Pearson Edexcel Level 3 Advanced GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015
131
CH
EMIS
TRY
A L
EVEL
PA
PER
3 M
AR
K S
CH
EME
Qu
esti
on
N
um
ber
A
nsw
er
Ad
dit
ion
al g
uid
ance
M
ark
1(a
) An
answ
er t
hat
mak
es r
efer
ence
to
the
follo
win
g po
ints
:
• st
ep 2
: in
solu
ble
impu
rities
are
rem
oved
by
filtr
atio
n of
the
hot
sol
utio
n
(1
)
• st
ep 4
: so
lubl
e im
puri
ties
rem
ain
in t
he s
olve
nt le
ft
afte
r fil
teri
ng t
he c
oole
d m
ixtu
re
(
1)
•
step
5:
the
solid
pro
duct
is w
ashe
d so
tha
t no
so
lubl
e im
puri
ties
for
m o
n th
e pr
oduc
t as
it d
ries
(1
)
3
1(b
) A d
escr
iption
tha
t m
akes
ref
eren
ce t
o th
e fo
llow
ing
poin
ts:
•
the
mel
ting
tem
pera
ture
is o
ver
a la
rger
ran
ge /
is
not
shar
p
(
1)
•
the
mea
sure
d m
elting
tem
pera
ture
is le
ss t
han
for
the
pure
sol
id
(1
)
2
(To
tal
Qu
esti
on
1 =
5 m
arks
)
Qu
esti
on
N
um
ber
A
nsw
er
A
dd
itio
nal
gu
idan
ce
Mar
k
2(a
) •
Fe2+
/ F
e(H
2O) 6
2+
(1
)
• C
r3+ / C
r(H
2O) 6
3+
(1)
Allo
w N
i2+ / N
i(H
2O) 6
2+
V3+
/ V
(H2O
) 63+
Igno
re n
ames
2
2(b
)(i)
C
r3+ / C
r(H
2O) 6
3+
(1
)
Ig
nore
nam
es
1
2(b
)(ii
) C
r(O
H) 3
/ C
r(H
2O) 3
(OH
) 3
(1
)
Igno
re n
ames
1
2(b
)(ii
i)
Cr(
OH
) 63-
(1)
Acc
ept
othe
r co
rrec
t sp
ecie
s Ig
nore
nam
es
(no
ecf
from
(b)
(i))
1
2(c
) Any
on
e fr
om:
• pu
rple
to
colo
urle
ss
• pu
rple
(so
lution
) de
colo
uris
ed
(
1)
Allo
w f
inal
col
our
of s
olut
ion
to b
e or
ange
Allo
w p
ink
for
purp
le
1
2(d
)(i)
C
l⁻
(1)
Rej
ect
Cl
Igno
re n
ames
1
Pearson Edexcel Level 3 Advanced GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015
132
Pearson Edexcel Level 3 Advanced GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015
133
(To
tal
Qu
esti
on
2 =
9 m
arks
)
Qu
esti
on
N
um
ber
A
nsw
er
A
dd
itio
nal
gu
idan
ce
Mar
k
2(d
)(ii
) An
expl
anat
ion
that
mak
es r
efer
ence
to
the
follo
win
g po
ints
: am
mon
ia s
olut
ion
cann
ot b
e us
ed b
ecau
se:
• am
mon
ia r
eact
s w
ith
the
iron
ions
to
form
a
prec
ipitat
e
(1
) o
r •
a pr
ecip
itat
e of
(Ir
on(I
I) h
ydro
xide
/ Fe
(OH
) 2/
Fe(H
2O) 4
(OH
) 2 f
orm
s
(1)
AN
D
• an
d so
obs
cure
s th
e di
ssol
ving
of
the
whi
te p
reci
pita
te
(1)
2
Qu
esti
on
n
um
ber
A
nsw
er
Ad
dit
ion
al g
uid
ance
M
ark
3(a
)(i)
•
eval
uation
of
num
ber
of m
oles
of
prop
anon
e
=
0.0
25 ×
2.0
= 0
.050
mol
(1
)
•
whi
ch is
gre
ater
tha
n th
e am
ount
of
iodi
ne,
whi
ch
is
0.05
0 ×
0.0
20 =
0.0
010
mol
(1)
2
3(a
)(ii
) •
mea
suri
ng c
