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Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level

CHEMISTRY 9701/11 Paper 1 Multiple Choice May/June 2017

MARK SCHEME

Maximum Mark: 40

Published

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2017 series for most Cambridge IGCSE®, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

9701/11 Cambridge International AS/A Level – Mark Scheme PUBLISHED

May/June 2017

© UCLES 2017 of 3

Question Answer Marks

1 C 1

2 D 1

3 A 1

4 C 1

5 B 1

6 A 1

7 D 1

8 C 1

9 B 1

10 C 1

11 B 1

12 A 1

13 B 1

14 A 1

15 C 1

16 C 1

17 A 1

18 C 1

19 A 1

20 B 1

21 D 1

22 B 1

23 B 1

24 C 1

25 D 1

26 C 1

27 A 1

28 C 1


May/June 2017

© UCLES 2017 of 3


29 A 1

30 B 1

31 B 1

32 D 1

33 D 1

34 A 1

35 D 1

36 A 1

37 B 1

38 D 1

39 B 1

40 A 1






MARK SCHEME

Maximum Mark: 40

Published



May/June 2017

© UCLES 2017 Page 2 of 3


1 B 1

2 B 1

3 C 1

4 D 1

5 A 1

6 D 1

7 C 1

8 C 1

9 A 1

10 A 1

11 C 1

12 B 1

13 B 1

14 C 1

15 D 1

16 C 1

17 D 1

18 D 1

19 B 1

20 D 1

21 D 1

22 C 1

23 A 1

24 D 1

25 D 1

26 B 1

27 D 1

28 A 1


May/June 2017



29 D 1

30 C 1

31 B 1

32 C 1

33 C 1

34 A 1

35 B 1

36 A 1

37 B 1

38 B 1

39 A 1

40 B 1






MARK SCHEME

Maximum Mark: 40

Published



May/June 2017



1 A 1

2 D 1

3 B 1

4 B 1

5 D 1

6 B 1

7 A 1

8 B 1

9 A 1

10 B 1

11 B 1

12 C 1

13 C 1

14 B 1

15 D 1

16 A 1

17 D 1

18 B 1

19 C 1

20 C 1

21 C 1

22 D 1

23 A 1

24 C 1

25 D 1

26 D 1

27 B 1

28 A 1


May/June 2017



29 A 1

30 C 1

31 C 1

32 B 1

33 A 1

34 B 1

35 A 1

36 A 1

37 A 1

38 D 1

39 C 1

40 B 1





CHEMISTRY 9701/21 Paper 2 AS Level Structured Questions May/June 2017

MARK SCHEME

Maximum Mark: 60

Published

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2017 series for most Cambridge IGCSE®, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.


May/June 2017



1(a) The mass of a molecule OR the (weighted) average / (weighted) mean mass of the molecules

1

Relative / compared to 112

(the mass) of an atom of carbon–12

OR on a scale in which a carbon–12 atom / isotope has a mass of (exactly) 12 (units)

1

1(b)(i) 3 1

1(b)(ii) 8 1

1(b)(iii) C3H8O + 4½O2 → 3CO2 + 4H2O 1

1(b)(iv)

AND propan–2–ol / 2–propanol

1

AND propan–1–ol / 1–propanol 1

Alternative answers (any two):

AND butan–1–ol / 1–butanol

AND butan–2–ol / 2–butanol

AND (2–)methylpropan–1–ol / (2–)methyl–1–propanol

AND (2–)methylpropan–2–ol / (2–)methyl–2–propanol


May/June 2017



1(b)(v) correct conversions of data to SI/consistent units p = 100 000 ; V = 20 × 10–6 ; T = 393

1

calculation of n ( = pVIRT) from M1 values 3 -6100×10 ×20×10=

8.31×393n

1

calculation of mass m ( = n × Mr) AND answer correct to 3sf m = 6.12 × 10–4 × 60 = 0.0367 (g) Alternative answer for using C4H10O: m = 6.12 × 10–4 × 74 = 0.0453 (g)

1

Total: 10


2(a) substance type of bonding type of lattice structure

copper metallic giant/metallic

ice covalent OR hydrogen(-bonding) / H(–bonding)

hydrogen-bonded / simple / molecular

silicon(IV) oxide covalent giant (molecular) / macromolecular

iodine covalent simple / molecular

sodium chloride ionic giant / ionic

1

1

1

1

1

2(b)(i) hydrogen bonding 1


May/June 2017



2(b)(ii) H-bond between O and H of different molecules 1

minimum three partial charges (in a row) over two H2O molecules, i.e.:

either δ–O––Hδ+ - - - - δ–O or Hδ+ - - - - δ–O––Hδ+

1

lone pair of electrons on O of H-bond, in line with H-bond 1

2(c)(i) X = liquid AND Z = solid 1

Y = liquid and solid OR ‘liquid / solid’ OR ‘liquid OR solid’ 1

2(c)(ii) (kinetic) energy reducing 1

motion slowing owtte 1

2(c)(iii) energy given out / released forming bonds / forming bonds exothermic 1

compensates for / counteracts heat loss / cooling owtte 1

Total: 15


May/June 2017



3(a)(i) A 1

3(a)(ii) H 1

3(a)(iii) G 1

3(a)(iv) B 1

3(a)(v) F 1

3(b)(i) (strong) heating 1

(to provide / overcome) high activation energy 1

3(b)(ii) white flame / white light / white smoke / white solid 1

3(b)(iii) →2 2 2Mg(s) + 2H O(l) Mg(OH) (s) + H (g) 2

3(c)(i) 2Mg(NO3)2 → 2MgO + 4NO2 + O2 1

3(c)(ii) →3 2CaCO CaO + CO 1

→2 2CaO + H O Ca(OH) 1

3(d)(i) reduce acidity in soil / increase pH of soil 1

(both) basic / base(s) 1

3(d)(ii) →+ 2+3 2 2CaCO + 2H Ca + CO + H O

OR →+ 2+

3 2 3CaCO + 2H Ca +H CO

1

Total: 16

970

© U

01/21

CLES 2017

Question

4(a)(i)

4(a)(ii)

4(a)(iii)

4(b)(i)

4(b)(ii)

4(b)(iii)

4(b)(iv)

4(c)(i)

(molecules

different str

sp2 ove

sp3 ove

sp2 = 116° –

sp3 = 106° –

(electrophil

bromine de

HOCH2CHB

CO2 / carbo

P = propana

Q = propan

Ca

/ isomers with) t

ructural / display

rlap of (2)s with

rlap of (2)s with

– 124°

– 112°

ic) addition

ecolourises / turn

BrCH2Br OR

on dioxide

al

one

ambridge Interna

the same molec

yed formulae / ar

two (2)p (atomi

all three (2)p (a

ns colourless / fa

ational AS/A LevPUBLISHED

Page 6 of 7

Answer

cular formula / sa

rangement of bo

c) orbitals

atomic) orbitals

des (from orang

vel – Mark Sche

r

ame number of a

onds

ge / brown)

me

atoms of each ellement

May/June 2

Marks

1

1

1

1

1

1

1

1

1

1

1

1

1

2017

970

© U

01/21

CLES 2017

Question

4(c)(ii)

4(d)(i)

4(d)(ii)

tr(i)iodomet

(molecules

Any two of: chir non diffe diffe

curly arrow

correct dipo

correct inte

Ca

thane / CHI3 /

/ isomers with) t

ral centre / C attan-super(im)posaberent spatial / 3Derent rotation of

w from lone pair o

ole on carbonyl

ermediate, includ

ambridge Interna

/

the same (molec

ached to four difble mirror image

D arrangement oplane-polarised

on :C≡N to C(δ+)

δ+C=Oδ– AND cu

ding C–O– AND

ational AS/A LevPUBLISHED

Page 7 of 7

Answer

cular and) struct

fferent groups / aes of atoms (owtte) light

urly arrow from b

curly arrow from

vel – Mark Sche

r

tural formula

atoms

bond to O(δ–)

m lone pair to H+

me

To

May/June 2

Marks

1

1

1

1

1

1

otal: 19

2017






MARK SCHEME

Maximum Mark: 60

Published



May/June 2017



1(a)

atomic number nucleon number number of electrons number of protons number of

neutrons symbol

6 3 3

58 326Fe +

2

1 1

1(b)(i) EITHER mass of an atom / isotope relative / compared to 1/12 (the mass) of (an atom of) C-12 OR on a scale in which a C-12 (atom / isotope) has (a mass of exactly) 12 (units) OR mass of one mol (of atoms) of an isotope relative / compared to 1/12 (the mass) of 1 mol of C-12 OR on a scale in which one mol C-12 (atom / isotope) has a mass of (exactly) 12 g

2 1 1

1(b)(ii) (10.0129 19.78) (80.22x) 10.8100

× +=

1

x = 10.9941 1

Total: 6


May/June 2017



2(a) strong triple bond 1

non-polar / no dipole 1

2(b)(i) Any 2 points covered correctly scores 2 marks Any 1 point covered correctly scores 1 mark • nitrogen (and oxygen) from the air / atmosphere (react): • high temperature (of internal combustion engine) / (engine) produces enough OR a lot of heat (energy) : • (so) breaks (strong) bond(s) in nitrogen (and oxygen) :

2

2(b)(ii) reduction / decomposition of NOx using a catalyst / catalytic convertor 1

