Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
www.maxpapers.com UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2010 question paper for the guidance of teachers 9701 CHEMISTRY 9701/11 Paper 1 (Multiple Choice), maximum raw mark 40 Mark schemes must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2010 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. Page 2 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 www.maxpapers.com Syllabus 9701 Question Number Key Question Number Key 1 2 D B 21 22 D A 3 4 5 B D A 23 24 25 B B D 6 7 D C 26 27 D B 8 9 10 B B B 28 29 30 C C D 11 12 C C 31 32 A C 13 14 15 A D D 33 34 35 A D A 16 17 D A 36 37 A D 18 19 20 B C C 38 39 40 B B D © UCLES 2010 Paper 11 www.maxpapers.com UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2010 question paper for the guidance of teachers 9701 CHEMISTRY 9701/12 Paper 1 (Multiple Choice), maximum raw mark 40 Mark schemes must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2010 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. Page 2 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 www.maxpapers.com Syllabus 9701 Question Number Key Question Number Key 1 2 D B 21 22 B C 3 4 5 C B D 23 24 25 B A D 6 7 A A 26 27 B A 8 9 10 D C B 28 29 30 B D D 11 12 C D 31 32 D B 13 14 15 B C D 33 34 35 C A A 16 17 C D 36 37 D A 18 19 20 B C C 38 39 40 C A D © UCLES 2010 Paper 12 www.maxpapers.com UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2010 question paper for the guidance of teachers 9701 CHEMISTRY 9701/13 Paper 1 (Multiple Choice), maximum raw mark 40 Mark schemes must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2010 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. Page 2 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 www.maxpapers.com Syllabus 9701 Question Number Key Question Number Key 1 2 B B 21 22 C D 3 4 5 D D A 23 24 25 D B B 6 7 C D 26 27 D B 8 9 10 A D B 28 29 30 C C D 11 12 B B 31 32 A A 13 14 15 C C D 33 34 35 C A D 16 17 A D 36 37 D A 18 19 20 B C A 38 39 40 B D B © UCLES 2010 Paper 13 www.maxpapers.com UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2010 question paper for the guidance of teachers 9701 CHEMISTRY 9701/21 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 must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2010 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. www.maxpapers.com Page 2 1 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 Paper 21 (a) the actual number of atoms of each element present (1) in one molecule of a compound (1) y (b) CХHУ + x + O2 4 [2] xCO2 + y H2O 2 xCO2 (1) y H2O (1) 2 [2] (c) (i) oxygen/O2 (1) (ii) carbon dioxide/CO2 (1) (iii) 10 cm3 (1) (iv) 20 cm3 (1) (d) CХHУ + 10 cm3 [4] y x + O2 4 20 cm3 xCO2 + 10 cm3 y H2O 2 1 mol of CxHy gives 1 mol of CO2 whence x = 1 (1) 1 mol of CxHy reacts with 2 mol of O2 whence and y x + 4 = 2 y = 4 (1) molecular formula is CH4 (1) [3] [Total: 11] © UCLES 2010 www.maxpapers.com Page 3 2 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 (a) N2 + 3H2 2NH3 (1) Paper 21 [1] (b) temperature between 300 and 550oC (1) correct explanation of effect of temperature on rate of formation of NH3 or on position of equilibrium (1) catalyst of iron or iron oxide (1) to speed up reaction or to reduce Ea (1) (c) manufacture of HNO3 or explosives or nylon or as a cleaning agent or as a refrigerant (1) [4] [1] (d) fertiliser in rivers causes excessive growth of aquatic plants/algae (1) when plants and algae die O2 is used up/fish or aquatic life die (1) (e) (i) CO NO (ii) CO NO (f) [2] by incomplete combustion of the hydrocarbon fuel (1) by reaction between N2 and O2 in the engine (1) toxic/effect on haemoglobin (1) toxic/formation of acid rain (1) [4] (i) platinum/Pt – allow palladium/Pd or rhodium/Rh (1) (ii) 2CO + 2NO → 2CO2 + N2 (1) [2] [Total: 14] © UCLES 2010 www.maxpapers.com Page 4 3 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 Paper 21 (a) (i) a compound which contains only carbon and hydrogen (1) (ii) separation of compounds by their boiling points (1) [2] (b) (i) high temperature and high pressure (1) high temperature and catalyst (1) (ii) C11H24 → C5H12 + C6H12 or C11H24 → C5H12 + 2C3H6 or C11H24 → C5H12 + 3C2H4 (1) [3] (c) (i) CH3 CH3CH2CH2CH2CH3 CH3CH2CHCH3 CH3CCH3 CH3 isomer B (1) isomer C CH3 isomer D (1) (1) (ii) the straight chain isomer (isomer B above) (1) it has the greatest van der Waals’ forces (1) because unbranched molecules have greater area of contact/ can pack more closely together (1) [6] (d) enthalpy change when 1 mol of a substance (1) is burnt in an excess of oxygen/air under standard conditions or is completely combusted under standard conditions (1) © UCLES 2010 [2] www.maxpapers.com Page 5 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 Paper 21 (e) (i) heat released = m c δT = 200 x 4.18 x 27.5 (1) = 22990 J = 23.0 kJ (1) (ii) 23.0 kJ produced from 0.47 g of E 2059 kJ produced from 0.47 x 2059 g of E (1) 23.0 = 42.08 g of E (1) allow ecf in (i) or (ii) on candidate’s expressions [4] (f) C3H6 = 42 E is C3H6 for ecf, E must be unsaturated and be no larger than C5 (1) [1] [Total: 18] 4 (a) reaction 1 reaction 2 reaction 3 reagent NaOH/KOH (1) solvent H2O/water/aqueous (1) reagent NH3/ammonia (1) solvent ethanol/C2H5OH/alcohol (1) reagent NaOH/KOH (1) solvent ethanol/C2H5OH/alcohol (1) [6] (b) with CH3CH2CH2CH2I rate would be faster (1) C-I bond is weaker than C-Br bond (1) C-I bond energy is 240 kJ mol 1, C-Br bond energy is 280 kJ mol 1 data must be quoted for this mark (1) (c) non-toxic volatile/low bp [3] non-flammable unreactive (any 2) © UCLES 2010 [2] www.maxpapers.com Page 6 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 Paper 21 (d) (i) when a covalent bond breaks the two electrons in the bond are shared between the two atoms (1) (ii) CCl2F2 → CClF2 + Cl (as minimum) allow CCl2F + F (1) [2] (e) they are flammable (1) [1] [Total: 14] 5 (a) NaBr/sodium bromide [1] (b) Br2/bromine or SO2/sulfur dioxide [1] (c) concentrated sulfuric acid is an oxidising agent or phosphoric(V) acid is not an oxidising agent [1] [Total: 3] © UCLES 2010 www.maxpapers.com UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2010 question paper for the guidance of teachers 9701 CHEMISTRY 9701/22 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 must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2010 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. www.maxpapers.com Page 2 1 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 Paper 22 (a) the actual number of atoms of each element present (1) in one molecule of a compound (1) y (b) CХHУ + x + O2 4 [2] xCO2 + y H2O 2 xCO2 (1) y H2O (1) 2 [2] (c) (i) oxygen/O2 (1) (ii) carbon dioxide/CO2 (1) (iii) 10 cm3 (1) (iv) 20 cm3 (1) (d) CХHУ + 10 cm3 [4] y x + O2 4 20 cm3 xCO2 + 10 cm3 y H2O 2 1 mol of CxHy gives 1 mol of CO2 whence x = 1 (1) 1 mol of CxHy reacts with 2 mol of O2 whence and y x + 4 = 2 y = 4 (1) molecular formula is CH4 (1) [3] [Total: 11] © UCLES 2010 www.maxpapers.com Page 3 2 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 (a) N2 + 3H2 2NH3 (1) Paper 22 [1] (b) temperature between 300 and 550oC (1) correct explanation of effect of temperature on rate of formation of NH3 or on position of equilibrium (1) catalyst of iron or iron oxide (1) to speed up reaction or to reduce Ea (1) (c) manufacture of HNO3 or explosives or nylon or as a cleaning agent or as a refrigerant (1) [4] [1] (d) fertiliser in rivers causes excessive growth of aquatic plants/algae (1) when plants and algae die O2 is used up/fish or aquatic life die (1) (e) (i) CO NO (ii) CO NO (f) [2] by incomplete combustion of the hydrocarbon fuel (1) by reaction between N2 and O2 in the engine (1) toxic/effect on haemoglobin (1) toxic/formation of acid rain (1) [4] (i) platinum/Pt – allow palladium/Pd or rhodium/Rh (1) (ii) 2CO + 2NO → 2CO2 + N2 (1) [2] [Total: 14] © UCLES 2010 www.maxpapers.com Page 4 3 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 Paper 22 (a) (i) a compound which contains only carbon and hydrogen (1) (ii) separation of compounds by their boiling points (1) [2] (b) (i) high temperature and high pressure (1) high temperature and catalyst (1) (ii) C11H24 → C5H12 + C6H12 or C11H24 → C5H12 + 2C3H6 or C11H24 → C5H12 + 3C2H4 (1) [3] (c) (i) CH3 CH3CH2CH2CH2CH3 CH3CH2CHCH3 CH3CCH3 CH3 isomer B (1) isomer C CH3 isomer D (1) (1) (ii) the straight chain isomer (isomer B above) (1) it has the greatest van der Waals’ forces (1) because unbranched molecules have greater area of contact/ can pack more closely together (1) [6] (d) enthalpy change when 1 mol of a substance (1) is burnt in an excess of oxygen/air under standard conditions or is completely combusted under standard conditions (1) © UCLES 2010 [2] www.maxpapers.com Page 5 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 Paper 22 (e) (i) heat released = m c δT = 200 x 4.18 x 27.5 (1) = 22990 J = 23.0 kJ (1) (ii) 23.0 kJ produced from 0.47 g of E 2059 kJ produced from 0.47 x 2059 g of E (1) 23.0 = 42.08 g of E (1) allow ecf in (i) or (ii) on candidate’s expressions [4] (f) C3H6 = 42 E is C3H6 for ecf, E must be unsaturated and be no larger than C5 (1) [1] [Total: 18] 4 (a) reaction 1 reaction 2 reaction 3 reagent NaOH/KOH (1) solvent H2O/water/aqueous (1) reagent NH3/ammonia (1) solvent ethanol/C2H5OH/alcohol (1) reagent NaOH/KOH (1) solvent ethanol/C2H5OH/alcohol (1) [6] (b) with CH3CH2CH2CH2I rate would be faster (1) C-I bond is weaker than C-Br bond (1) C-I bond energy is 240 kJ mol 1, C-Br bond energy is 280 kJ mol 1 data must be quoted for this mark (1) (c) non-toxic volatile/low bp [3] non-flammable unreactive (any 2) © UCLES 2010 [2] www.maxpapers.com Page 6 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 Paper 22 (d) (i) when a covalent bond breaks the two electrons in the bond are shared between the two atoms (1) (ii) CCl2F2 → CClF2 + Cl (as minimum) allow CCl2F + F (1) [2] (e) they are flammable (1) [1] [Total: 14] 5 (a) NaBr/sodium bromide [1] (b) Br2/bromine or SO2/sulfur dioxide [1] (c) concentrated sulfuric acid is an oxidising agent or phosphoric(V) acid is not an oxidising agent [1] [Total: 3] © UCLES 2010 www.maxpapers.com UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2010 question paper for the guidance of teachers 9701 CHEMISTRY 9701/23 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 must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2010 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. www.maxpapers.com Page 2 1 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 Paper 23 (a) atoms of the same element / with same proton (atomic) number / same number of protons (1) different numbers of neutrons / nucleon number / mass number (1) [2] (b) isotope no. of protons no. of neutrons no. of electrons 24 12 12 12 26 12 14 12 Mg Mg each correct row (1) (c) Ar = = [2] 24 × 78.60 + 25 × 10.11 + 26 × 11.29 (1) 100 1886.40 + 252.75 + 293.54 100 gives 24.33 to 4 sig fig (same as data in question) do not credit wrong number of sig figs or incorrect rounding up/down (1) (d) Mg + Cl2 → MgCl2 (1) (e) (i) n(Sb) = [2] [1] 2.45 = 0.020 (1) 122 (ii) mass of Cl in A = 4.57 – 2.45 = 2.12 g (1) n(Cl) = 4.57 2.45 35.5 = 2.12 = 0.06 35.5 allow ecf as appropriate (1) (iii) Sb : Cl = 0.02 : 0.06 = 1:3 empirical formula of A is SbCl3 (1) (iv) 2Sb + 3Cl2 → 2SbCl3 (1) (f) [5] (i) ionic (1) (ii) covalent (1) not van der Waals’ forces [2] [Total: 14] © UCLES 2010 www.maxpapers.com Page 3 2 (a) 1 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 Paper 23 S + O2 → SO2 (1) 2 2SO2 + O2 3 SO3 + H2O → H2SO4 or SO3 + H2SO4 → H2S2O7 (1) (b) condition 1 condition 2 2SO3 equation (1) equilibrium sign (1) [4] 400 – 600 °C (650 – 900K) (1) 1–10 atm/just above atmospheric pressure allow equivalent pressure units (1) vanadium pentoxide/vanadium(V) oxide/V2O5 (1) [3] (c) fertilisers/phosphates/ammonium sulfate or lead/acid batteries or paints/pigments or dyestuffs or steel pickling or metal treatment or detergents or explosives (1) [1] condition 3 (d) (i) 2H2S + 3O2 → 2SO2 + 2H2O (1) (ii) H2S –2 SO2 +4 S 0 all three (1) SO2 because the oxidation number of S is reduced (1) [3] (e) (i) 2NO + O2 → 2NO2 (1) SO2 + NO2 → SO3 + NO (1) SO3 + H2O → H2SO4 final product must be H2SO4 (1) (ii) corrosion of buildings or dissolving of Al 3+ ions from soil or pollution of rivers/killing aquatic life or making soil acidic/killing trees/corrosion of metals (1) (f) it is a reducing agent/inhibits oxidation (1) [4] [1] [Total: 16] © UCLES 2010 www.maxpapers.com Page 4 3 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 Paper 23 (a) (i) order of atoms must be C-C-O (1) linear (1) (ii) a molecule or atom with an unpaired electron or a species formed by the homolytic fission of a covalent bond (1) (iii) molecule has 2 bond pairs and one lone pair (1) and one unpaired electron (1) these may be shown in a diagram (b) (i) [5] H CN H CN —C—C—C—C— H H H H allow the structural formula —CH2CH(CN)CH2CH(CN)— (1) (ii) addition (1) [2] (c) (i) CH3CHO (1) (ii) O H2C O CH2 or H H O H H or (1) [2] (d) reagent product Br2 in an inert solvent BrCH2CHBrCHO NaCN + dil. H2SO4 CH2=CHCH(OH)CN allow CH2=CHCH(OH)CO2H Tollens’ reagent CH2=CHCO2H or CH2=CHCO2 NaBH4 CH2=CHCH2OH (4 × 1) [4] [Total: 13] © UCLES 2010 www.maxpapers.com Page 5 4 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 29.3 5.7 65.0 : : (1) 12 1 79.9 = 2.44 : 5.7 : 0.81 = 3 : 7 : 1 (1) C3H7Br = (3 × 12) + (7 × 1) + 79.9 = 122.9 use of 122.9 or 123 to prove molecular formula must be C3H7Br (1) (a) C : H : Br Syllabus 9701 Paper 23 = [3] (b) (i) mechanism must be SN2 dipole on C-Br bond or central C atom shown with δ+ (1) attack on C atom by lone pair of OH not from negative charge (1) transition state formed with negative charge shown (1) Br leaves/NaBr formed (1) (ii) C2H4/ethane (1) (iii) ethanol/C2H5OH (1) (iv) elimination (1) (c) (i) [7] H H H H HO—C—C— C—C—OH H H H H (1) (ii) must be skeletal or (1) [2] [Total: 12] 5 (a) AgCl/silver chloride (1) [1] (b) white (1) [1] (c) 1-iodobutane (1) [1] (d) C-I bond is weaker/longer than the other C-halogen bonds (1) C-I bond energy is 240 kJ mol 1 or covalent radius of I is 0.133 nm (1) [2] [Total: 5] © UCLES 2010 www.maxpapers.com UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2010 question paper for the guidance of teachers 9701 CHEMISTRY 9701/31 Paper 3 (Advanced Practical Skills), maximum raw mark 40 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 must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2010 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. www.maxpapers.com Page 2 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 Paper 31 Question Sections Indicative material Mark 1 PDO layout I Volume given for Rough titre and accurate titre details tabulated. 