 1 dm3 of 1 mol dm-3 aqueous sulfuric acid. D It contains the same number of atoms as 28 g of nitrogen gas. 2 The following are flight paths of charged particles as they pass through an electric field at the same speed. Which of the following correctly identifies X, Y and Z? X Y Z A 14N‾ 16O2+ 28Si2+ B 14N‾ 14C+ 28Si4+ C 15O+ 14C+ 14N+ D 15O‾ 14C+ 28Si+ X Y Z Topic: Mole Concept No of Al atoms in 27 g = 27 27.0 x 6.02 x 1023 = 6.02 x 1023 A No of C atoms in 1 12.0 g of 12C = 1 12 12.0 x 6.02 x 1023 = 4.18 x 1021 B No of Kr atoms in 24 dm3 of Kr = 24 24.0 x 6.02 x 1023 = 6.02 x 1023 N2  2 C H2SO4  2H+ No of H+ ions in 1 dm3 of aq H2SO4 = 1 x 1 x 2 x 6.02 x 1023 = 1.20 x 1024 D N2  2N No of N atoms in 28 g of N2 = 28 28.0 x 2 x 6.02 x 1023 = 1.20 x 1024 Answer: B

3 9729/01 CJC JC2 Preliminary Examination 2022 [Turn over 3 Use of the Data Booklet is relevant to this question. Species containing one or more unpaired electrons can be attracted by an external magnetic field and are said to be paramagnetic. Which of the following species is paramagnetic? 1 Cr3+ 2 Cu+ 3 Ni2+ A 3 B 1 and 2 C 1 and 3 D 1, 2 and 3 Concept: Angle of deflection particles 14N‾ 14C+ 28Si4+ m/z 1/14 1/14 4/28 = 1/7 14N‾ is negatively charged, so it is attracted to anode while 14C+ and 28Si4+ are attracted to cathode. Since angle of defle