chemistry a level paper 3 mark scheme · question number answer additional guidance mark 3(a)(i)...

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


132


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)


134


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)


136


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)


138


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)


140


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)


142


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


144


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


146


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


148


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

)

chemistry a level paper 3 mark scheme · question number answer additional guidance mark 3(a)(i)...

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