SRJC_H1_CHEM_P1 QNS ANS

1 SERANGOON JUNIOR COLLEGE General Certificate of Education Advanced Level Higher 1 CHEMISTRY 8872/01 PRELIMINARY EXAMINATION 28 August 2009 Paper 1 50 min Additional Materials: Data Booklet Multiple Choice Answer Sheet READ T HESE INSTRUCTIONS FIRST Write your name, index number on the OMS Sheet in the spaces provided. Write in soft pencil. There are thirty questions in this paper. Answer all questions. For each question, there are four possible answers labelled A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the OMR answer sheet. Read very carefully the instructions on the OMR answer sheet. You are advised to fill in the OMR Answer Sheet as you go along; no additional time will be given for the transfer of answers once the examination has ended. 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 question paper. This document consists of 14 printed pages and 0 blank page. SRJC 2009 8872/01/PRELIM/2009 [Turn Over

2 Answer ALL Questions 1 0.5 g of zinc powder was found to reduc e an acidified solution of 25.50 cm 3 of 0.200 mol dm-3 VO2 +. Which one of the following is the reduced product of VO2 +? A VO3 - B VO2+ C V3+ D V2+ Solution: D Amount of Zn = 30.5 7.645 1065.4  mol Amount of VO2 + = 325.5 0.200 5.1 101000   mol 2 3 3 VO 7.65 10 3 5.1 10 2 Znn n      Reduction: Zinc is a reducing agent, henc e it will undergo oxidat ion. From the data booklet, we choose the following half-equation: Zn(s) → Zn2+(aq) + 2e- Given the 3 mol of Zn reacts with 2 mol of VO 2 +, we can deduce that the 2 mol of VO 2 + takes in (2e x 3 =) 6 mol of e- from Zn. Hence, 1 mol of VO2 + will take in 3 mol of e-. Oxidation state of V in VO2 + = + 5 Hence, the final oxidation state of V = +5 – 3 = +2 2 3.920 g of an oxide of formula MO was completely dissolved in 30.0 cm 3 of 2.00 mol dm -3 sulphuric acid. The resulting solution was made up to 100 cm 3. 25.0 cm 3 of this solution wa s neutralised by 27.5 cm 3 of 0.100 mol dm-3 sodium hydroxide. What is the relative molecular mass of M? A 48.6 B 54.9 C 55.9 D 101.0 Solution: C amount of NaOH = 327.5 0.1 2.75 101000   mol amount of H2SO4 in 25.0 cm3 of solution = ½(2.75 x 10─3) = 1.375 x 10─3 mol SRJC 2009 8872/01/PRELIM/2009 [Turn Over

3 amount of H2SO4 in 100 cm3 of solution = 3100 1.375 10 0.005525.0   mol amount of initial amount of H2SO4 = 30.0 2.0 0.061000 mol Therefore, the amount of H2SO4 that reacts with MO = 0.06 – 0.0055= 0.0545 mol Given that 1 mol of H 2SO4 reacts with 1 mol of MO (M has an oxidation state of +2), therefore, amount of MO in