2NO2 + 4CO → 4CO2 + N2 OR 2NO + 2CO → 2CO2 + N2

1

2(b)(iii) (acts as a homogeneous) catalyst OR oxidising agent 1

SO2 + NO2 → SO3 + NO 1

NO + ½O2 → NO2 OR SO3 + H2O → H2SO4 1

2(b)(iv) 2NO2 + H2O → HNO2 + HNO3 OR 4NO2 + 2H2O + O2 → 4HNO3

1

2(c) fertiliser / nitrates dissolve in (river water) OR fertiliser / nitrates are washed / leached out / flows into (river water)

1


May/June 2017



algal bloom / promote algal growth / explosion of plant growth AND EITHER sunlight is blocked out (preventing photosynthesis) / plants can no longer carry out photosynthesis (and die) OR bacteria break down or decay dead organisms / plants / algae

1

drop in oxygen (concentration) 1

Total: 13


May/June 2017



3(a) (+) 103 1

3(b)(i) general shape of the curve and peak are displaced to right of original and starts at origin 1

the peak is lower and curve crosses once only finishing above original 1

3(b)(ii) rate increases AND correct explanation in terms of ‘more collisions’ 1

at higher T area above Ea is greater / more molecules with E ⩾ Ea 1

higher frequency of successful collisions OR more successful collisions per unit time / higher chance of successful collisions per unit time / higher proportion of successful collisions per unit time

1

3(b)(iii) increases (%) decomposition (of HBr) 1

(increasing T) shifts equilibrium to the right / in the forward direction / endothermic direction / towards H2 + Br2 1

to oppose the change or oppose the increase in temperature OR to absorb (additional) energy / heat OR to decrease the temperature

1

3(b)(iv) H-I bond strength less than H-Br OR less energy needed to break H-I ora

1

I (atom) is big(ger) (than Br) OR I (atom) has more shielding (than Br) ora

1

Br (atom) has greater (%) orbital / outer shell overlap OR attraction (of nucleus in iodine) for shared (pair of) electrons is weak(er) OR attraction (of nucleus in iodine) for bonding pair (or electrons) is weak(er) ora

1


May/June 2017



3(c)(i) H2 = 0.015 (mol) 1

HCl = 0.27 (mol) 1

3(c)(ii) HCl = 9/10 AND xH2 = 1/20 AND Cl2 = 1/20 OR HCl = 0.9(0) AND H2 = 0.05 AND Cl2 = 0.05

1

3(d)(i) (Kp =) 2 2

2H C

HCp p

p× l

l

1

3(d)(ii) equal number of moles (of gas) on either side (of equation) / (total) pressure cancels 1

3(d)(iii) 4.649 × 10–3 1

Total: 18


May/June 2017



4(a)(i) (A = )

1

4(a)(ii) (A / straight chain) has strong(er) (temporary dipole-) induced dipole (attractions) ora 1

(because A / straight chain has) bigger (surface) area / more (points of) contact (in unbranched isomer) ora OR (so) more energy required to break the intermolecular forces ora

1

4(a)(iii) CH3CHCHCH3 OR CH3CH=CHCH3 1

4(a)(iv) No rotation / restricted / limited rotation of C=C / (carbon) double bond 1

One (of the two) methyl groups / one (of the two) H (atoms) is on each C (of C=C) 1

4(a)(v)

arrow from the C=C double bond drawn to the bromine

1

dipole on Br2 in correct orientation AND arrow from the Br-Br bond to the Brδ– 1

correct carbocation / bromonium ion from the structure with C=C drawn 1

Br– with lone pair, negative charge AND arrow from lone pair to the carbon atom of intermediate OR using both arrows shown (in alternative diagram)

1

4(a)(vi) electrons in pi bond induce it (the dipole) OR (high) electron density in pi bond / double bond / C=C repels electrons (away from nearest Br) OR polarised by (high) electron density in pi bond / double bond / C=C

1


May/June 2017



4(b)(i) C = (2-)methylpropan-2-ol / (CH3)3COH / any unambiguous structure 1

D = (2-)methylpropan-1-ol / (CH3)2CHCH2OH / any unambiguous structure 1

E = (2-)methylpropanoic acid /(CH3)2CHCO2H / any unambiguous structure 1

4(b)(ii) 2C4H8O2 + Na2CO3 → 2C4H7O2Na + H2O + CO2 1

4(c)(i) triiodomethane 1

4(c)(ii) F = CH3CH2CH2COCH3 1

G = C2H5CH(CH3)CHO 1

4(c)(iii) a (tetrahedral) atom with four different groups / atoms / substituents attached OR a carbon (atom) with four different groups / atoms / substituents attached

1

4(d)(i) H C=O (group / bond) AND O–H (group / bond) 1

I C=O (group / bond) AND C–H (group / bond) 1


May/June 2017



4(d)(ii) H = ethanoic acid 1

I = methyl methanoate 1

Total: 23






MARK SCHEME

Maximum Mark: 60

Published



May/June 2017



1(a) (molecules / isomers with) the same molecular formula / same number of atoms of each element 1

different structural / displayed formulae / different arrangement of bonds 1

1(b)(i) 4 1

1(b)(ii) 6 1

1(b)(iii) molecular = C4H8 1

empirical = CH2 using alternative supplied data molecular = C6H12 empirical = CH2

1


May/June 2017



1(b)(iv)

1

alternative using supplied data: any two

1

1(b)(v) correct conversions of data to SI / consistent units P = 100 000; V = 25 × 10–6; T = 310

1

calculation of n ( = pV / RT)

−× × ×

×

3 6100 10 25 10n =8.31 310

1

calculation of mass m ( = n × Mr) AND answer correct to 3sf m = 9.705 × 10–4 × 56 = 0.0543 (g)

1

Alternative answer for using C6H12: m = 9.705 × 10−4 × 84 = 0.0815 (g)

Total: 11


May/June 2017



2(a)(i) halogen colour state

chlorine yellow / green gas

bromine red / brown / orange liquid

iodine grey / black solid

2

2(a)(ii) increasing number of electrons 1

(gives) increasing strength of van der Waals’ / id-id forces / London / dispersion forces 1

2(b) oxidising power decreases down the group. ora 1

ability to accept electrons decreases (down the group) ora 1

because (outer shell experiences) more shielding OR increased distance from nucleus (to outer shell) (outweighs the increasing nuclear charge down the group) ora

1

2(c)(i) solid sodium chloride: steamy / misty / white fumes 1

solid sodium iodide: purple fumes 1

2(c)(ii) (conc sulfuric) not powerful enough oxidising agent (to oxidise chloride) OR chloride not powerful enough reducing agent (to reduce sulfuric acid)

1

iodide reduces sulfuric acid OR iodide / I– is oxidised OR sulfuric acid oxidises iodide

1


May/June 2017



2(c)(iii) 2NaBr + 2H2SO4 Br2 + SO2 + Na2SO4 + 2H2O OR NaBr + H2SO4 → NaHSO4 + HBr AND 2HBr + H2SO4 → Br2 + SO2 + 2H2O OR 2NaBr + H2SO4 → Na2SO4 + 2HBr AND 2HBr + H2SO4→ Br2 + SO2 + 2H2O

2

2(d)(i) AgI (and AgCl solid) / silver ions reacting with iodide ions 1

2(d)(ii) AgCl (precipitate) dissolves (in ammonia) owtte 1

Total: 15


3(a)(i) (enthalpy / energy change) when one mole of a compound is formed 1

from its elements in their standard states / standard conditions 1

3(a)(ii) (∆Hr = ∑∆Hf products – Σ∆Hf reactants) –196 =2∆Hf SO3 – (2 × −296.8) 2∆Hf SO3 = –196 + (2 × −296.8) = –789.6

1

∆Hf SO3 = –394.8 (kJ mol–1) 1

3(b)(i) Mark to right of original Ea 1


May/June 2017



3(b)(ii) 2 marks for any two points: • Benefit of using a catalyst in terms of increasing rate or economic benefit i.e. (less heat required) • Creates alternative pathway with lower Ea • More molecules with E > Ea

2

3(b)(iii) (rate) increases AND correct explanation in terms of ‘more collisions’ 1

more successful collisions per unit time / higher chance of successful collisions per unit time / higher proportion of successful collisions per unit time

1

(yield) increases and shifts equilibrium to the right / in the forward direction / towards SO3 / towards the product / in exothermic direction

1

to oppose the change or oppose the increase in pressure / fewer molecules on RHS so eqm moves to right (to oppose change)

1

3(c)(i) SO2 = 0.01 (mol) AND SO3 = 0.99 (mol)

1

3(c)(ii) nTOT = 1.505 1

pO2 = 1.50 × 105 × (0.505 / 1.505) = 5.03 × 104 (Pa) 1

3(d)(i) ( )pK =

×

23

22 2

pSOpO pSO

1

3(d)(ii) 0.1946737305 1

Pa–1 1

Total: 17


May/June 2017



4(a) cracking 1

4(b) In any order CH2=CHCH2CH3 / CH2CHCH2CH3 / CH2CHC2H5 AND CH3CH=CHCH3 / CH3CHCHCH3 AND (CH3)2C=CH2 / (CH3)2CCH2

1

4(c)(i) (different) molecules with the same (molecular and) structural formula 1

(due to) different arrangement in space caused by C=C / double bond 1

4(c)(ii)

CH3 C C CH3

H H

Br

Hδ+

δ−

CH3 C+ C CH3

H H

H

Br-

arrow from the C=C double bond drawn to the H

1

dipole on H–Br in correct orientation AND arrow from the H-Br bond to the Brδ– 1

correct carbocation from the structure with C=C drawn 1

Br – with lone pair, negative charge AND arrow from lone pair to the positively charged carbon atom of intermediate 1


May/June 2017



4(d)(i) a (tetrahedral) atom with four different groups / atoms / substituents attached OR a carbon (atom) with four different groups / atoms / substituents attached

1

4(d)(ii) but–1–ene 1

4(d)(iii)

C CH3

CH2

Br

CH3

H

CCH3

CH2

Br

CH3

H One 3D structure of 2–bromobutane which must have 2 bonds shown the same and two different, i.e. three bond types altogether, e.g. two solid lines, one wedge and one dash. If two bonds are drawn in the plane of the paper, i.e. single solid lines, they must not be at 180 degrees to each other.

