Paper-5
(All Variants Yearly worked Solutions)
(2010-2017)
Article No. 235
Author:
Niaz Hussain
Zafar Iqbal
(Lahore Grammar School)
(Beaconhouse, LGS, KIMS, Roots)
Review Board:
o Shehbaz Ahmad
(LGS, Beaconhouse)
o Umar Zaman Khattak
(GACS)
o Rasheed Ahmed
(BDC)
o M. Luqman
(Aitchison College)
o Hafiz Farooq
(KIMS)
o Shahzad Zia
(GHA)
o Niaz Ahmed Awan
(Beaconhouse, LACAS, SICAS,
KIMS)
o Tanvir Gill
(BDC, City School)
o Mubashar Sulehri
(LGS, Bloomfield Hall, Roots)
o Shahzad Khokhar
(BDC)
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted, in
any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior
written permission of the Publisher.
Title
Chemistry A-Level Paper-5 Yearly Work Solutions (Article#235)
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CONTENTS
1
October/November 2017 Paper 52
--------------------------------------------------------------
2
2
October/November 2017 Paper 51
--------------------------------------------------------------
9
3
May/June 2017 Paper 52
--------------------------------------------------------------
18
4
May/June 2017 Paper 51
--------------------------------------------------------------
25
5
October/November 2016 Paper 52
--------------------------------------------------------------
33
6
October/November 2016 Paper 51
--------------------------------------------------------------
41
7
May/June 2016 Paper 52
--------------------------------------------------------------
50
8
May/June 2016 Paper 51
--------------------------------------------------------------
58
9
October/November 2015 Paper 52
--------------------------------------------------------------
65
10
October/November 2015 Paper 53
--------------------------------------------------------------
71
11
May/June 2015 Paper 52
--------------------------------------------------------------
77
12
May/June 2015 Paper 51
--------------------------------------------------------------
83
13
October/November 2014 Paper 52
--------------------------------------------------------------
89
14
May/June 2014 Paper 52
--------------------------------------------------------------
95
15
October/November 2013 Paper 51&52------------------------------------------------------------
100
16
October/November 2013 Paper 53
--------------------------------------------------------------
107
17
May/June 2013 Paper 52
--------------------------------------------------------------
113
18
October/November 2012 Paper 52
--------------------------------------------------------------
119
19
May/June 2012 Paper 52
--------------------------------------------------------------
125
20
October/November 2011 Paper 52
--------------------------------------------------------------
133
21
May/June 2011 Paper 52,51&53
--------------------------------------------------------------
140
22
October/November 2010 Paper 52&53------------------------------------------------------------
147
23
October/November 2010 Paper 51
--------------------------------------------------------------
154
24
May/June 2010 Paper 52
--------------------------------------------------------------
161
25
May/June 2010 Paper 51
--------------------------------------------------------------
168
2
Chemistry A-Level P-5
November 2017/P52
October/November 2017 Paper 52
1.
O/N 17/P52/Q1
Verdigris is a green pigment that contains both copper(II) carbonate, CuCO3, and copper(II)
hydroxide, Cu(OH)2, in varying amounts.
Both copper compounds react with dilute hydrochloric acid.
CuCO3(s) + 2HCl(aq)  CuCl2(aq) + CO2(g) + H2O(l)
Cu(OH)2(s) + 2HCl(aq)  CuCl2(aq) + 2H2O(l)
(a)
You are to plan an experiment to determine the percentage of copper(II) carbonate
in a sample of verdigris. Your method should involve the reaction of verdigris with
excess dilute hydrochloric acid.
You are provided with the following materials.

0.5 g of verdigris

10.0 mol dm–3 hydrochloric acid, HCl(aq)

commonly available laboratory reagents and equipment.
You may assume that any other material present in verdigris is unaffected by heating
and is not acidic or basic.
(i)
Explain why a titration would not be a suitable method to determine the
percentage of copper(II) carbonate in a sample of verdigris
[1]
–3
(ii)
The 10.0 mol dm HCl(aq) is corrosive and too concentrated for use in the
experiment.
Describe how you would accurately prepare 250cm 3 of 0.500mol dm–3
hydrochloric acid from the 10.0 mol dm –3 HCl(aq) provided.
Include details of any apparatus, including their capacities in cm 3, you would
use.
[2]
(iii)
Identify a dependent variable that you could measure to determine the
percentage of copper(II) carbonate in verdigris.
Your answer should be based on a difference that you can identify between
the reactions of copper(II) carbonate and copper(II) hydroxide with excess
dilute hydrochloric acid.
[1]
(iv)
Draw a diagram to show how you would set up apparatus and chemicals to
measure the dependent variable identified in (iii).
Label your diagram.
[2]
(v)
Using the axes below, sketch a graph to show how the dependent variable you
identified in (iii) would change during your experiment.
Extend the graph beyond the point at which the reaction is complete.
Label both axes.
0
[2]
(vi)
(vii)
A student carries out this experiment once.
Describe how this one experiment should be carried out to ensure that the
results are as accurate as possible.
[1]
A student suspected that their 0.5 g sample of verdigris only contained CuCO 3.
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
3
Chemistry A-Level P-5
(b)
(c)
November 2017/P52
Calculate the minimum volume, in cm 3, of 0.500 mol dm –3 HCl(aq) that would
be needed for the complete reaction of the sample if the student was correct.
[Mr: CuCO3 = 123.5]
Volume of 0.500 mol dm–3 HCl(aq) = ………… cm3 [2]
The following information gives some of the hazards associated with the chemicals
used in the procedure.
Copper(II) carbonate hydroxide
The solid is classified as health hazard and is
harmful if swallowed. Dispose of by reacting no
more than 60 g in 1 dm 3 of warm 1 mol dm–3
ethanoic acid before pouring down a foul-water
drain.
Hydrochloric acid
Solutions equal to or more concentrated than
6.5moldm–3 are classified as corrosive;
solutions equal to or more concentrated than
2.7 mol dm–3 but less concentrated than 6.5
mol dm–3 are classified as moderate hazard
and are irritant.
Describe one relevant precaution, other than eye protection and a lab coat, that should
be taken to keep the risk associated with the chemicals used to a minimum. Explain
your
answer.
[1]
Azurite is a blue copper-containing mineral. The copper compound in azurite has the
formula Cu3(CO3)2(OH)2. This copper compound reacts with sulfuric acid according to
the reaction shown.
Cu3(CO3)(OH)2(s) + 3H2SO4(aq)  3CuSO4(aq) + 2CO2(g) + 4H2O(l)
A student performed a series of titrations on 1.50 g samples of solid azurite using 0.400
mol dm–3 sulfuric acid.
It can be assumed that any other material present in azurite does not react with sulfuric
acid.
The titration data is given in the table.
experiment
final reading /
cm 3
initial reading /
titre/cm3
cm 3
rough
1
2
25.50
24.05
32.70
0.00
0.15
8.30
25.50
23.90
24.40
The indicator for the titration was bromophenol blue.
The student concluded that 24.15cm 3 of 0.400mol dm–3 sulfuric acid completely
neutralized 1.50 g of azurite.
(i)
Using the student’s value of 24.15cm3, calculate the percentage by mass of
Cu3(CO3)2(OH)2 in the sample of azurite.
Write your answer to three significant figures.
[Mr: Cu3(CO3)2(OH)2 = 344.5]
percentage by mass of Cu3(CO3)2(OH)2 in the sample of azurite = ……………%
[3]
(ii)
Identify two possible problems with the student’s titration and suggest
improvements to it.
Problem 1
Improvement 1
Problem 2
Improvement 2
[3]
Solution:
(a)
(i)
(ii)
CuCO3 and Cu(OH)2 both react with HCl to form CuCl2, so one of these compound
cannot be determined.
[1]
Acid provided (c1v2) = concentration to be prepared (c2v2)
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
4
Chemistry A-Level P-5
November 2017/P52
10 × v1 = 0.5 × 250
v1 =
(iii)
0.5  250
10
Take 12.5 cm3 of 10 mol dm–3 HCl, with the help of pipette or Burette add it into
measuring flask, and add distilled water upto mark.
[2]
Carbon dioxide gas will evolved, will be dependent variable, measure the volume of
gas will give the amount of CaCO3
[1]
(iv)
Gas Syringe
Thistle Funnel
Conical Flask
l
Verdigris
[2]
(v)
Volume
of gas
3
(cm )
0
(vi)
(vii)
Sealed all the apparatus carefully in order to avoid from gas loss.
mol of CuCO3 = 0.5 g/123.5 = 4.05 × 10–3 mol.
CuCO3 + 2HCl  CuCl2 + CO2 + H2O
1:2
mole of HCl
= 2 × 4.05 × 10–3
= 8.10 × 10–3 mole
Volume of HCl =
(b)
Time (sec)
[2]
[1]
8.10  103
 1000
0.5
volume of 0.500 mol dm–3 HCl(aq) = 16.2 cm3 [2]
10 mol
HCl is corrosive, so wear gloves also used mask in order to avoid from fumes
of HCl and CuCO3 being swallowed.
[1]
dm–3
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
5
Chemistry A-Level P-5
(c)
moles of H2SO4 in 24.15 cm3
(i)
November 2017/P52
=
0.4  24.15
1000
= 9.66 × 10–3 mol.
9.66  103
= 3.22 × 10–3 mol.
3
moles of Cu3(CO3)2(OH)2
=
mass of Cu3(CO3)2(OH)2
= 3.22 × 10–3 × 344.5
= 1.11 g.
% by mass
=
1.11
 100
1.50
= 74.0 %
percentage by mass of Cu3(CO3)2(OH)2 in the sample of azurite = 74.0 %
(ii)
Problem 1: Titers may not be concordant.
Improvement 1: Repeat titration to get titer difference of 0.1 cm 3.
Problem 2: Colour change of indicator may be marked with the colour of the
titrant.
Improvement 2: Student should try other indicator.
2.
[3]
[3]
O/N 17/P52/Q2
Activated charcoal is a form of carbon with a very high surface area. It can be used to remove
impurities from mixtures. It does this by a process called adsorption, where particles of the
impurity bond (adsorb) to the activated charcoal surface.
A student wanted to determine the ability of activated charcoal to absorb a blue dye (the
impurity) from aqueous solution.
The equation that links the mass of activated charcoal with the amount of blue dye adsorbed is
shown.
Y
c m
 
X

D
log    A b log[ X ]
m
difference in concentration of dye (in g dm –3) before and after
adsorption.
m
=
mass of activated charcoal (in g)
[X]
=
final concentration of dye (in g dm –3) after adsorption
A and b are constants.
The student used the following procedure to investigate the ability of activated charcoal to
adsorb a blue dye from an aqueous solution.

Place a 50.0 cm3 sample of a 25.00 g dm–3 solution of blue dye in a conical flask.

Add a weighed mass of activated charcoal to the flask.

Stir the contents of the flask for three minutes and then leave for one hour.

Filter the mixture.

Determine the final concentration of the blue dye, [X], by colorimetry.

Repeat the procedure using different masses of activated charcoal.
(a)
The final concentrations of blue dye after carrying out the procedure, [X], are shown in
the table.
(i)
Process the results to complete the table.
Record your data to two decimal places.
D
=
mass of
activated
charcoal, m
/g
Initial
concentration
of blue dye
/g dm–3
final
concentration
of blue dye, [X]
/g dm–3
0.20
25.00
0.25
0.30
difference in
concentration
of blue dye, D
/g dm–3
D
m
D
log  
m
0.96
120.20
2.08
25.00
0.69
97.24
1.99
25.00
0.60
81.33
1.91
P-5 (Planning, Analysis and Evaluation)
Log
[X]
Yearly Worked Solutions
6
Chemistry A-Level P-5
November 2017/P52
0.35
25.00
0.41
70.26
1.85
0.40
25.00
0.33
61.68
1.79
0.45
25.00
0.27
54.96
1.74
0.50
25.00
0.23
49.54
1.69
0.55
25.00
0.20
45.09
1.65
0.60
25.00
0.17
41.38
1.62
(ii)
[3]
By considering the data in the first three columns, state the effect of increasing
the mass of activated charcoal, m, on the amount of adsorption that occurs.
Explain this effect.
[2]
Plot a graph on the grid to show the relationship between log   and log [X].
m
Use a cross (×) to plot each data point. Draw the straight line of best fit.
D
(b)
2.15
2.10
2.05
2.00
1.95
1.90
D
log m
1.85
1.80
1.75
1.70
1.65
1.60
1.55
–0.90
–0.80
P-5 (Planning, Analysis and Evaluation)
–0.70
–0.60
–0.50
–0.40
log [X]
–0.30
–0.20
–0.10
0.00
Yearly Worked Solutions
7
Chemistry A-Level P-5
(c)
(d)
November 2017/P52
[2]
Circle the most anomalous point on the graph.
Suggest a reason why this anomaly may have occurred during the experimental
procedure.
[2]
(i)
Use the graph to determine the gradient of the best-fit line. State the
co-ordinates of both points you used in your calculation.
Determine the value of the constant b.
co-ordinates 1
co-ordinates 2
gradient = …………..………….
b = …………………. [2]
(ii)
Use the graph to determine a value for A.
A = …………………………. [1]
Solution:
(a)
(i)
mass of
activated
charcoal,
m/g
Initial
concentration
of blue dye/
g dm–3
final
concentration
of blue dye,
[X]/g dm–3
difference in
concentration
of blue dye, D
/g dm–3
D
m
D
log  
m
Log
[X]
0.20
25.00
0.96
24.04
120.20
2.08
–0.02
0.25
25.00
0.69
24.31
97.24
1.99
–0.16
0.30
25.00
0.60
24.40
81.33
1.91
–0.22
0.35
25.00
0.41
24.59
70.26
1.85
–0.39
0.40
25.00
0.33
24.67
61.68
1.79
–0.48
0.45
25.00
0.27
24.73
54.96
1.74
–0.57
0.50
25.00
0.23
24.77
49.54
1.69
–0.64
0.55
25.00
0.20
24.80
45.09
1.65
–0.70
0.60
25.00
0.17
24.83
41.38
1.62
–0.77
(ii)
[3]
Greater the mass of activated charcoal, greater would be the adsorption.
It will provide greater surface area for adsorption, so greater amount will be
adsorbed.
[2]
(b)
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
8
Chemistry A-Level P-5
November 2017/P52
2.15
2.10
2.05
2.00
1.95
1.90
log D
m
1.85
1.80
1.75
1.70
1.65
1.60
1.55
–0.90
(c)
(d)
–0.80
–0.70
–0.60
–0.50
–0.40
log [X]
–0.30
–0.20
–0.10
0.00
[2]
At (–0.22, 1.91), this is due to, less mass of activated charcoal taken, or less surface are so
less adsorption.
[2]
(i)
Co-ordinate 1
(1.69, 2.08)
Co-ordinates 2
(–0.02, –0.64)
m
y 2  y1
2.08  1.69

x2  x1 0.02  ( 0.64)
= 0.629
(ii)
P-5 (Planning, Analysis and Evaluation)
gradient = 0.629
b = 0.629
[2]
A = 2.09 [1]
Yearly Worked Solutions
9
Chemistry A-Level P-5
November 2017/P51
October/November 2017 Paper 51
1.
O/N 17/P51/Q1
Iodide ions, I–, and persulfate ions, S2O8–2, react according to the following equation.
2I–(aq) + S2O8 –2(aq)  I2(aq) + 2SO4 2–(aq)
The rate of reaction between these ions can be determined from the time it takes for a certain
amount of iodine, I2(aq), to be produced.