ylin
der/
bure
tte
(1)
1
3(a
)(ii
i)
• pi
pett
e
(1)
1
3(b
)
An
expl
anat
ion
that
mak
es r
efer
ence
to
the
follo
win
g po
ints
:
• th
e or
der
with
resp
ect
to io
dine
is z
ero
(1)
• be
caus
e th
e gr
aph
is a
str
aigh
t lin
e, s
how
ing
that
th
e ch
ange
in io
dine
con
cent
ration
is c
onst
ant
(1
)
2
3(c
) An
answ
er t
hat
mak
es r
efer
ence
to
the
follo
win
g po
ints
:
• 1st
ord
er w
ith
resp
ect
to H
+ a
nd p
ropa
none
(1)
•
H+ a
nd p
ropa
none
invo
lved
in r
eact
ion
befo
re r
ate
dete
rmin
ing
step
(th
eref
ore
1st
orde
r)
(1)
•
iodi
ne in
volv
ed a
fter
rat
e de
term
inin
g st
ep /
slo
w
step
(th
eref
ore
zero
ord
er)
(1)
3
(T
ota
l Q
ues
tio
n 3
= 9
mar
ks)
Pearson Edexcel Level 3 Advanced GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015
134
Pearson Edexcel Level 3 Advanced GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015
135
Qu
esti
on
n
um
ber
A
nsw
er
Ad
dit
ion
al g
uid
ance
M
ark
4(a
) •
axes
: co
rrec
t w
ay r
ound
, la
belle
d, s
uita
ble
scal
e
(1)
•
all p
oint
s pl
otte
d co
rrec
tly,
with
best
fit s
trai
ght
line
(1)
• ca
lcul
atio
n of
gra
dien
t of
str
aigh
t lin
e
(
1)
•
use
of g
radi
ent
= -
E a /
R t
o ca
lcul
ate
E a (
in k
J m
ol-1
)
(1
)
Plot
ted
poin
ts m
ust
cove
r at
leas
t ha
lf th
e gr
aph
pape
r on
eac
h ax
is
Allo
w ±½
a s
quar
e G
radi
ent
= (
-) 5
970
Allo
w ±
200
Act
ivat
ion
ener
gy =
597
0 x
8.
31 /
100
0
=
+49
.6 (
kJ m
ol-1
) Fi
nal a
nsw
er m
ust
be p
ositiv
e.
4
*4
(b)
This
que
stio
n as
sess
es a
stu
dent
’s a
bilit
y to
sho
w a
co
here
nt a
nd lo
gica
lly s
truc
ture
d an
swer
with
linka
ges
and
fully
-sus
tain
ed r
easo
ning
. M
arks
are
aw
arde
d fo
r in
dica
tive
con
tent
and
for
how
the
an
swer
is s
truc
ture
d an
d sh
ows
lines
of
reas
onin
g.
The
follo
win
g ta
ble
show
s ho
w t
he m
arks
sho
uld
be
awar
ded
for
indi
cative
con
tent
. N
umbe
r of
indi
cative
m
arki
ng p
oint
s se
en in
an
swer
Num
ber
of m
arks
aw
arde
d fo
r in
dica
tive
m
arki
ng p
oint
s 6
4 5–
4 3
3–2
2 1
1 0
0
Gui
danc
e on
how
the
mar
k sc
hem
e sh
ould
be
app
lied:
Th
e m
ark
for
indi
cative
con
tent
sho
uld
be
adde
d to
the
mar
k fo
r lin
es o
f re
ason
ing.
Fo
r ex
ampl
e, a
n an
swer
with
five
indi
cative
m
arki
ng p
oint
s, w
hich
is p
artial
ly
stru
ctur
ed w
ith
som
e lin
kage
s an
d lin
es o
f re
ason
ing,
sco
res
4 m
arks
(3
mar
ks f
or
indi
cative
con
tent
and
1 m
ark
for
part
ial
stru
ctur
e an
d so
me
linka
ges
and
lines
of
reas
onin
g).