1

Second structure either mirror of first OR all bonds drawn the same with position of two groups swapped. 1

4(d)(iv) intermediate / (secondary carbo) cation from X is more stable ora OR charge density of C+ (of the intermediate of X) is reduced

1

(due to) electron-releasing character / (positive) inductive effect of alkyl groups / / due to electron releasing alkyl group

1

4(e)(i) (2–)methylpropene / (2–)methylprop–1–ene 1

4(e)(ii)

CC

CC

BrH

H

H

HH

H

H

HH

C

C

C C

Br

H

H

H

HH

H H

HH 2

Total: 17





CHEMISTRY 9701/31 Paper 3 Advanced Practical Skills 1 May/June 2017

MARK SCHEME

Maximum Mark: 40

Published



May/June 2017



1(a) I: All the following data is recorded • rough titration: both burette readings and the titre • initial and final burette readings for two (or more) accurate titrations

Headings and units are not required for this mark

1

II: Titre values recorded for accurate titrations, and Appropriate headings and units in the accurate titration table

• initial / start (burette) reading / volume • final / end (burette) reading / volume • titre or volume used / added (not “difference”) • unit: / cm3 or (cm3) or in cm3 (for each heading)

or cm3 unit given for each volume recorded

1

III: All accurate burette readings are recorded to the nearest 0.05 cm3. The requirement to record to 0.05 applies to burette readings, including 0.00 cm3 (if this was the initial reading), but it does not apply to the titre. 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

IV: Final uncorrected titre is within 0.10 cm3 of any previous uncorrected accurate titre. 1


May/June 2017



Examiner rounds any accurate burette readings to the nearest 0.05 cm3, check subtractions and then select the “best” titres using the hierarchy: • identical titres then • accurate titres within 0.05 cm3, then • accurate titres within 0.10 cm3, etc.

These best titres should be used to calculate the mean titre, expressed to nearest 0.01 cm3. Examiner compares candidate’s tire value with that of the Supervisor.

Award V, VI and VII if δ ⩽ 0.30 (cm3) 1

Award V and VI if 0.30 < δ ⩽ 0.50 1

Award V, only, if 0.50 < δ ⩽ 0.80 1

1(b) Candidate calculates the mean correctly.

• Candidate must take the average of two (or more) titres that are within a total spread of not more than 0.20 cm3. • Working / explanation must be shown or ticks must be put next to the two (or more) accurate readings selected. • The mean should be quoted to 2 dp, and be rounded to nearest 0.01 cm3. • (e.g. 26.667 cm3 must be rounded to 26.67 cm3)

1


May/June 2017



1(b) Two special cases, where the mean need not be to 2 dp: • Allow mean expressed to 3 dp only for 0.025 or 0.075 (e.g. 26.325 cm3) • Allow mean if expressed to 1 dp, if all accurate burette readings were given to 1 dp and the mean is exactly

correct. • (e.g. 26.0 and 26.2 = 26.1 is allowed) • (e.g. 26.0 and 26.1 = 26.1 is wrong – should be 26.05)

Do not award this mark if:

• The rough titre was used to calculate the mean. • The candidate did only one accurate titration. • Burette readings were incorrectly subtracted to obtain any of the accurate titre values. • All burette readings used to calculate the mean were recorded as integers

Note: the candidate’s mean will sometimes be marked correct even if it was different from the mean calculated by the Examiner for the purpose of assessing accuracy.

1(c)(i) No of moles of thiosulfate used = 0.110 × mean titre / 1000 (expressed to 3 or 4 sig fig) 1

1(c)(ii) + (iii) Equation balanced I2 + 2Na2S2O3 → Na2S4O6 + 2NaI and no of moles of I2 = 0.5 × ans. in (i)

1

1(c)(iv) Correct answer, No of moles of copper(II) ions = 2 × answer (iii) (expressed to 3 or 4 sig fig)

1

1(c)(v) Mr = 26.0 / ans (iv) × 25 / 1000 1

Total: 12


May/June 2017



2(a) I: Table of data, to include: • Unit “covering” all weighings, or given for each weighing • No repeat headings (i.e. not two lists of weighings) • Appropriate headings for the three weighings: Mass of crucible and lid Mass of crucible, lid and FA 5 (or “contents before heating”) Mass of crucible, lid and residue / CuO / contents after heating

1

II: Weighings recorded • Six weighings recorded in the space provided. • All weighings recorded to same number of decimal places (one or more) • Label/heading to indicate which is Expt 1 and Expt 2

1

III: Both masses of FA 5 and residue, correctly subtracted • Masses of FA 5 used recorded on page 4, correctly subtracted • Masses of FA 5 used were between 2.5 – 3.0 g and 1.5 – 2.0 g • Masses of residue recorded on page 4, correctly subtracted

1

For assessment of accuracy, examiner must check and correct (if necessary) the masses of FA 5 used and of CuO obtained by the supervisor and by the candidate for Experiment 1.

• Examiner works out the ratio mass of FA5 / mass of CuO for the supervisor (2 dp) • Examiner works out the ratio (mass FA 5: mass CuO) for the candidate (2 dp) • Examiner calculates δ the difference between these two ratios.

Award IV and V if δ ⩽ 0.08 Award IV if 0.08 < δ ⩽ 0.15

2

VI: Observations made during heating Solid goes black / black residue (formed) or reference to blue/green flame

1

2(b)(i) • No of moles CuO = mass of residue / 79.5

• Answer must be correct and expressed to 3 or 4 sig fig 1


May/June 2017



2(b)(ii) • No of moles of FA 5 = answer (i) / 2 • Mr = mass of FA 5 used / no of moles of FA 5

1

2(b)(iii) Mr = mass of FA5 used in Expt 2 × 79.5 × 2/mass of residue (CuO) 1

2(b)(iv) Mr of FA 5 calculated from Ar values = 239 1

2(b)(v) Candidate should • correctly calculate the 2.5% of Mr in (iv) = 5.98 / 6.0, and • make a correct statement about the accuracy of the accepted formula, based on their result(s).

or correctly calculate % difference for their result(s) from Mr in (iv) and correct comment

1

2(c)(i) • heat (crucible and residue) to constant mass • heat more gently for longer period • cool in a desiccator

1

• to ensure that decomposition (of FA 5) is complete or to ensure that all the residue is CuO • to prevent escape of dust / smoke / solid (during heating)

1

2(c)(ii) Larger masses have lower percentage error in weighing 1

Total: 14


May/June 2017



FA 6 is Cu(NO3)2; FA 7 is FeCl3

3(a)(i) • melts or dissolves or blue liquid / solution formed • condensation or steam / vapour produced • black residue / solid • brown gas / fumes • gas / oxygen relights a glowing spill

4 or 5 observations correct = 2 marks 2 or 3 observations correct = 1 mark

2

3(a)(ii) FA 6 is Cu(NO3)2 1

3(b)(i) • with KI, FA 7 gives a brown / red-brown / red / orange solution • with starch, blue / blue-black / dark colour

1

• with FA 6, blue precipitate (formed) • on heating, (blue precipitate) turns black • With FA 7, red-brown / brown / rust ppt. (formed )

1

• With FA 6, no reaction / no change / no ppt. • With FA 7, white precipitate formed

1

• With FA 6, (pale) blue precipitate, then • deep/dark blue (solution) with excess • With FA 7, red-brown / brown / rust precipitate (forms)

1

Mg test Both observations correct With FA 6, brown / black precipitate / solid formed or blue colour fades / disappears With FA 7, fizzing / bubbling / effervescence

1

Test for hydrogen: (gas) “pops” with lighted splint 1


May/June 2017



3(b)(ii) FA 7 is acidic, because it fizzes / produces hydrogen with magnesium 1

3(b)(iii) Fe3+(aq) + 3OH–(aq) → Fe(OH)3(s) 1

3(b)(iv) Redox because iodine was produced (from iodide ions) 1

3(b)(v) You can’t be certain about the colour of the precipitate (with AgNO3) due to the coloured solution / colour of FA 7. or You can’t be sure whether the precipitate with AgNO3 is white / AgCl or cream / AgBr

1

3(b)(vi) Ammonia would react with the Fe3+ ions in FA 7 (masking the effect of ammonia on AgCl) or The cation in FA 7 gives a precipitate with ammonia (so the precipitate of AgCl would not appear to dissolve).