A mixture of solutions is prepared, containing known volumes of
○
aqueous ammonium persulfate, (NH4)2S2O8(aq),
○
aqueous sodium thiosulfate, Na2S2O3(aq),
○
starch indicator.

A known volume of aqueous potassium iodide, KI(aq), is added to this mixture and a
timer is started.

After the reactants are mixed, they react slowly to produce iodine, I2(aq).

Any iodine initially produced is removed by a reaction with thiosulfate ions.
I2(aq) + 2S2O32–(aq)  2I–(aq) + S4O62–(aq)

Iodine, I2(aq), is continuously removed until all of the thiosulfate ions have been used
up.

After that time any I2(aq) that is produced turns the starch indicator blue.

The time of the first appearance of the blue colour is recorded.

This procedure is repeated with different volumes of reactants, keeping the total volume
of the reaction mixture constant by adding the required volume of distilled water.
You are to plan a series of experiments to determine the effect of changing the concentration
of iodide ions on the rate of reaction.
You are provided with the following materials.
Solid ammonium persulfate, (NH4)2S2O8(s)
0.20 mol dm–3 aqueous KI, a source of I–(aq)
0.0050 mol dm–3 aqueous Na2S2O3, a source of S2O32–(aq)
Starch indicator
(a)
Calculate the mass of (NH4)2S2O8(s) that would be required to prepare 250 cm 3
of a standard solution of concentration 1.00 mol dm–3.
[Ar, values: N, 14.0; H, 1.0; S, 32.1; O. 16.0]
mass of (NH4)2S2O8(s) = ……………. g [1]
Describe how, after weighing the mass calculated in (i), you would prepare this
standard solution for use in your experiment.
Give the name and capacity, in cm3, of any apparatus used.
[1]
Explain how the use of starch solution improves the accuracy of the
experiment.
[1]
(i)
(ii)
(iii)
(b)
A student planned five experiments to investigate the effect of iodide concentration,
[I–], on the rate of reaction. The table shows the volumes used in experiment 1.
Complete the table for experiment 2 to 5.
experiment
volume of 1.00
mol dm–3
(NH4)2S2O8(aq)
/cm3
volume of
0.20 mol
dm–3
KI(aq)
/cm3
volume of
water
/cm3
volume of
0.0050 mol
dm–3
Na2S2O3(aq)
/cm3
volume of
starch
solution
/cm3
1
25.0
10.0
0.0
5.0
1.0
2
5.0
1.0
3
5.0
1.0
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solution
10
Chemistry A-Level P-5
(c)
(d)
(e)
November 2017/P51
4
5.0
1.0
5
5.0
1.0
[3]
In a different experiment, a student mixed the following solutions and measured the
time taken for the reaction.

10.0 cm3 of 1.00 mol dm–3 (NH4)2S2O8(aq)

5.0 cm3 of 0.0050 mol dm–3 (Na2S2O3(aq)

5.0 cm3 of 0.20 mol dm–3 KI(aq)

1.0 cm3 of starch indicator
(i)
The time taken for the blue colour to appear was 134 seconds (to the nearest
second). Calculate the rate of production of moles of I2, in mold m–3 s–1.
rate of production of moles of I2 = ………………… mold dm–3 s–1 [3]
(ii)
What should the student have done to make sure that the results were reliable?
[1]
(iii)
The 5.0 cm3 of 0.0050 mol dm –3 Na2S2O3(aq) was measured using a 50cm 3
burette which had graduations every 0.1 cm 3.
Calculate the maximum percentage error in the measured volume of this
solution.
percentage error = ……………….% [1]
A second student tried to perform the same experiment but found that the reaction
mixture turned blue immediately after KI(aq) was added.
State what error the student had made.
[1]
The following information gives some of the hazards associated with the chemicals
used in the procedure.
Ammonium persulfate
Solid is oxidizing and hazardous to the environment.
Contact with combustible material may cause fire. It is
classified as health hazard, is harmful if swallowed and
is irritating to eyes, respiratory system and skin.
Solution equal to or more concentrated then 0.2 mol dm –
3 should be labelled health hazard and hazardous to
the environment. Solutions equal to or more
concentrated than 0.05 mol dm –3 but less concentrated
than 0.2 mol dm–3 should be labelled health hazard.
Potassium iodide
All solutions are low hazards.
Sodium thiosulfate
All solutions are low hazards.
Describe one relevant precaution, other than eye protection and a lab coat that should
be taken to keep the risk associated with the chemicals used to a minimum. Explain
your
answer.
[1]
Solution:
(a)
(i)
n=
mass
250
×
Mr
1000
mass =
(ii)
(iii)
1 228.5  250
1000
mass of (NH4)2S2O4(s) = 57.12 g [1]
Dissolve all Na2S2O3 in 100 cm3 distilled water in a beaker. Now add this solution in
measuring flask and add distilled water in it to make the volume upto mark. Shake
the solution to homogenise the solution.
[2]
Starch solution give the sharp and point of reaction.
[1]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solution
11
Chemistry A-Level P-5
November 2017/P51
(b)
experiment
volume of 1.00
mol dm–3
(NH4)2S2O8(aq)
/cm3
volume of
0.20 mol
dm–3
KI(aq) /cm3
volume
of
water
/cm3
volume of
0.0050 mol
dm–3
Na2S2O3(aq)
/cm3
volume of
starch
solution
/cm3
1
25.0
10.0
0.0
5.0
1.0
2
25.0
8.0
2.0
5.0
1.0
3
25.0
6.0
4.0
5.0
1.0
4
25.0
4.0
6.0
5.0
1.0
5
25.0
2.0
8.0
5.0
1.0
[3]
(c)
(i)
moles of S2S3–2 =
0.005
5
1000
= 2.5 × 10–5 mole
moles of I2
=
2.5  105
 1.25  105 mol
2
concentration of I2 =
1.5  105
 1000
21
= 5.94 × 10–4 mol dm3
Rate
(ii)
(iii)
=
5.94  104
134
rate of production of moles of I2 = 4.44 × 10–6 mol dm–3 s–1
Repeal experiments to get average.

[3]
[1]
2  0.05
 100
5
percentage error = 2 %
(d)
(e)
[1]
Thiosulphate not added yet.
[1]
Wear glove (skin irritant, wear face mask, so not be swallowed, keep combustible materials
away.
[1]
O/N 17/P51/Q2
2.
Dilute sulfuric acid, H2SO4(aq), can be electrolyzed using platinum electrodes and a direct
current. Hydrogen gas is produced at the cathode and oxygen gas is produced at the anode.
The two gases are collected separately in burettes filled with dilute sulfuric acid placed over
each electrode.
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solution
12
Chemistry A-Level P-5
November 2017/P51
reaction at electrode in burette 1: 2H+(aq) + 2e–  H2(g)
reaction at electrode in burette 2: H2O(l) 
1
O2(g) + 2H+(aq) + 2e–
2
The production of hydrogen gas over time can be measured, and the data used to determine
the charge of one mole of electrons, known as the Faraday constant, F.
(a)
The volumes of hydrogen gas produced during the electrolysis process are recorded
in the table.
Process the results to calculate the volume of hydrogen gas produced, in cm 3, and the
charge passed, in coulombs, C.
charge (C) = current (A) × time (s)
The current was kept constant at 0.80 A.
(b)
time/s
reading on
burette 1/cm3
volume of hydrogen
gas produced / cm3
0
46.20
0.00
50
41.20
100
36.20
150
31.45
200
25.80
250
20.80
300
16.40
350
11.45
400
6.80
450
1.50
charge
passed / C
[2]
Plot a graph on the grid to show the relationship between volume of hydrogen gas
produced and charge passed.
Use a cross (×) to plot each data point. Draw the straight line of best fit.
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solution
13
Chemistry A-Level P-5
November 2017/P51
(c)
Do you think the results obtained (a) are reliable? Explain your answer.
(d)
(i)
(ii)
(iii)
[2]
[1]
The gradient of the line of best fit gives the volume of hydrogen gas produced
per coulomb.
Use the graph to determine the gradient of the line of best fit.
State the co-ordinates of both points you used in your calculation.
co-ordinates 1 ……………
co-ordinates 2: …………………
gradient = ……………… cm 3 C–1 [2]
Calculate the number of moles of hydrogen gas produced per coulomb.
If you were unable to obtain an answer for (d)(i), you may use the value 0.148
cm2 C–1, but this is not the correct answer.
[The molar volume of gas = 24.0 dm3 at room temperature and pressure.]
……………mol C–1 [1]
Use your answer to (ii) and the half-equation for the production of H2(g) to
calculate a numerical value for the Faraday constant (the charge of 1 mole of
electrons).
2H+(aq) + 2e–  H2(g)
……………. C mol–1 [1]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solution
14
Chemistry A-Level P-5
(e)
(i)
(ii)
(f)
November 2017/P51
The graph below shows the relationship between volume H 2(g) produced at
the cathode and time, in a similar experiment.
Draw a line on the graph to show the relationship between volume of O 2(g)
produced at the anode and time in this experiment.
[1]
Suggest why the volume of O2(g) measured in this experiment might be less
than that shown by your drawn line.
Assume that no gas is lost from leaks.
[1]
In these experiments, the pressure of the gas inside the burette is assumed to be
atmospheric pressure, Patm.
However, the pressure of water vapour and the mass of the solution in the burette
change the pressure of the gas to Pnew .
The expression below shows the relationship between Pnew and Patm.
Pnew = Patm – 2.81 – (9.81 × height of solution in burette)
(i)
Use the expression to sketch a graph on the axes below to show the
relationship between Pnew and the height of solution in the burette.
[1]
Show how Pnew changes the value of the Faraday constant calculated at Pnew
in (d) (iii).
Explain your answer.
[1]
A student’s teacher suggested it would be cheaper to use copper rather than platinum
electrodes in the electrolysis of dilute sulfuric acid.
(ii)
(g)
half-equation
E o /V
2H+(aq) + 2e–  H2(g)
0.00
Cu2+(aq) + 2e–  Cu(s)
+0.34
1
O2(g) + 2H+(aq) + 2e–  H2O(l)
2
+1.23
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solution
15
Chemistry A-Level P-5
November 2017/P51
Using the information in the table, suggest what effect, if any, the use of copper
electrodes would have on the volume of gas produced at each electrode. Explain your
answer.
cathode:
anode
[3]
Solution:
(a)
time/s
reading on
burette 1/cm3
volume of hydrogen
gas produced / cm3
charge
passed / C
0
46.20
0.00
0
50
41.20
5.00
40
100
36.20
10.00
80
150
31.45
14.75
120
200
25.80
20.40
160
250
20.80
25.40
200
300
16.40
29.80
240
350
11.45
34.75
280
400
6.80
39.40
320
450
1.50
44.70
360
[2]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solution
16
Chemistry A-Level P-5
November 2017/P51
(b)
44.00
40.00
36.00
2
y
volume of hydrogen gas produced / cm3
32.00
28.00
24.00
20.00
16.00
12.00
y1
8.00
4.00
0.00
0
(c)
(d)
40
80
X1
120
160
200
charge passed / C
240
280
X2

y 2  y1
x2  x2

35.4  10
280  80
co-ordinates 2
(x1 , x2) (280, 35)
gradient = 0.127 cm3 C-1 [2]
5.2 × 10–6 mol C–1
[1]
(ii)
(iii)
360
[2]
[1]
Yes, as all the points lies very close to the line of best fit.
(i)
co-ordinates 1
(y1 , y2) (80, 10)
320
=
1
gradient × 2
=
1
5.2  106  2
96.153 C mol–1 [1]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solution
17
Chemistry A-Level P-5
(e)
November 2017/P51
(i)
20.0
H2(g)
16.0
volume of
12.0
gas produced
8.0
/ cm3
4.0
0.0
0
50
100
150
200
250
300
time /s
(ii)
Some oxygen may be soluble in water.
[1]
[1]
Pnew
(f)
(ii)
(g)
height of solution
(i)
[1]
Pressure decrease as volume become larger and Faraday constant will be lower
than calculated.
[1]
cathode: As E o value of copper is larger and do not get reduce, in comparison with H+.
anode: If copper anod is used, so it began to dissolve.
[3]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solution
18
Chemistry A-Level P-5
June 2017/P52
May/June 2017 Paper 52
1.
M/J 17/P52/Q1
In 1804 the chemist John Dalton put forward the following idea. It is sometimes called “Dalton’s
Law’.
‘When two elements combine with each other to form more than one compound, the ratios of
the masses of one element that combine with a fixed mass of the other element are simple
whole numbers.’
A student used the apparatus shown to find out if Dalton’s Law is true for three oxides of lead.
Methane gas reduced the heated lead oxides to lead.
porcelain boat
containing
lead oxide A
porcelain boat
containing
lead oxide B
porcelain boat
containing
lead oxide C
excess methane
burning
methane
gas
heat
heat
heat
Lead and oxides of lead are harmful by inhalation and if swallowed. They are very toxic to
aquatic organisms and may cause long-term damage in the aquatic environment.
(a)
(b)
State two hazards associated with experimenting with lead oxides.
For each hazard, state a precaution (other than eye protection) that the student could
take to make sure that the experiment is carried out safely.
hazard 1
precaution
hazard 2
precaution
[2]
The student used the following procedure for the experiment.

Three clean, dry porcelain boats were weighed when empty.

Each boat was filled with a different lead oxide, labelled A, B or C and
reweighed.

The boats were placed in the apparatus and methane gas passed through.

All three samples were heated strongly until they were reduced to lead.

The boats were allowed to cool completely with the methane gas still passing
over them before they were re-weighed.