1 MMO Collection II In the correct spaces, records Initial and final burette readings for Rough titre and; Initial and final burette readings and, volume of FB 2 added recorded for each accurate titre Headings should match readings. Do not award this mark if: 50(.00) is used as an initial burette reading; More than one final burette reading is 50.(00); Any burette reading is greater than 50.(00) 1 MMO Decisions III Has two uncorrected, accurate titres within 0.1 cm3 Do not award this mark if having performed two titres within 0.1 cm3 a further titration is performed which is more than 0.10 cm3 from the closer of the initial two titres, unless a fourth titration, within 0.1 cm3 of the third titration or of the first two titres has also been carried out. 1 PDO Recording IV All accurate burette readings (initial and final) recorded to nearest 0.05 cm3. Assessed on burette readings only. 1 MMO Quality V, VI and VII Round any burette readings to the nearest 0.05 cm3 Check and correct subtractions in the titre table. Select the “best”titre using the hierarchy: two identical; titres within 0.05 cm3, titres within 0.10 cm3 etc. (a) 3 Award V, VI and VII for a difference to Supervisor within 0.20 cm3 Award V and VI only for a difference of 0.20+ cm3 – 0.40 cm3 Award V only for a difference of 0.40+ cm3 - 0.80 cm3 If the selected “best” titres are > 0.50 cm3 apart, cancel one of the Q marks awarded. © UCLES 2010 [7] www.maxpapers.com Page 3 (b) Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 ACE Interpretation (c) ACE Interpretation Paper 31 Calculates the mean, correct to 2 decimal places (third decimal place maybe rounded to the nearest 0.05 cm3) from any accurate titres within 0.20 cm3. A mean of exactly .x25 or .x75 is allowed but the candidate may round up or down to the nearest 0.05 cm3. If ALL burette readings are given to 1 decimal place then the mean can be given to 1 decimal place if numerically correct without rounding. Mean of 24.3 and 24.4 = 24.35 () Mean of 24.3 and 24.4 = 24.4 () Mean of 24.3 and 24.5 = 24.4 () Titres to be used in calculating the mean must be clearly shown – in an expression or ticked in the titration table. 1 No additional factor/expression is allowed in any step If an answer, with no working, is given in any section allow if correct. I Uses 15.0/248.2 only in step (i) If no working shown accept only the following evaluated answers: (0.060, 0.0604 or 0.06044) 1 [1] Uses answer (i) × cand average titre/1000 in step (ii) and answer (iv) × 1000/25 in step (v) 1 III Uses answer (ii) × ½ in step (iii), and answer (iii) × 2 in step (iv) 1 IV Appropriate working shown in a minimum of three sections. To include equations as steps for the working mark; In (iii) must see x2 or x0.5. In (iv) must see multiplication or division by 6, 1.2 or 2. 1:6 for IO3 /6H+, 1:1.2 for 5I /6H+, 1:2 for 6H+/3I2 1 V 3 to 5 significant figures in final answers to all sections attempted – minimum of three final answers required to qualify for the award of this mark. 1 II PDO Display Syllabus 9701 © UCLES 2010 1 [5] www.maxpapers.com Page 4 (d) Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 ACE Interpretation Syllabus 9701 Gives 0.1(0) cm3 as the maximum error in (i). Ignore any sign and 0.1 the expression /cand titre in (b) × 100 in (ii) 0.06 Evaluates /25.0 × 100 in step (iii) Accept only 0.240 or 0.24, or rounded to 0.2 provided 0.24 has been seen in the working. Paper 31 1 1 [2] [Total: 15] © UCLES 2010 www.maxpapers.com Page 5 2 (a) Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 Paper 31 PDO Layout I Records at least four different balance readings and at least one mass of solid/gas Accept 0.0(0X) g as the mass of the empty tube or a statement that the tube is tared. 1 PDO Recording II Gives all appropriate headings and units when recording results. Do not accept mass of empty tube as 0.0(00)g here unless tube is described as tared. (minimum of three pieces of information) 1 III All recorded balance readings consistent to at least 1 decimal place. (minimum of three balance readings) 1 MMO Decisions IV Evidence of reheating to “constant” mass. For balances reading to 1 d.p. two masses must be identical For 2 or 3 d.p.balances, two masses must be within 0.05 g 1 MMO Quality V and VI Check and correct all subtractions in the results table. mass heated Calculate /mass of residue to 3 significant figures. Compare to Supervisor standard or standard value of 1.45. 2 Award V and VI for a difference up to 0.15 Award V only for a difference of 0.15+ to 0.30 Where a candidate repeats the experiment use cumulative masses of FA 3 and residue. Where masses of FA 3 and residue cannot be checked, accept candidate values to calculate the ratio. (b) ACE Interpretation [6] 1 Evaluates cand mass loss from (a) /cand mass of FA 3 correct to 2–4 significant figures. Where mass loss or mass of FA 3 is not given in (a), check, from balance readings, the values. A candidate who incorrectly describes the mass of the residue as the mass loss in tabulated results in (a) may “correct” the error and use the correct mass loss here. © UCLES 2010 [1] www.maxpapers.com Page 6 (c) (d) Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 ACE Conclusions ACE Improvements Syllabus 9701 Paper 31 Uses Mr (values) of CO2 or H2O to justify how the ratio of CuCO3 to Cu(OH)2 affects the mass loss. If % loss is too high – more CuCO3 If % loss is too low – more Cu(OH)2 1 Draws apparatus showing the collection of carbon dioxide in a syringe or in a burette or measuring cylinder inverted over water. Allow use of an inverted tube if graduations are shown or it is suitably labelled. All apparatus should be recognisable from the drawing or appropriately labelled. 1 Shows, in the diagram, an effective method of removing water vapour. Named reagent; e.g. (concentrated H2SO4, CaCl2, silica gel, (CaO), anhydrous CuSO4. or stated purpose of an un-named reagent given. Allow also a suitable reflux arrangement, returning water to the heated tube. or a statement that water vapour condenses in a water bath. Do not accept a diagram showing the gas bubbling through water without some written indication that the water is a condenser. 1 [1] [2] [Total: 10] © UCLES 2010 www.maxpapers.com Page 7 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 FA 4 is Al2(SO4)3(aq); 3 (a) MMO Collection FA 5 is ZnSO4(aq); Syllabus 9701 Paper 31 FA 6 is Pb(NO3)2(aq); FA 7 is MgSO4(aq) 1 mark for correct observations in each of the vertical columns. or 1 mark for correct observations in each of the horizontal rows (i), (ii) and (iii). 3 mark maximum Mark the section by the method which gives the better mark. 4 [4] observations test (i) (ii) (iii) FA 4 FA 5 FA 6 FA 7 addition of NaOH white ppt white ppt white ppt white ppt further addition of NaOH ppt soluble ppt soluble ppt soluble ppt insoluble addition of NH3 white ppt white ppt white ppt white ppt further addition of NH3 ppt insoluble ppt soluble ppt insoluble ppt insoluble no ppt, no reaction, colourless or yellow solution no ppt, no reaction, colourless or yellow solution yellow ppt no ppt, no reaction, colourless or yellow solution addition of KI Minimum evidence required in observations for the ion identity marks I, II and III in (b) In some cases, identification may be allowed from incomplete observations. There must, however, be no observations that are contrary to those expected with any “correctly” identified ion. The same criteria will be applied to “candidate’s supporting evidence in awarding mark IV. Candidates are not permitted to introduce (from the Qualitative Analysis Notes) supporting evidence that is not given in the observations. Precipitate colour need not be mentioned in supporting evidence. Al3+ Zn2+ Pb2+ Mg2+ (white) precipitate, soluble in (excess) NaOH, if yellow ppt with KI (white) precipitate, soluble in (excess) NH3(aq) Yellow precipitate with KI (white) precipitate, insoluble in (excess) NaOH © UCLES 2010 www.maxpapers.com Page 8 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 FA 4 is Al2(SO4)3(aq); FA 5 is ZnSO4(aq); (b) ACE Conclusions (c) MMO Decisions Syllabus 9701 Paper 31 FA 6 is Pb(NO3)2(aq); FA 7 is MgSO4(aq) Do not accept any ion other than Al 3+, Zn2+, Pb2+or Mg2+ in any section. Marks I to III Ions must be correct, including charge, if a symbol has been given. – no ecf in this section. 1 Award I only if one ion only is identified from correct observations. 1 Award I and II if two ions only are identified from correct observations. 1 Award I, II and III if all four cations are identified from correct observations. The 4th cation may be identified by elimination from incomplete supporting evidence. 1 Award mark IV if the supporting evidence fits the ion identified and the practical performed for at least three of the four ions. 1 Allow ecf on ion order on mark IV. [4] Selects sodium or potassium chromate(VI), sulfuric acid or hydrochloric acid soln containing one of the following named ions or formula given followed by (aq): CrO42 , SO42 , Cl , Br but not I , soln containing CrO42 ions, H2SO4, HCl, [1] © UCLES 2010 www.maxpapers.com Page 9 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 Paper 31 FA 8 is CuSO4(aq) (d) MMO Collection I Records blue colour of solution fading/disappearing on adding zinc powder in (i) If no reaction with Zn(s) is reported do not allow blue to light blue solution. 1 II Records a temperature rise in (i) Accept reaction is exothermic/produces heat 1 III Records a red-brown, orange-brown, brown or black solid in (i) 1 IV Observes a green, lime green, fluorescent green or yellow-green solution in (ii) 1 V Observes solution turning blue, or blue solution in (iii) if solution green in (ii) or solution going towards blue in colour on adding water in (iii) 1 If solution is not mentioned in (ii) or (iii) but colours are correct – award point V only. (e) ACE Conclusions Completes the equation: → Cu(s) + Zn2+(aq) State symbols required [5] 1 [1] [Total: 15] © UCLES 2010 www.maxpapers.com UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2010 question paper for the guidance of teachers 9701 CHEMISTRY 9701/33 Paper 3 (Advanced Practical Skills), maximum raw mark 40 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 must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2010 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. www.maxpapers.com Page 2 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 Paper 33 Question Sections Indicative material Mark 1 PDO layout I Volume given for Rough titre and accurate titre details tabulated. 1 MMO Collection II In the correct spaces, records Initial and final burette readings for Rough titre and; Initial and final burette readings and, volume of FB 2 added recorded for each accurate titre Headings should match readings. Do not award this mark if: 50(.00) is used as an initial burette reading; More than one final burette reading is 50.(00); Any burette reading is greater than 50.(00) 1 MMO Decisions III Has two uncorrected, accurate titres within 0.1 cm3 Do not award this mark if having performed two titres within 0.1 cm3 a further titration is performed which is more than 0.10 cm3 from the closer of the initial two titres, unless a fourth titration, within 0.1 cm3 of the third titration or of the first two titres has also been carried out. 1 PDO Recording IV All accurate burette readings (initial and final) recorded to nearest 0.05 cm3. Assessed on burette readings only. 1 MMO Quality V, VI and VII Round any burette readings to the nearest 0.05 cm3 Check and correct subtractions in the titre table. Select the “best”titre using the hierarchy: two identical; titres within 0.05 cm3, titres within 0.10 cm3 etc. (a) 3 Award V, VI and VII for a difference to Supervisor within 0.20 cm3 Award V and VI only for a difference of 0.20+ cm3 – 0.40 cm3 Award V only for a difference of 0.40+ cm3 - 0.80 cm3 If the selected “best” titres are > 0.50 cm3 apart, cancel one of the Q marks awarded. © UCLES 2010 [7] www.maxpapers.com Page 3 (b) Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 ACE Interpretation (c) ACE Interpretation Paper 33 Calculates the mean, correct to 2 decimal places (third decimal place maybe rounded to the nearest 0.05 cm3) from any accurate titres within 0.20 cm3. A mean of exactly .x25 or .x75 is allowed but the candidate may round up or down to the nearest 0.05 cm3. If ALL burette readings are given to 1 decimal place then the mean can be given to 1 decimal place if numerically correct without rounding. Mean of 24.3 and 24.4 = 24.35 () Mean of 24.3 and 24.4 = 24.4 () Mean of 24.3 and 24.5 = 24.4 () Titres to be used in calculating the mean must be clearly shown – in an expression or ticked in the titration table. 