1

Total: 14





CHEMISTRY 9701/32 Paper 3 (Advanced Practical Skills 2) May/June 2017

MARK SCHEME

Maximum Mark: 40

Published



May/June 2017



1(a) I Initial and final burette readings and volume added recorded for rough titre and accurate titre details tabulated. [minimum 2 × 2 ‘boxes’ with relevant information]

1

II Initial and final burette readings recorded and volume of FB 2 added recorded for each accurate titration. Headings and units correct for accurate titrations Headings: initial / final (burette) reading / volume or reading / volume at start / finish and volume / FB 2 added/used or titre and Units: (cm3) or / cm3 or in cm3 [or cm3 by every entry]

1

III All accurate burette readings are recorded to the nearest 0.05 cm3 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

IV The final accurate titre recorded is within 0.10 cm3 of any other accurate titre. 1


May/June 2017



1(a) For assessment of accuracy (Q) marks, each Examiner should round any accurate burette readings to the nearest 0.05 cm3, check subtractions and then select the “best” titres for supervisor and candidate using the hierarchy: two identical; titres within 0.05 cm3; titres within 0.1 cm3; etc. These best titres should be used to calculate the mean titre, expressed to the nearest 0.01 cm3. The candidate’s titre is compared to the supervisor’s titre and δ calculated.

V, VI and VII Award V, VI and VII for δ ⩽ 0.20 cm3 Award V and VI for 0.20 cm3 < δ ⩽ 0.30 cm3 Award V for 0.30 cm3 < δ ⩽ 0.50 cm3

3

1(b) Check mean titre is correctly calculated from clearly selected values (ticks or working). • Candidate must average two (or more) titres where the total spread is ⩽ 0.20 cm3. • Working must be shown or ticks must be put next to the two (or more) accurate readings selected. • The mean should normally be quoted to 2 dp rounded to the nearest 0.01. [e.g. 26.667 must be rounded to 26.67] Two special cases where the mean may not be to 2 dp: allow mean to 3 dp only for 0.025 or 0.075 e.g. 26.325; allow mean to 1 dp if all accurate burette readings were given to 1 dp and the mean is exactly correct. [e.g. 26.0 and 26.2 = 26.1 is correct but 26.0 and 26.1 = 26.1 is incorrect.] Do not award this mark if: • the rough titre was used to calculate the mean; • candidate carried out only 1 accurate titration; • burette readings were incorrectly subtracted to obtain any of the accurate titre values; • all burette readings (resulting in titre values used in calculation of mean) are integers. Note: the candidate’s mean will sometimes be marked as correct even if it is different from the mean calculated by the examiner for the purpose of assessing accuracy.

1

1(c)(i) Correctly calculates 0.100 x 25

1000 = 2.5(0) × 10–3

1


May/June 2017



1(c)(ii) Correctly calculates

( )0.0025 x 1000

b to 3 or 4 sf

1

1(c)(iii) Correct expression 12.6 ÷ (ii) 1

1(c)(iv) Anion in FB 1 = CHCl2COO– (allow ecf: for candidate’s answer to (iii)) CH3COO– : ⩽ 77 CH2ClCOO– : 77.5 – 111.5 CHCl2COO– : 112 – 146 CCl3COO– : ⩾ 146.5

1

1(d)(i) Conc NaOH lower = > titre smaller = > smaller Mr 1

1(d)(ii) No effect on identification unless closer to smaller mass acid or (different Mr may lead to the) identification of a different acid with matching / close to Mr

1

Total: 14


2(a) I Unambiguous headings and correct units tabulated for all 6 thermometer readings, mean temps, and ∆Ts 1

II All thermometer readings recorded to 0.5 °C and ∆Ts correctly calculated and Mean temperatures correctly calculated to nearest .5 °C or to 1 or 2 dp

1

Award III if candidate ∆Ts within 1.5 °C 1

Award III and IV if candidate ∆Ts within 1.0 °C 1

2(b)(i) Correctly calculates n acid = 0.05(0) mol and n NaOH = 0.045 mol

1


May/June 2017



2(b)(ii) Correctly calculates 50 × 4.2 × ∆T1 to minimum 2 sf

1

2(b)(iii) Correct expression ( )

1000 x 0.045ii

1

2(b)(iv) +(v)

2 × mol NaOH in (i) or 0.09(0) in (iv) and

Correctly uses ( )( )∆100 x 4.2 x T 2

1000 x iv in (v)

1

Negative signs shown in (iii) and (v) and final answers to 2–4 sf in (ii), (iii) & (v) 1

2(c)(i) % error in vol of FB 3 = 0.5 x 100

50 = 1.(0)%

1

% error in vol of FB 4 = 2 x 0.25 x 100

25 = 2.(0)%

1

2(c)(ii) Use a burette / pipette for volume measurements / instead of a measuring cylinder or Add a lid to reduce heat loss (by convection) / to reduce convection or Use thermometer reading to 0.2 °C / smaller divisions / calibrations / more sensitive

1

Total 12


May/June 2017



FB 5 is CH3COOH; FB 6 is HCl; FB 7 is HNO3; FB 8 is CuSO4(aq); FB 9 is Na2edta

3(a)(i) Selects Na2CO3 / Mg 1

Effervescence / bubbling / fizzing greater / faster with FB 6 1

FB 5 is the weak acid (ora) with some evidence 1

3(a)(ii) no reaction / no ppt / no change with Ag+ and ‘not needed’ with Ba2+ (do not allow ‘no change’ unless there is no evidence of ammonia in 2nd test) Effervescence alone is not evidence so would expect ‘no change’.

1

Effervescence / gas / NH3 and turns (damp red) litmus blue 1

FB 7 is nitric acid from some evidence (can be effervescence in (ii)) 1


May/June 2017



3(b) (i) – (vi)

See below 6

Expected observations

test observation mark

(i) + Na2CO3

(pale) blue ppt (Allow blue-green / green-blue / turquoise / cyan) 1

(ii) + KI then

Yellow-brown / brown (not orange) 1

Na2S2O3 white / off-white ppt and soluble in excess 1

(iii) + c.HCl and

(blue) (solution) turns green (shade greener)

(iv) + H2O (green) (solution) turns (pale) blue (shade bluer)

1

(v) + NH3 (in excess) forming dark/deep blue solution or solution much darker than (iv)

1

(vi) + edta (solution) more blue / darker blue than (iv) 1


May/June 2017



3(vii) FB 8 contains Cu2+/copper(II) 1

3(viii) Cu2+(aq) + CO32–(aq) → CuCO3(s)

Allow 2Cu2+(aq) + 4I–(aq) → 2CuI(s) + I2(aq or s) Allow Cu2+(aq) + 2OH–(aq) → 2Cu(OH)2(s)

1

Total: 14


May/June 2017


Mark allocation

Skill Minimum mark

allocation

Breakdown of marks Question1

Question2

Question3

Total mark

Statement Minimum Marks

Manipulation, measurement and observation (MMO)

12 marks [17]

Successful collection of data and observations C 8 1 9 10

Quality of measurements and observations Q 2 3 2 5

Decisions relating to measurements of observations

De 2 1 1 2

Presentation of data and observations (PDO)

6 marks [7]

Recording data or observations R 2 1 1 2

Display of calculation and reasoning Di 2 1 2 3

Data layout L 2 1 1 2

Analysis, conclusions and evaluation (ACE)

10 marks [16]

Interpretation of data or observations and identifying sources of error

I 4 3 5 1 9

Drawing conclusions Con 5 3 3 6

Suggesting improvements Imp 1 1 1

Total 14 12 14 40






MARK SCHEME

Maximum Mark: 40

Published



May/June 2017



1(a) I Correct headings The following data are recorded in the space provided • mass of container with FA 2 • mass of (empty) container • mass of FA 2 ‘Mass’ must be stated for each piece of data. Unit / g (etc.) must be given for each piece of data. Subtraction for mass of FA 2 used must be correct.

1

II All the following data are recorded • two burette readings and titre for the rough titration • initial and final burette readings for two (or more) accurate titrations

1

III Titre values recorded for accurate titrations, and Appropriate headings and units in the accurate titration table • initial / start (burette) reading / volume • final / end (burette) reading / volume • titre or volume / FA 1 and used / added • unit: / cm3 or (cm3) or in cm3 (for each heading)

or cm3 unit given for each volume recorded

1

IV All accurate burette readings are recorded to the nearest 0.05 cm3. The requirement to record to 0.05 applies to burette readings, including 0.00 cm3 (if this was the initial reading), but it does not apply to the titre. This mark is not awarded 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

V The final accurate titre recorded is within 0.10 cm3 of any other accurate titre. • Do not include a reading if it is labelled “rough”. • Do not award the mark if any ‘accurate’ burette readings (apart from initial 0 cm3) are given to zero dp.

1


May/June 2017



For assessment of accuracy (Q) marks, each Examiner should round any burette readings to the nearest 0.05 cm3, check subtractions and then select the “best” titres using the hierarchy: • two (or more) accurate identical titres (ignoring any that are labelled “rough”), then • two (or more) accurate titres within 0.05 cm3, then • two (or more) accurate titres within 0.10 cm3, etc. These best titres should be used to calculate the mean titre, expressed to nearest 0.01 cm3. Calculate the candidate’s ratio to 1 dp, as shown below. Ratio = correct mean titre ÷ correct mass Calculate the difference (δ) between the candidate’s ratio and the supervisor’s ratio. Accuracy marks are awarded as follows.