The results are shown in the table.
lead
oxide
Mass of
porcelain
boat /g
Mass of
boat +
lead oxide
/g
Mass of
boat +
lead after
heating /g
A
5.26
9.31
9.04
B
5.12
8.96
8.48
C
5.23
10.52
10.06
Mass of
lead/g
Mass of
oxygen
/g
Mass of lead
that was
combined
with 1.0g
oxygen in the
lead oxide /g
Complete the table. Record the mass of lead that was combined with 1.0 g of oxygen
in the lead oxide to one decimal place.
Use the space below for any necessary calculations.
[2]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
19
Chemistry A-Level P-5
(c)
(i)
June 2017/P52
Use the values of mass of lead that was combined with 1.0 g oxygen in the
lead oxide to calculate the ratio of mass of lead in each compound.
mass of lead in A :
mass of lead in B :
mass of lead in C
……………………
…………………….
…………………..
[1]
(ii)
If Dalton’s Law is true there should be a simple whole number ratio of the
masses of lead in each compound combined with a fixed mass of oxygen.
Use your answer to (i) to state and explain whether the student’s experimental
results support Dalton’s Law.
[1]
In this experiment, identify
the independent variable.
the dependent variable
[2]
(i)
Another sample of lead oxide B was found to contain 3.067 g lead and 0.474
g oxygen.
Calculate the empirical formula of B. Show your working.
[Ar: Pb, 207.2; O, 16.0]
Empirical formula of B =……………. [1]
(ii)
What additional piece of information is required to calculate the molecular
formula of B?
[1]
Before the porcelain boats containing the lead were weighed, they were allowed to cool
completely with the methane gas still passing over them.
Apart from the hazards associated with handling hot apparatus, explain why this
procedure is essential to ensure that the results are reliable.
[1]
The student thought that not all of the lead oxide C had been reduced.
What should the student do to make sure all the lead oxide C had been reduced? [1]
(d)
(e)
(f)
(g)
Solution:
(a)
hazard 1:
precaution:
hazard 2:
precaution:
Lead compounds are toxic for aquatic life.
Do not dispose of lead and lead compounds in water.
Harmful vapours for inhalation.
Carry out heating in fume cupboard or well ventilated room.
[2]
(b)
lead
oxide
Mass of
porcelain
boat /g
Mass
of boat
+ lead
oxide
/g
Mass of
boat +
lead after
heating /g
Mass
of
lead/g
Mass of
oxygen
/g
Mass of lead that
was combined
with 1.0g oxygen
in the lead oxide
/g
A
5.26
9.31
9.04
3.78
0.27
14
B
5.12
8.96
8.48
3.36
0.48
7
C
5.23
10.52
10.06
4.83
0.46
10.5
[2]
(c)
(i)
mass of lead in
A:
mass of lead in
B:
mass of lead in
C
14
………………..
7
……………..
10.5
…………..
[1]
(ii)
14 7 10.5
: :
7 7 7
2 : 1 : 1.5
4:2:3
P-5 (Planning, Analysis and Evaluation)
All are in simple whole number ratio [1]
Yearly Worked Solutions
20
Chemistry A-Level P-5
(d)
(e)
Different lead oxide.
Mass of Lead combine with 1 g of oxygen.
(i)
(ii)
(f)
(g)
2.
June 2017/P52
3.067 (Pb)
0.0148
1 :
2
Molecular mass.
[2]
: 0.474 (O2)
: 0.0296
empirical formula of B = PbO2 [1]
[1]
In order to prevent the re-oxidation.
[1]
Again heat the lead oxide and re-weight sample, until no further loss of mass is
noticed.
[1]
M/J 17/P52/Q2
When light passes through solutions of chemical compounds some of the light may be
absorbed. The quantity of light absorbed is called the absorbance and it is measured by a
spectrophotometer. A simplified diagram of a spectrophotometer is shown. A glass cuvette is a
rectangular vessel.
sample of solution
light detecto r
light of one
wavelength
glass cuvette
(a)
(i)
A chemist placed distilled water in the glass cuvette. This was then put into the
spectrophotometer and a reading taken.
Explain why this reading was taken.
[1]
(ii)
Light passes through opposite sides of the cuvette. These two sides must be
wiped with a cloth to ensure they are clean and dry.
Explain why this procedure makes the readings more accurate.
[1]
Manganese is added to steel to increase its strength. A spectrophotometer can be used
to analyses the manganese content of steel. This is done by comparing the absorbance
of a solution of MnO4–(aq) prepared from a sample of steel, with the absorbance of
solutions of known concentrations of MnO4–(aq).
(b)
1.0 dm3 of a standard solution of 0.0300 mol dm –3 MNO4–was prepared by a chemist
using solid potassium manganite (VII), KMnO4, measured using a two decimal place
balance.
(i)
Calculate the mass of KMnO4 required to prepare this standard solution.
[Ar: K, 39.1; Mn, 54.9; O, 16.0]
mass of KMnO4 = ………….. g [1]
(ii)
Describe how the chemist should accurately prepare this standard solution
using a sample of KMnO4 of mass calculated in (i). There is a 1.0 dm 3
volumetric flask available.
[2]
–4
–3
(iii)
The chemist distilled this standard solution 3.0 × 10 mol dm for use in the
experiments.
Explain why the chemist did not prepare a solution of this concentration
directly, by dissolving the required mass of KMnO 4 in 1.0 dm3 of water.
[1]
(c)
The chemist needed to determine which wavelength of light was most absorbed by a
solution of MnO4–(aq). The clean, dry cuvette was filled with 3.0 × 10–4 MnO4– and
different wavelengths of light were passed through the solution. A graph of the results
was plotted.
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
21
Chemistry A-Level P-5
June 2017/P52
0.760
0.750
0.740
0.730
0.720
0.710
absorbance
0.700
0.690
0.680
0.670
0.660
0.650
0.640
0.630
0.620
0.610
0.600
510
515
520
525
530
535
540
545
550
wavelength / nm
(d)
Use the graph to estimate the wavelength of light that is most absorbed by the MnO4–
solution.
Wavelength of light most absorbed = ………….. nm [1]
The spectrophotometer was then set to the wavelength that is most absorbed by the
MnO4–(aq) solution.
The chemist measured the absorbance of solutions of known concentrations of MnO4–
(aq). The results are shown in the table.
(i)
concentration of
MnO4–(aq) / mol dm–3
absorbance
3.00 × 10–4
0.748
2.70 × 10–4
0.680
2.40 × 10–4
0.610
2.10 × 10–4
0.530
1.80 × 10–4
0.440
1.50 × 10–4
0.378
1.20 × 10–4
0.315
0.90 × 10–4
0.230
0.60 × 10–4
0.150
Plot a graph on the grid to show the relationship between the absorbance and
the concentration of MnO4–(aq).
Use a cross (×) to plot each data point. Draw a line of best fit.
[2]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
22
Chemistry A-Level P-5
June 2017/P52
0.800
0.700
0.600
0.500
absorbance
0.400
0.300
0.200
0.100
0.000
0.00
0.50 × 10–4 1.00 × 10–4 1.50 × 10–4 2.00 × 10–4 2.50 × 10–4 3.00 × 10–4 3.50 × 10–4
concentration of MnO 4–(aq) / mol dm –3
(ii)
(iii)
(e)
(f)
State the relationship between absorbance and concentration of MnO4–(aq).
Explain your answer with reference to particles.
[2]
Do you consider the results obtained to be reliable? Explain your answer. [1]
The chemist used the MnO4–(aq) solution of concentration 3.00 × 10–4 mol dm–
3 to prepare the solutions in the table on page 10.
Calculate the volume of 3.00 × 10–4 mol dm–3 MnO4–(aq) solution and the
volume of distilled water required to prepare a 25.00 cm3 solution of 2.70 × 10–
4 mol dm –3 MnO –(aq).
4
Give your answers to two decimal places.
Volume of 3.00 × 10–4 mol dm–3 MnO4–(aq) solution = …….. cm3
Volume of distilled water = ………. cm 3 [1]
(ii)
The volumes of the two solutions given in (e)(i) could be measured using the
same type of apparatus.
Name a suitable piece of apparatus which could be used to measure these
volumes.
[1]
The chemist dissolved a known mass of steel, containing manganese, in acid. The
manganese was then oxidized to manganite (VII) ions, MnO4–, using a very strong
oxidizing agent. The resulting solution was made up to 100.0 cm 3 in a volumetric flask.
A small sample of the solution of MnO4–(aq) prepared from the steel sample was placed
into a clean, dry cuvette and its absorbance measured using the spectrophotometer.
(i)
The absorbance of the MnO4–(aq) solution was 0.630.
Use the graph you have drawn in (d) (i) to determine the concentration of
MnO4–(aq) in this solution.
Give your answer to three significant figures.
concentration of MnO4–(aq) = …………. mol dm–3 [1]
(i)
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
23
Chemistry A-Level P-5
June 2017/P52
(ii)
(g)
(h)
Calculate the mass of manganese present in the steel sample. Show your
working
[Ar: Mn, 54.9]
Mass of manganese = …………… g [1]
The steel sample that the chemist used had a mass of 1.209 g.
Use the mass of manganese you calculated in (f)(ii) to calculate the percentage of
manganese by mass that was present in the steel sample.
(if you were unable to calculate an answer to (f)(ii) you may use 0.00143 g as the mass
of manganese. This is not the correct answer.
Percentage of manganese in the steel sample = …………………. % [1]
Another way of analyzing the manganese content of the steel sample is by titration.
The steel sample is prepared in the same way as previously. It is dissolved in acid and
then oxidized using a very strong oxidizing agent. The MnO 4–(aq) ions produced are
titrated with a solution of iron(II) ions. The equation for this reaction is shown.
MnO4- (aq) + 8H +(aq) + 5Fe2+ (aq)  Mn2+ (aq) + 5Fe3+ (aq) + 4H2O(I)
Explain why it is essential to remove the strong oxidizing agent used to prepare the
solution of steel sample before carrying out the titration.
[1]
Solution:
(a)
(b)
(c)
(d)
(i)
(ii)
(i)
Distilled water used to calibrate the spectrophotometer.
[1]
The water along with dirt on sides may absorb same light.
[1]
Mass = c × v × Mr
= 0.03 × 1 × 158
= 4.74
Mass of KMnO4 = 4.74 g [1]
3
(ii)
Dissolve 4.74 gram of KMnO4 in 200 cm distilled in a beaker. Now add this solution
into 1 dm3 measuring flask. Now add distilled in it with the help of same beaker, upto
the mark.
[2]
(iii)
This concentration required very less mass of KMnO 4, which cannot be weigh
accurately.
[1]
wavelength of light must absorbed = 529 nm
[1]
(i)
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
24
Chemistry A-Level P-5
June 2017/P52
0.800
0.700
0.600
0.500
absorbance
0.400
0.300
0.200
0.100
0.000
0.50× 10 –4 1.00 × 10 –4 1.50 × 10 –4 2.00× 10 –4 2.50 × 10 –4 3.00× 10 –4 3.50 × 10–4
0.00
concentration of MnO 4–(aq) / mol dm–3
(ii)
[2]
The concentration of MnO4–2(aq) is directly proportional to absorbance i.e. more ions
more will be absorbance.
[2]
Yes. All the points lies very close to each other.
[1]
C1 V1 = C2 V2
3 ×10–4 × V1 = 25 × 2.70 × 10–4
volume of 3.00 × 10–4 mol dm–3 MnO4–(aq) solution = 22.5 cm 3
volume of distilled water = 2.5 cm 3 [1]
Burette
[1]
(i)
concentration of MnO4–(aq) = 2.55 × 10–4 mol dm–3
(ii)
(e)
(f)
(iii)
(i)
 2.50  104  54.9 
(ii)
[1]
100
1000
mass of manganese = 1.37 × 10–3 g [1]
(g)
(h)
1.37  10
1.209
3
 100
percentage of manganese in the steel sample = 0.113 % [1]
Excess strong oxidizing agent may oxidize the Fe 2+ to Fe3+.
[1]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
25
Chemistry A-Level P-5
June 17/P-51
May/June 2017 Paper 51
1.
M/J 17/P51/Q1
The pain of muscle strains and swellings can be eased by using heat packs. As a source of
heat, some heat packs use the energy released when anhydrous calcium chloride dissolves in
water.
water
CaCl2 (s) 
 Ca2+ (aq)+ 2Cl- (aq)
A heat pack consists of a bag of water, inside which a smaller bag contains anhydrous calcium
chloride. When pressure is applied to the heat pack, the smaller bag bursts releasing the
anhydrous calcium chloride into the water. The heat pack is shaken to speed up dissolving.
Energy is released which warms the heat pack.
A student carried out an experiment to determine the enthalpy change when anhydrous calcium
chloride dissolves in distilled water. The results the student obtained are plotted on the graph
on next page.
(a)
By considering the graph of results, draw a labelled diagram of the experimental setup that the student could have used to produce the graph shown.
Label the apparatus and chemicals required to measure the two variables.
[2]
40
38
36
34
32
temperature
/°C
30
28
26
24
22
20
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0 10.0
time / minutes
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
Chemistry A-Level P-5
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
26
June 17/P-51
Explain why the student took readings between 0.0 minutes and 2.5 minutes.
[1]
Explain why the student did not take a reading at 3.0 minutes.
[1]
Explain why the temperature continued to increase between 3.5 minutes and 4.5
minutes.
[1]
Draw two straight lines of best fit on the grid. Extrapolate these lines to estimate the
theoretical temperature rise at 3.0 minutes. Give your answer to one decimal place.
theoretical temperature rise at 3.0 minutes = …………….. °C [2]
One of the results is anomalous. This occurred because the student took the
thermometer out of the solution and then replaced it just before the reading was taken.
The time at which the anomalous reading was taken was ……….. minutes.
Explain why these actions led to the anomalous point.
[1]
Explain why stirring the mixture would make this experiment more reliable.
[1]
Anhydrous calcium chloride is classified as a moderate health hazard. It is an irritant.
Apart from wearing eye protection, state one other relevant safety precaution the
student should have taken.
[1]
The student found the value for the enthalpy change of solution of anhydrous calcium
chloride to be –82.5 kJ mol–1.
A manufacturer produces a heat pack that contains 75.0 g of water.
Calculate the mass of anhydrous calcium chloride the manufacturer must use in the
inner bag to produce a rise in temperature of 30.0°C.
The specific heat capacity of water, c = 4.18 J g–1 K–1.
[Ar: Ca, 40.1, Cl, 35.5]
mass of anhydrous CaCl2 = ………… g [2]
Solution:
(a)
Stopwatch
Time
P-5 (Planning, Analysis and Evaluation)
Temprature
Yearly Worked Solutions
27
Chemistry A-Level P-5
3
June 17/P-51
40
o
8 .7 C
38
36
34
32
temperature
/C
30
28
26
24
22
o
5C
20.1
20
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0 10.0
time / minutes
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
It ensure the constant temperature during this time at room temperature.
At this time calcium chloride is added.
Calcium chloride is continue to dissolve in this time.
theoretical temperature rise at 3.0 minutes = 38.06°C
The time at which the anomalous reading was taken was 8.5 minutes.
The temperature was noted earlier or.
Stirring it will heat the solution uniformly.
Wears gloves or wear mask.
q=
[2]
[1]
[1]
[1]
[2]
[1]
[1]
[1]
75 × 4.8 × 30
= 9.405 kg
1000
=
9.405
 111.1
82.5
mass of anhydrous CaCl2 = 12.7 g [2]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
28
Chemistry A-Level P-5
2.
June 17/P-51
M/J 17/P51/Q2
Sucrose is a sugar. The concentration of a solution of sucrose can be measured by the optical
rotation, , of a sucrose solution. The more concentrated the solution, the greater the optical
rotation of the solution.
A polarimeter is used to measure optical rotation. Light is passed through a sample of the
sucrose solution in a glass cell, and the observed angle of rotation, obs, is measured.
A simplified diagram of a polarimeter is shown.
eyepiece
eye of
observer
light
glass cell
rotated light
If a glass cell of length 10 cm is filled with a solution of sucrose of concentration 1 g cm –3 the
measured angle of rotation is known as the specific rotation, [].
The observed angle of rotation obs, measured by the polarimeter is related mathematically to
the concentration of the sucrose solution by the equation shown.
obs = [] c
obs is the observed angle of rotation using a 10 cm cell.
[] is the specific rotation of sucrose solution.
c is the concentration of sucrose in g cm –3
A student wanted to determine the specific rotation of sucrose, []. Solutions of different
concentrations of sucrose at 20°C were placed in a polarimeter and the observed angle of
rotation, obs, recorded. The ‘+’ sign is used to show that the rotation is in a clockwise direction.
(a)
(i)
(ii)
(iii)
Concentration of
sucrose, c
/g cm–3
Observed angle of
rotation obs
0.0750
+5.05
0.0700
+4.70
0.0650
+4.40
0.0600
+4.00
0.0500
+3.30
0.0450
+2.55
0.0350
+2.30
0.0300
+1.95
0.0250
+1.68
0.0200
+1.40
Plot a graph on the grid on next page to show the relationship between
concentration of sucrose, c, and observed angle of rotation obs.
Use a cross (×) to plot each data point. Draw a line of bests fit.
[2]
Circle the most anomalous point on your graph.
[1]
Use the graph to determine the specific rotation, [], of sucrose.
Give this value to two decimal places.
State the co-ordinates of both points you used in your calculation.
co-ordinates 1 ……………………
co-ordinates 2 ………………………
specific rotation of sucrose, [] = ……….. [2]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
29
Chemistry A-Level P-5
June 17/P-51
5.5
5.0
4.5
4.0
3.5
3.0
α obs
2.5
2.0
1.5
1.0
0.5
0.0
0.000
0.0100
0.0200
0.0300
0.0400
0.0500
0.0600
0.0700
0.0800
c / g cm–3
(b)
(c)
You are asked to write instructions for another student to follow so they can prepare a
standard solution of 250 cm3 0.0750 g cm–3 sucrose. The student is provided with solid
sucrose and a 250 cm 3 volumetric flask.
(i)
Calculate the mass, in g, of sucrose the student would need to use.
mass of sucrose = …………… g [1]
(ii)
Describe how the student should accurately prepare the standard solution
using a sample of sucrose of mass calculated in (i).
[2]
(i)
The student used the standard solution prepared in (b) to prepare the solutions
in the table on the next page.
Calculate the volume of standard solution of concentration 0.0750 g cm –3 and
the volume of distilled water needed to prepare 15.00 cm 3 of sucrose solution
of concentration 0.0350 g cm –3.
Give your answers to two decimal places.
volume of standard solution = ……………. cm 3
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
Chemistry A-Level P-5
(d)
(e)
(f)
(g)
30
June 17/P-51
volume of distilled water = ………… cm 3
[1]
The volumes of the two solutions given in (c)(i) could be measured using the
same type of apparatus.
(ii)
Name a suitable piece of apparatus which could be used to measure these
volumes.
[1]
(iii)
In (a)(ii) you circled an anomalous point. This was caused by the student
incorrectly making one of the sucrose solutions.
Suggest the error made by the student that caused this anomaly.
[1]
The student recorded the observed angle of rotation, obs, for a sucrose solution of
unknown concentration as +3.75.
Determine the concentration of this sucrose solution in mol dm–3.
[Mr, sucrose: 342]
concentration of sucrose = ……………. mol dm –3
[3]
The glass cell of 10 cm length is expensive, so one cell is used for all the solutions that
are placed in the polarimeter.
Suggest how you would ensure that the concentration of solution in the cell is accurate
each time the cell is used for the different sucrose solutions.
[1]
Concentration of sucrose is the independent variable in this polarimeter experiment.
The glass cell of 10 cm length is replaced by a glass cell of 20 cm length. The 20 cm
glass cell is filled with 0.0750 g cm –3 sucrose solution.
Predict the value for the observed angle of rotation obs, for the sucrose solution of
concentration 0.0750 g cm–3 when the 20 cm cell is used. Explain your answer.
predicted value =
explanation
[2]
Before the angles of rotation of the sucrose solutions are measured, the glass cell is
first filled with distilled water and the angle of rotation measured.
Explain why this measurement is taken.
[1]
Solutions
(a)
(i)
(ii)
P-5 (Planning, Analysis and Evaluation)
[2]
[1]
Yearly Worked Solutions
31
Chemistry A-Level P-5
June 17/P-51
5.5
5.0
4.5
4.0
3.5
3.0
α obs
2.5
2.0
1.5
1.0
0.5
0.0
0.000
0.0100
0.0200
0.0300
0.0400
c /g cm
(b)
(iii)
co-ordinates 1
(i)
= 250 × 0.075
(ii)
(c)
(i)
(ii)
(iii)
(4.00)
0.0500
0.0600
0.0700
0.0800
–3
co-ordinates 2 (0059,0)
specific rotation of sucrose, [] = 67.66 [2]
mass of sucrose = 18.75 g [1]
Dissolve 18.75 g of sucrose in 100 cm 3 distilled water. Now add this solution
is the measuring flask of 250 cm 3. Now add the distilled water in the
measuring flask to make up the volume up to mask.
[2]
c1 v1 = c2 v2
0.0750 × 15 = 0.350 × v2
volume of standard solution = 7 cm 3
volume of distilled water = 8 cm 3 [1]
Pipette or burette
[1]
Solution is more diluted that expected value or
Value of obs is take an early.
[1]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
32
Chemistry A-Level P-5
(d)
At 3.75 is 0.056 g/cm 3
Concentration is 0.056 × 1000
= 56 g dm–3
=
(e)
(f)
(g)
June 17/P-51
56
342
concentration of sucrose = 0.164 mol dm –3 [3]
Wash the cell with solution to ensure the concentration.
[1]
predicted value = 5.05 × 2 = 10.10
Explanation:
The plane of polarized light passed or encounter with double
molecules.
[2]
This blank measurement is taken to caliberate the polarimeter to 0 degree
[1]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
33
Chemistry A-Level P-5
November 2016/P52
October/November 2016 Paper 52
1.
O/N 16/P52/Q1
Titrations using ethylene diaminetetra acetic acid (EDTA) can be used to determine the
concentration of metal ions in solution, such as Zn2 (aq).
A solution of EDTA is usually prepared from the hydrated disodium salt, Na2H2Y.2H2O . The
anion of EDTA is H2 Y 2- , where Y represents the organic part of the ion.
The equation for the reaction between Zn2+(aq) and EDTA is shown.
Zn2+ (aq) + H2Y2- (aq)  ZnY2- (aq)+2H+ (aq)
The indicator for the reaction is Solochrome Black, which changes colour at the endpoint from
purple to blue. The indicator only works at pH 10, so a buffer solution is added to the metal ion
solution to maintain the pH.
(a)
Explain why the pH would change during the titration if the buffer were not present.[1]
(b)
You are to plan a titration experiment to determine the concentration of zinc ions in a
solution of zinc sulfate of concentration approximately 0.1 mol dm –3.
You are provided with the following materials.
20.0 g of hydrated disordium EDTA, Na2H2Y.2H2O (Mr = 372.2)
(c)
aqueous zinc sulfate of approximate concentration 0.1 mol dm –3
buffer solution, pH 10
Solochrome Black indicator solution.
(i)
Name three pieces of volumetric apparatus you would use, with their
capacities in cm3.
[2]
(ii)
Calculate the mass of hydrated disordium EDTA that would be required for the
preparation of a standard solution of concentration 0.100 mol dm –3, using the
apparatus you have specified in (i).
mass of hydrated disordium EDTA = ……… g [1]
(iii)
Describe how you would prepare this standard solution for use in your
titration.
[2]
(iv)
After you have performed a rough titration, how would you ensure that your
next titration is accurate?
[1]
(v)
How would you ensure that your titration result is reliable?
[2]
The term hard water is used to describe water containing the dissolved metal ions,
Ca2+(aq) and Mg2+(aq). Both of these metal ions react with EDTA anions, H2Y2.
Ca2+(aq) + Mg2+(aq) + 2H2Y2- (aq)  CaY2- (aq) + MgY2- (aq) + 4H (aq)
In an experiment to determine the concentration of each of these metal ions, two
separate titration with EDTA need to be performed.
For titration 1, a 25.0 cm3 sample of hard water is titrated with 0.0100 mol dm–3 EDTA
solution using Solochrome Black solution as indicator.
For titration 2, another 25.0 cm 3 sample of the same hard water is first treated with
excess 2 mol dm–3 NaOH(aq) which precipitates all of the Mg2+(aq) ions as Mg(OH)2(s).
After this treatment, no Mg2+(aq) ions remain in solution, leaving only dissolved
Ca2+(aq) ions in solution. This solution is then titrated with 0.0100 mol dm –3 EDTA
solution using Solochrome Black solution as indicator.
The following information gives some of the hazards associated with the chemicals
used in the procedure.
Sodium hydroxide
Solutions equal to or more concentrated than 0.5 mol
dm–3 are classified as corrosive.
Solochrome Black
Solid Solochrome Black is classified as health hazard
and is irritating to eyes, respiratory system and skin. All
solutions are made up in ethanol and so are classified
as flammable and health hazard.
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
34
Chemistry A-Level P-5
(i)
(ii)
November 2016/P52
Identify one hazard that must be considered when planning the experiment
and describe a precaution, other than eye protection, that should be taken to
keep risks from this hazard to a minimum.
hazard:
precaution
[1]
Results obtained from this experiment are shown.
titre 1, 22.70 cm3
titre 2, 16.60 cm3
Use the results of the titrations to determine the concentrations of Ca2+(aq) and
Mg2+(aq) in the hard water.
concentration of Ca2+(aq) ……………………… mol dm–3
concentration of Mg2+(aq) …………………….. mol dm–3
[4]
Solution:
(a)
(b)
Reaction produces H+ ions, so pH will decrease.
(i)
1.
Volumetric flask 250 cm3.
2.
Pipette 25 cm3.
3.
Burette 50 cm3.
(ii)
Moles required to dissolved in 250 cm 3 =
[1]
[2]
0.1
 250
1000
Moles =
Mass =
(iii)
(c)
(iv)
(v)
(i)
(ii)
0.025
0.025 ×372.2
mass of hydrated disodium EDTA = 9.305 g [1]