If
the
re a
re n
o lin
kage
s be
twee
n po
ints
, th
e sa
me
five
indi
cative
mar
king
poi
nts
wou
ld y
ield
an
over
all s
core
of
3 m
arks
(3
mar
ks f
or in
dica
tive
con
tent
and
no
mar
ks
for
linka
ges)
.
6
Qu
esti
on
n
um
ber
A
nsw
er
Ad
dit
ion
al g
uid
ance
M
ark
*4
(b)
Co
nt.
Th
e fo
llow
ing
tabl
e sh
ows
how
the
mar
ks s
houl
d be
aw
arde
d fo
r st
ruct
ure
and
lines
of
reas
onin
g.
Ind
icat
ive
con
ten
t:
• ac
tiva
tion
ene
rgy
(EA)
for
the
form
atio
n of
A is
low
er
than
tha
t fo
r B
(E B
) •
henc
e at
40
ºC m
ore
colli
sion
s ex
ceed
EA t
han
exce
ed
E B
• so
A is
for
med
mor
e qu
ickl
y th
an B
at
40 º
C
• at
160
ºC
mor
e co
llisi
ons
exce
ed E
B (
and
E A)
than
at
40 º
C
• th
eref
ore
both
isom
ers
are
form
ed
• bu
t th
e re
action
s ar
e re
vers
ible
and
B is
the
mor
e st
able
isom
er,
ther
efor
e A
will
con
vert
to
B
N
umbe
r of
mar
ks
awar
ded
for
stru
ctur
e of
ans
wer
and
su
stai
ned
line
of
reas
onin
g
Ans
wer
sho
ws
a co
here
nt a
nd
logi
cal s
truc
ture
with
linka
ges
and
fully
sus
tain
ed li
nes
of
reas
onin
g de
mon
stra
ted
thro
ugho
ut.
2
Ans
wer
is p
artial
ly s
truc
ture
d w
ith
som
e lin
kage
s an
d lin
es o
f re
ason
ing.
1
Ans
wer
has
no
linka
ges
betw
een
poin
ts a
nd is
uns
truc
ture
d.
0
(To
tal
Qu
esti
on
4 =
10
mar
ks)
Pearson Edexcel Level 3 Advanced GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015
136
Pearson Edexcel Level 3 Advanced GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015
137
Qu
esti
on
N
um
ber
A
nsw
er
Ad
dit
ion
al g
uid
ance
M
ark
5(a
) •
subs
titu
tion
into
ΔS o
equ
atio
n
(1)
• ev
alua
tion
of Δ
S o
(1
)
• su
bstitu
tion
into
ΔG
o =
ΔH
o −
TΔ
S o ,
usi
ng Δ
S o
in
kJ
K-1
mol
-1
(
1)
• co
rrec
t an
swer
to
3 sf
(
1)
•
sinc
e Δ
G o
is n
egat
ive,
the
rea
ctio
n is
fea
sibl
e
(1)
Exam
ple
of c
alcu
lation
Δ
S o
= (
2 x
193)
- 1
92 -
(3 x
131
)
=
−19
9 J
K-1
mol
-1 /
−
0.1
99 k
J K
-1 m
ol-1
Δ
G o
= −
92.
0 -
(298
x −
0.1
99)
= −
32.
7 kJ
mol
-1 /
– 3
2 70
0 J
mol
-1
Th
e fir
st f
our
mar
king
poi
nts
can
be a
war
ded
for
a co
rrec
t an
swer
to
3 sf
with
no w
orki
ng
5
5(b
) An
expl
anat
ion
that
mak
es r
efer
ence
to
the
follo
win
g po
ints
:
• (a
s te
mpe
ratu
re in
crea
ses)
TΔ
S
beco
mes
mor
e ne
gative
(1)
• (e
vent
ually
) Δ
H ‒
TΔ
S
beco
mes
pos
itiv
e
(
1)
2
Qu
esti
on
N
um
ber
A
nsw
er
Ad
dit
ion
al g
uid
ance
M
ark
5(c
) •
Cor
rect
exp
ress
ion
for
Kp
(1)
•
Cal
cula
tion
of
mol
e fr
action
s fo
r N
2, H
2 an
d N
H3
(
1)
•
Cal
cula
tion
of
part
ial p
ress
ures
for
N2,
H2
and
NH
3
(1)
• Sub
stitut
ion
and
eval
uation
of
Kp
(1
)
• U
nits
: at
m−
2
(
1)
Exam
ple
of c
alcu
lation
(t
otal
num
ber
of m
oles
= 1
8)
mf
N2
=
2.8
8 ÷
18
= 0
.16
mf
H2
=
8.6
4 ÷
18
= 0
.48
mf
NH
3 =
6.4
8 ÷
18
= 0
.36
pN2
=
0.1
6 x
200
= 3
2 at
m
pH2
=
0.4
8 x
200
= 9
6 at
m
pNH
3 =
0.3
6 x
200
= 7
2 at
m
Kp
=
!!!"!(!)