1 No additional factor/expression is allowed in any step If an answer, with no working, is given in any section allow if correct. I Uses 15.0/248.2 only in step (i) If no working shown accept only the following evaluated answers: (0.060, 0.0604 or 0.06044) 1 [1] Uses answer (i) × cand average titre/1000 in step (ii) and answer (iv) × 1000/25 in step (v) 1 III Uses answer (ii) × ½ in step (iii), and answer (iii) × 2 in step (iv) 1 IV Appropriate working shown in a minimum of three sections. To include equations as steps for the working mark; In (iii) must see x2 or x0.5. In (iv) must see multiplication or division by 6, 1.2 or 2. 1:6 for IO3 /6H+, 1:1.2 for 5I /6H+, 1:2 for 6H+/3I2 1 V 3 to 5 significant figures in final answers to all sections attempted – minimum of three final answers required to qualify for the award of this mark. 1 II PDO Display Syllabus 9701 © UCLES 2010 1 [5] www.maxpapers.com Page 4 (d) Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 ACE Interpretation Syllabus 9701 Gives 0.1(0) cm3 as the maximum error in (i). Ignore any sign and 0.1 the expression /cand titre in (b) × 100 in (ii) 0.06 Evaluates /25.0 × 100 in step (iii) Accept only 0.240 or 0.24, or rounded to 0.2 provided 0.24 has been seen in the working. Paper 33 1 1 [2] [Total: 15] © UCLES 2010 www.maxpapers.com Page 5 2 (a) Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 Paper 33 PDO Layout I Records at least four different balance readings and at least one mass of solid/gas Accept 0.0(0X) g as the mass of the empty tube or a statement that the tube is tared. 1 PDO Recording II Gives all appropriate headings and units when recording results. Do not accept mass of empty tube as 0.0(00)g here unless tube is described as tared. (minimum of three pieces of information) 1 III All recorded balance readings consistent to at least 1 decimal place. (minimum of three balance readings) 1 MMO Decisions IV Evidence of reheating to “constant” mass. For balances reading to 1 d.p. two masses must be identical For 2 or 3 d.p.balances, two masses must be within 0.05 g 1 MMO Quality V and VI Check and correct all subtractions in the results table. mass heated Calculate /mass of residue to 3 significant figures. Compare to Supervisor standard or standard value of 1.45. 2 Award V and VI for a difference up to 0.15 Award V only for a difference of 0.15+ to 0.30 Where a candidate repeats the experiment use cumulative masses of FA 3 and residue. Where masses of FA 3 and residue cannot be checked, accept candidate values to calculate the ratio. (b) ACE Interpretation [6] 1 Evaluates cand mass loss from (a) /cand mass of FA 3 correct to 2–4 significant figures. Where mass loss or mass of FA 3 is not given in (a), check, from balance readings, the values. A candidate who incorrectly describes the mass of the residue as the mass loss in tabulated results in (a) may “correct” the error and use the correct mass loss here. © UCLES 2010 [1] www.maxpapers.com Page 6 (c) (d) Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 ACE Conclusions ACE Improvements Syllabus 9701 Paper 33 Uses Mr (values) of CO2 or H2O to justify how the ratio of CuCO3 to Cu(OH)2 affects the mass loss. If % loss is too high – more CuCO3 If % loss is too low – more Cu(OH)2 1 Draws apparatus showing the collection of carbon dioxide in a syringe or in a burette or measuring cylinder inverted over water. Allow use of an inverted tube if graduations are shown or it is suitably labelled. All apparatus should be recognisable from the drawing or appropriately labelled. 1 Shows, in the diagram, an effective method of removing water vapour. Named reagent; e.g. (concentrated H2SO4, CaCl2, silica gel, (CaO), anhydrous CuSO4. or stated purpose of an un-named reagent given. Allow also a suitable reflux arrangement, returning water to the heated tube. or a statement that water vapour condenses in a water bath. Do not accept a diagram showing the gas bubbling through water without some written indication that the water is a condenser. 1 [1] [2] [Total: 10] © UCLES 2010 www.maxpapers.com Page 7 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 FA 4 is Al2(SO4)3(aq); 3 (a) MMO Collection FA 5 is ZnSO4(aq); Syllabus 9701 Paper 33 FA 6 is Pb(NO3)2(aq); FA 7 is MgSO4(aq) 1 mark for correct observations in each of the vertical columns. or 1 mark for correct observations in each of the horizontal rows (i), (ii) and (iii). 3 mark maximum Mark the section by the method which gives the better mark. 4 [4] observations test (i) (ii) (iii) FA 4 FA 5 FA 6 FA 7 addition of NaOH white ppt white ppt white ppt white ppt further addition of NaOH ppt soluble ppt soluble ppt soluble ppt insoluble addition of NH3 white ppt white ppt white ppt white ppt further addition of NH3 ppt insoluble ppt soluble ppt insoluble ppt insoluble no ppt, no reaction, colourless or yellow solution no ppt, no reaction, colourless or yellow solution yellow ppt no ppt, no reaction, colourless or yellow solution addition of KI Minimum evidence required in observations for the ion identity marks I, II and III in (b) In some cases, identification may be allowed from incomplete observations. There must, however, be no observations that are contrary to those expected with any “correctly” identified ion. The same criteria will be applied to “candidate’s supporting evidence in awarding mark IV. Candidates are not permitted to introduce (from the Qualitative Analysis Notes) supporting evidence that is not given in the observations. Precipitate colour need not be mentioned in supporting evidence. Al3+ Zn2+ Pb2+ Mg2+ (white) precipitate, soluble in (excess) NaOH, if yellow ppt with KI (white) precipitate, soluble in (excess) NH3(aq) Yellow precipitate with KI (white) precipitate, insoluble in (excess) NaOH © UCLES 2010 www.maxpapers.com Page 8 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 FA 4 is Al2(SO4)3(aq); FA 5 is ZnSO4(aq); (b) ACE Conclusions (c) MMO Decisions Syllabus 9701 Paper 33 FA 6 is Pb(NO3)2(aq); FA 7 is MgSO4(aq) Do not accept any ion other than Al 3+, Zn2+, Pb2+or Mg2+ in any section. Marks I to III Ions must be correct, including charge, if a symbol has been given. – no ecf in this section. 1 Award I only if one ion only is identified from correct observations. 1 Award I and II if two ions only are identified from correct observations. 1 Award I, II and III if all four cations are identified from correct observations. The 4th cation may be identified by elimination from incomplete supporting evidence. 1 Award mark IV if the supporting evidence fits the ion identified and the practical performed for at least three of the four ions. 1 Allow ecf on ion order on mark IV. [4] Selects sodium or potassium chromate(VI), sulfuric acid or hydrochloric acid soln containing one of the following named ions or formula given followed by (aq): CrO42 , SO42 , Cl , Br but not I , soln containing CrO42 ions, H2SO4, HCl, [1] © UCLES 2010 www.maxpapers.com Page 9 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 Paper 33 FA 8 is CuSO4(aq) (d) MMO Collection I Records blue colour of solution fading/disappearing on adding zinc powder in (i) If no reaction with Zn(s) is reported do not allow blue to light blue solution. 1 II Records a temperature rise in (i) Accept reaction is exothermic/produces heat 1 III Records a red-brown, orange-brown, brown or black solid in (i) 1 IV Observes a green, lime green, fluorescent green or yellow-green solution in (ii) 1 V Observes solution turning blue, or blue solution in (iii) if solution green in (ii) or solution going towards blue in colour on adding water in (iii) 1 If solution is not mentioned in (ii) or (iii) but colours are correct – award point V only. (e) ACE Conclusions Completes the equation: → Cu(s) + Zn2+(aq) State symbols required [5] 1 [1] [Total: 15] © UCLES 2010 www.maxpapers.com UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2010 question paper for the guidance of teachers 9701 CHEMISTRY 9701/34 Paper 3 (Advanced Practical Skills), maximum raw mark 40 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 must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2010 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. www.maxpapers.com Page 2 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 Paper 34 Question Sections Indicative material Mark 1 PDO layout I Volume given for Rough titre and accurate titre details tabulated. 1 MMO Collection II In the correct spaces, records initial and final burette readings for Rough titre and; Initial and final burette readings and, volume of FB 2 added recorded for each accurate titre Headings should match readings. Do not award this mark if: 50(.00) is used as an initial burette reading; More than one final burette reading is 50.(00); Any burette reading is greater than 50.(00) 1 MMO Decisions III Has two uncorrected, accurate titres within 0.1 cm3 Do not award this mark if having performed two titres within 0.1 cm3 a further titration is performed which is more than 0.10 cm3 from the closer of the initial two titres, unless a fourth titration, within 0.1 cm3 of the third titration or of either of the pair has also been carried out. 1 PDO Recording IV All accurate burette readings (initial and final) recorded to nearest 0.05 cm3. Assessed on burette readings only. 1 MMO Quality V, VI and VII Round any burette readings to the nearest 0.05 cm3. Check and correct subtractions in the titre table. Select the “best” titre using the hierarchy: two identical; titres within 0.05 cm3, titres within 0.10 cm3 etc. (a) 3 Award V, VI and VII for a difference to Supervisor within 0.15 cm3 Award V and VI only for a difference of 0.15+ cm3 – 0.25 cm3 Award V only for a difference of 0.25+ cm3 – 0.40 cm3 If the selected “best” titres are > 0.40 cm3 apart, cancel one of the Q marks awarded. © UCLES 2010 [7] www.maxpapers.com Page 3 (b) Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 ACE Interpretation Syllabus 9701 Calculates the mean, correct to 2 decimal places (third decimal place rounded to the nearest 0.05 cm3) from any accurate titres within 0.20 cm3. A mean of exactly .x25 or .x75 is allowed but the candidate may round up or down to the nearest 0.05 cm3. If ALL burette readings are given to 1 decimal place then the mean can be given to 1 decimal place if numerically correct without rounding. Mean of 24.3 and 24.4 = 24.35 () Mean of 24.3 and 24.4 = 24.4 () Titres to be used in calculating the mean must be clearly shown – in an expression or ticked in the titration table. Paper 34 1 [1] ACE Interpretation No additional factor/expression is allowed in any step If an answer, with no working, is given in any section allow if correct. I Uses 2.00/158.0 in step (i) and answer (i) × cand titre/1000 in step (ii) PDO Display II Uses answer (ii) × 5 in step (iii) and answer (iii) × 1000/25 in step (iv) 1 III Uses answer (iv) × 151.9 in step (v), and answer (v) × 100/21.50 in step (vi) 1 IV Appropriate working shown in a minimum of four sections. 1 V 3 to 5 significant figures in final answers to all sections attempted – minimum of four final answers required 1 (c) 1 [5] [Total: 13] © UCLES 2010 www.maxpapers.com Page 4 2 (a) Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 Paper 34 PDO Layout I Records at least four different balance readings and at least one mass of solid/gas Accept 0.0(0X) g as the mass of the empty tube or a statement that the tube is tared. 1 PDO Recording II Gives all appropriate headings and units when recording results. Do not accept mass of empty tube as 0.0(00)g here unless tube is described as tared. (minimum of three pieces of information) 1 III All recorded balance readings consistent to at least 1 decimal place. (minimum of three balance readings) 1 MMO Decisions IV Evidence of reheating to “constant” mass. For balances reading to 1 d.p. two masses must be identical For 2 or 3 d.p. balances, two masses must be within 0.05 g 1 MMO Quality V and VI checks and corrects if necessary all subtractions in the results table. mass heated Calculate /mass of residue to 3 significant figures. Compare to supervisor standard or standard value of 1.40. 2 Award V and VI for a difference up to 0.10 Award V only for a difference of 0.10+ to 0.20 Where a candidate repeats the experiment use cumulative masses of FB 3 and residue. Where masses of FB 3 and residue cannot be checked, accept candidate values to calculate the ratio. (b) ACE Interpretation ACE Conclusions Calculates 2.71, (2.710, 2.7097) 1 and (ii) Has: cand value in (i) x mass loss from table in (a) If no mass loss is recorded in the table, check the value used. (iii) Ticks the appropriate box for the experiment 1 and makes some comparison between mass of NaHCO3 and the mass of FB 3 used If mass of NaHCO3 calculated in (ii) ≥ mass of FB 3, ignore any ticked box but award the mark for any statement that the mass is not possible. [6] (i) © UCLES 2010 [2] www.maxpapers.com Page 5 (c) (d) Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 ACE Improvements ACE Interpretation Syllabus 9701 Paper 34 (i) No mass change with Na2CO3 (on heating). (ii) Evidence for no gas produced, e.g.: limewater unaffected, no gas collected in a gas syringe If there is reference to measuring mass and to measuring volume but the absence of change is not mentioned, award one of the two marks available. 1 1 Max errors of 0.05, 0.005 and 0.0005 respectively for balances A, B and C. Calculates: 1.11% error for balance A 0.25% error for balance B 0.20% error for balance C Allow ecf on % errors only if: (i) Max errors given are 0.1, 0.01 and 0.001 respectively for balances A, B and C and % errors are 2.22%, 0.50% and 0.40% (ii) All max errors are incorrect by a factor 10 e.g. 0. 5, 0. 05 and 0. 