AQA Jun 2017 Paper 2 Q2 (with explained solutions) |

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AQA Jun 2017 Paper 2 Q2

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0 2 An experiment was carried out to determine the relative molecular mass (Mr) of a volatile hydrocarbon X that is a liquid at room temperature. A known mass of X was vaporised at a known temperature and pressure and the volume of the gas produced was measured in a gas syringe. Data from this experiment are shown in Table 1. Table 1 Mass of X Temperature Pressure Volume 194 mg 373 K 102 kPa 72 cm3 0 2 . 1 Calculate the relative molecular mass of X. Show your working. Give your answer to the appropriate number of significant figures. The gas constant, R = 8.31 J K1 mol1 [5 marks] *03* Relative molecular mass Turn overIB/M/Jun17/7404/2<br />
Analysis of a different hydrocarbon Y shows that it contains 83.7% by mass of carbon. Calculate the empirical formula of Y. Use this empirical formula and the relative molecular mass of Y (Mr = 86.0) to calculate the molecular formula of Y. [4 marks] Empirical formula Molecular formula 0 2 . 2 *04* IB/M/Jun17/7404/2<br />

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MARK SCHEMEAS-LEVEL CHEMISTRY7404/2JUNE 2017 Question Marking Guidance Mark Comments 02.1 Stage 1 M1=10210372106 8.31373 M2 M3 = 0.0024 / 0.00237 / 0.002369 / 0.0023693Stage 2 M4(=) = 0.194M5 = 82 (2sf only) 1 1 1 1 As this is an extended response question, each separate step of correct working is required in M1-M5 Correct answer with no working scores 2 marks M1 - If expression not written out, M1 could score from a correct expression for M2 (even if unit conversions are not correct for M2) M2allow an expression that gives correct value for M3 M3 should be at least 2sf (do not allow 0.0023 but do allow 0.00236) M4 must show 0.194 or 194 x 10-3 in working to score M5 must be 2sf ECF:No ECF within either stage 1 or stage 2 (except for transcription errors)Allow ECF from stage 1 into stage 2, i.e for M4 and M5 based on incorrect M3, (but if expression for M4 is inverted, cannot score M5)(Note that if 72 x 10-3 used in M2, then M3 = 2.4, M5 = 0.082) Ignore units for M3 and M5 Note that if T = 273 + 373 = 646, M5 = 140 (2sf) 12 of 26<br />
MARK SCHEMEAS-LEVEL CHEMISTRY7404/2JUNE 2017 Marking Guidance Mark Comments M1 dividing %s by relative atomic masses C = 83.7/12(.0), H = 16.3/1(.0) M2 converting (C:H 6.975:16.3) to 3:7 M3 empirical formula = C3H7 M4 molecular formula = C6H14 1 1 M1 & M2 are for working M3 for C3H7 only, marked independently M4 for C6H14 only, marked independently (ignore additional correct structures) Formulae with no working cannot score M1 or M2 Alternative method: M1 working that shows 83.7% of 86 is 72 M2 idea of 72/12 gives 6 C atoms Alternative method: working that shows that C6H14 (or C3H7) contains 83.7% C scores M1 & M2 13 of 26 Question 02.2<br />

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