1(a) Award VI, VII and VIII if δ ⩽ 0.2 (cm3 g–1) 1

Award VI and VII if 0.2 < δ ⩽ 0.4 1

Award VI, only, if 0.4 < δ ⩽ 0.6 1

• Spread penalty: if the two “best” (corrected) titres used by the Examiner were ⩾ 0.50 cm3 apart, maximum 2 accuracy marks.

• If only a rough titration is shown, award Q marks based on that, maximum 2 accuracy marks.


May/June 2017



1(b) Candidate calculates the mean correctly. • Candidate must take the average of two (or more) titres that are within a total spread of not more than 0.20 cm3. • Working / explanation must be shown or ticks must be put next to the two (or more) accurate readings selected. • The mean should be quoted to 2 dp, and be rounded to nearest 0.01 cm3. (e.g. 26.665 cm3 must be rounded to 26.67 cm3) Two special cases, where the mean need not be to 2 dp: • Allow mean expressed to 3 dp only for 0.025 or 0.075 (e.g. 26.325 cm3) • Allow mean if expressed to 1 dp, if all accurate burette readings (apart from initial 0) were given to 1 dp and the

mean is exactly correct. (e.g. 26.0 and 26.2 = 26.1 is allowed) (e.g. 26.0 and 26.1 = 26.1 is wrong – should be 26.05) This mark is not awarded if: • The rough titre was used to calculate the mean. • The candidate did only one accurate titration. • Burette readings were incorrectly subtracted to obtain any of the accurate titre values. • All burette readings used to calculate the mean were recorded as integers Note: the candidate’s mean will sometimes be marked correct even if it was different from the mean calculated by the Examiner for the purpose of assessing accuracy.

1

1(c)(i) No of moles of H2SO4 used = 0.05(0) × (b) / 1000 to minimum 2 sf 1

1(c)(ii) and

1(c)(iii)

2NaHCO3 + H2SO4 Na2SO4 + 2CO2 + 2H2O and No of moles of NaHCO3 = 2 × answer (i)

1


May/June 2017



1(c)(iv) Mass of NaHCO3 = answer (iii) × 10 × 84 1

1(c)(v) % = answer (iv) / mass of FA 2 used × 100 1

All answers attempted in (i), (iii), (iv) & (v) are shown to 3 or 4 sf Minimum 3 answers attempted to gain the mark

1

1(c)(vi) Any one of the following answers. • the impurity does not react with (sulfuric) acid / FA 1 / NaHCO3 • the impurity is not alkaline / acidic • the impurity is neutral

1

1(c)(vii) % error (= 0.1 / 250 × 100) = 0.04% 1

Total: 16


May/June 2017



2(a) I Four weighings recorded and correct headings given and mass of FA 4 used and mass of residue recorded • (Mass of) crucible, (lid) • (Mass of) crucible, (lid) and FA 4 (or 'contents before heating) • (Mass of) crucible, (lid) and contents / residue / FA 4 after (first) heating • (Mass of) crucible, (lid) and contents / residue / FA 4 after re-heating • (Mass of) FA 4 • (Mass of) residue / FA 5 / contents after heating If ‘mass’ not written then ‘g’ must be with each entry. Use of lid must be consistent.

1

II • All weighings recorded to same decimal places (one or more). • Third and fourth weighings are within 0.05 g of each other (or both equal if a one decimal place balance was used) • Mass of FA 4 and FA 5 / residue must be correctly subtracted.

1


May/June 2017



2(a) III and IV: • For assessment of accuracy, examiner must check and correct (if necessary) the masses of FA 4 used and of

residue (smaller mass) obtained by the supervisor and by the candidate. • Work out ratio mass of FA4/mass of residue for the supervisor (2 dp) • Work out ratio mass of FA4/mass of residue for candidate (2 dp) • Calculate the difference (δ) between these two ratios. Award III and IV if δ ≤ 0.05 Award III if 0.05 < δ ≤ 0.10

2

2(b)(i) and

2(b)(ii)

(i) Mass NaHCO3 = (% purity from 1(c)(v) / 100) × mass of FA 4 used and (ii) Mass impurity = mass of FA 4 – answer (i) or mass impurity = % impurity / 100 x mass FA 4

1

2(b)(iii) Mass of decomposition solid = mass of residue (FA 5) from table – mass of impurity (ii) and expressed to 2, 3 or 4 sig fig or mass of decomposition solid = mass of NaHCO3 – mass lost on heating [(i) – (mass FA 4 – mass FA 5)]

1

2(b)(iv) Mass of residue obtained = answer (iii) × 84 / answer (i) 1


May/June 2017



2(b)(v) If correct, (84 g) NaHCO3 would give 40 g residue / NaOH (owtte) or mole ratio 1:1.3 (so not 1:1) or Answers could refer to mass / moles of CO2

1

2(c)(i) Lid reduces / stops absorption of water (vapour) by solid / residue / FA 5 while cooling 1

2(c)(ii) Repeat the experiment and ignore anomalous results / to obtain concordant / consistent results or cool in a desiccator or use larger mass of FA 4 / contents / solid

1

2(d)(i) Any two observations required • fizzing / effervescence / bubbling • gas turns limewater milky / chalky / cloudy white / white ppt • solid dissolves / colourless solution forms • rapid/brisk effervescence = 2 observations

1

2(d)(ii) FA 5 contains carbonate ion / CO32–

and reference to fizzing (with acid) or to CO2 liberated (with acid) or positive limewater test or correct equation 1

2(d)(iii) 2NaHCO3(s) → H2O(g) + CO2(g) + Na2CO3(s) 1

2(d)(iv) (From equation) 84 g NaHCO3 should give 0.5 × 106 g residue (= 53 g) and gives a (sensible) comment based on student’s 52.3 g

1

Total: 14


May/June 2017



FA 6 is MnCl 2; FA 7 is Al 2(SO4)3

3(a)(i) Ba2+ test: all observations correct • FA 6 – no change / no reaction / no ppt / solution stays colourless with both • FA 7 – white precipitate with Ba2+ and • white ppt (remains) / insoluble / no reaction with HNO3

1

AgNO3 test: both observations correct • FA 6 – white precipitate • FA 7 – no change / no reaction / solution stays colourless / no ppt

1

Na2CO3 test: both observations correct • FA 6 – no reaction / solid does not dissolve / no effervescence • FA 7 – fizzing / bubbling / effervescence / or gas / CO2 turns limewater milky / chalky / cloudy white / (forms) white ppt

1

3(a)(ii) FA 7 has lower pH and gas / CO2 given off / it fizzes (more rapidly if fizzing with both) with sodium carbonate

1


May/June 2017



3(b) Reagents: NaOH and NH3 (names or correct formulae) 1

Observations – (3 × 1 mark) • FA 6 + NaOH : off-white / buff / beige / light brown ppt • FA 6 + NH3 : off-white / buff / beige / light brown ppt

1

• FA 6 : both ppts insoluble in excess and darken / turn brown with either 1

• FA 7 + NaOH : white ppt and soluble in excess • FA 7 + NH3 : white ppt and insoluble in excess

1

3(c) Conclusions (one mark for each). • FA 6 is MnCl 2 • FA 7 is Al 2(SO4)3

2

Total: 10






MARK SCHEME

Maximum Mark: 40

Published



May/June 2017



1(a) I Mass (of Mg) with correctly displayed unit and all temperatures recorded Initial T must be between 10–45 °C

1

II All temperature readings to .5 ºC with at least one ending in .0 °C and at least one ending in .5 °C 1

Round any thermometer readings to the nearest .5 °C Calculate ∆T from T at 2 minutes to T max from the table. Compare with supervisor ∆T. Award III if ∆T within 2 °C of supervisor Award III and IV if ∆T within 1 ºC of supervisor

2

1(b) I Axes labelled (T on y-axis & t on x-axis). Scale chosen so that plotted points (and 10 °C extra on y-axis) occupy more than half the available space in both directions.

1

II Points plotted to within half a small square. Points that should be on lines must be on the line and points that should not be on lines must not be on lines.