005. % errors are 11.1%, 2.5% and 2.0% 1 [2] 1 [2] [Total: 12] © UCLES 2010 www.maxpapers.com Page 6 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 FB 4 is MnSO4(aq); 3 (a) FB 5 is MgSO4(aq); MMO Collection Syllabus 9701 Paper 34 FB 6 is Al2(SO4)3(aq); FB 7 is (NH4)2SO4(aq) Give one mark for each of the following: I for FB 4 – tests (i) and (iv) II for FB 5 – tests (i) and (iv) III for FB 6 – tests (i) and (iv) IV for FB 7 – tests (i), (iii) and (iv) V Give one mark for any change/darkening of the initial precipitate in test (ii) for FB 4 to a qualified brown. The darkening may be described in test (i) or in test (iv) VI Describes the test on gas for ammonia in test (iii) for any solution that has no precipitate in either part test of (i) and is warmed. The test for ammonia is expected with FB 7 Do not award (VI) if the test is carried out with a solution in which a precipitate had formed at any stage or If a solution in which no precipitate is formed is not warmed with sodium hydroxide 1 1 1 1 1 1 [6] Results required with NaOH(aq) and NH3(aq) for the award of marks I to IV in 3(a) observations test FB 4 addition of NaOH (i) Do not accept cream or equivalent colour precipitates further addition of NaOH (iii) (iv) off-white, pale brown, buff or beige precipitate precipitate insoluble FB 5 FB 6 FB 7 No precipitate or no change white precipitate white precipitate precipitate insoluble precipitate soluble warming solution with NaOH Do not accept clear on its own as an observation; clear solution is acceptable no precipitate or no change (may be left blank) any reference to a gas being evolved or reference to red litmus turning blue addition of NH3 as NaOH as NaOH as NaOH as NaOH further addition of NH3 as NaOH as NaOH precipitate insoluble as NaOH © UCLES 2010 www.maxpapers.com Page 7 (b) Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 ACE Conclusions Syllabus 9701 Paper 34 Do not accept any ion other than Mn2+, Mg2+, Al3+or NH4+ in any section. Marks I to III Ions must be correct, including charge, if a symbol has been given. – no ecf in this section. Award I only if one ion only is identified from correct observations. 1 Award I and II if two ions only are identified from correct observations. 1 Award I, II and III if all four cations are identified from correct observations. The 4th cation may be identified by elimination from incomplete supporting evidence. 1 A deduction of Mn2+ is allowed from a cream ppt with NaOH(aq) and NH3(aq) 1 IV Award this mark if the supporting evidence fits the ion identified and the practical performed for at least three of the four ions Allow ecf on ion order for mark IV. (Mg2+ and Al 3+ are most likely to be interchanged depending on “solubility in excess” observations. [4] Minimum evidence required in observations for the ion identity marks I, II and III. In some cases, identification may be allowed from incomplete observations. There must, however, be no observations that are contrary to those expected with any “correctly” identified ion. The same criteria will be applied to “candidate’s supporting evidence in awarding mark IV. Candidates are not permitted to introduce (from the Qualitative Analysis Notes) supporting evidence that is not given in the observations. Mn2+ off-white precipitate with each reagent, or off-white precipitate turning brown with either of the reagents identification of the ion is allowed from an incorrect observation of a cream or yellow-white precipitate – one ion is known to be Mn2+ Mg2+ white precipitate, insoluble in (excess) NaOH Al 3+ white precipitate, soluble in (excess) NaOH NH4 + no precipitate/no change with either reagent or ammonia, alkaline gas or gas turning red litmus blue evolved © UCLES 2010 www.maxpapers.com Page 8 (c) (d) (e) (f) Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 Paper 34 MMO Collection Records no precipitate/no reaction with each of the reagents. 1 ACE Conclusions States that Pb2+/lead(II) would give similar results. Award this mark providing there are no contrary observations for the solution identified as containing Al 3+ 1 Records a white ppt in (i) Records a yellow precipitate or precipitate turning yellow in (ii). 1 1 Award one mark for any attempt to describe replacement of Cl by I in the ppt. 1 MMO Collection ACE Conclusions [1] [1] [2] [1] [Total: 15] © UCLES 2010 www.maxpapers.com UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2010 question paper for the guidance of teachers 9701 CHEMISTRY 9701/35 Paper 3 (Advanced Practical Skills), maximum raw mark 40 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 must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2010 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. www.maxpapers.com Page 2 Mark Scheme: Teachers’ version GCE A / AS LEVEL – October/November 2010 Syllabus 9701 Paper 35 Question Sections Indicative material 1 PDO Layout I MMO Collection II Follows instructions – initial and final burette readings recorded for Rough titre and initial and final burette readings and volume of FA 2 added recorded for each accurate titre and headings should match readings. Do not award this mark if: 50(.00) is used as an initial burette reading; more than one final burette reading is 50.(00); any burette reading is greater than 50.(00) 1 MMO Decisions III Has two uncorrected, accurate titres within 0.1 cm3 Do not consider the Rough even if ticked. Do not award this mark if having performed two titres within 0.1 cm3 a further titration is performed which is more than 0.10 cm3 from the closer of the initial two titres, unless a fourth titration, within 0.1 cm3 of the third titration has also been carried out. 1 PDO Recording IV All accurate burette readings (initial and final) recorded to nearest 0.05 cm3 Assess this mark on burette readings only 1 MMO Quality V, VI and VII Round any burette readings to the nearest 0.05 cm3. Check and correct subtractions in the titre table. Select the “best” titre using the hierarchy: two identical; titres within 0.05 cm3; titres within 0.1 cm3; etc. Award V, VI and VII for a difference from Supervisor within 0.20 cm3 3 (a) Volume given for Rough titre. and accurate titre details tabulated. Mark 1 Award V and VI only for a difference of 0.20+ cm3 – 0.30 cm3 Award V only for a difference of 0.30+ - 0.50 cm3 If the “best” titres are ≥ 0.50 cm3 apart cancel one of the Q marks. © UCLES 2010 [7] www.maxpapers.com Page 3 (b) Mark Scheme: Teachers’ version GCE A / AS LEVEL – October/November 2010 ACE Interpretation Syllabus 9701 Calculates the mean, correct to 2 decimal places from any accurate titres within 0.20 cm3. The third decimal place may be rounded to the nearest 0.05 cm3. A mean of exactly .x25 or .x75 is allowed but the candidate may round up or down to the nearest 0.05 cm3. If ALL burette readings are given to 1 decimal place then the mean can be given to 1 decimal place if numerically correct without rounding. Mean of 24.3 and 24.4 = 24.35 () Mean of 24.3 and 24.4 = 24.4 () Paper 35 1 Titres to be used in calculating the mean must be clearly shown – in an expression or ticked in the titration table. (c) ACE Interpretation I Correctly evaluates II Uses answer (i) × 10 .00 = 0.25(0) 40 mean titre in step (ii) 1000 [1] 1 1 and 1000 in step (iii) 10 If an answer, with no working, is given in any section allow if correct. answer (ii) x Total [2] [Total: 10] © UCLES 2010 www.maxpapers.com Page 4 2 (a) Mark Scheme: Teachers’ version GCE A / AS LEVEL – October/November 2010 PDO Recording I Syllabus 9701 Has correct headings (minimum three) and units in the weighing table in (2)(a) and correct units in the titration table in (2)(b) Paper 35 1 Acceptable units are /g, (g), mass in grams, mass in g; similarly /cm3, II All three balance reading are read with constant precision (same no of decimal places) and to at least 1 decimal place 1 [2] On Supervisor script scale the titre for 3.00 g of FA 3 added to the acid. Calculate 8 × (3.00 – mass of FA 3 used) and subtract from the titre obtained. Mass of FA 3 used = (mass tube + FA 3) – (mass tube + residue) If (mass tube + residue) < mass of empty tube then use (mass tube + FA 3) – (mass tube). (b) MMO Quality Award I and II if the difference between candidate and Supervisor scaled titres is within 0.40 cm3 1 Award I only if the difference is between 0.50+ cm3 and 0.80 cm3 1 (c) There is no mark available for this section. (d) ACE Interpretation I Uses mean titre × 0.280 in step (i) 1000 [2] 1 and uses answer (i) × 250 in step (ii) 25 II Correctly evaluates PDO Display 0.5 × 250 = 0.125 in step 1000 1 III Uses answer (iv) × 0.5 × 100 in step (v) 1 IV Working shown in a minimum of three sections Working should be a step in the right direction: step (i) 0.28 × titre volume (in cm3/dm3) step (ii) Use of 25 & 250 or 10 step (iii) 0.5 and 250 step (iv) the correct two numbers step (v) would need to include 2 (0.5) and 100 step (vi) must be correct 1 V 3 to 5 significant figures in final answers to all sections attempted – minimum of three final answers required 1 © UCLES 2010 [5] www.maxpapers.com Page 5 (e) Mark Scheme: Teachers’ version GCE A / AS LEVEL – October/November 2010 ACE Conclusions Syllabus 9701 Explains one of the following: Paper 35 1 If 5.5 g of CaCO3 had been used the titre would be too small/not enough HCl remains for the titration (not ‘all the acid has reacted’) or Difficult/takes too long to dissolve 5.5 g of solid/it will not all dissolve in 150 cm3 (of acid) or Excessive/too fast effervescence/fizzing/rate of gas evolved or Acid spray (f) (g) ACE Interpretation (i) If balance displays to 1 decimal place: error in balance reading is ±0.05 g or ±0.1(0) g error in mass of FA 3 is ±0.1 g or ±0.2 g If balance displays to 2 decimal places: error in balance reading is ±0.005 g or ±0.01 g error in mass of FA 3 is ±0.01 g or ±0.02 g If balance displays to 3 decimal places: error in balance reading is ±0.0005 g or 0.001g error in mass of FA 3 is ±0.001 g or ±0.002 g [1] 1 (ii) Correctly evaluates to at least 2 significant figures: candidate' s error in mass of FA 3 × 100 mass of FA 3 used 1 (i) Gives correct equation for the thermal decomposition of calcium carbonate including state symbols 1 (ii) Outlines: ACE weigh container Improvements weigh container + solid (heating and) weighing again repeated (heating and) weighing to constant mass or weigh container weighing container + solid (heating and) measuring gas volume when no further increase and cooled to room temperature / use of pV = nRT / PV = constant T 1 ACE Conclusions Total [2] [2] [14] © UCLES 2010 www.maxpapers.com Page 6 Mark Scheme: Teachers’ version GCE A / AS LEVEL – October/November 2010 FA 7 is Fe2(SO4)3(aq); 3 (a) FA 8 is CrCl3(aq); Syllabus 9701 FA 9 is ZnI2(aq) [ZnCl2 + KI] PDO Layout I (Tabulates) observations clearly, showing: observation when each reagent is first added and observation when reagent added to excess (if there is a ppt) MMO Collection II, III and IV 1 mark for correct observations in each of the columns or rows representing FA 7, FA 8 and FA 9 or 1 mark for correct observations in the row or column representing a reagent added (initial and excess count as one row/column) ACE Conclusions Paper 35 1 3 Award V only if one ion only is correctly identified 1 Award V and VI if all three ions are correctly identified from candidate’s observations. Allow ecf* 1 Minimum for observations marks: Solution NaOH NH3 FA 7 red-brown/brown/rust ppt insoluble (in excess) red-brown ppt insoluble (in excess) (suitable qualified brown) FA 8 grey-green ppt soluble/dissolves (in excess) giving a dark green solution grey-green ppt insoluble (in excess) FA 9 White/milky white ppt soluble/dissolves (in excess) White/milky white ppt soluble/dissolves (in excess) Minimum for conclusions marks: (with incomplete but not CON observations) FA 7 FA 8 FA 9 red-brown ppt with either; grey-green ppt with either/(dark) green solution with excess NaOH; white ppt soluble in excess NH3. * ecfs allowed FA 8 FA 9 FA 9 FA 9 allow Fe2+ if green ppt insoluble in excess NaOH (no grey-green ppts) allow Al3+ and Pb2+ if white ppt insoluble in excess NH3 allow Ba2+ and NH4+ if no ppt with either allow Mg2+ if white ppt insoluble in excess of both © UCLES 2010 [6] www.maxpapers.com Page 7 (b) Mark Scheme: Teachers’ version GCE A / AS LEVEL – October/November 2010 Syllabus 9701 Selects barium chloride or barium nitrate for the test in step (i) Do not allow Ba2+ alone Ba2+(aq) or soln containing Ba2+ (ions) is acceptable Paper 35 MMO Decisions I 1 MMO Collection II Records white/off-white precipitate with only FA 7 MMO Decisions III Selects silver nitrate or lead nitrate in (ii) to add to the solutions (that do not contain sulfate) Do not allow Ag+ or Pb2+ alone Aqueous ions or solutions containing the ion are acceptable as above MMO Collection IV Appropriate observations FA 8 white ppt with Ag+/white ppt or no ppt with Pb2+ FA 9 yellow ppt with either Ignore observations with any solution candidate has identified as sulfate 1 ACE Conclusions V FA 8 is chloride, FA 9 is iodide Credit if the supporting evidence fits the ion identified and the practical performed for FA 8 and FA 9 provided there is no CON observation in (i) Do not credit if Ag+ gives a ppt with FA 7 1 1 1 Marks IV and V may be awarded from FA 8 white ppt chloride (IV) FA 9 yellow ppt iodide (V) [5] Other possibilities: Two white ppts with aqueous Ba2+ then remaining solution tested with aqueous Ag+/Pb2+ This would score marks I, III and may score one of IV or V Aqueous Ba2+ gives positive result with solution other than FA 7 and tests with aqueous Ag+/Pb2+ performed (This would score marks I and III) Ignore observation and conclusion with FA 7 Award correct observation and valid conclusion for third ion thus scoring one of IV or V Aqueous Ba2+ gives positive result with all three solutions Award mark I, and mark III may be awarded for selection of aqueous Ag+/Pb2+ or statement that no further testing is required but no other marks can be awarded in this section. © UCLES 2010 www.maxpapers.com Page 8 Mark Scheme: Teachers’ version GCE A / AS LEVEL – October/November 2010 Syllabus 9701 Paper 35 FA 10 is NaNO3(s); FA 11 is NaNO2(s) (c) (i) MMO Collection I Solid/FA 10 melts/to a liquid/solution (on heating) 1 II Observes bubbles of gas in liquid/solution or Liquid/solution turns yellow/pale yellow 1 MMO Decisions III Describes an appropriate test in either (i) or (ii) for any of the following gases: O2, CO2, NH3 or SO2 There must be a reference to gas being evolved before this mark can be awarded. 