specimen ms - paper 1 edexcel chemistry as-level
DESCRIPTION
chemestry question paper from edexcellTRANSCRIPT
!"#$%&'()*"+,"-(."/"-(0(1*/#',"*(234%5*5#$6(78)(5'(89":5%;$62#:<-"(1%%"%%:"';(=#;"$5#-%(>(?%%3"(@(>(A"4$3#$6(BC@D(E(!"#$%&'()*3,#;5&'(.5:5;"*(BC@D
00
CH
EMIS
TRY
AS
PA
PER
1 M
AR
KS
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
:
• at
om
s w
ith
the
sam
e nu
mbe
r of
pro
tons
(1
)
• w
ith
diff
eren
t nu
mbe
rs o
f ne
utro
ns
(1)
Rej
ect
‘Ele
men
ts w
ith
the
sam
e…’
Igno
re r
efer
ence
s to
the
sam
e nu
mbe
r of
el
ectr
ons
Igno
re ‘a
tom
s of
the
sam
e el
emen
t th
at d
iffer
on
ly in
mas
s nu
mbe
r’
2
1(b
) C
1
1(c
) •
calc
ulat
ion
of %
30Si a
nd s
ubst
itut
ion
into
ex
pres
sion
sho
win
g su
m o
f ab
unda
nce
x m
ass
num
ber
÷ t
otal
abu
ndan
ce
(1)
• ev
alua
tion
of
corr
ect
answ
er t
o 3
s.f.
(
1)
Exam
ple
of c
alcu
lation
: (9
2.2
x 28
) +
(4.
67 x
29)
+
( 3
.13
x 30
)
10
0
(=
28.
1093
) =
28.
1
C
orre
ct a
nsw
er w
ith
no w
orki
ng t
o 3.
s.f
scor
es
2 m
arks
Ig
nore
any
uni
ts
2
1(d
)
• ca
lcul
atio
n of
num
ber
of m
oles
of
mol
ecul
es
pres
ent
(
1)
•
use
of A
voga
dro
num
ber
to c
onve
rt t
o nu
mbe
r of
m
olec
ules
(
1)
Exam
ple
of c
alcu
lation
: nu
mbe
r of
mol
es o
f m
olec
ules
= 5
.67
÷ 1
70.1
=
0.0
3333
...
num
ber
of m
olec
ules
= 0
.033
33..
.x 6
.02
x 10
23
= 2
.01
x 10
22
Allo
w 2
x 1
022
Cor
rect
ans
wer
no
wor
king
sco
res
2 m
arks
2
(To
tal
for
Qu
esti
on
1 =
7 m
arks
)
Qu
esti
on
N
um
ber
A
nsw
er
A
dd
itio
nal
Gu
idan
ce
Mar
k
2(a
) An
answ
er t
hat
mak
es r
efer
ence
to
the
follo
win
g po
ints
: O
(at
om)
• ha
s m
ore
prot
ons
/ ha
s gr
eate
r nu
clea
r ch
arge
(1)
•
has
smal
ler
(ato
mic
) ra
dius
(th
an C
ato
m)
(1)
Igno
re r
efer
ence
s to
shi
eldi
ng
Allo
w j
ust
‘sm
alle
r’
Allo
w r
ever
se a
rgum
ent
for
carb
on
2
2(b
) An
expl
anat
ion
that
mak
es r
efer
ence
to:
(o
nly
carb
on d
ioxi
de is
non
-pol
ar)
•
beca
use
only
car
bon
diox
ide
is s
ymm
etri
cal /
line
ar
(1
)
O
R
bond
pol
aritie
s ar
e ve
ctor
s /
vect
or q
uant
itie
s
• an
d th
eref
ore
the
bond
pol
aritie
s ca
ncel
(1
)
2
!"#$%&'()*"+,"-(."/"-(0(1*/#',"*(234%5*5#$6(78)(5'(89":5%;$62#:<-"(1%%"%%:"';(=#;"$5#-%(>(?%%3"(@(>(A"4$3#$6(BC@D(E(!"#$%&'()*3,#;5&'(.5:5;"*(BC@D
0F
!"#$%&'()*"+,"-(."/"-(0(1*/#',"*(234%5*5#$6(78)(5'(89":5%;$62#:<-"(1%%"%%:"';(=#;"$5#-%(>(?%%3"(@(>(A"4$3#$6(BC@D(E(!"#$%&'()*3,#;5&'(.5:5;"*(BC@D
