NYJC_Prelim_CHEM_P2 ANS

NANYANG JUNIOR COLLEGE JC 2 PRELIMINARY EXAM Higher 1 CANDIDATE NAME CLASS TUTOR’S NAME CHEMISTRY 8872/02 Paper 2 23 Sep 2009 Candidates answer Section A on the Question Paper. 2 hours Candidates answer Section A on the Question Paper Additional Materials: Answer Paper Data Booklet READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. For Examiner’s Use A1 A2 A3 B1 B2 B3 Section A Answer all questions. Section B Answer two questions on separate answer paper. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. Total This document consists of 14 printed pages. [Turn over

2 H1 Chemistry 8872/02 NYJC J2/09 Prelim For examiner’s use only Section A Answer all questions in this section in the spaces provided. 1(a) A sample of carbon containing different isotopes was artificially enriched. The relative isotopic masses and its relative abundances determined in a mass spectrometer are given in the table below. Relative isotopic mass Relative abundance 12C 85.0 % 13C 10.0 % 14C 5.0 % Calculate the relative atomic mass of carbon in the sample to 1 decimal place. [1] A r = 12 x 0.850 + 13 x 0.100 + 14 x 0.050 = 12.2 [1] (b) A 0.100 mol of a hydrocarbon W was burnt completely to produce 26.4 g of carbon dioxide gas and 10.8 g of water. The same mass of W when burnt under a container with 300 g of water at 30 oC was found to bring the water to boil. The process was known to be only 70% efficient. The specific heat capacity of water is 4.2 J g –1 K–1. (i) Determine the molecular formula of the hydrocarbon W. [2] CxHy + (x + ¼ y)O 2  xCO 2 + ½ y (H 2O) Amt of CO2 formed = 26.4 / 44.0 = 0.600 mol Amt of H2O formed = 10.8 / 18.0 = 0.600 mol C xHy CO2 H2O n 0.100 0.600 0.600 Lowest ratio 1 6 6 Mol ratio 1 x ½ y Comparing coefficient x = 6 6 = ½ y  y = 12 Molecular formula T is C6H12 [1] [1] (ii) Calculate the enthalpy change of combustion of W. [2] Heat lost from combustion = heat gained by water [1] [1]

3 H1 Chemistry 8872/02 NYJC J2/09 Prelim For examiner’s use only c -1 -1 m c T 300 x 4.2 x (100-30) 70H = = - n 0.100 100 = -1260 000 J mol = -1260 kJ mol  (c) Arranging the following in order of increasing boiling points. Explain your choice in terms of structure and bonding. H 2O, NH 3, NaCl, CH 4 [5]  H 2O, NH3, CH4 and C6H14 have simple molecular structure consisting of molecules held together by weak forces.