1

III Two lines of best fit drawn – one up to 2 minutes and the other after the reaction has occurred. 1

IV Both lines extrapolated to 2½ minutes and vertical line drawn at 2½ minutes 1

V Examiner to calculate ∆T from candidate graph and award mark if within 0.5 °C of candidate’s ∆T 1

1(c)(i) Correctly calculates energy evolved = 25 × 4.2 × ∆T and answer to 2 – 4 sf. 1

1(c)(ii) Correct use of moles of magnesium = mass Mg from (a)/24.3 1

Correct use of ∆H =

( )( )n Mg x 1000

i and answer must be negative

1

1(d) 2 masses, 4 thermometer readings and 2 temperature rises with correct units and unambiguous headings shown 1

Examiner to calculate ∆T longer piece/ ∆T shorter piece to 2 dp Award 2 marks if 1.80 to 2.20 Award 1 mark if 1.70 to 2.30

2


May/June 2017



1(e) • correct (larger) ∆T from thermometer readings and correct (larger) mass (from balance readings) • correct expression of 25 × 4.2 × ∆T • correct expression for division by number of moles of Mg • answer with negative sign and evidence of division by 1000 and answer to 2 – 4 sf ∆H = – 25 × 4.2 × ∆T × 24.3 ÷ [m(Mg) × 1000] 3 points correct = 1 mark 4 points correct = 2 marks

2

1(f)(i) Either yes because the reaction is faster so less heat is lost or no because a catalyst does not alter ∆H / ∆T

1

1(f)(ii) No effect because the acid is in excess / magnesium is the limiting reagent / all the Mg reacts or ∆T would be larger because the reaction is faster as acid is diprotic (owtte) so less heat lost

1


May/June 2017



1(g)(i) Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g) Chemical symbols = 1 mark Correct balancing and state symbols = 1 mark

2

1(g)(ii) Answer = + 1.9 = 2 marks Answer = – 1.9 / 1.9 / + 3.8 = 1 mark Some working must be shown to score both marks

2

1(h)(i) and

1(h)(ii)

(i) & (ii) together Allow any two correct statements • a stronger acid or correct identification provides a greater concentration of H+ / more hydrogen ions (ora) • (some) energy required to break O–H bond (allow OH bond) • –I effect/increased electronegativity of Cl increases strength of (trichloroethanoic) acid / makes it easier to release

H+ (compared to ethanoic acid)

2

Total: 25


May/June 2017



FB 5 is HCl; FB 6 is H2SO4; FB 7 is HNO3; FB 8 is KI(aq)+Na2CO3(aq)

2(a)(i) AgNO3 observations correct 1

Ba(NO3)2 observations correct 1

Na2CO3 observations correct 1

Gas / CO2 / fizz turned limewater milky / chalky / cloudy white / formed white ppt with limewater in at least one box 1

Test FB 5 FB 6 FB 7

AgNO3 White ppt No reaction / no change / no ppt

No reaction / no change / no ppt

NH3 (ppt) soluble No reaction / no change / no ppt

No reaction / no change / no ppt

(not ‘no observation’ or ‘–‘)

Ba(NO3)2 No reaction / no change / no ppt

White ppt No reaction / no change / no ppt

HNO3 No reaction / no change / no ppt

(ppt) insoluble No reaction / no change / no ppt

Na2CO3 Effervescence / fizz / bubbles Effervescence / fizz / bubbles Effervescence / fizz / bubbles

Positive limewater test – see above


May/June 2017



2(a)(ii) H+ / hydrogen ion 1

2(a)(iii) Adds named reactive metal (or symbol) (Mg or Zn, allow Al, Fe) / named suitable acid-base indicator 1

Effervescence / fizz / bubbles / gas / H2 pops with lighted splint / correct final colour (chosen indicator must change colour in the pH range < 7)

1

2(a)(iv) FB 5 Cl – FB 6 SO42– FB 7 unknown

Allow names of ions 3 correct scores 2 2 correct scores 1

2

2(a)(v) Test: Name / correct formula of strong acid (and warm) or (acidified) potassium manganate(VII) / KMnO4 No (brown) gas or not decolourised Conclusion: FB 7 is NO3

– / nitrate

1

2(b) see expected observations table 4

Ions present I– and CO32–

1

Total: 15


May/June 2017


Expected observations

Test Observation

HCl Fizz / etc. or gas / CO2 turns limewater milky / etc. and

H2O2 Brown / yellow (darker yellow if yellow with HCl) / red-brown / orange-brown / yellow-brown (solution) and

Starch Blue-black / black / dark blue (not purple) colour [1]

NaOH No reaction / no ppt / solution remains colourless [1]

CuSO4 Blue/green/brown range of coloured ppt and

HCl Brown colour [1]

Na2S2O3 White / cream / off-white / pale grey and solid / residue / ppt [1]






MARK SCHEME

Maximum Mark: 40

Published



May/June 2017



1(a) I Constructs a table for results showing volume of FA 1, volume of water, reaction time, reaction rate for all experiments carried out

1

II Appropriate headings and units for recorded data given. Volumes in cm3 or / cm3 or (cm3). Time in seconds or / s or (s) All volumes except zero given to .00.

1

III All times recorded to the nearest second. 1

IV 3 additional volumes chosen intervals not less than 2.00 cm3 and all volumes of FA 1 ⩾ 6.00 cm3 and one volume of FA 1 ⩽ 8.00 cm3

1

V In all 3 additional experiments water is added to make a total of 20.(00) cm3 1

VI + VII Compare time for 20.00 cm3 of FA 1 with that of supervisor. 2 marks for ± 3 s 1 mark for ± 5 s

2

VIII Compare ratio of time for 10.00 cm3 of FA 1 / time for 20.00 cm3 of FA 1. 1 mark for ratio between 1.8 – 2.2

1

IX All rates correctly calculated using 500 / time (minimum 2 sf and 1 dp)

1

X Units for rate given as s–1 1


May/June 2017



1(b) I Rate on y-axis and volume on x-axis. Axes clearly labelled and suitable linear scales. 1

II Scale chosen to use more than half of each axis for origin and plotted points 1

III All points plotted correctly to within half a square and in the correct square. 1

IV Draws a line of best fit. This may be a straight line or a smooth curve with anomalous points indicated. 1

1(c) Rate is (directly) proportional to concentration of peroxodisulfate or comment suitable to shape of graph 1

1(d)(i) Reads rate from graph correct to one small square and shows use of this number in calculation 1

Shows use of 500 / rate 1

1(d)(ii) Correctly calculates (0.5 / time for expt 1) × 100 to 2 or more sf 1

1(d)(iii) The student is correct as the reaction time would be longer and so the (percentage) error reduced. 1

1(d)(iv) There is so much thiosulfate that all the iodide reacts so there is no iodine to turn the starch blue-black. 1


May/June 2017



1(e)(i) Record time to nearest second with units of s 1

Candidate’s time compared with that from Expt 1. 1 mark for ± 3 s

1

1(e)(ii) Estimates a time as 4 × ans (i) 1

Time / rate related to concentration of S2O32– / FA 3

Increased concentration of FA 3 increases time of reaction / time longer / decreases rate of reaction / rate lower / smaller / reaction slower.

1

Total: 24


May/June 2017



FA 4 is (NH4)2Fe(SO4)2 FA 5 is KAl(SO4)2 FA 6 is Na2SO3 FA 7 is H2SO4 FA 8 is NaNO2

2(a)(i)

test observation

mark FA 4 FA 5

+ NaOH green ppt white ppt 1

insoluble in excess soluble in excess 1

then warm

gas / ammonia turns (damp red) litmus blue

no reaction / litmus stays red

1

+ NH3 green ppt white ppt 1

and turning brown (in air) in either alkali test

insoluble in excess insoluble in excess 1

5


May/June 2017



2(a)(ii) FA 4 contains NH4 + and Fe 2+

FA 5 contains Al 3+ 2 marks for all three correct 1 mark for any two correct

2

2(b) Selects BaCl2(aq) or Ba(NO3)2(aq) followed by appropriate acid (acid must be named) OR Selects acidified potassium manganate(VII) OR Selects named acid and tests gas with acidified potassium manganate(VII)

1

White ppt that is soluble in acid OR Decolourises (potassium manganate(VII))

1

SO3 2‒ 1

2(c)(i) + Mg Effervescence / fizzing / bubbles 1

Gas / H2 / fizz pops with a lighted splint 1

+ FA 8 Brown (yellow / orange) fumes or gas turns blue litmus red/bleached or blue solution 1

2(c)(ii) H2SO4 1

NaNO2 1

2(c)(iii) Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g) 1

Total: 16





CHEMISTRY 9701/41 Paper 4 A Level Structured Questions May/June 2017

MARK SCHEME

Maximum Mark: 100

Published



May/June 2017



1(a) solubility increases down the group 1

∆Hlatt and ∆Hhyd both decrease or ∆Hlatt and ∆Hhyd both become less exothermic / more endothermic

1

∆Hlatt decreases / changes more (than ∆Hhyd as OH– being smaller than M2+) 1

∆Hsol becomes more exothermic / more negative / less endothermic / less positive 1

1(b)(i) ∆Hr1 – (538 + 2x230 + 394) = –(1216 + 286) ∆Hr1 – 1392 = –1502

1

∆Hr1 = –110 1

1(b)(ii) let ∆Hf(HCO3–(aq)) = y

2y – 538 = –1216 – 394 – 286 – 26

1

y = –692 1

1(b)(iii) ∆Hr3 –538 – 2(230 + 394) = –538 – 2(692) ∆Hr3 = –136

1

1(b)(iv) ∆Hr3 will be identical to ∆Hr4, / unchanged 1

as the reaction is the same, or: 2OH–(aq) + 2CO2(g) → 2HCO3

–(aq) or metal ions stay in solution/metal ions are unchanged / are spectators

1


May/June 2017



1(c) more gaseous moles are being consumed (in reaction 3) or more CO2 moles are being consumed (in reaction 3)

1

∆S is therefore expected to be more negative/less positive for reaction 3. 1

Total: 13


2(a)(i)

H O C N H N OC1 + 1

16 electrons on each diagram 1

2(a)(ii) HNC = 115–125° AND NCO = 180° 1

2(a)(iii) cyanic acid, because it’s a stronger / higher bond enthalpy / triple / C≡N / more electrons involved bond 1

2(b)(i) [H+] = √([HNCO]Ka) = √(0.1 × 1.2 × 10–4) or 3.46 × 10–3 1

pH = log [H+] = 2.5 (2.46) 1

2(b)(ii) Na2CO3 + 2(NH2)2CO → 2NaNCO + CO2 + 2NH3 + H2O 1

2(c)(i) (n(OH–) at start = (2 × 0.1 × 30) / 1000 = 6 × 10–3 mol) (n(OH–) reacted = (0.1 × 20) / 1000 = 2 × 10–3 mol) n(OH–) remaining = (6–2) × 10–3 = 4 × 10–3 mol, (in 50 cm3)