0D
Qu
esti
on
N
um
ber
A
nsw
er
A
dd
itio
nal
Gu
idan
ce
Mar
k
2(c
)
•
lone
pai
r of
ele
ctro
ns o
n O
of
one
mol
ecul
e
(1
)
• δ+
sym
bol o
n on
e re
leva
nt H
ato
m A
ND
δ–
sym
bol
on o
ne r
elev
ant
O a
tom
(1)
If
no
repr
esen
tation
of
a hy
drog
en b
ond
(by
dash
ed li
ne o
r si
mila
r),
then
onl
y on
e of
the
se m
arks
can
be
awar
ded
No
pena
lty
for
show
ing
both
pos
sibl
e hy
drog
en
bond
s
2
(To
tal
for
Qu
esti
on
2 =
6 m
arks
)
Q
ues
tio
n
Nu
mb
er
An
swer
Ad
dit
ion
al G
uid
ance
M
ark
3(a
) B
1
3
(b)(
i)
An
expl
anat
ion
that
mak
es r
efer
ence
to
the
follo
win
g po
ints
: •
use
a fu
me
cupb
oard
(1)
•
as c
hlor
ine
is t
oxic
/ p
oiso
nous
(1
)
2
3(b
)(ii
) An
expl
anat
ion
that
mak
es r
efer
ence
to
the
follo
win
g po
ints
: •
cool
the
rea
ctio
n ve
ssel
/ s
urro
und
the
flask
with
cold
wat
er
(1)
•
in o
rder
to
prev
ent
subl
imat
ion
(of
PCl 5)
/ to
pr
even
t th
e PC
l 5 t
urni
ng in
to a
gas
(1)
2
3(b
)(ii
i)
• PC
l 3 +
Cl 2 →
PC
l 5
(1
)
Igno
re s
tate
sym
bols
, ev
en if
inco
rrec
t
1
!"#$%&'()*"+,"-(."/"-(0(1*/#',"*(234%5*5#$6(78)(5'(89":5%;$62#:<-"(1%%"%%:"';(=#;"$5#-%(>(?%%3"(@(>(A"4$3#$6(BC@D(E(!"#$%&'()*3,#;5&'(.5:5;"*(BC@D
0G
!"#$%&'()*"+,"-(."/"-(0(1*/#',"*(234%5*5#$6(78)(5'(89":5%;$62#:<-"(1%%"%%:"';(=#;"$5#-%(>(?%%3"(@(>(A"4$3#$6(BC@D(E(!"#$%&'()*3,#;5&'(.5:5;"*(BC@D
0H
Qu
esti
on
N
um
ber
A
nsw
er
A
dd
itio
nal
Gu
idan
ce
Mar
k
3(c
) •
calc
ulat
ion
of
mol
es P
Cl 5
(=
mol
es P
OC
l 3)
(
1)
• m
oles
HC
l = m
oles
PC
l 5 x
5
(
1)
•
volu
me
HC
l = m
oles
HC
l x 2
4 dm
3
(1)
Allo
w e
cf f
or s
teps
in c
alcu
lation
, ig
nore
si
gnifi
cant
fig
ures
in f
inal
ans
wer
exc
ept
one
sign
ifica
nt f
igur
e C
orre
ct a
nsw
er w
ith
no w
orki
ng s
core
s 3
mar
ks
Exam
ple
of c
alcu
lation
M
oles
PC
l 5 =
4.1
7
= 0
.02(
00)
(mol
)
20
8.5
M
oles
HC
l = 5
x 0
.02(
00)
= 0
.1(0
0) (
mol
)
Vol
ume
HC
l = 0
.1 x
24
= 2
.4 (
dm3 )
3
(To
tal
for
Qu
esti
on
3 =
9 m
arks
)
Qu
esti
on
N
um
ber
A
nsw
er
A
dd
itio
nal
Gu
idan
ce
Mar
k
4(a
) D
1
4(b
) •
calc
ulat
ion
of
mol
es A
gCl
(1)
•
tota
l mol
es o
f Ag
in 5
00 c
m3
(w
here
mol
es A
g =
mol
es A
gCl)
(1)
• ca
lcul
atio
n of
tot
al m
ass
of A
g
(1)
•
eval
uation
of
corr
ect
answ
er t
o 3.
s.f
usin
g %
by
mas
s of
Ag
= M
ass
Ag
x 1
00%
(1)
5.