1 MMO Collection IV Positive identification of oxygen gas in (i): glowing splint rekindles/relights/glows brighter (gas evolved rekindles a glowing splint would gain marks III and IV) (‘glowing splint rekindles’ would gain mark III not IV) 1 V On adding acid to residue to FA 11, observes brown/yellow-brown gas (not yellow, orange or red-brown) or blue solution (allow greenish blue) 1 (ii) Total [5] [16] © UCLES 2010 www.maxpapers.com UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2010 question paper for the guidance of teachers 9701 CHEMISTRY 9701/36 Paper 3 (Advanced Practical Skills), maximum raw mark 40 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 must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2010 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. www.maxpapers.com Page 2 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 Paper 36 Question Sections Indicative material Mark 1 PDO layout I Volume given for Rough titre and accurate titre details tabulated. 1 MMO Collection II Follows instructions - dilutes 45.50 – 46.50 cm3 FB 1 and initial and final burette readings recorded for Rough titre and initial and final burette readings and volume of FB 2 added recorded for each accurate titre Headings should match readings. Do not award this mark if: 50(.00) is used as an initial burette reading; more than one final burette reading is 50.(00); any burette reading is greater than 50.(00) 1 MMO Decisions III Has two uncorrected, accurate titres within 0.1 cm3 Do not consider the Rough even if ticked. Do not award this mark if having performed two titres within 0.1 cm3 a further titration is performed which is more than 0.10 cm3 from the closer of the initial two titres, unless a fourth titration, within 0.1 cm3 of the third titration (or first two) has also been carried out. 1 PDO Recording IV All accurate burette readings (initial and final) recorded to nearest 0.05 cm3 (Accurate titration & dilution tables) Assess this mark on burette readings only 1 (a) For candidates and Supervisor scale titre for 46.00 cm3 FB 1 diluted. 46.00 Calculate titre × volume of FB1 added Calculate difference in Supervisor and candidate scaled values and award “quality” marks as below. [If candidate has not recorded a volume diluted, use 46.00 cm3] © UCLES 2010 www.maxpapers.com Page 3 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 MMO Quality Syllabus 9701 V, VI and VII Round any burette readings to the nearest 0.05 cm3. Check and correct subtractions in the titre table. Select the “best” titre using the hierarchy: two identical; titres within 0.05 cm3; titres within 0.1 cm3; etc. Paper 36 3 Award V, VI and VII for a difference from Supervisor within 0.20 cm3 Award V and VI only for a difference of 0.20+ cm3 – 0.30 cm3 Award V only for a difference of 0.30+ – 0.50 cm3 If the “best” titres are ≥ 0.50 cm3 apart cancel one of the Q marks. (b) (c) ACE Interpretation ACE Interpretation PDO Display [7] Calculates the mean, correct to 2 decimal places (third decimal place rounded to the nearest 0.05 cm3) from any accurate titres within 0.20 cm3. A mean of exactly .×25 or .×75 is allowed but the candidate may round up or down to the nearest 0.05 cm3. If ALL burette readings are given to 1 decimal place then the mean can be given to 1 decimal place if numerically correct without rounding. Mean of 24.3 and 24.4 = 24.35 () Mean of 24.3 and 24.4 = 24.4 () Titres to be used in calculating the mean must be clearly shown – in an expression or ticked in the titration table. 1 I Expression correct in step (i) volume diluted × 0.125 1000 1 II Uses answer to (i) × III Uses answer to (ii) × 2 in step (iii) and 1000 answer to (iii) × in step (iv) titre If an answer, with no working, is given in any section allow if correct. 25 in step (ii) 250 [1] 1 1 IV Appropriate working shown in a minimum of 3 sections. 1 V 1 3 to 5 significant figures in final answers to all sections attempted – minimum of 3 final answers required to qualify for the award of this mark. © UCLES 2010 [5] www.maxpapers.com Page 4 (d) (e) Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 Paper 36 For Student A explains that final burette reading was also 0.05 cm3 greater than the true value (“error” in same direction) Ignore parallax error Not errors cancel – reason needed 1 (ii) For Student B explains that final burette reading was 0.05 cm3 greater than the true value (“error” in opposite direction) Not errors compound each other/add up 1 ACE Conclusions (i) Explains that carbon dioxide is acidic (and its absorption reverses the colour change of the indicator) 1 ACE Improvements (ii) Puts acid/FB 3 in burette and pipettes NaOH/ FB 2 into flask or Heat the solution/Use hot solution 1 ACE Interpretation (i) [2] [2] [Total: 17] © UCLES 2010 www.maxpapers.com Page 5 2 (a) (b) (c) (d) Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 Paper 36 PDO Recording I Records results in a single table for both experiments. No repetition of headings 1 MMO Quality II Titre for either Flask A or B within 0.50 cm3 of Supervisor 1 III Titre for either Flask A or B within 0.30 cm3 of Supervisor 1 IV Titre for both Flask A and B within 0.30 cm3 of Supervisor 1 ACE Interpretation ACE Conclusions ACE Conclusions [4] (i) Calculates a volume of 200 cm3 in step (i) 1 (ii) Correctly calculates titre x 5 for each flask 1 Mark consequentially to practical results Chooses expt with lower titre – less remains (or converse argument) or higher value in (b)(iii) Allow ecf 1 Comparison of candidate’s Kc values Judgement on constancy or otherwise Supports/does not support equilibrium 1 [2] [1] [1] [Total: 8] © UCLES 2010 www.maxpapers.com Page 6 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 FB 7 is Fe(NH4)2(SO4)2(aq); 3 (a) (b) FB 8 is Na2SO4(aq); Syllabus 9701 Paper 36 FB 9 is CaCl2(aq) MMO Decisions I Selects sodium hydroxide as reagent (Not if + Al) and describes (warming the solution and) testing any gas evolved with red litmus/pH paper 1 MMO Collection II Records positive test for ammonia gas with FB 7 only Must link gas/NH3 with positive test (Allow even if Al mentioned in I) 1 MMO Decisions III Selects barium chloride or nitrate together with HCl or HNO3 Do not accept Ba2+ as a reagent Accept Ba2+(aq) or a solution containing Ba2+ ions 1 MMO Collection IV White ppt, persisting in acid with FB 7 and with FB 8 Allow from unspecified strong acid provided there is no ppt with FB 9. 1 MMO Conclusions V Mark consequentially to observations for solutions containing NH4+ and SO42 ecf allowed here but not with other identities Allow from strong acid or from H2SO4 if clearly added after Ba2+(aq) 1 [5] PDO Layout I (Tabulates) observations clearly, showing: observation when each reagent is first added and observation when reagent added to excess if there is a ppt 1 MMO Collection II, III and IV 1 mark for correct observations in each of the columns or rows representing FB 7, FB 8 and FB 9 or 1 mark for correct observations in the row or column representing a reagent added (initial and excess count as one row/column) 3 [4] Minimum observations Solution FB 7 NaOH Green ppt insoluble (in excess) NH3 Green ppt insoluble (in excess) FB 8 no reaction/no change/no ppt Not “–“ words needed (Only penalise once) colourless soln/no reaction/no change/no ppt © UCLES 2010 FB 9 White ppt insoluble (in excess) No reaction/no change/no ppt www.maxpapers.com Page 7 (c) Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 ACE Conclusions Syllabus 9701 One mark for FB 7 and FB 9 containing Fe2+ and Ca2+ respectively provided no CON obs in (a) or (b) No ecf Ignore FB 8, ignore supporting evidence Paper 36 1 [1] FB 10 is CuCO3(s) (d) (i) (ii) I observes the solid turning black in step (i) 1 II observes fluidity in solid layer in step (i) Allow description of fluidised solid as “liquid” 1 MMO Decisions III describes an appropriate test for any of the following gases: O2, CO2, NH3 or SO2 (gas or O2/etc needed) 1 MMO Collection IV lime water turns milky/cloudy/chalky Gas or CO2 turns limewater milky. scores III and IV 1 V 1 MMO Collection on adding acid to residue from FB 10, observes green solution (on warming) Ignore any residual solid Allow blue-green or bluish green Allow if (qualified) green solution turns blue on cooling May award either III or IV here but only for gas tests for CO2 or SO2 or limewater observations [5] [Total: 15] © UCLES 2010 www.maxpapers.com UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2010 question paper for the guidance of teachers 9701 CHEMISTRY 9701/41 Paper 4 (A2 Structured Questions), maximum raw mark 100 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 must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2010 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. www.maxpapers.com Page 2 1 Mark Scheme: Teachers’ version GCE A LEVEL – October/November 2010 Syllabus 9701 Paper 41 (a) PCl5 + 4H2O → H3PO4 + 5HCl (1) SiCl4 + 2H2O → SiO2 + 4HCl (or giving H2SiO3, Si(OH)4 etc.) (1) (b) bond energies: S-S Cl-Cl S-Cl = 264 kJ mol = 244 kJ mol = 250 kJ mol [2] 1 1 1 ∆H = 8 × 264 + 8 × 244 – 16 × 250 = +64 kJ mol 1 (2) [2] (c) (i) +2 (1) (ii) (half) the sulfur goes up by +2, (1) (the other half) goes down by –2 (1) (iii) HCl (can be read into (iv)) (1) (iv) 2SCl2 + 2H2O → S + SO2 + 4HCl (1) (v) (+ AgNO3) (+ K2Cr2O7) white ppt. (1) solution turns green (1) [7] [Total: 11] 2 (a) (i) A ligand is a species that contains a lone pair of electrons, or that can form a dative bond (to a transition element) (1) (ii) species OH NH4+ CH3OH CH3NH2 can be a ligand cannot be a ligand (4 × ½) [3] (b) (i) C is [Cu(NH3)6]2+ SO42 (allow [Cu(NH3)4]2+ SO42 (1) D is CuO (1) E is Na2SO4 (1) F is BaSO4 (1) (ii) acid-base or neutralisation (1) [5] (c) (i) any two from: brown fumes or vapour evolved / gas relights glowing splint / black solid formed (2) (ii) 2Cu(NO3)2 → 2CuO + 4NO2 + O2 (1) [3] [Total: 11 max 10] © UCLES 2010 www.maxpapers.com Page 3 3 Mark Scheme: Teachers’ version GCE A LEVEL – October/November 2010 Syllabus 9701 Paper 41 (a) (i) Cu(s) – 2e → Cu2+(aq) allow electrons on RHS (1) (ii) Eo for Ag+/Ag is +0.80V which is more positive than +0.34V for Cu2+/Cu, (1) so it’s less easily oxidised (owtte) (1) (iii) Eo for Ni2+ is –0.25V, (1) Ni is readily oxidised and goes into solution as Ni2+(aq) (1) [Mark (ii) and (iii) to max 3] (iv) Cu2+(aq) + 2e → Cu(s) (1) (v) Eo for Zn2+/Zn is negative / = –0.76V, so Zn2+ is not easily reduced. (1) (vi) The blue colour fades because Cu2+(aq) is being replaced by Zn2+(aq) or Ni2+(aq) or [Cu2+] decreases (1) [7] (b) amount of copper = 225/63.5 = 3.54(3) mol (1) amount of electrons needed = 2 × 3.54 = 7.08/9 (7.087) mol (1) no. of coulombs = 20 × 10 × 60 × 60 = 7.2 × 105 C no. of moles of electrons = 7.2 × 105/9.65 × 104 = 7.46 mol (1) percentage “wasted” = 100 × (7.461 – 7.087)/7.461 = 5.01 (5.0)% (accept 4.98–5.10) (1) [4] (c) Eo data: Ni2+/Ni = –0.25V Fe2+/Fe = –0.44V (1) Because the Fe potential is more negative than the Ni potential, the iron will dissolve (1) [2] [Total: 13] 4 (a) (i) SnO2 (ii) PbO Can be read into equation (1) 2NaOH + SnO2 → Na2SnO3 + H2O (1) Can be read into equation (1) PbO + 2HCl → PbCl2 + H2O (1) [4] (b) moles of oxygen = 9.3/16 = 0.581 mol moles of lead = 90.7/207 = 0.438 mol (both 3 s.f.) (1) so formula is Pb3O4 (1) [2] (c) (i) Ksp = [Pb2+][Cl ]2 (1) units = mol3 dm 9 (1) (ii) if [Pb2+] = x, Ksp = 4x3, so x = 3√{Ksp/4} [Pb2+] = 3√{2 × 10 5/4} = 1.71 × 10–2 mol dm 3 (1) (iii) [Pb2+] = 2 × 10 5/(0.5)2 = 8.0 × 10–5 mol dm 3 (1) (iv) common ion effect, or increased [Cl ] forces solubility equilibrium over to the left (1) [Max 4] [Total: 10] © UCLES 2010 www.maxpapers.com Page 4 5 Mark Scheme: Teachers’ version GCE A LEVEL – October/November 2010 Syllabus 9701 Paper 41 (a) (i) ester (1) (ii) H is nitrobenzene – structure needed here (1) J is phenyldiazonium chloride – structure needed here (1) (iii) step 2 step 3 step 4 step 5 Sn/Zn + HCl / H2 + named cat / NaBH4 / LiAlH4 / Na + ethanol (1) HNO2/NaNO2 + HCl at T = 10°C or less (1) heat/warm to T > 10°C (1) CH3COCl / CH3COCOCOCH3 (1) [7] (b) (i) compounds that have the same molecular formula, but different structures (1) (ii) phenol (NOT hydroxy) (1) (methyl) ketone or carbonyl (1) (iii) K is 4-ethanoylphenol, HO-C6H4-COCH3 (must be 1,4- disubstituted isomer) (1) (iv) K I2 + NaOH(aq) NaOH(aq) Br2(aq) O CO2 ((1) for CO2 ; (1) for –O ) In any positions Br OH Br COCH3 (2 × Br needed) O COCH3 (anion needed) [4] [8 max 7] [Total: 14] © UCLES 2010 www.maxpapers.com Page 5 6 Mark Scheme: Teachers’ version GCE A LEVEL – October/November 2010 (a) Syllabus 9701 Paper 41 O * * (1) for each centre – more than 2 centres shown deduct 1 mark [2] LiAlH4 or NaBH4 or Na + ethanol or H2 + Ni (1) heat with Al2O3 / porous pot or conc. H2SO4 / H3PO4 (1) (b) (i) step 1 step 2 (ii) M (1) L (1) (letters may be reversed) [4] (c) (i) M (no mark) (ii) CO2H O P i.e. 3,7-dimethyl-6-oxo-octanoic acid (1) (iii) 2,4-DNPH (1) orange ppt. with P (none with N) (1) Mark ecf from candidates’ P (d) (+) H () Cl H + [3] H Cl Cl 2 curly arrows (1) carbocation intermediate + Cl (1) lone pair on Cl and last curly arrow (1) [3] [Total: 12] © UCLES 2010 www.maxpapers.com Page 6 7 Mark Scheme: Teachers’ version GCE A LEVEL – October/November 2010 Syllabus 9701 Paper 41 (a) (i) Disulfide bond / group / bridge (1) (ii) The tertiary structure (1) (iii) The substrate will no longer bond to / fit into the active site (1) or shape of active site