Dissolve 9.305 gm of EDTA in a Beaker of 100 cm 3 distilled water.

Add it into volumetric flask, of 250 cm 3, add distilled water in it to make up to
volume upto the mask.
[2]
Add the solution dropwise into flask, according to your calculated end volum
[1]
Repeat it three times to get concordant readings.
[1]
hazard:
Sodium hydroxide is corrosive in action.
precaution:
Wear gloves, and keep Solochrome Black away from
flame.
[1]
Titration for Ca+2
=
= 6.64 × 10–3 mol dm–3
Ca2+
Titration for Ca+2 and mg+2
Concentration of
2.
0.01 16.60
25
mg+2
=
0.01 22.7
25
= 9.08 ×10–3
= 9.08 × 10–3 – 6.64 × 10–3
concentration of Ca2+ (aq) 6.64 × 10–3 mol dm–3
concentration of Mg2+(aq) 2.44 × 10–3 mol dm–3
[4]
O/N 16/P52/Q2
Benzenediazonium chloride, C6H5N2Cl is readily hydrolysed at temperature above 5°C, forming
phenol, nitrogen gas and hydrochloric acid.
C6H5N2Cl(aq) + H2O(l)  C6H5OH(aq) + N2(g) + HCl (aq)
The progress of the reaction can be monitored by measuring the volume of gas produced over
time. The volume of gas produced V, after time, t, is proportional to the amount of
benzenediazonium chloride that has been hydrolysed. The final volume of gas produced, Vfinal,
is proportional to the original concentration of benzenediazonium chloride.
The order of reaction can be determined from these results.
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
35
Chemistry A-Level P-5
(a)
(i)
November 2016/P52
The experimentally determined volumes of gas produced during the hydrolysis
of benzenediazonium chloride at 50°C are recorded below.
Process the results to allow you to plot graph of (Vfinal – V) against time, t.
Vfinal = 252 cm3
time/s
Volume, V/cm3
0
0
150
32
300
62
450
87
600
110
750
129
900
146
1050
160
1200
173
1350
184
1500
193
(Vfinal – V)/ cm3
[1]
(ii)
Plot a graph to show how (Vfinal – V) /
varies with time/s.
Use a cross (×) to plot each data point. Draw the curve of best fit.
cm3
260
240
220
200
180
160
140
(Vfinal – V ) / cm3
120
100
80
60
40
20
0
0
200
400
600
800
time / s
1000
1200
1400
1600
[2]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
36
Chemistry A-Level P-5
(iii)
(iv)
(b)
November 2016/P52
Do you think the results obtained in (i) are reliable? Explain your answer. [1]
Use the graph to determine the half-life, t 12 , of this reaction.
State the co-ordinates of both points you used in your calculation.
co-ordinates 1
co-ordinates 2
half-life = ………………. s [2]
A student set up an experiment to determine the order of the reaction in (a). Part of the
experimental set-up is shown below:
boiling tube
water bath
benzenediazonium
chloride
anti-bumping
granules
(i)
(c)
Complete the diagram above to show the experimental set-up the student
could have used to collect and measure the volume of gas evolved by the
reaction.
[2]
(ii)
The water bath was set at 60°C.
At a reaction temperature 60°C, the measurements made would be less
accurate than measurements made at room temperature.
State why the measurements made at a higher temperature are less accurate.
State the effect this will have on the values of Vfinal –V.
[2]
The graph below shows the results obtained from another benzenediazonium chloride
hydrolysis reaction performed at a different temperature.
(i)
(ii)
The point at time = 450 s is considered to be anomalous.
Suggest what caused the anomaly.
[1]
The rate of reaction at different times can be calculated by drawing tangents
to the best-fit line. The gradient of the tangent is equal to the rate of reaction,
in cm3 s–1.
Use the graph in (c) to read the value of Vfinal –V at time t = 200 s and write this
value in the table below.
Draw a tangent to the curve at time t = 200 s. Use the tangent to determine the
gradient at time t = 200 s.
State the co-ordinates of both points you used in your calculation.
Co-ordinates 1
Co-ordinates 2
gradient at 200 s = …………… cm 3 s-1
Use your gradient to complete the table.
(Vfinal –V) / cm3
rate of reaction / cm 3
500
104
– 0.143
600
91
– 0.127
900
59
– 0.0867
1000
52
– 0.0720
1400
30
– 0.0417
time / s
200
[4]
(iii)
The concentration of benzenediazonium chloride is directly proportional to
(Vfinal –V).
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
37
Chemistry A-Level P-5
November 2016/P52
Use the date in the table in (ii) to calculate the order of reaction with respect to
benzenediazonium chloride.
You must show your working.
[2]
Solution:
(a)
(i)
time/s
Volume, V/cm3
(Vfinal – V) cm3
0
0
252
150
32
220
300
62
190
450
87
165
600
110
142
750
129
123
900
146
106
1050
160
92
1200
173
79
1350
184
68
1500
193
59
[1]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
38
Chemistry A-Level P-5
(ii)
November 2016/P52
260
y 1 240
220
200
180
160
140
(V final – V ) / cm3
y2
120
100
80
60
40
20
0
0
(iii)
(iv)
200
400
600
800
time / s
1000
1200
Yes most of the points come on line.
co-ordinates 1 (180, 220)
co-ordinates 2: (860, 110)
1400
1600
[2]
[1]
half-life = 680 s [2]
(b)
(i)
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
39
Chemistry A-Level P-5
November 2016/P52
boiling tube
water bath
benzenediazonium
chloride
anti-bumping
granules
(ii)
[2]
Rate of reaction would be higher at higher temperature and measurement would be
difficult and volume of greater.
The value of Vfinal –V will be lower.
[2]
(c)
250
200
7. 5
15
150
(Vfinal – V) /cm3
100
×
×
×
×
×
50
×
×
×
0
0
200
X1
P-5 (Planning, Analysis and Evaluation)
400
600
800
1000
1200
1400
time /s
Yearly Worked Solutions
40
Chemistry A-Level P-5
(i)
(ii)
Less reading is taken.
co-ordinates 1 (200, 50)
November 2016/P52
[1]
co-ordinates 2 (680, 157.5)
gradient at 200 s = –0.223 cm3 s–1
time / s
(Vfinal –V) / cm3
rate of reaction / cm 3
200
157.5
– 0.223
500
104
– 0.143
600
91
– 0.127
900
59
– 0.0867
1000
52
– 0.0720
1400
30
– 0.0417
[4]
(iii)
1st
half life
time
2nd half life
= 200 to 100
= 540
= 1020 – 540
= 520
half life for both is same, so, it is first order reaction.
P-5 (Planning, Analysis and Evaluation)
[2]
Yearly Worked Solutions
41
Chemistry A-Level P-5
November 2016/P51
October/November 2016 Paper 51
1.
O/N 16/P51/Q1
When hydrated barium chloride, BaCl2.xH2O, dissolves in water, Ba2+(aq) and Cl–(aq) ions are
formed.
The concentration of chloride ions in solution can be determined by titration with aqueous silver
nitrate of known concentration.
Ag+(aq) + Cl–(aq)  AgCl(s)
The indicator for the reaction is aqueous potassium chromate(VI), K2CrO4(aq). At the endpoint
of the titration, it forms a red precipitate in the presence of excess silver ions.
(a)
The solubilities, in g dm –3, of different ionic compounds at 20°C are given in the table
below.
anion
(b)
cation
Cl–
CrO42–
SO42–
Ag+
0.0019
0.022
293
Ba2+
358
0.0028
0.00245
With reference to these data, where relevant, answer the following questions.
(i)
Name the red precipitate and give an equation for its formation.
name:
equation:
[2]
2+
Sulfuric acid must be added to the solution to prevent the Ba (aq) ions from
interfering with the action of the potassium chromate(VI) indicator.
(ii)
How would Ba2+(aq) ions interfere with the action of this indicator?
[1]
(iii)
How does the addition of sulfuric acid prevent Ba2+(aq) ions from interfering
with the action of this indicator?
[1]
In an initial rough titration, excess silver nitrate solution is added so that the endpoint
is exceeded.
Draw a sketch graph to show how the mass of silver chloride varies with the volume of
silver nitrate added.
Label both axes.
0
(c)
[2]
You are to plan to titration experiment to determine the value of x in BaCl2.xH2O.
You are provided with the following materials.
300 g of hydrated barium chloride, BaCl2.xH2O
0.050 mol dm–3 aqueous silver nitrate
1.0 mol dm–3 potassium chromate(VI) solution
1.0 mol dm–3 sulfuric acid
(i)
Name three pieces of volumetric apparatus you would use, with their
capacities in cm3.
1.
2.
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
42
Chemistry A-Level P-5
November 2016/P51
3.
(d)
[2]
(ii)
Describe how you would make a solution of barium chloride that is suitable for
use in your titration.
[2]
(iii)
A known volume of barium chloride solution is transferred to a conical flask.
In what order should the other three solutions then be added to the flask?
first
second
third
[1]
(iv)
How would you ensure that your titration result is reliable?
[1]
(v)
In another experiment, a student dissolved 3.13 g of hydrated barium chloride.
BaCl2.xH2O, in distilled water to give 1.00 dm 3 of solution.
It was calculated that the concentration of Ba2+(aq) ions was 0.0128 mol dm –3.
Determine the value of x in BaCl2.xH2O
[Ar: Ba, 137.3; Cl, 35.5; H, 1.0; O, 16.2]
x = …………. [2]
The following information gives some of the hazards associated with the chemicals
used in the procedure.
Barium
chloride
Solid barium chloride is classified as toxic. Solutions equal to or
more concentrated than 0.4 mol dm –3 are classified as moderate
hazard and are harmful if swallowed. Solutions less concentrated
than 0.4 mol dm–3 are classified as non-hazardous.
Potassium
chromate(VI)
All solutions more concentrated than 0.9 mol dm –3 are classified as
health hazard. They may cause skin, eye and respiratory irritation.
Silver nitrate
Solutions equal to or more concentrated than 0.18 mol dm –3 are
classified as corrosive. Solutions equal to or more concentrated
than 0.06 mol dm–3 but less than 0.18 mol dm –3 are classified as
moderate hazard and cause skin and eye irritation. Solutions less
concentrated than 0.06 mol dm –3 are classified as non-hazardous.
Identify one hazard that must be considered when planning the experiment and
describe a precaution, other than eye protection, that should be taken to keep risks
from this hazard to a minimum.
hazard:
precaution:
[1]
Solution:
(a)
(i)
name:
Silver chromate
equation:
2Ag+ + K 2CrO4  Ag2CrO4 + 2K +
aq
aq
aqpH
aq
[2]
1
(ii)
(iii)
Barrium chromate would form, which is insoluble and could interfere the
reaction.
[1]
H2SO4 form BaSO4, which is insoluble.
[1]
(b)
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
43
November 2016/P51
Mass of Precipitate
Chemistry A-Level P-5
0
Volume of AgNo 3
[2]
(c)
(i)
(ii)
(iii)
(iv)
(v)
cm3
1.
Volumetric flask
250
2.
Pipette
25 cm3
3.
Burette
50 cm3