!!!!.!! !
!(!)
Kp
=
!"!
!" ! !"!
=
1.
83 x
10−
4 a
tm−
2
Alter
native
met
hod
for
calc
ulat
ion:
!.!"
!
!.!" ! !.!"!
(=
7.3
2421
875)
K
p =
!.!"
!
!.!" ! !.!"!
x
!!""!
=
1.8
3 x
10−
4 a
tm-2
C
orre
ct a
nsw
er w
ith
no w
orki
ng w
ith
units
scor
es 5
mar
ks
5
(To
tal
Qu
esti
on
5 =
12
mar
ks)
Pearson Edexcel Level 3 Advanced GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015
138
Pearson Edexcel Level 3 Advanced GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015
139
Qu
esti
on
N
um
ber
A
nsw
er
Ad
dit
ion
al g
uid
ance
M
ark
6(a
) An
answ
er t
hat
mak
es r
efer
ence
to
the
follo
win
g:
• su
itab
le v
olum
es o
f et
hano
l and
wat
er
(
1)
•
evid
ence
of
calc
ulat
ion
to s
how
one
com
pone
nt o
f m
ixtu
re in
exc
ess
(1
)
• m
ixed
tog
ethe
r in
sim
ple
calo
rim
eter
/ p
olys
tyre
ne
cup
with
lid
(1)
•
stir
and
mea
sure
max
imum
tem
pera
ture
cha
nge
(
1)
•
calc
ulat
e en
ergy
cha
nge
usin
g Q
=m
c∆T
(1)
•
calc
ulat
e st
reng
th o
f hy
drog
en b
ond
per
mol
e by
sc
alin
g up
fro
m t
he a
mou
nt o
f lim
itin
g co
mpo
nent
of
mix
ture
(i.e
. co
mpo
nent
not
in e
xces
s)
(1)
6
6(b
)(i)
•
eval
uation
of
num
ber
of m
oles
of
2-hy
drox
yben
zoic
aci
d us
ed
(1)
•
calc
ulat
ion
of m
ass
of a
spir
in a
t 10
0% y
ield
(1)
•
calc
ulat
ion
of m
ass
of a
spir
in a
t 65
% y
ield
(1
)
Exam
ple
of c
alcu
lation
2
/ 13
8 =
0.0
145
mol
0.
0145
x 1
80 =
2.6
1 g
2.61
/100
x 6
5 =
1.7
0 g
C
orre
ct a
nsw
er w
ith
no w
orki
ng s
core
s 3
mar
ks
3
Qu
esti
on
N
um
ber
A
nsw
er
Ad
dit
ion
al g
uid
ance
M
ark
6(b
)(ii
) Th
e m
ark
for
each
rea
son
mus
t be
link
ed w
ith
the
corr
ect
impr
ovem
ent.
