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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
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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]
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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
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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
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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
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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]
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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]
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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]
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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
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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
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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]
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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]
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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
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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
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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
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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]
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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
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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]
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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]
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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
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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]
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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
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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
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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
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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
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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
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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
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[2]
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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