1

so [OH–]end = (4 × 10–3 × 1000) / 50 = 0.08 mol dm–3 1


May/June 2017



2(c)(ii) [H+] = Kw / [OH–] = (1 × 10–14) / 0.08 = 1.25 × 10–13 mol dm–3 1

so pH = –log(1.25 × 10–13) = 12.9 1

2(c)(iii) curve starts at 2.46 / 2.5 1

vertical portion (end point) at vol added = 10.0 cm3 1

finishes at pH = 12.9 1

2(d)(i) monodentate: (a species that) forms one dative / coordinate bond 1

ligand: a species that uses a lone pair of electrons to form a dative / coordinate bond to a metal atom / metal ion 1

2(d)(ii) [Ag(NCO)2]– or [Ag(OCN)2]– correct formula 1

correct charge 1

2(e)(i) n(BaCO3) =1.66 / 197.3 = 8.4(1) × 10–3 mol 1

2(e)(ii) n(RNCO) = 8.41 × 10–3 mol, so Mr = 1 / (8.41 × 10–3) = 119 1

2(e)(iii) molecular formula = C7H5NO 1


May/June 2017



2(e)(iv)

1

Total: 23


3(a)(i) +3 or Co3+ 1

3(a)(ii) oxidation 1

ligand displacement / replacement / exchange / substitution 1


May/June 2017



3(a)(iii)

ClCo

ClH3N

H3N

NH3

NH3

NH3

CoNH3H3N

H3N

Cl

Cl

cis trans

NH3

CoClH3N

Cl

NH3

NH3

orCl

CoNH3H3N

H3N

NH3

Cl

or

1 + 1

geometrical or cis-trans 1

3(b)(i) The number of bonds / atoms bonded to an atom / ion / species / metal 1

3(b)(ii) C 6 [Cr(CN)6] – D – [Ni(NH2CH2CH2NH2)3] 2+/+2 E 4 [PtCl4] – F 6 – 3–/–3

6

3(c)(i) Kstab(1) = [FeSCN2+]/([Fe3+][SCN–]) mol–1 dm3 Kstab(2) = [FeCl4–]/([Fe3+][Cl –]4) mol–4 dm12

3

3(c)(ii) Keq(3) = Kstab(1) / Kstab(2) 1

3(c)(iii) Keq(3) = 1750 1

mol3 dm–9 1

Total: 19


May/June 2017



4(a)(i) optical, because it contains a / one chiral C-atom or chiral C-atoms or chiral atom / centre or C* indicated or C with 4 different groups

1

4(a)(ii) C10H14O + 3H2 → C10H20O correct formulae 1

balancing 1

4(b)(i) electrophilic substitution 1

4(b)(ii) step 3 reduction 1

step 5 substitution / hydrolysis 1

4(b)(iii) step 1 (CH3)2CHCl + Al Cl3 / Al Br3 / FeCl3 / FeBr3 1 + 1

step 2 HNO3 + H2SO4 conc (T < 55 °C) 1

step 3 Sn + HCl 1

step 4 HNO2 (or NaNO2 + HCl ) (at T < 10 °C) 1

the two temperatures for steps 2 and 4 1

4(c)(i) H2 + Pt or H2 + Ni + heat or pressure 1


May/June 2017



4(c)(ii) HH

H CH(CH3)2

OH

CH3

(CH3)2CH, CH3 and OH on the correct ring atoms i.e. structure is correct

1

all Hs on the same side of the ring 1

Total: 15


5(a)

J K L M

amine methyl ketone

aromatic amine aldehyde

amine methyl ketone amide

J and L correct 1 + 1

K correct 1 + 1

M correct 1

5(b)(i) hydrolysis 1

5(b)(ii) P is C6H5NH2 1

Q is CH3CH2CO2Na 1


May/June 2017



5(c) 1

1

1

1

K&L only: two chiral atoms shown 1

5(d) W is C6H5CO2Na 1

Total: 14

J isNH

O

orNHCH3

O

NH2or

O

CHO

NH2

K is

L is

NH2

O

HN

OM is


May/June 2017



6(a) Any of the three methods possible. Any 4 of the 5 points for each method available for maximum 4 marks. Method 1 1 Ensure both solutions (A and B) at 40 °C before mixing 2 mix known volumes of A and B and start the clock 3 at known time take out a sample / X and add it to ice-cold solvent 4 titrate against HCl 5 repeat at time at known time intervals Method 2 1 Ensure both solutions (A and B) at 40 °C before mixing 2 mix known volumes of A and B and start the clock 3 at known time pour into ice-cold solvent or pour ice-cold solvent in 4 titrate against HCl 5 repeat with different concentrations of either A or B, or repeat using different times Method 3 1 Ensure both solutions (A and B) at 40 °C before mixing 2 mix known volumes of A and B and start the clock and add pH meter 3 at a known time . . . . 4 . . . . record the pH 5 repeat pH readings at known time intervals

4

6(b)(i) from 1 and 3: when [RCl ] is trebled, so is rate, so order w.r.t. [RCl ] = 1 1

from 1 and 2: when both concentrations are doubled, rate doubles so [OH–] has no effect on rate, so order w.r.t.[OH–] = 0 1

6(b)(ii) rate = k[RCl ] AND units: sec–1 1 / s 1

6(b)(iii) relative rate = 2.0 1


May/June 2017



6(c)(i)

C-Cl dipole and first curly arrow

1

intermediate cation 1

OH– with lone pair and curly arrow 1

6(c)(ii) Beginning with candidate’s mechanism in (c)(i): If SN1: racemate / mixture of / two optical isomers will be formed, because: the intermediate is planar / has a plane of symmetry / OH– can approach from top or bottom or from any direction If SN2: one optical isomer because attack always from fixed direction / from same side / the “configuration” always inverts / there is an asymmetric transition state

1


May/June 2017



6(d)(i)

δ value number of H atoms group splitting result with D2O

1.4 3 CH3 / methyl doublet peak remains 2.7 1 OH / hydroxyl / alcohol singlet peak disappears 4.0 1 CH quartet peak remains

the three groups are in their correct places wrt the δ values 1

no. of H atoms for each peak agrees with group column 1

splitting patterns doublet, singlet and quartet are assigned to correct groups 1

peak identified as OH disappears with D2O, no other peak disappears 1

Total: 16






MARK SCHEME

Maximum Mark: 100

Published



May/June 2017



1(a)(i) increases down the group 1

radius / size of (cat)ion/M2+ increases 1

less polarisation / distortion of anion / carbonate ion / CO32– 1

1(a)(ii) Na+ has smaller ionic charge and larger ionic radii OR the charge density of the Na+ is lower

1

1(b)(i) 2KHCO3 → K2CO3 + CO2 + H2O 1

1(b)(ii) NaHCO3 because Na+ is smaller OR charge density Na+ is larger 1

1(c)(i) LE = ∆Hf – 2(∆Hat + IE) – ½(O=O) – (EA1 + EA2) = –361 – 2(89) – 2(418) – 496/2 – (–141+798) = –2280 (kJ mol–1) correct answer scores [3]

3 1 1 1

1(c)(ii) LE of Na2O will be more negative AND as Na(+) is smaller / larger charge density / smaller radii AND so greater attraction (between the ions) OR (ionic) bonds will be stronger

1

Total: 10


May/June 2017



2(a) Add AgNO3 Cl – gives a white ppt and I– gives a yellow ppt. 1

Add NH3(aq); ppt dissolves and ppt is insoluble 1

2(b)(i) conductivity decreases during the reaction, AND number of Na+ / I– / ions are decreased / used up (from solution)

1

2(b)(ii) (Equilibrate) solutions at 40 °C / with a water bath (cannot be after mixing) mix known volumes and start the clock / timing clearly mentioned/implied measure conductance / conductivity at regular intervals / every measured time [method A]

OR measure the time for conductance to go to zero / a specific value / to be constant [method B] prepare a curve of conductance vs. time [related to method A]

OR prepare a curve of conductance vs. concentration [related to method A] OR repeating the experiment at different concentrations [related to method A and B]

any 3 points

3

2(c)(i) [R-Cl ]: rate increases by 5 / 3 when concentration increases by 10 / 6 (5 / 3), so order = 1

1

[I–]: rate increases by 5 / 3 when concentration increases by 5 / 3, so order = 1

1

2(c)(ii) rate = k[I–][CH3CH2CHClCH3] AND units of k = dm3 mol–1 s–1 1

2(c)(iii) relative rate = 5 / 5.3 1


May/June 2017



2(d)(i) either SN1 or SN2 mechanism

C-Cl dipole AND C-Cl curly arrow 1

intermediate cation OR 5-valent transition state (charge essential) 1

I– with lone pair AND other curly arrow 1

2(d)(ii) If SN1 in 2(d)(i) mixture of / two optical isomers will be formed, AND the intermediate can be formed by the I– approaching from top or bottom plane If SN2 in 2(d)(i) one optical isomer AND attack always from fixed direction / opposite side