• Im
prov
emen
t: s
wap
the
wat
er in
flow
and
ou
tflo
w in
the
con
dens
er
(1)
• R
easo
n : t
o im
prov
e ef
ficie
ncy
of c
onde
nsin
g pr
oces
s
(
1)
•
Impr
ovem
ent:
add
ant
i-bu
mpi
ng g
ranu
les
to
flask
(1)
• R
easo
n: t
o pr
omot
e sm
ooth
boi
ling/
to p
reve
nt
mat
eria
l esc
apin
g fr
om t
op o
f co
nden
ser
(
1)
•
Impr
ovem
ent:
inse
rt c
onde
nser
into
nec
k of
fla
sk
(1
)
• R
easo
n: t
o pr
even
t es
cape
of
reag
ents
(
1)
6
(To
tal
for
qu
esti
on
6 =
15
mar
ks)
Pearson Edexcel Level 3 Advanced GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015
140
Pearson Edexcel Level 3 Advanced GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015
141
Qu
esti
on
N
um
ber
A
nsw
er
Ad
dit
ion
al g
uid
ance
M
ark
7(a
)(i)
•
(sat
urat
ed)
pota
ssiu
m n
itra
te
(
1)
Allo
w p
otas
sium
chl
orid
e
1
7(a
)(ii
) An
expl
anat
ion
that
mak
es r
efer
ence
to
the
follo
win
g po
ints
:
• to
com
plet
e th
e ci
rcui
t
(1)
•
by a
llow
ing
mov
emen
t of
(po
sitive
and
neg
ativ
e) io
ns
(1
)
2
7(b
)(i)
• co
ntai
ner/
beak
er c
onta
inin
g th
e si
de a
rm a
nd s
ilver
, bo
th d
ippi
ng in
to s
ilver
nitra
te s
olut
ion
(1
)
• co
nnec
ting
wir
e fr
om s
ilver
and
cal
omel
ele
ctro
de t
o co
mpl
ete
the
circ
uit
(1)
• (h
igh
resi
stan
ce/d
igital
) vo
ltm
eter
(1)
3
7(b
)(ii
) •
solu
tion
con
cent
ration
1.0
mol
dm
-3
(1)
• te
mpe
ratu
re 2
98 K
(1)
Igno
re m
ention
of
pres
sure
2
Qu
esti
on
N
um
ber
A
nsw
er
Ad
dit
ion
al g
uid
ance
M
ark
7(c
)(i)
•
corr
ect
equa
tion
(1)
• em
f =
0.8
0 –
(+)0
.27
=
(+
)0.5
3 (V
)
(1
)
No
sign
in a
nsw
er s
core
s 1
mar
k, a
min
us s
ign
give
n sc
ores
0 m
arks
C
orre
ct a
nsw
er w
ith
no w
orki
ng s
core
s 2
mar
ks
2
7(c
)(ii
) •
(+)0
.03
(V)
(1
)
1
7(c
)(ii
i)
• Fe
2+(a
q) +
Ag+
(aq)
→ F
e3+(a
q)+
Ag(
s)
(
1)
Allo
w r
ever
sibl
e ar
row
s
1
7(c
)(iv
) •
rear
rang
emen
t an
d su
bstitu
tion
into
equ
atio
n
(1)
• ev
alua
tion
of
ln K
and
con
vers
ion
to K
(
1)
Exam
ple
of c
alcu
lation
ln
K =
__
_289
2__
= (
+)1
.167
8
8.3
1 x
298
K
= 3
.215
02
= 3
.22
Ig
nore
sf
exce
pt 1
N
ote
if ln
K =
1.1
678
is u
sed
answ
er is
3.2
1 C
orre
ct a
nsw
er w
ith
no w
orki
ng s
core
s 2
mar
ks
2
(To
tal
for
qu
esti
on
7 =
14
mar
ks)
Pearson Edexcel Level 3 Advanced GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015
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143
Qu
esti
on
N
um
ber
A
nsw
er
Ad
dit
ion
al g
uid
ance
M
ark
8(a
) (i
)
• ci
nnam
on o
il up
per
laye
r
(1)
• se
para
ting
fun
nel
(1)
Funn
el m
ust
be s
uita
ble
for
a st
oppe
r
2
8(a
)(ii
) •
addi
tion
of
(anh
ydro
us)
sodi
um s
ulfa
te /
cal
cium
ch
lori
de /
mag
nesi
um s
ulfa
te
(1)
1
8
(a)(
iii)
•
from
clo
udy
to c
lear
(
1)
D
o no
t ac
cept
col
ourl
ess
in p
lace
of
clea
r 1
8(b
)(i)
•
cont
ains
C=
C/a
lken
e
(
1)
1
8(b
)(ii
) •
cont
ains
car
bony
l gro
up/
alde
hyde
or
keto
ne
(1)
C=
O a
lone
1
8(b
)(ii
i)
• al
dehy
de/
-CH
O
(1)
1
Qu
esti
on
N
um
ber
A
nsw
er
Ad
dit
ion
al g
uid
ance
M
ark
8(c
)(i)
•
77
C6H
5+
(1
)
• 10
3 C
8H7+
(1)
Mus
t sh
ow a
cha
rge
but
only
pen
alis
e om
issi
on
once
Allo
w s
truc
tura
l, di
spla
yed
or s
kele
tal f
orm
ulae
2
8(c
)(ii
)
Allo
w n
on-d
ispl
ayed
ben
zene
C-H
s 2
8(c
)(ii
i)
• Th
e pe
ak is
due
to
the
pres
ence
of
an a
tom
of
a (13
C)
isot
ope
(1)
Allo
w r
efer
ence
to
othe
r na
med
isot
ope
of H
or
O
1
Pearson Edexcel Level 3 Advanced GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015
144
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145
Qu
esti
on
N
um
ber
A
nsw
er
Ad
dit
ion
al g
uid
ance
M
ark
8(d
) •
initia
l mol
es o
f N
aOH
in 2
50 c
m3
(1)
•
exce
ss m
oles
of
NaO
H in
25.