is changed [3] (b) (i) Acid-base / proton donor / neutralisation / salt formation (1) (ii) The ability of the –CO2H group to form hydrogen bonds (1) and ionic interactions (1) The –CO2H/–CO2 group is no longer able to interact with –NH2/–NH3+ (1) The Ag+ forms a strong bond with –COO (1) [5] max [4] (c) (i) 8 but allow 4O2 if specified as molecules (1) (ii) Dative / co-ordinate (1) (iii) Octahedral / 6 co-ordinate (1) [3] [Total: 10] 8 (a) Protons (1) in NMR, energy is absorbed due to the two spin states (1) Electrons (1) in X-ray crystallography, X-rays are diffracted (by regions of high electron density) (1) [4] (b) (i) 1 – no mark The spectrum of alcohol / Y contains different peaks Alcohol / Y contains different chemical environments Spectrum 2 contains only one peak (1) (ii) Spectrum 2 only shows 1 peak so Z must be a ketone (1) Hence Y must be a 2° alcohol (1) Number of carbon atoms present 0.6 × 100 3 (1) 17.6 × 1.1 Thus Z must be CH3COCH3 (1) Hence Y must be propan-2-ol, CH3CH(OH)CH3 (1) (iii) H │ Y is CH3 – C – CH3 │ OH (1) (iv) All of the protons in Z are in the same chemical environment (1) [8] max [7] [Total: 11] © UCLES 2010 www.maxpapers.com Page 7 9 Mark Scheme: Teachers’ version GCE A LEVEL – October/November 2010 Syllabus 9701 Paper 41 (a) (i) A few nanometres (accept 0.5–10 nm) (1) (ii) Graphite/graphene (1) (iii) van der Waals’ (1) Carbon atoms in the nanotubes are joined by covalent bonds (1) (as are the hydrogen atoms in a hydrogen molecule) or no dipoles on C or H2 or the substances are non-polar [4] (b) More hydrogen can be packed into the same space/volume (1) [1] (c) If a system at equilibrium is disturbed, the equilibrium moves in the direction which tends to reduce the disturbance (owtte) (1) When H2 is removed the pressure drops and more H2 is released from that adsorbed (1) The equilibrium H2adsorbed H2gaseous (1) Equilibrium shifts to the right as pressure drops (1) [4] [Total: 9] © UCLES 2010 www.maxpapers.com UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2010 question paper for the guidance of teachers 9701 CHEMISTRY 9701/42 Paper 4 (A2 Structured Questions), maximum raw mark 100 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 must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2010 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. www.maxpapers.com Page 2 1 Mark Scheme: Teachers’ version GCE A LEVEL – October/November 2010 Syllabus 9701 Paper 42 (a) PCl5 + 4H2O → H3PO4 + 5HCl (1) SiCl4 + 2H2O → SiO2 + 4HCl (or giving H2SiO3, Si(OH)4 etc.) (1) (b) bond energies: S-S Cl-Cl S-Cl = 264 kJ mol = 244 kJ mol = 250 kJ mol [2] 1 1 1 ∆H = 8 × 264 + 8 × 244 – 16 × 250 = +64 kJ mol 1 (2) [2] (c) (i) +2 (1) (ii) (half) the sulfur goes up by +2, (1) (the other half) goes down by –2 (1) (iii) HCl (can be read into (iv)) (1) (iv) 2SCl2 + 2H2O → S + SO2 + 4HCl (1) (v) (+ AgNO3) (+ K2Cr2O7) white ppt. (1) solution turns green (1) [7] [Total: 11] 2 (a) (i) A ligand is a species that contains a lone pair of electrons, or that can form a dative bond (to a transition element) (1) (ii) species OH NH4+ CH3OH CH3NH2 can be a ligand cannot be a ligand (4 × ½) [3] (b) (i) C is [Cu(NH3)6]2+ SO42 (allow [Cu(NH3)4]2+ SO42 (1) D is CuO (1) E is Na2SO4 (1) F is BaSO4 (1) (ii) acid-base or neutralisation (1) [5] (c) (i) any two from: brown fumes or vapour evolved / gas relights glowing splint / black solid formed (2) (ii) 2Cu(NO3)2 → 2CuO + 4NO2 + O2 (1) [3] [Total: 11 max 10] © UCLES 2010 www.maxpapers.com Page 3 3 Mark Scheme: Teachers’ version GCE A LEVEL – October/November 2010 Syllabus 9701 Paper 42 (a) (i) Cu(s) – 2e → Cu2+(aq) allow electrons on RHS (1) (ii) Eo for Ag+/Ag is +0.80V which is more positive than +0.34V for Cu2+/Cu, (1) so it’s less easily oxidised (owtte) (1) (iii) Eo for Ni2+ is –0.25V, (1) Ni is readily oxidised and goes into solution as Ni2+(aq) (1) [Mark (ii) and (iii) to max 3] (iv) Cu2+(aq) + 2e → Cu(s) (1) (v) Eo for Zn2+/Zn is negative / = –0.76V, so Zn2+ is not easily reduced. (1) (vi) The blue colour fades because Cu2+(aq) is being replaced by Zn2+(aq) or Ni2+(aq) or [Cu2+] decreases (1) [7] (b) amount of copper = 225/63.5 = 3.54(3) mol (1) amount of electrons needed = 2 × 3.54 = 7.08/9 (7.087) mol (1) no. of coulombs = 20 × 10 × 60 × 60 = 7.2 × 105 C no. of moles of electrons = 7.2 × 105/9.65 × 104 = 7.46 mol (1) percentage “wasted” = 100 × (7.461 – 7.087)/7.461 = 5.01 (5.0)% (accept 4.98–5.10) (1) [4] (c) Eo data: Ni2+/Ni = –0.25V Fe2+/Fe = –0.44V (1) Because the Fe potential is more negative than the Ni potential, the iron will dissolve (1) [2] [Total: 13] 4 (a) (i) SnO2 (ii) PbO Can be read into equation (1) 2NaOH + SnO2 → Na2SnO3 + H2O (1) Can be read into equation (1) PbO + 2HCl → PbCl2 + H2O (1) [4] (b) moles of oxygen = 9.3/16 = 0.581 mol moles of lead = 90.7/207 = 0.438 mol (both 3 s.f.) (1) so formula is Pb3O4 (1) [2] (c) (i) Ksp = [Pb2+][Cl ]2 (1) units = mol3 dm 9 (1) (ii) if [Pb2+] = x, Ksp = 4x3, so x = 3√{Ksp/4} [Pb2+] = 3√{2 × 10 5/4} = 1.71 × 10–2 mol dm 3 (1) (iii) [Pb2+] = 2 × 10 5/(0.5)2 = 8.0 × 10–5 mol dm 3 (1) (iv) common ion effect, or increased [Cl ] forces solubility equilibrium over to the left (1) [Max 4] [Total: 10] © UCLES 2010 www.maxpapers.com Page 4 5 Mark Scheme: Teachers’ version GCE A LEVEL – October/November 2010 Syllabus 9701 Paper 42 (a) (i) ester (1) (ii) H is nitrobenzene – structure needed here (1) J is phenyldiazonium chloride – structure needed here (1) (iii) step 2 step 3 step 4 step 5 Sn/Zn + HCl / H2 + named cat / NaBH4 / LiAlH4 / Na + ethanol (1) HNO2/NaNO2 + HCl at T = 10°C or less (1) heat/warm to T > 10°C (1) CH3COCl / CH3COCOCOCH3 (1) [7] (b) (i) compounds that have the same molecular formula, but different structures (1) (ii) phenol (NOT hydroxy) (1) (methyl) ketone or carbonyl (1) (iii) K is 4-ethanoylphenol, HO-C6H4-COCH3 (must be 1,4- disubstituted isomer) (1) (iv) K I2 + NaOH(aq) NaOH(aq) Br2(aq) O CO2 ((1) for CO2 ; (1) for –O ) In any positions Br OH Br COCH3 (2 × Br needed) O COCH3 (anion needed) [4] [8 max 7] [Total: 14] © UCLES 2010 www.maxpapers.com Page 5 6 Mark Scheme: Teachers’ version GCE A LEVEL – October/November 2010 (a) Syllabus 9701 Paper 42 O * * (1) for each centre – more than 2 centres shown deduct 1 mark [2] LiAlH4 or NaBH4 or Na + ethanol or H2 + Ni (1) heat with Al2O3 / porous pot or conc. H2SO4 / H3PO4 (1) (b) (i) step 1 step 2 (ii) M (1) L (1) (letters may be reversed) [4] (c) (i) M (no mark) (ii) CO2H O P i.e. 3,7-dimethyl-6-oxo-octanoic acid (1) (iii) 2,4-DNPH (1) orange ppt. with P (none with N) (1) Mark ecf from candidates’ P (d) (+) H () Cl H + [3] H Cl Cl 2 curly arrows (1) carbocation intermediate + Cl (1) lone pair on Cl and last curly arrow (1) [3] [Total: 12] © UCLES 2010 www.maxpapers.com Page 6 7 Mark Scheme: Teachers’ version GCE A LEVEL – October/November 2010 Syllabus 9701 Paper 42 (a) (i) Disulfide bond / group / bridge (1) (ii) The tertiary structure (1) (iii) The substrate will no longer bond to / fit into the active site (1) or shape of active site is changed [3] (b) (i) Acid-base / proton donor / neutralisation / salt formation (1) (ii) The ability of the –CO2H group to form hydrogen bonds (1) and ionic interactions (1) The –CO2H/–CO2 group is no longer able to interact with –NH2/–NH3+ (1) The Ag+ forms a strong bond with –COO (1) [5] max [4] (c) (i) 8 but allow 4O2 if specified as molecules (1) (ii) Dative / co-ordinate (1) (iii) Octahedral / 6 co-ordinate (1) [3] [Total: 10] 8 (a) Protons (1) in NMR, energy is absorbed due to the two spin states (1) Electrons (1) in X-ray crystallography, X-rays are diffracted (by regions of high electron density) (1) [4] (b) (i) 1 – no mark The spectrum of alcohol / Y contains different peaks Alcohol / Y contains different chemical environments Spectrum 2 contains only one peak (1) (ii) Spectrum 2 only shows 1 peak so Z must be a ketone (1) Hence Y must be a 2° alcohol (1) Number of carbon atoms present 0.6 × 100 3 (1) 17.6 × 1.1 Thus Z must be CH3COCH3 (1) Hence Y must be propan-2-ol, CH3CH(OH)CH3 (1) (iii) H │ Y is CH3 – C – CH3 │ OH (1) (iv) All of the protons in Z are in the same chemical environment (1) [8] max [7] [Total: 11] © UCLES 2010 www.maxpapers.com Page 7 9 Mark Scheme: Teachers’ version GCE A LEVEL – October/November 2010 Syllabus 9701 Paper 42 (a) (i) A few nanometres (accept 0.5–10 nm) (1) (ii) Graphite/graphene (1) (iii) van der Waals’ (1) Carbon atoms in the nanotubes are joined by covalent bonds (1) (as are the hydrogen atoms in a hydrogen molecule) or no dipoles on C or H2 or the substances are non-polar [4] (b) More hydrogen can be packed into the same space/volume (1) [1] (c) If a system at equilibrium is disturbed, the equilibrium moves in the direction which tends to reduce the disturbance (owtte) (1) When H2 is removed the pressure drops and more H2 is released from that adsorbed (1) The equilibrium H2adsorbed H2gaseous (1) Equilibrium shifts to the right as pressure drops (1) [4] [Total: 9] © UCLES 2010 www.maxpapers.com UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2010 question paper for the guidance of teachers 9701 CHEMISTRY 9701/43 Paper 4 (A2 Structured Questions), maximum raw mark 100 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 must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2010 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. www.maxpapers.com Page 2 1 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 Paper 43 (a) (i) P2O5 + 3H2O → 2H3PO4 (or similar) or P4O10 + 6H2O → 4H3PO4 (1) SO2 + H2O → H2SO3 (1) (ii) 2NO2 + H2O → HNO2 + HNO3 (1) (iii) 2ClO2 + 2NaOH → NaClO2 + NaClO3 + H2O or ionic eqn (1) [4] (b) (i) 2CH4 + C2H6 + H2S + 9O2 → 4CO2 + SO2 + 8H2O Formulae (1), balanced (1) (ii) (The SO2 produced) causes acid rain (1) or consequence of acid rain – defoliation etc. – or respiratory problem (iii) 1000 dm3 contains 50 dm3 of H2S this is 50/24 (= 2.083 moles) (1) Mr(ethanolamine) = 24 + 7 + 14 + 16 = 61 therefore mass = 2.083 × 61 = 127(.1)g (1) (or ecf) (iv) acid-base (1) (v) ∆H = ∆Hf(rhs) – ∆Hf(lhs) = {(3 × 11 – 2 × 242)}{–}{(2 × –21 – 297)} –1 for each { } in which there is an error = –451 + 339 = –112 (kJ mol 1) (2) [8] [Total: 12] 2 (a) any three from: d-orbitals / sub-shells / energy levels are split or equivalent * (1) colour due to absorption of light (1) when e promoted to higher orbital * (1) ∆E = hf or hυ or h /λ (marks * could be in labelled diagram) (1) (b) blue is [Cu(H2O)6]2+ (or full correct name of ion) (1) ligand exchange/displacement/replacement (1) ((NH4)2CuCl4 contains) [CuCl4]2 (1) CuSO4 is white as it has no ligands (1) [3] [max 3] (c) n(thio) = 0.02 × 19.5/1000 = 3.9 × 10 4 mol (1) n(thio) = n(Cu2+), so n(Cu2+) in 50 cm3 = 3.9 × 10 4 mol so [Cu2+] = 3.9 × 10 4 × 1000/50 = (7.8 × 10–3 (mol dm 3)) (1) {or all-in-one-line: n(thio) = n(Cu2+), so [Cu2+] = 0.02 × 19.5/50 = (7.8 × 10–3 mol dm 3)} (2) in 100 cm3, there will be 7.8 × 10 4 mol, which is 63.5 × 7.8 × 10 4 = 0.049 – 0.050% (1) Allow ecf on 2nd and 3rd marks 0.5 gets 2 marks only [3] [Total: 9] © UCLES 2010 www.maxpapers.com Page 3 3 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 (a) reaction I: reduction or hydrogenation (1) reaction II: oxidation or redox (1) (b) thymol: or or menthol: or menthone: Br2(aq) (1) NaOH(aq) (1) FeCl3(aq) (1) Cr2O72 /H+ (1) Lucas test or ZnCl2/HCl (1) 2,4-DNPH/Brady’s reagent (1) Paper 43 [2] decolourises or white ppt (1) dissolves (1) violet/purple (colour) (1) orange → green (1) cloudy or white ppt (1) orange ppt (1) [6] [Total: 8] 4 reaction I: reaction II: reaction III: reaction IV: reaction V: reaction VI: reaction VII: Cl2 + light (1) (not aq) Br2 + Al Br3 or Fe or FeBr3 (1) (not aq) NaOH, heat in ethanol (1) (allow aqueous EtOH) HNO3 + H2SO4 (1) conc and < 60°C (1) (2 marks) KMnO4 + H+/OH + heat (1) Sn + HCl (1) HNO2 + HCl, < 10°C (1) X is N N OH (1) allow –N2— and –ONa [max 8] [Total: 8] © UCLES 2010 www.maxpapers.com Page 4 5 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 Paper 43 (a) (i) 2H2O – 4e → 4H+ + O2 (1) (ii) 2Cl – 2e → Cl2 (1) [2] (b) (i) Eo = (1.23 – (–0.83)) = 2.06V (1) (ii) Eo = (1.36 – (–0.83)) = 2.19V (1) (in (i) if (a)(i) as 4(OH ) – 4e → 2H2O + O2 ecf is 0.4 – (–0.83) = 1.23 (1) – needs working shown) [2] (c) (i) no change (because [H2O] does not change) (1) smaller/less positive (1) (ii) The (overall) Eo for Cl2 production will decrease, (whereas that) for O2 production will stay the same. (answer could be in terms of 1st Eo decreasing and becoming lower than 2nd)(or Eo for Cl2 becomes less than for O2) (1) [3] (d) (i) Cl + 3H2O → ClO3 + 3H2 (1) (ii) n(C) = 250 × 60 × 60 = (9 × 105 C) (1) n(e ) = 9 × 105/96500 = 9.33 mol n(NaClO3) = 9.33/6 = (1.55 mol) – allow ecf (1) Mr(NaClO3) = 106.5 mass (NaClO3) = 1.55 × 106.5 = 165.5 g (1) (165 – 166 gets 3 marks, 993 gets 2 marks as ecf) [4] [Total: 11] © UCLES 2010 www.maxpapers.com Page 5 6 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 Paper 43 (a) (i) Br2 (ignore solvent, but do not credit AlCl3 or HCl or light) (1) (ii) curly arrow from C=C to Br (1) another one breaking Br-Br bond. (1) correct intermediate cation and Br produced (not Brδ ) (1) [max 3] (b) B is NH2CH2CH2NH2 (1) C is NCCH2CH2CN (1) E is ClCOCH2CH2COCl (1) [3] (Allow (CH2)2 or C2H4. Allow correct atoms in any order on LHS but order must be correct on RHS) (c) reaction II: heat, dilute H+(aq) or HCl(aq) or HCl(conc) or H2SO4(aq) (1) reaction III: H2 + Ni (or other named catalyst) or LiAlH4 or Na in ethanol (1) [2] (d) NH4+ (1) [1] (e) (i) [-NHCH2CH2CH2CH2NH-COCH2CH2CO-] (1) (allow (CH2)4 and (CH2)2) (not dimer, needs bonds both ends) (ii) HCl (1) (f) (i) [H+] = 10 [2] pH = 10 2.6 = 2.51 × 10 3 (mol dm 3) (1) (ii) Ka = [H+]2/c = 6.31 × 10 5 (mol dm 3) (allow ecf from (i)) (1) [2] [Total: 13] 7 (a) NH2CH2CH2CH2NH2 + HCl → NH2CH2CH2CH2NH3+ Cl (1) NH2CH2CH2CH2NH3+ Cl + HCl → Cl NH3+CH2CH2CH2NH3+ Cl (1) (Deduct 