1


May/June 2017



2(e)(i) 4 peaks 1

2(e)(ii) 1 + 1

number of peaks = 2 number of peaks = 3 1

Total: 18 Question Answer Marks

3(a)

four shared pairs: S=O and 2 × S-Cl 1

all (9) lone pairs 1

3(b)(i) NaOH + HCl → NaCl + H2O 1

2NaOH + SO2 → Na2SO3 + H2O 1

CH3C

CH3CH3

Cl

CH3C

CH3H

CH2Cl


May/June 2017



3(b)(ii) moles (at start) = 0.5 × 60 / 1000 = 3 × 10–2 AND moles (at end) = 0.5 × 10.8 / 1000 = 5.4 × 10–3

1

moles reacted (= (30–5.4) × 10–3 =) 2.5 × 10–2 correct ans. scores [2] 1

3(b)(iii) moles of RCO2H = 2.46 × 10–2/3 = 8.2–8.3 × 10–3 mole 1

3(b)(iv) Mr = 1.00 / (8.2 × 10–3) = 121.95 (=122) 1

3(b)(v) C7H6O2 OR C6H5CO2H 1

3(c)(i) LiAl H4 1

3(c)(ii)

angelic acid T U

CO2HCO2H

NH2

3

3(c)(iii) angelic acid: geometrical OR cis-trans compound T: optical

1

Total: 14


May/June 2017



4(a)(i) Mr = 52 + 6 × 18 + 3 × 35.5 = 266.5 1

4(a)(ii) 1.00g = 1 / 266.5 OR 3.75 × 10–3 moles (of complex in 1g) for A, n=2 AND [Cr(H2O)4Cl2]Cl.2H2O for B, n=1 AND [Cr(H2O)5Cl ]Cl2.H2O for C, n=0; AND [Cr(H2O)6]Cl3

2

4(b)(i) Geometric(al) / cis-trans 1

4(b)(ii) 1

4(b)(iii) isomer 2 AND dipoles do not cancel OR CN– are on the same side of the molecule

1

Total: 6


May/June 2017



5(a)(i) bidentate: (a species that) forms two dative bonds / donates two lone pairs 1

ligand: a species that uses a lone pair to form a dative bond to a metal atom / metal ion 1

5(a)(ii)

N

CrNN

N

Cl

Cl

Cl

CrClN

N

N

N

N

CrNCl

Cl

N

N

each structure [1] x 3

3

5(b)(i) Kstab1 = [Cu(NH3)42+]/[Cu2+][NH3]4 1

Kstab2 = [Cu(en)22+]/[Cu2+][en]2 1

mol–4 dm12 AND mol–2 dm6 1

5(b)(ii) Keq3 = Kstab2 / Kstab1 1

5(b)(iii) Keq3 = Kstab2 / Kstab1 = 4.4(2) × 106 1

mol2 dm–6 1

5(c)(i) (∆Seq1 is negative as) more / 5 moles of reactants are forming (one mole of) the complex OR (∆Seq2 is positive as) fewer / 3 moles of reactants are forming (one mole of) the complex

1

5(c)(ii) ∆Geq2 = –100 – 298 × 40 / 1000 OR ∆G =∆H – T∆S = –112 or –111.9 (kJ mol–1) correct answer [2]

2 1 1


May/June 2017



5(c)(iii) Since (∆Geq2) is more negative (than ∆Geq1) AND equilibrium 2 is more feasible

1

5(c)(iv) ∆H(3) = –8 (kJ mol–1) 1

5(c)(v) ligand exchange / replacement / substitution / displacement 1

Total: 17


6(a)(i) the lower / smaller the pKa, the stronger the acid 1

6(a)(ii) pKa = –log(Ka) or pKa = –lg(Ka) or Ka = 10–pka 1

6(a)(iii) (stronger than ethanoic acid because) Cl is electron-withdrawing 1

and so stabilises the RCO2– anion / conjugate base

or weakens O-H bond (so H+ is more easily released) 1

6(b)(i) NH3+CH2CO2

– → NH2CH2CO2– + H+

OR NH3+CH2CO2

– + H2O → NH2CH2CO2– + H3O+

1

6(b)(ii) Ka = 10–9.87 = 1.35 × 10–10 [H+] = √(Ka.c) = 3.67 × 10–6

1

pH = 5.4 (5.43–5.44) min 2sf 1


May/June 2017



6(b)(iii) curve starts at 5.4 and continuous 1


finishes at pH = 12.5 at 20 cm3 (and does not increase in pH)

1

Total: 10 Question Answer Marks

7(a)

W X Y Z

acyl chloride / COCl

methyl ketone / CH3CO group aryl chloride

aldehyde / CHO chloro(alkane) / RCl

Alkene / C=C phenol / C6H5OH aryl chloride

0–1 [0]; 2 [1]; 3 [2]; 4 [3]; 5 [4]; 6–8 [5]

5


May/June 2017



7(b)(i)

1 + 1

1 + 1

7(b)(ii)

OR any chiral atom correctly labelled

1

Total: 10


8(a)(i) step 1 electrophilic substitution ignore acylation 1

step 2 nucleophilic addition 1

8(a)(ii) hydrolysis 1

CH2COCl

orCOCl

CH3

COCH3

Cl

Cl

W X

CHO ZCH=CH2

Cl

HOY

CHO

CH2Clor


May/June 2017



8(a)(iii) step 1 Cl CH2CHO (allow Br, I for Cl ) 1

Al Cl3 1

step 2 HCN + NaCN 1

step 3 heat in H3O+ / heat H+(aq) 1

step 5 NH3 under pressure (+ heat) or heat NH3 in a sealed tube 1

8(a)(iv) 1 + 1

1

1

8(b)(i) P is tyr 1

tyr is 2– AND it is small / has a small Mr 1

with NaOH(aq)

NH2

OCO2 [2]

with HCl(aq)

NH3

HOCO2H

[1]

with Br2(aq) NH3

HOCO2

Br

Br

NH2

HOCO2H

Br

Br

or

[1]


May/June 2017



8(b)(ii) (dipeptide / phe-tyr) 2– is about double the Mr / mass of (phe) 1 OR mass / charge ratios are about the same for each (for dipeptide / phe-tyr and phe)

1

Total: 15






MARK SCHEME

Maximum Mark: 100

Published



May/June 2017



1(a) solubility increases down the group 1

∆Hlatt and ∆Hhyd both decrease or ∆Hlatt and ∆Hhyd both become less exothermic / more endothermic

1

∆Hlatt decreases / changes more (than ∆Hhyd as OH– being smaller than M2+) 1

∆Hsol becomes more exothermic / more negative / less endothermic / less positive 1

1(b)(i) ∆Hr1 – (538 + 2x230 + 394) = –(1216 + 286) ∆Hr1 – 1392 = –1502

1

∆Hr1 = –110 1

1(b)(ii) let ∆Hf(HCO3–(aq)) = y

2y – 538 = –1216 – 394 – 286 – 26

1

y = –692 1

1(b)(iii) ∆Hr3 –538 – 2(230 + 394) = –538 – 2(692) ∆Hr3 = –136

1

1(b)(iv) ∆Hr3 will be identical to ∆Hr4, / unchanged 1

as the reaction is the same, or: 2OH–(aq) + 2CO2(g) → 2HCO3

–(aq) or metal ions stay in solution/metal ions are unchanged / are spectators

1


May/June 2017



1(c) more gaseous moles are being consumed (in reaction 3) or more CO2 moles are being consumed (in reaction 3)

1

∆S is therefore expected to be more negative/less positive for reaction 3. 1

Total: 13


2(a)(i)

H O C N H N OC1 + 1

16 electrons on each diagram 1

2(a)(ii) HNC = 115–125° AND NCO = 180° 1

2(a)(iii) cyanic acid, because it’s a stronger / higher bond enthalpy / triple / C≡N / more electrons involved bond 1

2(b)(i) [H+] = √([HNCO]Ka) = √(0.1 × 1.2 × 10–4) or 3.46 × 10–3 1

pH = log [H+] = 2.5 (2.46) 1

2(b)(ii) Na2CO3 + 2(NH2)2CO → 2NaNCO + CO2 + 2NH3 + H2O 1

2(c)(i) (n(OH–) at start = (2 × 0.1 × 30) / 1000 = 6 × 10–3 mol) (n(OH–) reacted = (0.1 × 20) / 1000 = 2 × 10–3 mol) n(OH–) remaining = (6–2) × 10–3 = 4 × 10–3 mol, (in 50 cm3)

1

so [OH–]end = (4 × 10–3 × 1000) / 50 = 0.08 mol dm–3 1


May/June 2017



2(c)(ii) [H+] = Kw / [OH–] = (1 × 10–14) / 0.08 = 1.25 × 10–13 mol dm–3 1

so pH = –log(1.25 × 10–13) = 12.9 1

2(c)(iii) curve starts at 2.46 / 2.5 1


finishes at pH = 12.9 1

2(d)(i) monodentate: (a species that) forms one dative / coordinate bond 1

ligand: a species that uses a lone pair of electrons to form a dative / coordinate bond to a metal atom / metal ion 1

2(d)(ii) [Ag(NCO)2]– or [Ag(OCN)2]– correct formula 1

correct charge 1

2(e)(i) n(BaCO3) =1.66 / 197.3 = 8.4(1) × 10–3 mol 1

2(e)(ii) n(RNCO) = 8.41 × 10–3 mol, so Mr = 1 / (8.41 × 10–3) = 119 1

2(e)(iii) molecular formula = C7H5NO 1