0 cm
3
(1)
•
expr
essi
on f
or m
oles
of
tota
l NaO
H r
eact
ed
(1
)
• ev
alua
tion
of
mol
es o
f ci
nnam
ic a
cid
(1
)
•
eval
uation
of
Mr o
f ci
nnam
ic a
cid
to 1
dp
(1)
Exam
ple
calc
ulat
ion
Initia
l mol
es o
f N
aOH
= (
250
÷ 1
000)
x 0
.500
=
0.1
25
Mol
es o
f ex
cess
NaO
H in
25.
0 cm
3
=
(28
.25
÷ 1
000)
x 0
.400
= 0
.011
3
Mol
es o
f N
aOH
rea
cted
= 0
.125
– (
10 x
0.0
113)
M
oles
of
cinn
amic
aci
d =
mol
es o
f N
aOH
rea
cted
= 0
.012
M
r of
cinn
amic
aci
d =
1.7
8 ÷
0.0
12 =
148
.3
Allo
w e
cf f
rom
2nd
mar
k
Cor
rect
ans
wer
to
1 dp
with
no w
orki
ng s
core
s 5
mar
ks
5
(To
tal
for
Qu
esti
on
8 =
17
mar
ks)
Qu
esti
on
N
um
ber
A
nsw
er
Ad
dit
ion
al g
uid
ance
M
ark
9(a
)(i)
An
expl
anat
ion
that
mak
es r
efer
ence
to
the
follo
win
g po
ints
:
• co
pper
for
ms
an io
n w
ith
an in
com
plet
e d-
sub-
shel
l / w
ith
a co
nfig
urat
ion
of 3
d9
(1
)
• bu
t th
e on
ly io
n fo
rmed
by
zinc
has
a c
ompl
etel
y fil
led
d-
sub-
shel
l
(1)
2
9(a
)(ii
) An
expl
anat
ion
that
mak
es r
efer
ence
to
the
follo
win
g po
ints
:
• (i
n br
ass)
the
laye
rs o
f po
sitive
ions
can
slid
e ov
er
one
anot
her
(1)
• an
d th
ere
are
(alw
ays)
ele
ctro
ns b
etw
een
the
laye
rs p
reve
ntin
g re
puls
ion
betw
e en
the
ions
in
one
laye
r an
d th
ose
in a
noth
er la
yer
(1
)
• (i
n so
dium
chl
orid
e) w
hen
a la
yer
of io
ns is
di
spla
ced,
ions
with
the
sam
e ch
arge
bec
ome
clos
e to
one
ano
ther
and
rep
el
(1
)
3
9(b
)(i)
•
brow
n fu
mes
(1)
• a
gree
n/bl
ue s
olut
ion
form
ing
(1)
2
Pearson Edexcel Level 3 Advanced GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015
146
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147
Qu
esti
on
N
um
ber
A
nsw
er
Ad
dit
ion
al g
uid
ance
M
ark
9(b
)(ii
) •
I 2(a
q) +
2S
2O32-
(aq)
→ 2
I- (aq
) +
S4O
62-(a
q)
(
1)
Sta
te s
ymbo
ls m
ust
be p
rese
nt
1
9(b
)(ii
i)
• am
ount
of
thio
sulfa
te
(1)
•
uses
am
ount
of
thio
sulfa
te =
am
ount
of
iodi
ne t
o de
term
ine
amou
nt o
f th
iosu
lfate
= a
mou
nt o
f co
pper
(II)
ions
in 2
5 cm
3 po
rtio
n
(
1)
• ev
alua
tion
of
num
ber
of m
oles
of
Cu
in s
ampl
e
(1
)
•
calc
ulat
es m
ass
of C
u
(1)
•
• pe
rcen
tage
cop
per
to 3
sf
(1)
Exam
ple
of c
alcu
lation
am
ount
of
thio
sulfa
te =
22.