00
allo
w e
cf f
or s
teps
in c
alcu
lation
; in
clud
ing
for
final
ans
wer
dep
ende
nt o
n ro
undi
ng in
ste
ps o
f th
e ca
lcul
atio
n.
corr
ect
answ
er t
o 3.
s.f
with
no w
orki
ng s
core
s 4
mar
ks
Exam
ple
of c
alcu
lation
M
oles
AgC
l =
0.
617
=
0.0
0430
.. (
mol
)
14
3.4
To
tal m
oles
Ag
=
0.0
0430
x 5
00
= 0
.043
0...
50.
0 M
ass
Ag
= 0
.043
0 x
107.
9 =
4.6
425.
.. =
4.6
4 (g
) %
by
mas
s of
Ag
= 4
.642
5…
x 10
0%
5.0
0
=
92.
9 %
4
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0I
!"#$%&'()*"+,"-(."/"-(0(1*/#',"*(234%5*5#$6(78)(5'(89":5%;$62#:<-"(1%%"%%:"';(=#;"$5#-%(>(?%%3"(@(>(A"4$3#$6(BC@D(E(!"#$%&'()*3,#;5&'(.5:5;"*(BC@D
0J
Qu
esti
on
N
um
ber
A
nsw
er
A
dd
itio
nal
Gu
idan
ce
Mar
k
4(c
)(i)
An
answ
er t
hat
mak
es r
efer
ence
to
the
follo
win
g po
ints
:
• (a
rea
ctio
n in
whi
ch a
n) e
lem
ent
(in
a sp
ecie
s)
(1
) •
is s
imul
tane
ousl
y ox
idis
ed a
nd r
educ
ed
/ fo
r w
hich
the
oxi
dation
num
ber
both
incr
ease
s an
d de
crea
ses
(in
the
sam
e re
action
)
(
1)
Rej
ect
‘ato
m’
2
4(c
)(ii
) C
1
(To
tal
for
Qu
esti
on
4 =
8 m
arks
)
Q
ues
tio
n
Nu
mb
er
An
swer
Ad
dit
ion
al G
uid
ance
M
ark
5(a
) A
1
5
(b)(
i)
C
1
5(b
)(ii
) SrC
O3(
s) →
SrO
(s)
+
CO
2(g)
• sp
ecie
s
(1
)
• st
ate
sym
bols
(
1)
2
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FC
!"#$%&'()*"+,"-(."/"-(0(1*/#',"*(234%5*5#$6(78)(5'(89":5%;$62#:<-"(1%%"%%:"';(=#;"$5#-%(>(?%%3"(@(>(A"4$3#$6(BC@D(E(!"#$%&'()*3,#;5&'(.5:5;"*(BC@D
F@
Q
ues
tio
n
Nu
mb
er
An
swer
Ad
dit
ion
al G
uid
ance
M
ark
*5
(b)(
iii)
Th
is q
uest
ion
asse
sses
a s
tude
nt’s
abi
lity
to s
how
a
cohe
rent
and
logi
cally
str
uctu
red
answ
er w
ith
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
appl
ied:
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 s
truc
ture
d w
ith
som
e lin
kage
s an
d lin
es o
f re
ason
ing,
sc
ores
4 m
arks
(3
mar
ks f
or in
dica
tive
co
nten
t an
d 1
mar
k fo
r pa
rtia
l str
uctu
re a
nd
som
e lin
kage
s an
d lin
es o
f re
ason
ing)
.
If t
here
are
no
linka
ges
betw
een
poin
ts,
the
sam
e fiv
e in
dica
tive
mar
king
poi
nts
wou
ld
yiel
d an
ove
rall
scor
e of
3 m
arks
(3
mar
ks f
or
indi
cative
con
tent
and
no
mar
ks f
or li
nkag
es).
