CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 202 series 970 CHEMISTRY 970/23 Paper 2 (AS Structured Questions), maximum raw mark 60 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the October/November 202 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.
Page 2 Mark Scheme: Teachers version Syllabus Paper In this question, numerical answers should be given to three significant figures. (a) (i) C 6 H 2 O 6 + 6O 2 6CO 2 + 6H 2 O () (ii) M r C 6 H 2 O 6 = 80 () 80 g C 6 H 2 O 6 6 mol CO 2 200 g C 6 H 2 O 6 6 200 mol CO 2 80 = 40.0 mol to 3 sf allow ecf on wrong equation and/or wrong M r () (iii) 6.82 0 9 people will produce 6.82 0 9 40.0 mol CO 2 = 2.728 0 mol CO 2 () 2.728 0 mol CO 2 2.728 0 44 =.20032 0 3 g =.20 0 7 tonnes CO 2 to 3 sf () [5] allow ecf on answer from (ii) (b) (i) 2C 8 H 8 + 25O 2 6CO 2 + 8H 2 O or C 8 H 8 + 2½O 2 8CO 2 + 9H 2 O () (ii) M r C 8 H 8 = (8 2) + (8 ) = 4 () mass of 4.00 dm 3 of octane = 4000 0.70 = 2800 g () n(c 8 H 8 ) = 2800 = 24.564035 mol in 4.00 dm 3 4 = 24.6 mol to 3 sf () (iii) 2 mol C 8 H 8 produce 6 44 g CO 2 24.6 mol C 8 H 8 produce 6 44 24.6 g CO2 2 = 8659.2 g CO 2 = 8660 g CO 2 to 3 sf () [5]
Page 3 Mark Scheme: Teachers version Syllabus Paper (c) 6.82 0 9 people produce.20 0 7 tonnes CO 2 per day 8660 g CO 2 produced when car travels 00 km when travelling km, car produces 8660 = 8.66 0 g 00 = 8.66 0 5 tonnes () to produce.20 0 7 tonnes CO 2 car must travel.20 0 7 8.66 0 5 =.38568293 0 =.39 0 km to 3 sf () [2] (d) possible pollutants and the damage they cause CO NO X NO NO2 SO 2 H 2 O C toxic toxic toxic toxic global respiratory respiratory global respiratory warming problems problems warming problems photochemical smog acid rain acid rain unburned C 8 H 8 respiratory problems compound () damage () [2] [Total: 4]
Page 4 Mark Scheme: Teachers version Syllabus Paper 2 (a) (i) white fumes/steamy fumes () (ii) NaCl + H 2 SO 4 NaHSO 4 + HCl or 2NaCl + H 2 SO 4 Na 2 SO 4 + 2HCl () (iii) an acid that is completely ionised in solution or an acid that is completely dissociated into H + ions in solution () [3] (b) (i) purple/violet vapour (I 2 ) or black/brown solid (I 2 ) or irritating/acrid gas (SO 2 ) or stinking gas (H 2 S) or yellow solid (S) () (ii) conc. H 2 SO 4 is an oxidising agent or HI is a reducing agent () which oxidises HI or which reduces H 2 SO 4 () [3] (c) (i) white ppt formed not creamy white or off white () which dissolves in NH 3 (aq) () (ii) NaCl (aq) + AgNO 3 (aq) AgCl(s) + NaNO 3 (aq) or Cl (aq) + Ag + (aq) AgCl(s) equation () all state symbols correct () AgCl(s) + 2NH 3 (aq) [Ag(NH 3 ) 2 ] + Cl (aq) or AgCl(s) + 2NH 3 (aq) [Ag(NH 3 ) 2 ] Cl (aq) equation () all state symbols correct () (iii) precipitate is yellow () precipitate does not dissolve () [8] [Total: 4]
Page 5 Mark Scheme: Teachers version Syllabus Paper 3 (a) manufacture of ammonia/haber process or hydrogenation of fats/oils or making margarine or hydrocracking () [] (b) (i) increasing the pressure equilibrium will move to LHS () fewer moles/molecules on LHS or more moles/molecules on RHS () (ii) decreasing the temperature equilibrium will move to LHS () forward reaction is endothermic () [4] (c) rate will increase () collisions will occur more frequently () [2] (d) (i) K c = [CO 2 ][H 2 ] [CO][H 2 0] () (ii) CO(g) + H 2 O (g) CO 2 (g) + H 2 (g) initial moles 0.40 0.40 0.20 0.20 equil moles (0.40 y) (0.40 y) (0.20 + y) (0.20 + y) equil concn./mol dm 3 (0.40 y) (0.40 y) (0.20 + y) (0.20 + y) K c = (0.20 + y) 2 = 6.40 0 () (0.40 y) 2 (0.20 + y) = 6. 40 0 = 0.8 (0.40 y) (0.20 + y) = 0.8 (0.40 y) 0.20 + y = 0.32 0.8 y.8 y = 0.2 gives y = 0.067 () at equilibrium n(co) = n(h 2 O) = (0.40 0.067) = 0.33 mol and n(co 2 ) = n(h 2 ) = (0.20 + 0.067) = 0.27 mol () allow ecf as appropriate [5] [Total: 2]
Page 6 Mark Scheme: Teachers version Syllabus Paper 4 (a) (i) reaction organic compound reagent structural formulae of organic product A CH 3 CH(OH)CH 3 NaBH 4 no reaction B CH 3 COCH 3 C CH 3 CO 2 CH(CH 3 ) 2 D (CH 3 ) 3 COH Tollens reagent warm KOH(aq) warm Cr 2 O 7 2 /H + heat under reflux no reaction CH 3 CO 2 K or CH 3 CO 2 + (CH 3 ) 2 CHOH no reaction E CH 3 COCH 3 NaBH 4 CH 3 CH(OH)CH 3 F (CH 3 ) 3 COH PCl 5 (CH 3 ) 3 CCl G CH 3 CH=CHCH 2 OH MnO 4 /H + heat under reflux CH 3 CO 2 H + HO 2 CCO 2 H each correct answer gets (9 ) (ii) reaction G colour at the beginning of the reaction purple colour at the end of the reaction colourless not clear ( + + ) [2] [Total: 2]
Page 7 Mark Scheme: Teachers version Syllabus Paper 5 (a) (i) H J K CH 2 =CHCH 2 CH 2 OH CH 3 CH 2 COCH 3 CH 3 CH 2 CH 2 CHO CH 3 CH=CHCH 2 OH CH 2 =CHCH(OH)CH 3 (ii) each correct answer gets (5 x ) (iii) () correct structure drawn fully displayed () chiral centre clearly shown by* () [8] [Total: 8]