specimen ms - paper 1 edexcel chemistry as-level

!"#$%&'()*"+,"-(."/"-(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

!"#$%&'()*"+,"-(."/"-(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

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

!"#$%&'()*"+,"-(."/"-(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

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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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

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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)

!"#$%&'()*"+,"-(."/"-(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

DB