1 only, if Cl omitted twice but allow with H+) (b) starts at 11.3 and finished as 1.6 (1) steep portions at 10 cm3 and 20 cm3 volume added (1) [2] [2] [Total: 4] © UCLES 2010 www.maxpapers.com Page 6 8 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 Paper 43 (a) (i) diagram to show tetrahedral arrangement (3D or bond angle marked) (1) (ii) 4 covalent bonds/bond pairs (with Cl) only or no lone pairs. (1) [2] (b) (i) steamy/white fumes/gas or heat evolved (1) (fumes are) HCl (from hydrolysis of Sn-Cl bonds) or exothermic reaction/bond breaking (1) (can award second mark for HCl (g) in eqn.) (ii) SnCl4 + 2H2O → SnO2 + 4HCl etc. (allow partial hydrolysis and with OHs) (1) [3] [Total: 5] 9 (a) Sugar/deoxyribose, phosphate, base (or better)(not ribose) (1) [1] (b) Diagram showing sugar-phosphate backbone (chain) (1) Bases on side-chain (1) Base paired – A-T or G-C (1) H-bonds shown and labelled (1) (c) mRNA, ribosome, tRNA [4] all three correct (2) (mRNA first allow 1 mark) [2] (d) (i) (4 × 4 × 4) = 64 (1) (ii) START (or Met) – ser – arg – leu – asp – val (2) (5 correct order score (1)) (iii) Amino acid leu is changed to pro (1) [4] [Total: 11] © UCLES 2010 www.maxpapers.com Page 7 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 Paper 43 10 (a) (i) Partition – substance is distributed between the stationary and mobile phase or has different solubility in each phase (1) Adsorption – substances form bonds of varying strength with or are attracted to or are held on to stationary phase. (1) (ii) Technique Separation method Paper chromatography Partition Thin-layer chromatography Adsorption Gas/liquid chromatography Partition 3 correct → (2) 2 correct → (1) (iii) %X = 44% (±2) %; %Y = 56% (±2%) (1) [5] (b) (i) They are largely composed of (carbon and) hydrogen which are active in the NMR (owtte) or protons/H+/H exist in different chemical environments (with characteristic absorptions) (1) (ii) 2 correct displayed formulae (1) In propanone all the protons are in a similar chemical environment (and hence there will be one proton peak.) (1) In propanal there are (three) different chemical environments and hence there will be (three) proton peaks or three different chemical environments or three proton peaks (1) [4] [Total: 9] © UCLES 2010 www.maxpapers.com Page 8 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Syllabus 9701 11 (a) Any two from: The drug can be localised in a part of the body (1) Smaller doses can be given reducing cost (1) Smaller doses can be given with fewer possible side effects (1) More immediate action / acts faster (1) Paper 43 [2] (b) (May circle whole functional group) Any 2 circles (2) [2] (c) (i) Must not react with the drug or must not breakdown too easily/quickly (1) (ii) The swelling/hydrolysis would begin in the stomach (and the drug would be released too soon) or stomach is acidic or has low pH (1) [2] (d) Addition, condensation (1) Suitable equation for addition (1) Suitable equation for condensation (1) (Addition equation must show polymeristion and balance – allow nX → X2n or Xn or Xn/2) (Condensation can be simple reaction e.g. to single ester or amide but must balance – 2 products) (If polymerisation RHS must show a repeat unit but can leave out other product – HCl etc.) [3] (e) Hydrolysis (1) [1] [Total: 11] © UCLES 2010 www.maxpapers.com UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2010 question paper for the guidance of teachers 9701 CHEMISTRY 9701/51 Paper 5 (Planning, Analysis and Evaluation), maximum raw mark 30 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 must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2010 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. www.maxpapers.com Page 2 Question 1 (a) (b) (c) Mark Scheme: Teachers’ version GCE A LEVEL – October/November 2010 Sections Syllabus 9701 Paper 51 Indicative material Mark Predicts a direct proportionality. Accept statements such as ‘no. of moles of the precipitate/PbCl2 will increase (as the number of moles of NaCl increases’). Equation shows a 1 to 2 molar ratio or wtte. (If a ‘plateau’ graph is described for the first mark allow a correctly explanation for the second mark i.e. all the lead nitrate has been used up) [1] PLAN Method & Problem & ACE All the lead nitrate was used up; moles or concentration of lead nitrate; total volume of lead nitrate. NOT amount A diagonal straight-line going through the origin. The line will abruptly change to a horizontal line Possible alternatives: • Diagonal line only, with +ve slope from the origin – 1 mark • Diagonal line not starting at the origin with a horizontal line – 1 mark • Curve from the origin with decreasing gradient and horizontal straight-line – 1 mark • Curve not from the origin with decreasing gradient and horizontal straight-line – 0 marks • Any lines showing an increase in gradient – 0 marks [1] PLAN Problem Independent variable – volume/mass/moles of NaCl Dependent variable – moles/mass of PbCl2 / ppt Other variables – temperature. NOT amount and NOT concentration of the NaCl. Three points correct – 2 marks Two points correct – 1 mark Any incorrect suggestions cancel correct suggestions PLAN Problem © UCLES 2010 [1] [1] [1] [2] www.maxpapers.com Page 3 Question (d) (i) Mark Scheme: Teachers’ version GCE A LEVEL – October/November 2010 Sections PLAN Method (ii) (e) (f) PLAN Method PLAN Method Syllabus 9701 Paper 51 Indicative material Calculating an appropriate mass of lead nitrate; this should be 8.275(8.3)g Dissolving the solid in water (or stirring) in a beaker or other appropriate vessel (volumetric/graduated/dilution flask) using less than 250 cm3 Adding to the volumetric/graduated/dilution flask and adding water to the 250 cm3 mark (if 250 cm3 of water are added directly to a volumetric/ graduated/dilution flask (containing the solid of course) allow 1 mark) Any dilution from a ‘given’ solution of lead nitrate gets 0 marks out of 3. Synthesis of the lead nitrate from lead and nitric acid scores zero. A student who uses 33.1 grams of lead nitrate to make up correctly 1 dm3 of solution can gain the first two marks, but to gain the third mark a measured 250 cm3 needs to be taken using an appropriate measuring vessel. Apparatus for volume measurement (burette/pipette/measuring cylinder)(not a syringe) used to measure 50 cm3 or less of lead nitrate and 100 cm3 or less of sodium chloride (mention of only one measuring vessel is enough for this mark) Method for drying the precipitate (adding propanone and allowing to evaporate/pressing with filter paper/warm oven/sun leaving out to dry. NOT heat or the use of a Bunsen or microwave. Table needs • volumes of lead nitrate, sodium chloride and mass/weight of lead chloride/ppt with appropriate data (volumes must total no more than 250 cm3 in each case) • moles of sodium chloride and lead chloride/ppt • correct units throughout (/cm3, /cm3 and /g [accept grams]; accept ( ) instead of / ) (these are for the items in bullet point one) Normally there will be 5 sets of volumes with the volume of lead nitrate being constant. (no figures are required for the mass of the ppt) (allow 4 sets if the volumes are different to those in (d)). Ignore; volumes of water; units in the mole columns; items such as filter paper; numbers in the mole columns. Do not allow; n for the number of moles; zero for the volume of lead nitrate or sodium chloride. The drying process should be repeated to constant mass Allow heat/reheat to constant mass/weight. Total Mark [1] [1] [1] [1] [1] [1] [1] [1] [1] [16] © UCLES 2010 www.maxpapers.com Page 4 Question 2 Mark Scheme: Teachers’ version GCE A LEVEL – October/November 2010 Sections Syllabus 9701 Paper 51 Indicative material Mark (a) ACE Data The required Mr is correct (180) (allow if given in the table.) Ignore units. [1] (b) ACE Data The required two column headings are correct. • Moles of glucose/solute; B/180; B/Mr; mass of glucose/Mr etc. IGNORE units. • Molality; D × 10; D/(A × 10 3) or equivalent; units mol kg 1 (allow Cm instead of molality) The calculations are correct and both columns are fully completed to 3 s.f. (allow ecf for the calculations, only if the calculation is absolutely clear) If only one column is given with correct header and numbers give one mark. If the molality column is given and it is fully correct give both marks (the formula in the heading needs to be something like (B × D × 10)/180 or D × 10 if the D column heading is correct as B/180 / or equivalent), if only the heading or the numbers are correct give one mark (even if column D is incorrect in some way, this means that a correct set of values for the molality, to 3sf could be worth a mark). If the first two values for the molality are 0.555 and 0.677 and the remainder are all correct award the mark. Table values at the end. [1] Give one mark for labelling the x-axis molality and the y-axis freezing point depression provided the plotted points and the origin if used cover at least half the scalings in both directions. Scales must allow all the points to be plotted; if less than nine points are plotted allow plotting for those actually plotted. (in other cases where correct scales are used all 9 points must be plotted, allow plotting to the nearest half a square in each direction) (If either of the axes have non-linear scales do not award the first two marks) Freezing point depression must have units /°C; ignore any molality units Give one mark for correctly plotting all the points plotted. Give one mark for drawing a ‘line of best fit’ passing through the origin. (again allow a half square accuracy) (If the origin is not used do not give this mark) These three points can stand alone as far as possible while allowing for ecf. Even if the axes are unlikely to access the mark, allow the other two marks. [1] (c) ACE Data © UCLES 2010 [1] [1] [1] www.maxpapers.com Page 5 Question (d) Mark Scheme: Teachers’ version GCE A LEVEL – October/November 2010 Sections ACE Data (e) (i) ACE data (ii) Syllabus 9701 Paper 51 Indicative material Mark One mark if the two anomalous points furthest from the line (one on each side) are circled. Allow only one anomaly if there is only one or all the anomalies are on the same side. In plotting the points, it is possible that some points will be a little way from the correctly drawn line. These in many cases are likely not to be ‘ringed’. If 5 or more points are ‘ringed’ do not award this mark but allow any subsequent correct discussion. For each of the two different anomalies an appropriate explanation gains one mark. If the graph is plotted correctly point 3 will be above the line and point 7 below the line. POINT 3; temperature measured after freezing was complete (i.e. late); too much glucose added; reduced amount of water. POINT 7; temperature recorded before freezing was complete; not all glucose dissolved; too much water. If two correct suggestions are given but not ‘tied’ to a particular point award one mark. If the comments are assigned to the wrong points NO marks. [1] Appropriately drawn lines on the graph with correctly deduced intercepts. For a correct calculation from the candidates figures with correct units (°C kg mol 1) Since the results produce a good linear graph, the procedure is OK. Accept ‘constant gradient’, and references such as ‘most of the points are close to the drawn line’ etc. ‘Line passing through the origin’ is irrelevant. [1] [2] [1] [1] (f) ACE Data The NaCl produces twice the number of particles. Allow the statement that NaCl produces Na+ and Cl . [1] (g) ACE Data Between answers to (a) and (f). [1] Total [14] © UCLES 2010 www.maxpapers.com Page 6 Mark Scheme: Teachers’ version GCE A LEVEL – October/November 2010 Syllabus 9701 Paper 51 F A B C D E mass of water /g mass of glucose /g 100 10.0 freezing point depression ∆Tf /°C 1.03 moles of glucose B 180 0.0556 molality D × 10 mol/kg 0.556 100 12.2 1.26 0.0678 0.678 100 18.0 2.09 0.100 1.00 100 23.3 2.40 0.129 1.29 100 27.7 2.86 0.154 1.54 100 30.9 3.22 0.172 1.72 100 33.1 3.31 0.184 1.84 100 38.6 3.98 0.214 2.14 100 42.3 4.37 0.235 2.35 © UCLES 2010 www.maxpapers.com UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2010 question paper for the guidance of teachers 9701 CHEMISTRY 9701/52 Paper 5 (Planning, Analysis and Evaluation), maximum raw mark 30 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 must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2010 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. www.maxpapers.com Page 2 Question 1 (a) (b) (c) Mark Scheme: Teachers’ version GCE A LEVEL – October/November 2010 Sections Syllabus 9701 Paper 52 Indicative material Mark Predicts a direct proportionality. Accept statements such as ‘no. of moles of the precipitate/PbCl2 will increase (as the number of moles of NaCl increases’). Equation shows a 1 to 2 molar ratio or wtte. (If a ‘plateau’ graph is described for the first mark allow a correctly explanation for the second mark i.e. all the lead nitrate has been used up) [1] PLAN Method & Problem & ACE All the lead nitrate was used up; moles or concentration of lead nitrate; total volume of lead nitrate. NOT amount A diagonal straight-line going through the origin. The line will abruptly change to a horizontal line Possible alternatives: • Diagonal line only, with +ve slope from the origin – 1 mark • Diagonal line not starting at the origin with a horizontal line – 1 mark • Curve from the origin with decreasing gradient and horizontal straight-line – 1 mark • Curve not from the origin with decreasing gradient and horizontal straight-line – 0 marks • Any lines showing an increase in gradient – 0 marks [1] PLAN Problem Independent variable – volume/mass/moles of NaCl Dependent variable – moles/mass of PbCl2 / ppt Other variables – temperature. NOT amount and NOT concentration of the NaCl. Three points correct – 2 marks Two