6
Qu
esti
on
N
um
ber
A
nsw
er
A
dd
itio
nal
Gu
idan
ce
Mar
k
*5
(b)(
iii)
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:
• C
loud
ines
s /
milk
ines
s /
form
atio
n of
whi
te p
pte
due
to r
eact
ion
betw
een
limew
ater
and
car
bon
diox
ide
•
The
shor
ter
the
tim
e (f
or l
imew
ater
to
go c
loud
y),
the
fast
er t
he r
ate
of d
ecom
posi
tion
•
Rat
e of
de
com
posi
tion
de
pend
s on
m
etal
io
n si
ze
and
cha
rge
/ ch
arge
den
sity
•
B f
aste
r th
an A
as
Mg2+
(ra
dius
) sm
alle
r th
an C
a2+
• B
fas
ter
than
D a
s ch
arge
den
sity
of
Mg2+
gre
ater
th
an L
i+ /
hig
her
char
ge o
f M
g2+ h
as m
ore
effe
ct
than
sm
alle
r ra
dius
of
Li+
• C
doe
s no
t de
com
pose
as
K+ h
as (
rela
tive
ly)
larg
e ra
dius
and
sm
all c
harg
e
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
!"#$%&'()*"+,"-(."/"-(0(1*/#',"*(234%5*5#$6(78)(5'(89":5%;$62#:<-"(1%%"%%:"';(=#;"$5#-%(>(?%%3"(@(>(A"4$3#$6(BC@D(E(!"#$%&'()*3,#;5&'(.5:5;"*(BC@D
FB
!"#$%&'()*"+,"-(."/"-(0(1*/#',"*(234%5*5#$6(78)(5'(89":5%;$62#:<-"(1%%"%%:"';(=#;"$5#-%(>(?%%3"(@(>(A"4$3#$6(BC@D(E(!"#$%&'()*3,#;5&'(.5:5;"*(BC@D
F0
(To
tal
for
Qu
esti
on
5 =
10
mar
ks)
Qu
esti
on
N
um
ber
A
nsw
er
A
dd
itio
nal
Gu
idan
ce
Mar
k
6(a
)
• ca
lcul
atio
n of
% b
y m
ass
of o
xyge
n
(1)
• ev
alua
tion
of
num
ber
of m
oles
of
C,
H,
N a
nd O
(
1)
•
conf
irm
atio
n of
rat
io 1
: 6
: 2
: 2
(1)
Exam
ple
of c
alcu
lation
(%
by
mas
s of
) O
= 4
1.03
(%)
C
:
H
:
N
:
O
1
5.38
:
7
.69
:
35
.90
:
41.0
3 1
2.0
1.0
14
.0
16.0
1.2
8
:
7.6
9
:
2.5
6
:
2.
56
1
.28
:
7
.69
:
2
.56
:
2.5
6 1
.28
1
.28
1
.28
1.2
8
=
1
:
6
:
2
:
2
3
Qu
esti
on
N
um
ber
A
nsw
er
A
dd
itio
nal
Gu
idan
ce
Mar
k
6(b
)(i)
•
H2N
CO
ON
H4 →
2N
H3
+ C
O2
(1
)
Igno
re s
tate
sym
bols
, ev
en if
inco
rrec
t 1
6(b
)(ii
) A
1
6
(b)(
iii)
• al
l ele
ctro
n pa
irs
corr
ectly
show
n fo
r C
=O
an
d C
—O
(1
)
•
corr
ect
elec
tron
pai
rs f
or C
—N
bon
d an
d t
he —
NH
2 gr
oup
and
the
lone
pai
r on
N
(1)
2
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FF
!"#$%&'()*"+,"-(."/"-(0(1*/#',"*(234%5*5#$6(78)(5'(89":5%;$62#:<-"(1%%"%%:"';(=#;"$5#-%(>(?%%3"(@(>(A"4$3#$6(BC@D(E(!"#$%&'()*3,#;5&'(.5:5;"*(BC@D
FD
Q
ues
tio
n
Nu
mb
er
An
swer
Ad
dit
ion
al G
uid
ance
M
ark
6(b
)(iv
) •
shap
e: t
rigo
nal p
lana
r
(1)
•
just
ifica
tion
:
C=
O t
reat
ed a
s a
sing
le b
ond
pair
of
elec
tron
s/(s
hape
of
ion)
bas
ed o
n th
ree
bond
pai
rs
of e
lect
rons
(ar
ound
cen
tral
C a
tom
)/(s
hape
of
ion
base
d on
) th
ree
area
s of
ele
ctro