7 x
0.25
100
0
=
5.6
75 x
10-3
(m
ol)
5.
675
x 10
-3 (
mol
) =
am
ount
of
copp
er(I
I)
io
ns in
25
cm3
port
ion
am
ount
of
Cu
in s
ampl
e =
5.6
75 x
10-3
x 1
0
= 5
.675
x 1
0-2 (
mol
)
m
ass
of C
u =
5.6
75 x
10-2
x 6
3.5
=
3.6
0362
5
perc
enta
ge c
oppe
r =
3.6
0362
5 x
100
/5.0
0
= 7
2.07
25 =
72.
1%
Allo
w e
cf f
rom
2nd
mar
k C
orre
ct a
nsw
er t
o 3
sf w
ith
no w
orki
ng s
core
s 5
mar
ks
5
9(b
)(iv
) •
calc
ulat
ion
of p
erce
ntag
e un
cert
aint
y fr
om b
alan
ce
= ±
0.00
5x 2
x10
0/5.
00
= 0
.2%
an
d
perc
enta
ge u
ncer
tain
ty in
mea
n titr
e fr
om b
uret
te
= 2
x±0.
05 x
100
/22
.7
= 0
.44%
(1)
• so
bur
ette
rea
ding
is m
ost
sign
ifica
nt
(1)
2
(To
tal
for
Qu
esti
on
9 =
15
mar
ks)
Qu
esti
on
N
um
ber
A
nsw
er
Ad
dit
ion
al g
uid
ance
M
ark
10
(a)
• C
6H5C
OO
H +
CaO
→ C
6H6
+ C
aCO
3
(1)
Acc
ept
C6H
5CO
OH
+ C
aO →
C6H
6 +
CaO
+ C
O2
1
10
(b)
An
answ
er t
hat
mak
es r
efer
ence
to
the
follo
win
g po
ints
: •
-2 in
cyc
lohe
xane
and
-1
in b
enze
ne
(1
)
• so
(ca
rbon
is)
oxid
ised
(1)
2nd
mar
k st
ands
alo
ne
2
10
(c)
spec
trum
1 is
met
hylb
enze
ne,
beca
use
• it c
onta
ins
an a
bsor
ptio
n at
296
2 –
2853
cm
-1
(1)
• ow
ing
to a
lkyl
C—
H s
tret
chin
g
(1
)
Iden
tific
atio
n un
qual
ified
gai
ns n
o m
arks
2
10
(d)
nC6H
5CH
=C
H2 →
( C
H—
CH
2 )
n
|
C
6H5
• co
rrec
t pr
oduc
t fo
rmul
a
(
1)
•
bala
nced
equ
atio
n
(
1)
2
10
(e)(
i)
Iron
/ ir
on(I
II)
brom
ide
Allo
w a
lum
iniu
m /
alu
min
ium
bro
mid
e Allo
w c
orre
ct f
orm
ulae
1
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149
Qu
esti
on
N
um
ber
A
nsw
er
Ad
dit
ion
al g
uid
ance
M
ark
10
(e)(
ii)
• re
agen
t fo
r st
ep 1
(1
)
• pr
oduc
t of
ste
p 1
(1)
• re
agen
t fo
r st
ep 2
(
1)
•
prod
uct
of s
tep
2
(1)
• re
agen
t fo
r st
ep 3
(
1)
•
cata
lyst
for
ste
p 3
(1)
6
Ex
ampl
e of
syn
thes
is:
(To
tal
for
Qu
esti
on
10
= 1
4 m
arks
)