Weigh 3 gram of BaCl2.xH2O in 50 cm3 of distilled water.

Add their solution into 250 cm 3 volumetric flask.

Now add distilled water in it to make up the volume upto mark.
first:
Sulphuric acid
second:
Barrium chloride
third:
Silver nitrate
Titration will be repeated to get three concordant readings.
0.0128 × 208.3
= 2.67 g of BaCl2
Mass of H2O
= 3.13 – 2.67
= 0.46 of H2O
x=
Mol of x
2.
hazard:
precaution:
[2]
[1]
[1]
0.46 g 0.0128
=
18 x
1
= 35.9 gm.
=
(d)
[2]
35.9
 1.99
18
x= 2
[2]
It is corrosive so wear gloves.
Potassium dichromate cause skin irritation, sulphuric also skin irritant.
[1]
O/N 16/P51/Q2
Sucrose, C12H22O11, is a naturally occurring sugar found in sugarcane and many fruits. It can
be hydrolysed in acid solution to give glucose and fructose. All three molecules are chiral and
will rotate the plane of polarized light. The degree of rotation is known as the optical rotation,

CH2OH
O
OH
HO
CH2OH
CH2OH
O
H+ catalyst
HO
O
OH
OH
sucrose
O
OH
+ H 2O
CH2OH
HO
+
OH
OH
glucose
CH2OH
O
HO
OH
CH2OH
OH
fructose
In the presence of excess water, the reaction can be considered to be first order with respect
to sucrose concentration.
The progress of the reaction can be monitored using a polarimeter, which measures the optical
rotation, , of the solution. The more concentration the solution, the greater the optical rotation
of the solution.
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
44
Chemistry A-Level P-5
November 2016/P51
The concentration of sucrose at time t can be represented as ( – final), where final is the
optical rotation of the solution after 6 hours.
The mathematical relationship is given by the following equation.
log10  α - αfinal  = A -
kt
2.30
A is a consent.
k is the rate consent.
(a)
The experimentally determined values of optical rotation during the hydrolysis of
sucrose at 298 K are recorded below.
Process the results to allow you to plot a graph of log10 (–final) against time t.
Calculate (–final) and record it to 1 decimal place.
Calculate log10 (–final) and record it to 2 decimal places.
(b)
(i)
(ii)
time/s
Optical
rotation, 
0
39.9
300
29.1
600
21.3
900
15.5
1200
10.6
1500
6.2
1800
2.4
2100
–0.3
2400
–2.5
2700
–4.5
final
–12.0
(–final)
log10 (–final)
[2]
Plot a graph on the grid on mentioned below graph to show how log10 (–final)
varies with time t.
Use a cross (×) to plot each data point. Draw the line of best fit.
[2]
State and explain whether the results and your graph confirm the relationship
log10  α - αfinal  = A -
P-5 (Planning, Analysis and Evaluation)
kt
.
2.30
[1]
Yearly Worked Solutions
45
Chemistry A-Level P-5
November 2016/P51
1.8
1.7
1.6
1.5
1.4
1.3
log 10( α – α final)
1.2
1.1
1.0
0.9
0.8
0.7
0.6
0
500
1000
1500
2000
2500
3000
time / s
(c)
(i)
(ii)
Determine the gradient of the graph.
State the co-ordinates of both points you used for your calculation.
Record the value of the gradient to three significant figures.
co-ordinates 1 ……………..
co-ordinates 2
……………….
gradient = …………… s–1
[2]
Use the gradient value to calculate a value for k in the expression shown.
log10  α - αfinal  = A -
kt
2.30
[2]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
46
Chemistry A-Level P-5
(d)
November 2016/P51
The graph below shows the results obtained from a second hydrolysis of sucrose
reaction performed at a different temperature.
45.0
40.0
35.0
30.0
25.0
α – α final
20.0
15.0
10.0
5.0
0.0
0
400
660
(i)
(ii)
(iii)
800
1200
660
1600
time / s
(v)
2400
2800
The point at time = 2000 s is considered to be anomalous.
Suggest what caused the anomaly.
[1]
Use the graph to determine the half-life, t1/2, of this reaction.
State the co-ordinates of both points you used in your calculation.
co-ordinates 1 ……………..
co-ordinates 2
……………….
half-life = …………… s [2]
For a first-order reaction, the following relationship exists.
half-life, t1/2 =
(iv)
2000
0.693
k
Use this relationship and your answer to (ii) to determine k´, the rate constant
for this second hydrolysis reaction.
If you have been unable to determine the half-life of the reaction in (ii), you
may use the value t1/2 = 500 s, though this is not the correct answer.
k´ = ……………………. s–1 [1]
State whether the temperature of the second reaction was higher or lower than
that of the first.
Explain your answer with reference to the answers you obtained in (c)(ii) and
(d)(iii).
If you have been unable to calculate a value for k in (c)(ii), you may use the
value k = 800 × 100–4, though this is not the correct answer.
[1]
Would the value of the half-life change if the reaction were repeated with twice
the initial concentration of sucrose? Give a reason for your answer.
[1]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
47
Chemistry A-Level P-5
November 2016/P51
Solution:
(a)
(b)
(i)
(ii)
time/s
0
300
600
900
1200
1500
1800
2100
2400
2700
Optical rotation, 
39.9
29.1
21.3
15.5
10.6
6.2
2.4
–0.3
–2.5
–4.5
final
–12.0
(–final)
51.9
41.1
33.3
27.5
22.6
18.2
14.4
11.7
9.5
7.5
log10(–final)
1.72
1.61
1.52
1.44
1.35
1.26
1.16
1.07
0.98
0.88
Yes, most of the points lies on the line, with few anomalies.
P-5 (Planning, Analysis and Evaluation)
[2]
[2]
[1]
Yearly Worked Solutions
48
Chemistry A-Level P-5
November 2016/P51
1.8
1.7
1.6
1.5
y2
1.4
1.3
log10( α – α final )
1.2
1.1
1.0
0.9
0.8
0.7
0.6-1.41
y1
0.6
0
x1
(c)
(i)
500
co-ordinates 1 (0, 1000)
1.41  0.6

1000  0
1000
x2
1500
2000
2500
3000
time /s
co-ordinates 2
0.6, 1.41
gradient = 8.1 × 10–4 s–1
[2]
+
(ii)
k = log10
k
= A - log10  α - αfinal 
2.36
    log10  α - αfinal   
  1.30
k= 


t


k = 8.1 × 104
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
49
Chemistry A-Level P-5
log10  α - αfinal   A 
November 2016/P51
-kT
2.3
(2.3 (log (–final) + A = –kT
log10  α - αfinal   A
2.3
T
k
k = –2.3 × 8.4 × 10–4
=1.932 × 10–3
[2]
(d)
45.0
40.0
35.0
30.0
25.0
α – α final
20.0
15.0
10.0
5.0
0.0
0
400
660
800
1200
660
1600
2000
(i)
(ii)
(–final) is recorded too early.
co-ordinates 1 (4.4, 2.2)
(0,660)
(iii)
(iv)
(v)
k = 1005 ×
Second value is less so temperature is lower.
Half-life remain constant because its under reaction.
P-5 (Planning, Analysis and Evaluation)
2400
2800
time /s
[1]
co-ordinates 2 (2.2, 1.1)
(660, 1320)
10–3
half-life = 660 s
= 1.05 × 10–3 s–1
[2]
[1]
[1]
[1]
Yearly Worked Solutions
50
Chemistry A-Level P-5
June 2016/P52
May/June 2016 Paper 52
1.
M/J 16/P52/Q1
A more reactive metal will displace a less reactive metal from a solution of its salt. This reaction
is exothermic. If the same reaction is set up in an electrochemical cell than, instead of an
enthalpy change, electrical energy is produced and a cell voltage can be measured.
You are to plan an investigation of the reaction of three different metals (magnesium, iron and
zinc) with aqueous copper(II) sulfate. You will plan to investigate whether there is a relationship
O
between their cell potential values. Ecell
, and their enthalpy changes of reaction Hr.
Mg(s) + Cu2 (aq)  Mg2 (aq) + Cu(s)
Fe(s) + Cu2 (aq)  Fe2 (aq) + Cu(s)
Zn(s) + Cu2 (aq)  Zn2 (aq) + Cu(s)
Copper(II) sulfate solution is classified as moderate hazard.
Zinc sulfate solution is classified as corrosive.
Iron(II) sulfate solution is classified as a health hazard.
(a)
(b)
(c)
(d)
O
Predict how Hr may change as Ecell
increases. Give a reason for your prediction. [1]
The first part of the investigation is to determine the enthalpy change, Hr, for the
reaction off the same number of moles of three powdered metals with 0.500 mol dm –3
Copper(II) sulfate.
When determining the Hr for the reaction of the metals listed above with aqueous
copper(II) sulfate.
the independent variable is,
the dependent variable is
[2]
3
You are provided with a sample of powdered metal and 50.0 cm of 0.500 mol dm–3
aqueous copper(II) sulfate.
(i)
Draw a fully labelled diagram to show how the apparatus should be set up to
allow you to determine the increase in temperature of aqueous copper(II)
sulfate.
You should use apparatus normally found in a school or college laboratory.
[1]
(ii)
State the measurements you would make in your experiment.
[2]
(iii)
Other than eye protection, state one precaution you would take to make sure
that the experiment proceeds safely.
[1]
(iv)
For the reaction with magnesium, calculate the mass of magnesium, in g, you
would use so that it is in a small excess. You must show your working.
[Ar: Mg, 24.3]
mass of Mg = g [2]
(v)
Explain why the metal used should be in powdered form rather than in strips.
[1]
(vi)
The aqueous copper (II) sulfate and metal mixture should be stirred
continuously. Explain why.
[1]
In one experiment, the increase in temperature when excess magnesium powder is
added to 50.0 cm 3 of 0.500 mol dm–3 aqueous copper(II) sulfate is 58.5°C.
Calculate the enthalpy change for this reaction, Hr, in kJ mol–1.
Assume the specific heat capacity, c, of the reaction mixture is 4.18 J g–1 K–1.
Assume 1.0 cm3 of 0.500 mol dm –3 aqueous copper (II) sulfate has a mass of 1.0 g.
Include a sign in your answer.
Mg(s) + Cu2+(aq)  Mg2+(aq) + Cu(s)
Hr = kJ mol–1 [2]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
51
Chemistry A-Level P-5
(e)
June 2016/P52
O
The second part of the investigation involves determining the cell potential, Ecell
, for the
three electrochemical cells.
cell reaction
Mg(s) +
Cu2+(aq)
 Mg2+(aq) + Cu(s)
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Fe(s) + Cu2+(aq)  Fe2+(aq) + Cu(s)
(f)
(g)
O
Draw a diagram of the apparatus you would use to measure the Ecell
for the
magnesium/copper cell. Your labels should include the names of the metals and the
names and concentrations of the solutions you would use.
[3]
Explain why the enthalpy change determination and cell potential determination should
be carried out at the same temperature as each other.
[1]
O
Accepted Ecell values are shown for the cell reactions.
cell reaction
O
/V
Ecell
1
Mg(s) + Cu2+(aq)  Mg2+(aq) + Cu(s)
+2.72
2
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
+1.10
3
Fe(s) + Cu2+(aq)  Fe2+(aq) + Cu(s)
+0.78
Hr
Use your prediction in (a), your answer to (d) and data from the table to predict Hr
values for reactions 2 and 3.
Complete the table with these values.
[1]
Solutions:
(a)
(b)
(c)
O
With the increase in Ecell
, increase, the Hr decreases or the reaction become more
exothermic.
[1]
the independent variable is
metal
the dependent variable is
increase in temperature.
[2]
(i)
Stirer
Thermometer
cuso 4
Metal
[1]
(ii)
(iii)
 Temperature before adding metal and after adding metal.
 Mass of metal before reaction and after reaction.
Zinc sulphate is corrosive so wear gloves.
 0.5 
(iv)
(v)
(vi)
(d)
[2]
[1]
50
 24.3
1000
mass of Mg = 0.6075 g [2]
Greater surface area, higher will be rate of reaction.
[1]
Uniform heating is ensured by stirring.
[2]
q
=
mC∆T
=
50 × 4.18 × 58.5
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
52
Chemistry A-Level P-5
q
June 2016/P52
12226.5
 103
0.025
=
Hr = –489 kJ mol–1 [2]
(e)
(f)
(g)
Both values can be compared easily under same conditions
1
cell reaction
O
/V
Ecell
Hr
Mg(s) + Cu2+(aq)  Mg2+(aq) + Cu(s)
+2.72
– 4.89