n de
nsity
(aro
und
cent
ral C
ato
m)/
(sha
pe o
f io
n ba
sed
on)
thre
e vo
lum
es o
f el
ectr
on d
ensi
ty (
arou
nd c
entr
al C
ato
m)
(1
) el
ectr
on p
airs
/ele
ctro
n re
gion
s re
pel t
o po
sition
s of
m
axim
um s
epar
atio
n/m
inim
um r
epul
sion
(1)
Rej
ect
‘ato
ms
repe
l’/‘b
onds
rep
el’/
Just
‘e
lect
rons
rep
el’
3
(To
tal
for
Qu
esti
on
6 =
10
mar
ks)
Qu
esti
on
N
um
ber
A
nsw
er
A
dd
itio
nal
Gu
idan
ce
Mar
k
7(a
) An
expl
anat
ion
that
mak
es r
efer
ence
to:
• bo
iling
tem
pera
ture
s in
crea
se f
rom
flu
orin
e to
io
dine
(1
)
• (a
s) m
ore
elec
tron
s pe
r (X
2) m
olec
ule
from
flu
orin
e
to io
dine
(1)
• (t
here
fore
the
) st
reng
th o
f Lo
ndon
for
ces
incr
ease
s fr
om f
luor
ine
to io
dine
(1)
Pl
us o
ne
from
:
• (s
o) m
ore
ener
gy r
equi
red
to s
epar
ate
mol
ecul
es
(1)
•
(so)
mor
e en
ergy
req
uire
d to
bre
ak t
he
inte
rmol
ecul
ar f
orce
s
(1
)
Allo
w m
olec
ules
incr
ease
in s
ize
/ m
ass
from
flu
orin
e to
iodi
ne
Allo
w ‘m
ore
Lond
on f
orce
s’ f
rom
flu
orin
e to
io
dine
Allo
w ‘m
ore
heat
’ nee
ded
to s
epar
ate
mol
ecul
es
Allo
w m
ore
ener
gy r
equi
red
to o
verc
ome
the
in
term
olec
ular
att
ract
ions
Rej
ect
‘mor
e en
ergy
req
uire
d to
bre
ak c
oval
ent
bond
s’
Allo
w r
ever
se a
rgum
ent
4
7(b
) D
1
7
(c)(
i)
• H
2SO
4 +
2H
+ +
2e–
→ S
O2
+ 2
H2O
o
r
• SO
42– +
4H
+ +
2e– →
SO
2 +
2H
2O
(1
)
Allo
w m
ultipl
es
Igno
re s
tate
sym
bols
, ev
en if
inco
rrec
t
1
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FG
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FH
Q
ues
tio
n
Nu
mb
er
An
swer
Ad
dit
ion
al G
uid
ance
M
ark
7(c
)(ii
) •
H2S
O4
+ 2
H+ +
2Br– →
SO
2 +
2H
2O +
Br 2
o
r
• SO
42– +
4H
+ +
2Br– →
SO
2 +
2H
2O +
Br 2
(1)
No
ecf
from
(c)
(i)
Allo
w m
ultipl
es
Igno
re s
tate
sym
bols
, ev
en if
inco
rrec
t
1
7(c
)(ii
i)
• re
duci
ng a
gent
/ele
ctro
n do
nor/
redu
ces
sulfu
ric
acid
/red
uces
H2S
O4
(1
)
1
7(d
)(i)
•
hydr
ogen
chl
orid
e /
HC
l
(
1)
Ig
nore
sta
te s
ymbo
ls
1
7(d
)(ii
) O
bser
vation
: •
blac
k so
lid /
gre
y so
lid /
pur
ple
vapo
ur
OR
pun
gent
gas
O
R y
ello
w s
olid
O
R g
as s
mel
ling
of r
otte
n eg
gs
(
1)
Equa
tion
s:
• N
aI +
H2S
O4 →
NaH
SO
4 +
HI
(1
)
• 2H
I +
H2S
O4 →
I 2
+
SO
2 +
2H
2O
OR
6H
I +
H2S
O4 →
3I
2 +
S +
4H
2O
OR
8H
I +
H2S
O4 →
4I
2 +
H2S
+ 4
H2O
(1)
2nd e
quat
ion
mus
t m
atch
obs
erva
tion
mad
e
Allo
w p
urpl
e so
lid
Allo
w 2
NaI
+ H
2SO
4 →
Na 2
SO
4 +
2I
Allo
w c
ombi
nation
s of
bot
h eq
uation
s fo
r bo
th
mar
ks
e.g.