points correct – 1 mark Any incorrect suggestions cancel correct suggestions PLAN Problem © UCLES 2010 [1] [1] [1] [2] www.maxpapers.com Page 3 Question (d) (i) Mark Scheme: Teachers’ version GCE A LEVEL – October/November 2010 Sections PLAN Method (ii) (e) (f) PLAN Method PLAN Method Syllabus 9701 Paper 52 Indicative material Calculating an appropriate mass of lead nitrate; this should be 8.275(8.3)g Dissolving the solid in water (or stirring) in a beaker or other appropriate vessel (volumetric/graduated/dilution flask) using less than 250 cm3 Adding to the volumetric/graduated/dilution flask and adding water to the 250 cm3 mark (if 250 cm3 of water are added directly to a volumetric/ graduated/dilution flask (containing the solid of course) allow 1 mark) Any dilution from a ‘given’ solution of lead nitrate gets 0 marks out of 3. Synthesis of the lead nitrate from lead and nitric acid scores zero. A student who uses 33.1 grams of lead nitrate to make up correctly 1 dm3 of solution can gain the first two marks, but to gain the third mark a measured 250 cm3 needs to be taken using an appropriate measuring vessel. Apparatus for volume measurement (burette/pipette/measuring cylinder)(not a syringe) used to measure 50 cm3 or less of lead nitrate and 100 cm3 or less of sodium chloride (mention of only one measuring vessel is enough for this mark) Method for drying the precipitate (adding propanone and allowing to evaporate/pressing with filter paper/warm oven/sun leaving out to dry. NOT heat or the use of a Bunsen or microwave. Table needs • volumes of lead nitrate, sodium chloride and mass/weight of lead chloride/ppt with appropriate data (volumes must total no more than 250 cm3 in each case) • moles of sodium chloride and lead chloride/ppt • correct units throughout (/cm3, /cm3 and /g [accept grams]; accept ( ) instead of / ) (these are for the items in bullet point one) Normally there will be 5 sets of volumes with the volume of lead nitrate being constant. (no figures are required for the mass of the ppt) (allow 4 sets if the volumes are different to those in (d)). Ignore; volumes of water; units in the mole columns; items such as filter paper; numbers in the mole columns. Do not allow; n for the number of moles; zero for the volume of lead nitrate or sodium chloride. The drying process should be repeated to constant mass Allow heat/reheat to constant mass/weight. Total Mark [1] [1] [1] [1] [1] [1] [1] [1] [1] [16] © UCLES 2010 www.maxpapers.com Page 4 Question 2 Mark Scheme: Teachers’ version GCE A LEVEL – October/November 2010 Sections Syllabus 9701 Paper 52 Indicative material Mark (a) ACE Data The required Mr is correct (180) (allow if given in the table.) Ignore units. [1] (b) ACE Data The required two column headings are correct. • Moles of glucose/solute; B/180; B/Mr; mass of glucose/Mr etc. IGNORE units. • Molality; D × 10; D/(A × 10 3) or equivalent; units mol kg 1 (allow Cm instead of molality) The calculations are correct and both columns are fully completed to 3 s.f. (allow ecf for the calculations, only if the calculation is absolutely clear) If only one column is given with correct header and numbers give one mark. If the molality column is given and it is fully correct give both marks (the formula in the heading needs to be something like (B × D × 10)/180 or D × 10 if the D column heading is correct as B/180 / or equivalent), if only the heading or the numbers are correct give one mark (even if column D is incorrect in some way, this means that a correct set of values for the molality, to 3sf could be worth a mark). If the first two values for the molality are 0.555 and 0.677 and the remainder are all correct award the mark. Table values at the end. [1] Give one mark for labelling the x-axis molality and the y-axis freezing point depression provided the plotted points and the origin if used cover at least half the scalings in both directions. Scales must allow all the points to be plotted; if less than nine points are plotted allow plotting for those actually plotted. (in other cases where correct scales are used all 9 points must be plotted, allow plotting to the nearest half a square in each direction) (If either of the axes have non-linear scales do not award the first two marks) Freezing point depression must have units /°C; ignore any molality units Give one mark for correctly plotting all the points plotted. Give one mark for drawing a ‘line of best fit’ passing through the origin. (again allow a half square accuracy) (If the origin is not used do not give this mark) These three points can stand alone as far as possible while allowing for ecf. Even if the axes are unlikely to access the mark, allow the other two marks. [1] (c) ACE Data © UCLES 2010 [1] [1] [1] www.maxpapers.com Page 5 Question (d) Mark Scheme: Teachers’ version GCE A LEVEL – October/November 2010 Sections ACE Data (e) (i) ACE data (ii) Syllabus 9701 Paper 52 Indicative material Mark One mark if the two anomalous points furthest from the line (one on each side) are circled. Allow only one anomaly if there is only one or all the anomalies are on the same side. In plotting the points, it is possible that some points will be a little way from the correctly drawn line. These in many cases are likely not to be ‘ringed’. If 5 or more points are ‘ringed’ do not award this mark but allow any subsequent correct discussion. For each of the two different anomalies an appropriate explanation gains one mark. If the graph is plotted correctly point 3 will be above the line and point 7 below the line. POINT 3; temperature measured after freezing was complete (i.e. late); too much glucose added; reduced amount of water. POINT 7; temperature recorded before freezing was complete; not all glucose dissolved; too much water. If two correct suggestions are given but not ‘tied’ to a particular point award one mark. If the comments are assigned to the wrong points NO marks. [1] Appropriately drawn lines on the graph with correctly deduced intercepts. For a correct calculation from the candidates figures with correct units (°C kg mol 1) Since the results produce a good linear graph, the procedure is OK. Accept ‘constant gradient’, and references such as ‘most of the points are close to the drawn line’ etc. ‘Line passing through the origin’ is irrelevant. [1] [2] [1] [1] (f) ACE Data The NaCl produces twice the number of particles. Allow the statement that NaCl produces Na+ and Cl . [1] (g) ACE Data Between answers to (a) and (f). [1] Total [14] © UCLES 2010 www.maxpapers.com Page 6 Mark Scheme: Teachers’ version GCE A LEVEL – October/November 2010 Syllabus 9701 Paper 52 F A B C D E mass of water /g mass of glucose /g 100 10.0 freezing point depression ∆Tf /°C 1.03 moles of glucose B 180 0.0556 molality D × 10 mol/kg 0.556 100 12.2 1.26 0.0678 0.678 100 18.0 2.09 0.100 1.00 100 23.3 2.40 0.129 1.29 100 27.7 2.86 0.154 1.54 100 30.9 3.22 0.172 1.72 100 33.1 3.31 0.184 1.84 100 38.6 3.98 0.214 2.14 100 42.3 4.37 0.235 2.35 © UCLES 2010 www.maxpapers.com UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2010 question paper for the guidance of teachers 9701 CHEMISTRY 9701/53 Paper 5 (Planning, Analysis and Evaluation), maximum raw mark 30 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 must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2010 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. www.maxpapers.com Page 2 Question 1 (a) (i) (ii) (b) (c)(i)(ii) (d) (i) Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Sections PLAN Problem Paper 53 Indicative material Mark Qualitative answer only for the first mark. Predicts any direct proportionality. E.g. Increasing concentration increases rate or doubling the molecules per unit volume (concentration) doubles the rate. Could gain this mark from next part. No ecf for opposite answer. [1] Increasing concentration/molecules per unit volume increases frequency of collisions. [1] PLAN Problem A diagonal straight-line of positive gradient going through the origin (+/– 1mm). No up or down curves or plateaux. If part (i) refers to some description of catalyst saturation, then accept a straight line horizontal plateau after the up straight line. [1] PLAN Problem Independent variable – concentration of H2O2. Do not allow volume/moles/amount. Has to be concentration. [1] Dependent variable – Time for the fixed volume of oxygen (or a stated volume of oxygen) OR time OR time taken. Not rate of reaction (negator) as that is a derived quantity. [1] Reaction vessel (conical flask) divided or with a small tube inside for the catalyst or any other internal separation device. Thistle or dropping funnels negate. Ignore heating [1] Syringe or equivalent apparatus (over water) connected to the reaction vessel in an airtight format. [1] The volume of the gas collecting apparatus shown and is not exceeding 1 dm3. [1] Completes the table. Correct units required for each column. 0 cm3 of H2O2 not allowed. Total volumes of solutions need not be constant. [1] If reaction vessel mark awarded, give mark for shaking to react (catalyst and solution present). If a thistle or dropping funnel used then mark is for adding the liquid reagent to the catalyst. In other situations give mark for adding solid catalyst to the solution (only) and closing the vessel. [1] Starting the reaction and a stopwatch simultaneously. Accept “start the clock” only if it is unambiguously related to the reaction start. [1] Recording the time taken to produce a chosen/fixed volume of gas (15 cm3 for example). [1] Starting the reaction and the stopwatch simultaneously is difficult. Accept any reaction starting process in conjunction with starting the clock. Accept closing the apparatus and starting the clock. Do not accept loss of oxygen whilst closing the bung (neutral). [1] PLAN Method PLAN Method (ii) (e) Syllabus 9701 PLAN Method © UCLES 2010 www.maxpapers.com Page 3 Question (f) Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Sections PLAN method Syllabus 9701 Paper 53 Indicative material Mark There are 6 marking points which are, Three columns, concentration of hydrogen peroxide, time and rate (or 1/time). Ignore other columns Each column needs a correct unit in correct format i.e. /mol dm–3, /s, s–1 or the use of brackets (s). Accept seconds, minutes (not sec, min or m), or not M or molarity. Two marks for 5 or 6 correct points. One mark for 3 or 4 correct points. No marks for 2 or less correct points. Total [2] [15] © UCLES 2010 www.maxpapers.com Page 4 2 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Question Sections (a) ACE Data Syllabus 9701 Paper 53 Indicative material Mark One of the two column headings correct in heading, unit and expression. [1] The calculations are correct in both columns (first two and last two in each) and both columns are fully completed (to 3SF). (One mark for each column). [2] If an expression is not given and all the data is totally correct then the last 2 marks are available. ECF data from incorrect expressions provided correctly calculated (and provided some attempt at a titration calculation is made). For incorrect expressions check calculate test data. Then the last 2 marks are available. If an expression is not given and all the data has been calculated correctly except it has not been divided by 2, then 1 mark is available. (b) ACE Data Give one mark for unambiguously labelling and scaling the x-axis and the y-axis provided the plotted points cover at least half the scalings in both directions. Plot may be either way round. Headings could be names or D or E [1] Give one mark for correctly plotting the first, last and anomalous points and those that deviate significantly from the line (+/– ½ square except where a grid line is involved). All 10 data points must be plotted on the grid. ECF plots of incorrect data. [1] [1] Give one mark for drawing a ‘line of best fit’ which must pass through the origin (+/– ½). (c) ACE Data The anomalies must be ringed and normally must include the two points furthest away from the drawn line on each side of the line (ignore other anomalies). If all the anomalies are on one side of the line – ring the furthest away (also ignore other anomalies). Accept only one anomaly if that is the situation where there is only one anomaly (the candidate may not have ringed all the anomalies). This mark negated if more than 5 anomalies. For each of the two different anomalies an appropriate explanation gains one mark. Explanations must be related to the particular point and include the nature of the deviation. Award 1 mark for two correct explanations not properly linked to a point. © UCLES 2010 [1] [2] www.maxpapers.com Page 5 Mark Scheme: Teachers’ version GCE A/AS LEVEL – October/November 2010 Question Sections (d) (i) ACE data (ii) Syllabus 9701 Paper 53 Indicative material Mark For appropriately drawn lines on the graph with correctly deduced intercepts (+/– ½ square except where a grid line is involved) give one mark. [1] Correctly calculates the value of the gradient. This should be in the order of 16.3/0.061. ECF incorrect intercepts. [1] Yes, Since the results produce a good linear/straight line graph, the procedure is OK. Normally a “no” answer is not acceptable. Do not accept an unjustified “yes” answer. [1] (e) ACE Data Any facilitation that takes the (succinic) acid into an aqueous phase will suffice e.g. to ensure that all the reactants are mixed in the aqueous layer / so the reactants are in solution in water for the neutralisation / makes the titration work because the acid is in the aqueous layer / extract or mix the acid into the aqueous phase to react / produce H+(aq) to react with the alkali. Answers must have the acid in an aqueous phase. [1] (f) ACE Data Low (titre) values. Thus % errors are high (consequential). Percentage error is required. [1] [1] Total [15] © UCLES 2010