2
Zn(s) +
+ Cu(s)
+1.10
– 240
3
Fe(s) + Cu2+(aq)  Fe2+(aq) + Cu(s)
+0.78
–130
Cu2+(aq)
Zn2+(aq)
[3]
[1]
[1]
2.
M/J 16/P52/Q2
The relative molecular mass, Mr, of volatile liquids can be determined using the apparatus
below.
steam in
gas syringe
self-sealing
cap
volatile
liquid
hypodermic
syringe
steam jacket
thermometer
steam out
A known mass of volatile liquid is injected into the gas syringe using a hypodermic syringe. The
injected volatile liquid vaporizes and the volume of vapour is recorded.
The experiment can be repeated using different samples of the sample volatile liquid. The
following mathematical relationship can be used to calculate the relative molecular mass if the
experiment is carried out at 100°C and 1.01 ×105 Pa.
 3.07  104 
V 
m
Mr


P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
53
Chemistry A-Level P-5
June 2016/P52
m is the mass of the volatile liquid in g.
V is the volume of the volatile liquid in cm 3 when vaporized.
A graph of V against m can be plotted.
A group of students is given a volatile liquid hydrocarbon, Y, and asked to find its relative
molecular mass in a series of experiments using this procedure.

A 100 cm3 gas syringe is placed in a steam jacket.

Approximately 5 cm 3 of air is pulled into the gas syringe.

The temperature is allowed to reach a constant 100°C.

Once the air in the gas syringe has stopped expanding, its volume is recorded.

The hypodermic syringe is filled with liquid Y.

The total mass of the hypodermic syringe and liquid Y is recorded.

A little liquid Y is injected into the hot gas syringe.

The total mass of the hypodermic syringe is recorded again.

The maximum volume of air and vapour in the gas syringe is recorded.

The mass of liquid Y injected into the gas syringe is calculated and recorded.
The results from the group of students are given in the table.
mass of
syringe +
liquid Y
before
injection /g
mass of
syringe +
liquid Y
after
injection /g
Volume of
air in gas
syringe
before
injection
/cm3
Volume of
air + vapour
Y in gas
syringe after
injection /
cm3
4.83
4.68
7
55
5.33
5.23
9
44
4.85
4.64
13
85
5.09
4.92
11
69
5.31
5.07
14
97
5.57
5.48
8
39
5.32
5.12
9
79
5.17
4.94
12
91
4.84
4.72
7
48
5.05
4.83
11
84
Mass of
liquid Y
used /g
Volume of
vapour Y /
cm3
(a)
Process the results in the table to calculate both the masses of volatile liquid Y used
and the volumes of vaporized Y.
[2]
(b)
Plot a graph on the grid to show the relationship between mass of liquid Y and volume
of vapour Y.
Use a cross (×) to plot each data point.
Draw the line of best fit.
[2]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
Chemistry A-Level P-5
(c)
54
June 2016/P52
Liquid Y evaporates easily, even at room temperature. This can cause anomalous
results giving points below the line of best fit.
(i)
Explain how such anomalies occur.
(ii)
With reference to the experimental procedure, explain how this source of error
could be minimized.
[1]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
55
Chemistry A-Level P-5
(d)
(e)
June 2016/P52
(i)
Determine the gradient of your graph. State the co-ordinates of both points you
used for your calculation. Record the value of the gradient to three significant
figures.
co-ordinates 1.
co-ordinates 2.
gradient = [2]
(ii)
Use the gradient value in (i) and the mathematical relationship on page 7 to
calculate the experimentally determined relative molecular mass of Y.
experimentally determined Mr of Y = [2]
Compound Y is a hydrocarbon that contains 85.7% carbon by mass.
The diagram shows the mass spectrum of compound Y.
Use all the information given to determine the molecular formula of Y.
molecular formula of Y
[2]
Solutions:
(a)
mass of
syringe +
liquid Y
before
injection /g
mass of
syringe +
liquid Y
after
injection /g
Volume
of air in
gas
syringe
before
injection
/cm3
Volume of
air + vapour
Y in gas
syringe
after
injection /
cm3
Mass of
liquid Y
used /g
Volume of
vapour Y /
cm3
4.83
4.68
7
55
0.15
48
5.33
5.23
9
44
0.10
35
4.85
4.64
13
85
0.21
72
5.09
4.92
11
69
0.17
58
5.31
5.07
14
97
0.24
83
5.57
5.48
8
39
0.09
31
5.32
5.12
9
79
0.20
70
5.17
4.94
12
91
0.23
79
4.84
4.72
7
48
0.12
41
5.05
4.83
11
84
0.22
73
[2]
(b)
75  35
40


0.21  0.096 0.114
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
56
Chemistry A-Level P-5
June 2016/P52
90
80
Y2
70
volume of
vapour Y
/ cm3
60
50
Y1
40
30
0.08
0.10
x 1 0.12
0.14
0.16
0.18
0.20
x 2 0.22
0.24
mass of liquid Y /g
[2]
(c)
(i)
(ii)
(d)
(i)
Y may evaporates from the syringe, or
Y evaporate before weighing or after injection
[1]
Keep the syringe cool as much as possible, to stop evaporation and close off the
needle end to stop vaporation.
[1]
co-ordinates 1: (x1 y1) (0.118, 41)
co-ordinates 2: (x2 y2) (0.22, 77)
gradient
=
77  41
0.22  0.118
=
36
0.102
gradient = 352.94
P-5 (Planning, Analysis and Evaluation)
[2]
Yearly Worked Solutions
57
Chemistry A-Level P-5
(ii)
Mr 
3.07  102
gradient

3.07  104
353
June 2016/P52
experimentally determined Mr of Y = 869
(e)
Mr from spectrum is 84.
C = 6 × 12
=
H = 12 × 1
=
[2]
72
12
84
Empirical formula mass 84
molecular formula of Y C6H12
P-5 (Planning, Analysis and Evaluation)
[2]
Yearly Worked Solutions
Chemistry A-Level P-5
58
June 16/P-51
May/June 2016 Paper 51
1.
M/J 16/P51/Q1
Lithium is a soft alkali metal which may be cut with a knife. It is usually stored under oil because
it reacts rapidly with moisture and oxygen in the air.
Lithium is corrosive and may cause burns.
Lithium is high flammable and in large amounts reacts violently with water.
2Li(s) + 2H2O(I)  2LiOH(aq) + H2(g)
This reaction can be used to determine the relative atomic mass of lithium by measuring the
volume of hydrogen produced from a small amount of lithium.
(a)
Draw the apparatus you could use to measure the volume of hydrogen produced, using
standard laboratory equipment.
Label the chemicals in your diagram and show how the reactants can be kept apart
until the reaction is started.
[3]
(b)
To successfully carry out this experiment a correct procedure must be followed. The
lithium you will use is stored as large pieces under oil.
(i)
Beginning with a large piece of lithium being removed from the oil, state how
you would prepare a small pieces of lithium for use in this experiment.
[1]
(ii)
By only observing the gas collecting apparatus, state how you would know the
reaction had stopped.
[1]
(iii)
Other than eye protection, state two precautions you would take to make sure
that the experiment proceeds safely.
1.
2.
[2]
(iv)
The relative atomic mass of lithium is known to be approximately 7.
What approximate volume of hydrogen gas would a 0.1 g mass of lithium
produce?
(1 mol of gas occupies 24.0 dm 3 at room temperature and pressure.)
Volume of H2(g) produced ………………. [1]
(v)
What would be the capacity (volume) of the gas collecting apparatus you would
use for the volume of hydrogen produced in (iv)?
volume of gas collecting apparatus ………….. [1]
(c)
Another method that can be used to determine the relative atomic mass of lithium is by
titration of the lithium hydroxide produced during its reaction with water.
The following experimental procedure may be used.
1.
Add 100.0 cm3 of distilled water to a clean beaker.
2.
Add a known mass of lithium to the distilled water.
3.
After the reaction is complete, transfer 25.0 cm 3 of the solution of
lithium hydroxide from the beaker to a clean conical flask.
4.
Titrate this with an acid of known concentration.
(i)
State how you would accurately measure the total volume of distilled
water in step 1.
[1]
(ii)
State how you would know that the reaction between lithium and distilled water
was complete.
[1]
(iii)
State how you would transfer 25.0 cm 3 of the solution of lithium hydroxide into
a clean conical flask in step 3.
[1]
(iv)
State how you would ensure that your titration result was reliable.
[1]
(d)
To make sure that the beaker and the conical flask used in the experimental procedure
in (c) are clean, a student decides to wash them out with some distilled water before
starting the experiment.
Some water remains in the beaker. State the effect, if any, this would have on the
calculated relative atomic mass of lithium. Explain your reasoning.
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
Chemistry A-Level P-5
59
June 16/P-51
Some water remains in the conical flask. State the effect, if any, this would have on the
calculated relative atomic mass of lithium. Explain your reasoning.
[2]
Solution:
(a)
(b)
(c)
(d)
2.
[3]
Cut small piece of lithium inside the oil with knife, then add into flask and remove oil.
[1]
(ii)
Bubbles are stop coming out, and gas volume in the syringe stop to change.
[1]
(iii)
1. Wear gloves, as it is corrosive in action.
2. Keep the lithium piece away from water and flame.
[2]
0.1
1
(iv)
[since 2 moles Lithium produce 1 mole = 24 dm3 of hydrogen]

 24 
7
2
volume of H2(g) produced 171 cm3 [1]
(v)
volume of gas collecting apparatus 250 cm 3 [1]
(i)
Burette is used to measure volume.
[1]
(ii)
All solid lithium will reacted and dissolve, no bubble arise.
[1]
(iii)
Pipette or Burette.
[1]
(iv)
Repeat the titrations to get concordant readings.
[1]
Solution get diluted, and greater Ar will appear.
Number of moles of LiOH put in the solution remain same, so no effect.
[2]
(i)
M/J 16/P51/Q2
Activation energy EA, is the minimum energy with which particles must collide so that a reaction
occurs. The activation energy for the reaction of magnesium with aqueous hydrogen ions can
be determined in the laboratory.
Mg(s) + 2H+(aq)  Mg2+(aq) + H2(g)
A magnesium strip is placed in dilute hydrochloric acid and the time taken (t), in seconds, for
1  1
the magnesium to disappear is measured. The initial rate of reaction is calculated as
time  t 
If the experiment is repeated at several different temperatures then the following mathematical
relationship can be used to calculate EA.
EA
 1
 1
log10   
 