2N
aI +
3H
2SO
4 →
2N
aHSO
4 +
I2
+ S
O2
+
2H2O
3
Qu
esti
on
N
um
ber
A
nsw
er
A
dd
itio
nal
Gu
idan
ce
Mar
k
7(d
)(ii
i)
An
expl
anat
ion
that
mak
es r
efer
ence
to
the
follo
win
g po
ints
:
• io
dide
ions
are
bet
ter
redu
cing
age
nts
(tha
n ch
lori
de io
ns)
(1)
•
beca
use
iodi
de io
ns lo
se e
lect
rons
mor
e re
adily
/
elec
tron
s in
iodi
de io
ns a
re le
ss s
tron
gly
held
by
the
nucl
eus
(1
)
Allo
w H
I is
a b
ette
r re
duci
ng a
gent
(th
an H
Cl)
Allo
w r
ever
se a
rgum
ent
2
(To
tal
for
Qu
esti
on
7 =
14
mar
ks)
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FI
!"#$%&'()*"+,"-(."/"-(0(1*/#',"*(234%5*5#$6(78)(5'(89":5%;$62#:<-"(1%%"%%:"';(=#;"$5#-%(>(?%%3"(@(>(A"4$3#$6(BC@D(E(!"#$%&'()*3,#;5&'(.5:5;"*(BC@D
FJ
Qu
esti
on
N
um
ber
A
nsw
er
A
dd
itio
nal
Gu
idan
ce
Mar
k
8(a
) •
corr
ect
calc
ulat
ion
of a
ll m
ean
titr
es (
23.1
5 an
d 22
.25
and
22.3
0 an
d 22
.70
and
22.2
0)
(1)
• co
ncor
dant
titre
s tick
ed (
2, 3
and
5)
and
cal
cula
tion
of
mea
n titr
e =
22.
25 (
cm3 )
(1
)
(i.e
. th
ose
that
agr
ee w
ithi
n ±
0.20
cm
3 )
2
8(b
)(i)
•
calc
ulat
ion
of n
umbe
r of
mol
es t
rich
loro
etha
noic
ac
id (
= n
umbe
r of
mol
es o
f N
aOH
)
(1)
•
rear
rang
emen
t an
d ev
alua
tion
of
tric
hlor
oeth
anoi
c
acid
con
cent
ration
in m
ol d
m-3
(
1)
•
eval
uation
of
Mr o
f tr
ichl
oroe
than
oic
acid
and
co
nver
sion
to
conc
entr
atio
n in
g d
m-3
, to
1 d
p
(
1)
Allo
w e
cf f
or s
teps
in c
alcu
lation
; in
clud
ing
for
final
ans
wer
dep
ende
nt o
n ro
undi
ng in
ste
ps o
f th
e ca
lcul
atio
n.
Cor
rect
ans
wer
with
no w
orki
ng t
o 1d
p sc
ores
3
mar
ks
Exam
ple
of c
alcu
lation
mol
es a
cid
= m
oles
NaO
H =
(0.
130
x 25
.0)
10
00
= 3
.25
x 10
–3 / 0
.003
25 (
mol
) co
ncen
trat
ion
of a
cid
= 3
.25
x 10
–3 x
100
0
22
.25
=
0.1
46…
mol
dm
-3
conc
entr
atio
n ac
id in
g d
m3 =
0.1
46…
x 1
63.5
=
23.
