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CAMBRIDGE INTERNATIONAL EXAMINATIONS
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 should be read in conjunction with the question paper and the Principal
Examiner Report for Teachers.
Cambridge will not enter into discussions about these mark schemes.
Cambridge is publishing the mark schemes for the May/June 2013 series for most IGCSE,
GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level
components.
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MARK SCHEME for the May/June 2013 series
s
er
GCE Advanced Subsidiary Level and GCE Advanced Level
Page 2
1
(a)
Mark Scheme
GCE AS/A LEVEL – May/June 2013
a base is a proton acceptor or
a lone pair donor
a weak base is not fully ionised
e.g. NH3 + H2O NH4+ + OH– or
B + H+ BH+ or equivalent
⇌ is necessary
(b) (i) stated pressure
stated temperature
named catalyst
greater than 1 atm up to 5 atm
400 to 500 oC
V2O5/vanadium(V) oxide
(ii) SO3 is dissolved in concentrated H2SO4
and then diluted with water
not 'SO3 dissolved in water' as the only statement
Syllabus
9701
Paper
22
(1)
(1)
(1)
(1)
(1)
(1)
(1)
(c) (i) with concentrated sulfuric acid
ClCH2CH=CHCl
(1)
with ammonia
H2NCH2CH(OH)CH2NH2
(1)
(ii) nucleophilic
substitution
[3]
(1)
(1)
[4]
[4]
[Total: 11]
© Cambridge International Examinations 2013
Page 3
2
Mark Scheme
GCE AS/A LEVEL – May/June 2013
25.0 × 1.00
= 0.025 mol
1000
16.2 × 2.00
n(NaOH) =
= 0.0324 mol
1000
0.0324
= 0.0162 mol
n(H2SO4) reacting with NaOH =
2
n(H2SO4) reacting with NH3 = 0.025 - 0.0162 = 0.0088 mol
n(NH3) reacting with H2SO4 = 2 x 0.0088 = 0.0176 mol
n(NaNO3) reacting = n(NH3) produced = 0.0176 mol
mass of NaNO3 that reacted = 0.0176 x 85 = 1.496 g
1.496 × 100
= 91.2195122 = 91.2
% of NaNO3 =
1.64
give one mark for the correct expression
give one mark for answer given as 91.2 – i.e to 3 sig. fig.
allow ecf where appropriate
(a) (i) n(H2SO4) =
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
Syllabus
9701
Paper
22
(1)
(1)
(1)
(1)
(1)
(1)
(1)
(1)
(1)
[9]
(b)
NaNO3 +5 and NH3 -3 both required
(1)
[1]
[Total: 10]
© Cambridge International Examinations 2013
Page 4
3
(a)
Mark Scheme
GCE AS/A LEVEL – May/June 2013
Syllabus
9701
Paper
22
penalise (-1) the use of names of elements or formulae of compounds
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Ca
O or N or C
C or N or S or F or Cl or Br
Si or Ge or B
Al or Si or P or S or H
Al
(1)
(1)
(1)
(1)
(1)
(1)
[6]
(b) (i)
element
Na
Mg
Al
Si
P
S
oxide
Na2O
MgO
Al2O3
SiO2
P2O5/P4O10
or
P2O3/P4O6
SO2
flame
yellow
or
orange
white
white
white
white
or
yellow
blue
formula of oxide
colour of flame
(1)
(1)
(ii)
chloride
NaCl
MgCl2
7
6.5 to
6.9
pH
AlCl3
or
Al2Cl6
SiCl4
PCl3
or
PCl5
SCl2
or
S2Cl2
1 to 4
formula of chloride
pH of solution formed
(1)
(1)
[4]
(c) (i)
I
Cl
(1)
(ii) intermolecular forces/van der Waals’ forces
are stronger or greater in ICl
ICl has most electrons or
has the largest permanent dipole
(iii) ICl
greatest difference in electronegativity is between I and Cl
(1)
(1)
(1)
(1)
[5]
[Total: 15]
© Cambridge International Examinations 2013
Page 5
4
Mark Scheme
GCE AS/A LEVEL – May/June 2013
Syllabus
9701
Paper
22
(a)
A
Br2 in an inert organic
solvent
B
PCl5
C
H2 and Ni catalyst
D
NaBH4
E
K2Cr2O7/H+,
heat under reflux
CH3CHBrCHBrCH2OH
CH3CH=CHCH2Cl
CH3CH2CH2CH2OH
NO REACTION
CH3CH=CHCO2H
give one mark for each correct answer
(5 × 1)
[5]
(1)
[1]
(b)
(c)
correct C4 with C=C in position 2
accept cis form
correctly shown –CO2H
allow ecf on candidate’s answer to E in (a)
© Cambridge International Examinations 2013
(1)
(1)
[2]
Page 6
Mark Scheme
GCE AS/A LEVEL – May/June 2013
(d) (i) reagent
Paper
22
observation
2,4-dinitrophenylhydrazine
Tollens' reagent
red/orange ppt.
silver mirror or
grey ppt. or
black ppt.
brick red ppt.
Fehling’s reagent
correct reagent
observation
(1)
(1)
(ii) reduction or nucleophilic addiction
(e)
Syllabus
9701
C : H : O =
(1)
[3]
73.7
12.3
14.0
:
:
12
1
16
= 6.14 : 12.3
= 7.01 : 14.1
: 0.875
: 1
gives C7H14O
formula must be given
(1)
(1)
[2]
[Total: 13]
© Cambridge International Examinations 2013
Page 7
5
(a)
Mark Scheme
GCE AS/A LEVEL – May/June 2013
Syllabus
9701
C4H8O2
Paper
22
(1)
[1]
(4 × 1)
[4]
(c) (i) CHO or aldehyde absent
(ii) >CO or carbonyl absent
(iii) CO2H or carboxylic acid present
(1)
(1)
(1)
[3]
(d) (i) CH3CO2H or ethanoic acid
(ii) Y above
(1)
(1)
[2]
(e)
(1)
[1]
(b)
HCO2CH2CH2CH3
HCO2CH(CH3)2
W
X
CH3CO2CH2CH3
CH3CH2CO2CH3
Y
Z
give one mark for each correct answer
none – no chiral carbon atoms present
[Total: 11]
© Cambridge International Examinations 2013