 t  0.0191  T 
T is the temperature measured in K.
1
is the initial rate of reaction in s–1.
t
1
 1
A graph of log10   against
can be plotted.
T
t 
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
60
Chemistry A-Level P-5
June 16/P-51
Experimental procedure
1.
25 cm3 of dilute hydrochloric acid is added to a boiling tube.
2.
The boiling tube is placed in a water bath until the dilute hydrochloric acid reaches a
constant temperature. This temperature is recorded.
3.
A magnesium strip of mass 0.10 g is added to the boiling tube, the mixture stirred and
the time taken for the magnesium to disappear is recorded.
4.
The temperature of the water bath is changed and the experiment is repeated.
(a)
The results of the experiment, carried out at different temperatures, are recorded in the
table below.
 1
 1
Process the results to calculate the reciprocal of temperature   and log10   . The
T 
t 
1
first value of
has been done for you.
T
1
Record
in standard form to three significant figures.
T
 1
 1
Record log10   to two decimal places. You should except log10   to be negative.
t 
t 
(b)
Temperature
/°C
Time t/s
Temperature
T/K
1
/ K 1
T
1 1
/s
t
15
83
288
3.47 × 10–3
1.20 × 10–2
20
58
293
1.72 × 10–2
27
36
300
2.78 × 10–2
30
28
303
3.57 × 10–2
34
18
307
5.56 × 10–2
38
19
311
5.26 × 10–2
40
15
313
6.67 × 10–2
43
12
316
8.33 × 10–2
48
9
321
1.11 × 10–1
55
8
328
1.25 × 10–1
 1
log10  
t 
[3]
Plot a graph on the grid on graph mentioned below to show the relationship between
1
 1
log10   and .
T
t 
Use a cross (×) to plot each data point.
Draw a line of best fit.
P-5 (Planning, Analysis and Evaluation)
[2]
Yearly Worked Solutions
61
Chemistry A-Level P-5
June 16/P-51
1
/K–1
T
3.0  10 –3
3.1  10 –3
3.2  10
–3
3.3  10 –3
3.4  10 –3
3.5  10 –3
3.6  10 –3
−0.8
−0.9
−1.0
−1.1
−1.2
−1.3
1
log10 ( )
t
−1.4
−1.5
−1.6
−1.7
−1.8
−1.9
−2.0
(c)
(d)
On your graph, circle the two points you consider to be the most anomalous. Label
each one with a different letter. Explain what may have caused each of the anomalies
you have identified, giving a different reason each time.
Make it clear in your answer to which point you are referring.
[2]
(i)
Determine the gradient of your graph. State the co-ordinates of both points you
used for your calculation.
Record the value of the gradient to three significant figures.
 co-ordinates 1
 co-ordinates 2
gradient = ……………….[2]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
62
Chemistry A-Level P-5
June 16/P-51
(ii)
(e)
(f)
(g)
(h)
Use your gradient from (i) and the mathematical relationship on next page to
calculate the activation energy EA, in kJ mol–1.
Include a sign in your answer.
EA = ……………….. kJ mol–1 [2]
State whether you consider the results to be reliable. Explain your answer.
[1]
Student X commented that data collected at higher temperatures in the experiment
may be less accurate than that collected at lower temperatures.
State whether student X is correct. Explain why.
[1]
If the magnesium strip is not stirred it floats to the surface of the hydrochloric acid.
State how this will affect the reaction time. Explain why.
[1]
The experiment (a) is repeated using dilute ethanoic acid instead of dilute hydrochloric
acid. The concentration of both acids is equal. The same temperatures are used as in
(a).
State the effect this change in acid will have on the initial rate values. Give a reason for
this.
[1]
Solution:
(a)
Temperature
/°C
Time t/s
Temperature
T/K
1
/ K 1
T
1 1
/s
t
 1
log10  
t 
15
83
288
3.47 × 10–3
1.20 × 10–2
–1.92
293
3.41 ×
10–3
1.72 ×
10–2
–1.76
3.33 ×
10–3
2.78 ×
10–2
–1.56
3.30 ×
10–3
3.57 ×
10–2
–1.45
3.26 ×
10–3
5.56 ×
10–2
–1.25
3.22 ×
10–3
5.26 ×
10–2
–1.28
3.19 ×
10–3
6.67 ×
10–2
–1.18
3.16 ×
10–3
8.33 ×
10–2
–1.08
3.12 ×
10–3
1.11 ×
10–1
–0.95
3.05 ×
10–3
1.25 ×
10–1
–0.90
20
27
30
34
38
40
43
48
55
58
36
28
18
19
15
12
9
8
300
303
307
311
313
316
321
328
[3]
(b)
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
63
Chemistry A-Level P-5
June 16/P-51
1
/ K–1
T
3.0× 10 –3
3.1× 10 –3
3.2 × 10 –3
3.3 × 10 –3
3.4 × 10 –3
3.5 × 10 –3
3.6 × 10 –3
−0.8
−0.9
−1.0
−1.1
−1.2
−1.3
1
log 10 ( )
t
−1.4
−1.5
−1.6
−1.7
−1.8
−1.9
−2.0
(c)
(d)
(e)
(f)
[2]
X, is point due to late time is stopped or timer was started early.
Y-point time was started late, or stopped early or Mg disappear earlier.
[2]
(i)
co-ordinates (1): (–1.46, –0.8)
co-ordinates (2): (3.3 × 10–3, 3.07 × 10–3)
gradient = 2869.56 [2]
Gradient × 0.0191
(ii)
– EA =
EA = 0.0548 kJ mol–1 [2]
1000
Most of the points lies close to the line, so the results are reliable.
[1]
Reaction time is very less at higher temperature so the uncertainty or percentage error is
greater X is corrected.
[1]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
Chemistry A-Level P-5
(g)
(h)
64
June 16/P-51
Area of contact is less, so reaction time is greater.
H+ ion concentration in case of Ethanoic acid is less, so initial rate is very slow.
P-5 (Planning, Analysis and Evaluation)
[1]
[1]
Yearly Worked Solutions
65
Chemistry A-Level P-5
November 2015/P52
October/November 2015 Paper 52
1.
O/N 15/P52/Q1
It is possible to determine the relative molecular mass, Mr, of a small sample of a volatile liquid
by measuring its mass and then heating to vaporize it to obtain its volume as a gas.
(a)
Explain how the relative molecular mass can be determined in this way.
[2]
(b)
(i)
The volume of the vaporized sample depends on its temperature and pressure.
In an experiment, a sample of volatile liquid of known mass was vaporized and
its volume recorded. The pressure was correctly recorded as 101 kPa but the
temperature was incorrectly recorded as 50°C. The correct temperature was
60°C.
By considering the effect of these different temperatures on the gas volume,
explain how the value of the calculated Mr would be affected.
[3]
(ii)
The temperature was maintained at 60°C but the pressure was increased to
110 kPa. Would this have given an answer that was nearer to the true values
of the relative molecular mass? Explain your answer.
[1]
In an experiment to determine the relative molecular mass of hexane, boiling point 69°C, a
specialist piece of apparatus called a Victor Meyer tube can be used. This consists of a long
tube with a bulb at the base in which a sample can be vaporized. The tube has a side arm to
allow the escape of gas from within the tube. The tube is surrounded by another which can be
used to heat the contents of the first tube.
A diagram of the apparatus is shown below.
stopper
small
sample tube
hexane
Victor Meyer tube
sand
A small sample tube containing the hexane is inserted at the top of the Vector Meyer tube. The
sample tube is small enough to fit inside the Victor Meyer tube and falls onto the hot sand
below. The sand will cushion its fall so that the sample tube does not break. The stopper is then
quickly replaced at the top of the Victor Meyer tube. The hot sand causes the hexane to
vaporize and expel air contained in the Victor Meyer tube.
(c)
Complete the diagram above to show:
 how the apparatus should be heated.
 a connection to further apparatus which would allow the air expelled from the Victor
Meyer tube when the sample of hexane is vaporized to be collected and measured.
[2]
(d)
Suggest one hazard associated with the use of hexane.
[1]
(e)
(i)
With the gas collection apparatus connected to the heated Victor Meyer tube,
expelled air will be collected before the hexane is introduced. Explain why. [2]
(ii)
At which stage of the experimental procedure should the sample tube be
dropped into the Victor Meyer tube?
[1]
(f)
State what measurements you would need to make in order to determine the relative
molecular mass of hexane.
[3]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
66
Chemistry A-Level P-5
November 2015/P52
Solution:
(a)
Note down the pressure exerted by the vapourised gasp. Note down the volume of gas
evaporated. Mass is also known. Use the formula pV = nRT to get “n” number of moles.
Then use Mr =
(b)
(i)
(ii)
m
to get Mr.
n
[2]
Volume of gas is lowered at lower temperature (i.e. of 50°C). This in turn causes the
calculated value of the Mr to be less than the actual one.
[3]
Yes the answer would be close to the true value as the increase in pressure would
cause the Mr to increase.
[1]
stopper
small
sample tube
Gas Syringe
hexane
Victor Meyer tube
stand to hold gas syringe
in place ubter to heat the
sample uniformly
sand
stand to hold gas syringe
in place ubter to heat
the sample uniformly
Tripod
stand
Bunsen
burner
(c)
(d)
(e)
(f)
2.
[2]


(i)
(ii)



Causes breathing difficulties.
Causes irritation to the skin.
The air would expand and will move towards the collection apparatus.
Tube syringe no longer moves.
Temperature and pressure.
The mass of tube with Hexane and mass of empty tube.
Syringe reading before addition of hexane and after additional
hexane.
[1]
[1]
[1]
of
[3]
O/N 15/P52/Q2
In an experiment, various masses of solid barium hydroxide are added to 60.0 cm 3 of a solution
of hydrochloric and contained in a polystyrene cup.
In each experiment a fresh sample of the acid is taken and its initial temperature is measured.
After the Barium hydroxide has been added, the acid is stirred and the maximum temperature
reached is noted.
The results of each experiment are recorded in the table below.
(a)
Complete the table below to give the temperature rise obtained from each experiment
to one decimal place and the amount of barium hydroxide used in mol to three
significant figures in each case.
The mass of 1 mol of barium hydroxide is 171 g.
Initial
temperature
of HCl/°C
mass of barium
hydroxide
added / g
Maximum
temperature
reached /°C
21.0
0.500
22.2
P-5 (Planning, Analysis and Evaluation)
Temperature
rise / °C
barium
hydroxide
added / mol
Yearly Worked Solutions
67
Chemistry A-Level P-5
(b)
(i)
November 2015/P52
20.6
1.00
23.0
21.2
1.50
24.9
21.8
2.00
26.5
20.5
3.00
27.8
21.4
4.00
31.1
21.2
5.00
31.6
21.0
6.00
31.4
20.8
8.00
31.2
[2]
Using the grid, plot a graph to show how the temperature rise varies with the
moles of barium hydroxide added.
[1]
12.0
11.0
10.0
9.0
temperature rise / C
8.0
7.0
6.0
5.0
4.0
3.0
2.0
1.0
0.0
0.000
0.010
0.020
0.030
0.040
0.050
barium hydroxide added / mol
(ii)
(c)
Draw two lines of best fit on your graph and state the value on the x-axis at
the point of intersection of the two lines.
Value on the x-axis at the point of intersection is ………..
[2]
Use the value on the x-axis at the point of intersection to calculate the concentration of
the hydrochloric acid in mol dm–3.
[2]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
68
Chemistry A-Level P-5
(d)
November 2015/P52
Explain the variation in temperature that takes place when barium hydroxide is added
to the hydrochloric acid.
[2]
(i)
When the experiment is done in the way described, the results are not very
accurate.
Apart from limitations due to the accuracy of the measuring equipment,
suggest why:
 all the temperature rises measured are less than theoretically should be
expected.
 the temperature rises are more inaccurate as they approach their
maximum value.
[2]
(ii)
What improvement would you make to achieve greater accuracy?
[1]
3
3
In another experiment, 60.0 cm of ethanoic acid is used instead of the 60.0 cm of
hydrochloric acid.
If the ethanoic acid has the same concentration as the hydrochloric acid, draw on your
graph another pair of lines to show the results you would expect to obtain.
Explain your answer.
[3]
(e)
(f)
Solution:
(a)
Initial
temperature
of HCl/°C
21.0
20.6
21.2
21.8
20.5
21.4
21.2
21.0
20.8
mass of barium
hydroxide
added/ g
0.500
1.00
1.50
2.00
3.00
4.00
5.00
6.00
8.00
Maximum
temperature
reached /°C
22.2
23.0
24.9
26.5
27.8
31.1
31.6
31.4
31.2
Temperature
rise / °C
1.2
2.4
3.7
4.7
7.3
9.7
10.4
10.4
10.4
barium
hydroxide
added / mol
0.00292
0.00585
0.00877
0.0117
0.0175
0.0234
0.0292
0.0351
0.0468
[2]
(b)
(i)
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
69
Chemistry A-Level P-5
November 2015/P52
12.0
11.0
10.0
9.0
temperature rise / C
8.0
7.0
With Ehanoic Acid
6.0
With HCl
5.0
4.0
3.0
2.0
1.0
0.0
0.000
0.010
0.020
0.030
0.040
0.050
barium hydroxide added / mol
(ii)
(c)
(d)
(e)
Value on the x-axis at the point of intersection is 0.0248.
[1]
[2]
Ba(OH)2 + 2HCl  BaCl2 + 2H2O
0.0248 × 2
= 0.496 mol
0.496
 1000
=
60
= 0.827 mol dm-3
[2]
The given reaction is exothermic as there is a rise in temperature. Upto a certain mass/moles
of Ba(HO)2 added there is a rise in temperature till HCl fully reacted after which
volume/concentration of HCl became a limiting factor, hence the reaction does not proceed.
Consequently the rise in temperature becomes constant.
[2]
(i)

Heat is lost to the surroundings.

As the reaction proceeds the reaction gets slower
[2]
(ii)

Use powdered Ba(OH)2.

Use lid to cover the polystyrene cup.
[1]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
Chemistry A-Level P-5
(f)
70
November 2015/P52
As ethanoic acid is a weak acid some of the heat energy which is released is used to ionize
ethanoic acid.
[3]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
Chemistry A-Level P-5
71
November 2015/P53
October/November 2015 Paper 53
1.
O/N 15/P53/Q1
The halogenoalkanes can react with hydroxide ions to form an alcohol and a halide ion.
(a)
The rate at which the reaction occurs depends on which of the halogenoalkanes is
chosen. The reaction is a nucleophilic attack by the hydroxide ion and the rate might
depend on:

the polarity of the carbon-halogen bond,

the bond strength of the carbon-halogen bond.
For (i) and (ii), chlorobutane, bromobutane and iodobutane should be considered.
(i)
If the rate of reaction was only controlled by the polarity of the carbon-halogen
bond, the order of reactivity (most reactive to least reactive) would be
Explain the variation in the polarity of the carbon-halogen bonds.
[2]
(ii)
If the rate of reaction was only controlled by the bond strength of the carbonhalogen bond, the order of reactivity (most reactive to least reactive) would be
Explain the variation in the bond strength of the carbon-halogen bonds.
[2]
(b)
An experiment can be carried out to compare the extent of the reaction between
aqueous hydroxide ions and chlorobutane, bromobutane and iodobutane. Samples of
the halogenoalkanes are reacted with sodium hydroxide for 2 minutes at 50°C. After
the reaction, addition of aqueous silver nitrate causes the formation of a silver halide
precipitate.
Some hazards associated with the use of halogenoalkanes include:

very hazardous in case of skin and particularly eye contact,

very hazardous if inhaled or ingested,

flammable.
To carry out this experiment, the following would be supplied.

usual laboratory apparatus

laboratory reagents including a suitable aqueous solution of sodium hydroxide
and aqueous silver nitrate

samples of each of the three liquid halogenoalkanes
(i)
Identity the independent variable and the dependent variable in this
experiment.
Independent variable
dependent variable
[1]
(ii)
The amount of each halogenoalkane liquid to use is most practically measured
by its volume. Usually equal volumes of the three halogenoalkanes are used.
Explain why this is not ideal and what change should be made to obtain a more
reliable comparison between the halogenoalkanes.
[2]
(iii)
Answer the following questions about the experiment.

Having measured the quantity of fhalogenoalkanes, what must be
ensured about the amount of sodium hydroxide used?

How would the reaction tubes be heated and the experiment be
started?

What reagent could be added which would ensure that after 2 minutes
the reaction was stopped?

After the reaction has been stopped and aqueous silver nitrate added,
the amount of precipitate formed could be determined by measuring
its height.
Why is it necessary to leave the tubes for some time before making
this measurement?
[5]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
72
Chemistry A-Level P-5
November 2015/P53
(iv)
(c)
Although the amount of silver halide formed can reasonably be obtained by
measuring the height of the precipitate in the reaction tube, this is not very
reliable.
Explain what should be done with the precipitate to obtain a more reliable
measurement of the amount of silver halide produced.
[2]
The use of halogenoalkanes is hazardous and both gloves and eye protection are
necessary. State one other essential precaution which should be taken when carrying
out the experiments.
[1]
Solution:
(a)
(b)
(c)
2.


Chlorobutane – Bromobutane – Iodobutane
Electronegativity decreases from chlorine to iodine. Higher electronegativity
results in higher reactivity.
[2]
(ii)

Iodobutane – bromobutane – chlorobutane

Bond strength decreases from the chlorobutane, to Iodobutene so reactivity
will increase.
[2]
(i)
Independent variable: Type of halogenoalkane
Dependent variable:
Amount of precipitate formed.
[1]
(ii)
Equal moles of the halogenoalkanes are not being used. For a more reliable
comparison, use equal moles of each halogenoalkane.
Or
Equal volumes do not contains equal number of moles so for reliable comparison
must take equal number of moles.
[2]
(iii)

Use same amount of NaOH for each halogenoalkane.

Place the tubes in a water bath. Heat the water bath to 50°C by a Bunsen
burner. Use thermometer to ensure that this temperature is achieved and
maintained. Mix the reagents and immediately start timer.

Nitric acid.