9 g
dm-3
3
Qu
esti
on
N
um
ber
A
nsw
er
A
dd
itio
nal
Gu
idan
ce
Mar
k
8(b
)(ii
) •
calc
ulat
ion
of n
umbe
r of
gra
ms
of
tric
hlor
oeth
anoi
c ac
id in
250
cm
3
(
1)
•
calc
ulat
ion
of %
pur
ity,
sho
win
g it is
< 9
9.9%
(1)
O
R •
conv
ersi
on o
f m
easu
red
mas
s in
to t
heor
etic
al
conc
entr
atio
n in
g d
m-3
(1)
• ca
lcul
atio
n of
% p
urity,
sho
win
g it is
< 9
9.9%
(1)
Exam
ple
of c
alcu
lation
: m
ass
acid
in 2
50 c
m3
= 2
3.9
x 2
50
= 5
.975
g
1000
pu
rity
= 5
.975
x
100
%
= 9
6.4
%,
whi
ch is
<99
.9%
6.
2
OR
th
eore
tica
l con
cent
ration
= 6
.2 x
100
0 =
24.
8 g
dm-3
2
50
puri
ty =
23.
9
x 10
0%
= 9
6.4
%,
whi
ch is
< 9
9.9%
2
4.8
Allo
w e
cf o
n va
lue
in (
b)(i
)
2
8(c
)(i)
• (e
ach
mas
s re
adin
g) =
1.6
1 %
an
d (e
ach
pipe
tte
read
ing)
= 0
.160
%
(1
)
Exam
ple
of c
alcu
lation
Ea
ch m
ass
read
ing:
(±
) 2
x 0.
05 x
100
%
6.
2
=
1.6
1%
Each
pip
ette
vol
ume:
±
0.0
4 x
100%
25
.0
= 0
.160
%
1
8(c
)(ii
) An
expl
anat
ion
that
mak
es r
efer
ence
to:
• To
tal %
err
or =
2.4
2%
(1)
• cl
aim
is n
ot c
orre
ct b
ecau
se 9
6.4
± 2
.42%
is
still
low
er t
han
the
man
ufac
ture
r’s
valu
e of
99
.9%
(1
)
ecf
on v
alue
obt
aine
d in
(c)
(i)
2
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DC
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D@
Q
ues
tio
n
Nu
mb
er
An
swer
Ad
dit
ion
al G
uid
ance
M
ark
8(d
) M
axim
um t
hre
e m
arks
for
issu
e id
entifie
d M
axim
um t
hre
e m
arks
for
impr
ovem
ent
iden
tifie
d w
hich
m
ust
be li
nked
with
asso
ciat
ed is
sue
iden
tifie
d
• is
sue:
mas
s of
(so
lid)
acid
not
acc
urat
ely
wei
ghed
ou
t
(1
) •
impr
ovem
ent:
wei
gh m
ass
of a
cid
by
diff
eren
ce/r
inse
out
the
wei
ghin
g bo
ttle
/use
a
bala
nce
read
ing
to 2
d.p
./us
e a
mor
e pr
ecis
e ba
lanc
e
(1
)
• is
sue:
som
e ac
id w
ill b
e le
ft in
the
bea
ker/
som
e ac
id
will
not
be
tran
sfer
red
to t
he v
olum
etri
c fla
sk
(1)
•
impr
ovem
ent:
rin
se o
ut t
he b
eake
r (i
n w
hich
the
so
lid a
cid
was
dis
solv
ed)
and
add
the
was
hing
s to
th
e vo
lum
etri
c fla
sk
(1)
•
issu
e: in
suff
icie
nt m
ixin
g of
the
so
lution
/con
cent
ration
of
the
solu
tion
will
not
be
unifo
rm
(
1)
•
impr
ovem
ent:
inve
rt t
he v
olum
etri
c fla
sk (
seve
ral
tim
es)
(1)
•
issu
e: b
uret
te n
ot r
inse
d
(1
)
• im
prov
emen
t: b
uret
te s
houl
d be
rin
sed
with
acid
so
lution
bef
ore
use
(1
)
Rej
ect
use
of a
‘mor
e ac
cura
te’ b
alan
ce
Allo
w p
ipet
te n
ot r
inse
d w
ith
sodi
um h
ydro
xide
6
(To
tal
for
Qu
esti
on
8 =
16
mar
k)
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DB