Allow the precipitate to settle.
[5]
(iv)
Filter the precipitates and dry them. Then weight them using an electronic balance.
[2]
Do not expose the halogenoalkanes directly to frames
OR
Wear a mask, us a fume cupboard, carry out experiment in a well ventilated
room.
[1]
(i)
O/N 15/P53/Q2
At 25°C, dinitrogen tetroxide, N2O4(g), forms an equilibrium mixture with nitrogen dioxide,
NO2(g),
N2O4(g)
 2NO2(g)
H = +57.2 kJ mol–1
As N2O4(g) is colourless and NO2(g) is brown, the composition of an equilibrium mixture can be
determined by its colour.
(a)
Write an expression for the equilibrium constant, Kc, for this equilibrium.
[1]
(b)
In an experiment, quantities of N2O4 are left to reach an equilibrium which contains
N2O4 and NO2. These are analysed to determine the concentrations of N 2O4(g) and
NO2(g) that are present. The results are listed in the first two columns of the table below.
Complete the third column of the table to give the value of [NO2(g)]2 for each of the
results of the experiment. Values should be given to three significant figures.
[N2O4(g)]/mol dm–3
[NO2(g)]/mol dm–3
0.900
0.0729
0.800
0.0687
0.700
0.0643
0.600
0.0595
P-5 (Planning, Analysis and Evaluation)
[NO2(g)]2/mol2 dm–6
Yearly Worked Solutions
73
Chemistry A-Level P-5
0.500
0.0548
0.400
0.0486
0.300
0.0390
0.200
0.0344
0.100
0.0243
November 2015/P53
[2]
(c)
(i)
The value of the equilibrium constant for N2O4(g)  2NO2(g) can be calculated
from a graph of [N2O4(g)] against [NO2(g)]2.
Use the grid to plot this graph and draw a line of best fit through the plotted
points.
0.006
0.005
[NO2(g)]2 / mol2 dm–6
0.004
0.003
0.002
0.001
0.000
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
[N2O 4(g)]/ mol dm –3
(ii)
[2]
Choose two suitable sets of values from your graph and use them to calculate
a value for the equilibrium constant, Kc. Give your answer to three significant
figures and give its units.
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
74
Chemistry A-Level P-5
November 2015/P53
co-ordinates of two points used ………………..
(d)
(e)
(f)
……………………
Kc = …………. units …………
[3]
Identify the result which is most anomalous and suggest a reason, other than a
calculation error, why this may have occurred.
[2]
(i)
On your graph, draw a line that would be obtained if the temperature of the
equilibrium mixture was raised.
(ii)
Explain the position of the line drawn in (i).
(iii)
What effect, if any, would the higher temperature have on the value of Kc? [1]
(iv)
How would your value for the equilibrium constant change if the pressure
applied to the equilibrium mixture was increased.
[1]
In the experiments, the result have been obtained by starting with pure N 2O4(g) and
then letting the equilibrium with NO2(g) form.
Calculate the starting concentration of pure N2O4(g) that would be required to produce
the mixture of 0.900 mol dm –3 of N2O4(g) and 0.0729 mol dm –3 of NO2(g) once
equilibrium had been established.
[1]
Solution:
NO2 
N2O4 
2
(a)
Kc =
[1]
(b)
[N2O4(g)]/mol dm–3
[NO2(g)]/mol dm–3
[NO2(g)]2/mol2 dm–6
0.900
0.0729
0.00531
0.800
0.0687
0.00472
0.700
0.0643
0.00413
0.600
0.0595
0.00354
0.500
0.0548
0.00300
0.400
0.0486
0.00236
0.300
0.0390
0.00152
0.200
0.0344
0.00118
0.100
0.0243
0.00590
[2]
(c)
(i)
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
75
Chemistry A-Level P-5
November 2015/P53
0.006
0.005
[NO 2(g)]2 /mol2 dm–6
0.004
0.003
0.002
0.001
0.000
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
–3
[N2O 4(g)]/mol dm
[2]
(ii)
Co-ordinates of two points used: (0.07, 0.65)
(K c ) Gradient 

(d)
(e)
(0.0004, 0.00385)
0.00385  0.00040
0.65  0.07
0.00345
0.58
= 0.00594
Kc
= 0.00594 units mol dm–3
[2]
–3
The result at (N2O4) = 0.3 mol dm is the most anomalous, it is because temperature might
have been lower than 25°C (too low) or the equilibrium might not have been established,
yet
[1]
(i)
On Graph
[1]
(ii)
Reaction is endothermic hence increasing temperature would give higher
concentration of NO2 i.e. steeper gradient.
[1]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
76
Chemistry A-Level P-5
(iii)
(iv)
(f)
November 2015/P53
Value of Kc increases.
No change.
[1]
[1]
N2O4(g)  2NO2(g)
At start
At equilibrium
x
0.900
x
0.0729
 0.900
2
x = 0.900 + 0.03645
= 0.0963 mol dm–3
P-5 (Planning, Analysis and Evaluation)
O
0.0729
[1]
Yearly Worked Solutions
77
Chemistry A-Level P-5
June 2015/P52
May/June 2015 Paper 52
1.
M/J 15/P52/Q1
This question concerns electrolysis of different compounds.
(a)
During the electrolysis of dilute sulfuric acid using a current of 0.75 A for 90 minutes,
the volume of oxygen gas collected was recorded and is shown in the graph below.
250
volume of oxygen
200
150
volume
of gas
/ cm 3
100
50
0
0
10
20
30
40
50
60
70
80
90
time / min
(i)
(ii)
(iii)
(b)
Give equations for the reactions that occur at each electrode in the electrolysis
of sulfuric acid.
[2]
On the graph above, use a ruler to draw and label a line (hydrogen) to predict
the volume of hydrogen that would be given off during same experiment [1]
On the graph above, use a ruler to draw and label a line (oxygen) to predict the
volume oxygen that would be produced if a current of 0.45A was used instead
of the 0.75A used in the original experiment.
[1]
During the electrolysis of potassium butanedioate, the following reaction occurs.
CH2COO-K +
2H2O + 
CH2COO-K +
 C2H4 + 2CO2 + H2 + 2K +OH
potassium
butanedioate
An experiment can be carried out to confirm the above equation. In order to do this, the
amounts of hydrogen, ethane and carbon dioxide produced need to be measured.
Hydrogen is produced at one electrode, ethane and carbon dioxide are produced at the
other. The carbon dioxide can be separated from the ethene by absorbing it in an alkali
before the volume of ethene is measured.
(i)
Using the power supply drawn below, draw a fully labelled circuit diagram and
apparatus which shows how:

the current could be measured,

the hydrogen produced could be collected and its volume measured.

the carbon dioxide could be removed using a named alkali,

the volume of ethene could be measured.
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
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Chemistry A-Level P-5
+
June 2015/P52
–
d.c. power
supply
(ii)
(iii)
(iv)
(v)
(vi)
State what measurements should be taken when carrying out the experiment.
C coulombs of electricity resulted in V cm3 of hydrogen gas being produced
during the electrolysis.
In terms of C and V, state the number of coulombs, N, that would be required
to produce 24 dm3 of hydrogen.
In terms of N, state the number of faradays of electricity that would be required
to produce 1 mol of hydrogen at room temperature and pressure.
(1 faraday of electricity = 96 500 coulombs)
Give the equation for the reaction that takes place when the carbon dioxide is
absorbed by the alkali.
Predict the organic product that would be obtained at the anode when a
solution of potassium hexanedioate is electrolyzed.
Solution:
(a)
(i)
4H+ (aq) + 4e  2H2 (g)
At cathode

4OHaq
 2H2( ) O + O2  4e 


At anode
[2]
(ii)
(II)
(III)
volume
of gas
3
/cm
[1]
(iii)
(b)
0.75 A : 250 cm 3 of O2
0.45 A : x
x = 150 cm3
[1]
(i)
P-5 (Planning, Analysis and Evaluation)
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Chemistry A-Level P-5
June 2015/P52
d.c.power
supply
Gas syringe to
measure volume 2
1+2
Gas syringe ensure
volume of Ethene
boiling tube with
KoH Solution to
absorb Cu 2
Cathode
Anode
Electrolyte
[5]
(ii)
(iii)
 Note down the current by Ammeter.
 Now down the volume of hydrogen gas evolved.
C = v cm3
N : 24
N=
(iv)
(v)
C
× 24 dm3
v × 10-3
Faraday =
[2]
[Ampere]
[1]
N
96500
2 KOH + CO2  K 2CO3 + H2O
(aq)
(g)
(aq)
O
[1]
( )
O
K O - C - (CH2 )4 - C - O-K + +2H2O  2K + + 2OH  2CO2 + H2 + C4H8
+ -
(vi)
aq
( )
aq
aq
(g)
(g)
CH3 - CH=CH-CH3
But–2–ene
2.
[1]
M/J 15/P52/Q2
In order to identify a monoprotic (monobasic) hydroxycarboxylic acid, HX, the following
experiments are carried out.
25.0 cm3 of an aqueous solution of HX is titrated against 0.0500 mol dm –3 aqueous sodium
carbonate. The end-point of the titration is reached when 25.0 cm 3 of the aqueous sodium
carbonate has been added.
(a)
(i)
Write the equation for the complete neutralisatiion of HX with sodium
carbonate.
[1]
(ii)
How does the equation show that the concentration of HX is 0.100 mol dm–3.
[1]
(b)
(i)
State the acid dissociation constant Ka, for the above reaction in terms of H+
and HX only.
[1]
(ii)
The pH of the aqueous solution of HX is 2.43.
Use the pH and the concentration of HX to show that the pKa of the acid is
3.86. All your working must be shown.
[2]
(c)
In an experiment various masses of the sodium salt of the acid, NaX, are added to
separate portions of 100 cm 3 of HX with stirring. After each addition the pH of the
solution obtained is measured. The results of the experiment are recorded in the table
below.
P-5 (Planning, Analysis and Evaluation)
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80
Chemistry A-Level P-5
(i)
June 2015/P52
Mass of naX added / g
pH
0.00
2.43
0.10
2.81
0.20
3.11
0.30
3.19
0.40
3.41
0.60
3.59
0.80
3.71
1.00
3.81
1.20
3.89
1.50
3.99
2.00
4.11
Plot a graph to show how the pH of the solution varies with the mass of NaX
added. Draw the curve of bests fit.
4.2
4.0
3.8
3.6
3.4
pH
3.2
3.0
2.8
2.6
2.4
0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
mass of NaX added / g
(ii)
(d)
(i)
[2]
Circle one anomalous point on your graph and give a reason for how this could
have occurred using the experimental procedure described.
[2]
The graph shows that a pH of 3.86 is obtained when 1.12 g of NaX is added to
100 cm3 of HX.
Remember that pKa of HX is also 3.86.
Use this information to calculate the relative molecular mass, Mr, of HX.
P-5 (Planning, Analysis and Evaluation)
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81
Chemistry A-Level P-5
(ii)
(e)
June 2015/P52
Show your working.
[Ar: H, 1.0; C, 12.0; O, 16.0; Na, 23.0]
[3]
The calculated Mr is subject to small experiment error. Suggest a structure for
the organic hydroxycarboxylic acid, HX, that best fits your Mr data.
If you have not calculated a value for the Mr, use the value of 104. This is not
the correct value.
[1]
Another method for determining the concentration of the acid HX could be to evaporate
a sample of the solution and weigh and solid that remains.
Suggest two reasons why this might not be a very good method of finding the mass of
solid HX in a sample of the solution.
[2]
Solution:
(a)
(i)
(ii)
(b)
(i)
Na2CO3 + 2HX  2 NaX + H2O + CO2
aq
aq
aq
( ) (g)
[1]
Molar ration of Na2CO3 and HX is 1:2. Since both have equal volume so
concentration of HX is 0.1 mol dm –3.
[1]
H    X  
Ka     
HX 
 sin ce H   X 
2
H  
Ka   
 HX 
(ii)
(c)
[1]
[H+]
[H+]
= 10–2.43
= 3.7 + 5 × 10–3 mol dm–3.
Ka
=
Ka
pKa
pKa
= 1.38 × 10–14
= –log (Ka)
= 3.86
3.715  103 
 0.1
2
[2]
(i)
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
82
Chemistry A-Level P-5
June 2015/P52
[2]
(d)
(ii)
(i)
Less mass of NaX is measured.
PH = pKa + log
3 8 6  3 8 6  log
Acid
0.01 
Salt
Mr 
0.1
x
1.12
Mr
1.12
0.1
l o g x  l o g 0.1
Mr = 112
mass of X
(salt) = 0.1 mol dm–3
NaX(n) = c × v
mass of HX
= 112 – 23
= 89
= 90
100
= 0.1
1000
= 0.01 moles
[3]
(ii)
H H
O
H H
or
H C C C O
H OH
(e)



H
O
HO C C C O
H H
H
HX may decomposed during heating.
Acid may got evaporate.
Acid may spit out
P-5 (Planning, Analysis and Evaluation)
[1]
[2]
Yearly Worked Solutions
83
Chemistry A-Level P-5
June 2015/P51
May/June 2015 Paper 51
1.
M/J 15/P51/Q1
A saturated aqueous solution of magnesium methanoate, Mg(HCOO) 2, has a solubility of
approximately 150 g dm –3 at room temperature. Its exact solubility can be determined by
titrating magnesium methanoate against aqueous potassium manganate(VII).
During the titration, the methanoate ion, HCOO –, is oxidised to carbon dioxide while the
manganate(VII) ion, MnO4–, is reduced to Mn2+.
You are supplied with:
a saturated aqueous solution of Mg(HCOO)2
aqueous potassium manganate(VII), KMnO4, of concentration 0.0200 mol dm –3.
(a)
(i)
Write the half equations for the oxidation of HCOO –(aq) to CO2(g) and the
reduction of MnO4–(aq) to Mn2+(aq) in acid solution.
[2]
(ii)
Using the approximate solubility above, calculate the concentration, in mol
dm–3, of the saturated aqueous magnesium methanoate and the concentration
of the methanoate ions present in this solution.
[Ar: H, 1.0; C, 12.0; 0.16.0; Mg, 24.3]
(iii)
[2]
In order to obtain a reliable titre value, the saturated solution of magnesium
methanoate needs to be diluted.
Describe how you would accurately measure a 5.0 cm 3 sample of saturated
magnesium methanoate solution and use it to prepare a solution fifty times
more dilute than the saturated solution.
[2]
(iv)
Before the titration is carried out, dilute sulfuric acid must be added to the
magnesium methanoate.
Explain why this is necessary and also whether the volume of sulfuric acid
chosen will affect the result of the titration.
[2]
(v)
The potassium manganate(VII) is added from a burette into the magnesium
methanoate in a conical flask.
Describe what you would see when you had reached the end-point of the
titration.
[1]
(vi)
1 mol of acidified MnO4– ions reacts with 2.5 mol of HCOO– ions.
25.0 cm3 of the diluted solution prepared in (iii) required 25.50 cm 3 of 0.0200
mol dm–3 potassium manganite(VII) solution to reach the end-point.
Use this information to calculate the concentration, in mol dm–3, of HCOO– ions
in the diluted solution.
………. mol dm–3 [1]
(vii)
Use your answer to (vi) calculate the concentration, in mol dm–3, of the
saturated solution of magnesium methanoate, Mg(HCOO)2. Give your answer
to three significant figures.
……….mol dm–3 [1]
(b)
The solubility of magnesium methanoate can be determined at higher temperatures
using the same titration.
In an experiment to determine how the concentration of saturated magnesium
methanoate varies with temperature, name the independent variable and the
dependent variable.
Independent variable
Dependent variable
(c)
[1]
The solubility of magnesium methanoate increases with temperature.
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
84
Chemistry A-Level P-5
June 2015/P51
What does this tell you about H for the process below?
Mg(HCOO)2(s)  Mg2+ (aq) + 2HCOO–(aq)
Explain your answer.
(d)
[2]
A student used the same titration method, this time to measure the concentration of a
saturated solution of barium methanoate.
Explain why the acidification of the solution with dilute sulfuric acid might make the
titration difficult to do.
[1]
Solution:
(a)
(i)
HCOO
(aq)

+
 CO2 (g) + H(aq)
+ 2e-  5
5e + 8H (aq) + MnO

(ii)


(aq)
2+
 Mn(aq)
+ 4H2O

2
[2]
Mg(HCOO)2  Mg2+ + 2HCOO–
Mr of Mg(HCOO)2 = 24.3 + (2 × 1.0) + (12.0 × 2) + (16.0 × 2 × 2)
= 114.3
Concentration of Mg(HCOO)2 =
(iii)
(iv)
(v)
150
= 1.312 mol dm–3
114.3
Concentration of Mg(HCOO)2 in mol dm–3 = 1.312
Concentration of HCOO– in mol dm–3 = 1.312 × 2 = 2.624 mol dm–3
[2]
Use a burette to accurately measure 5 cm 3 of magnesium methanoate. Pour it into
a volumetric flask and fill it up to the 250 cm 3 mark with distilled water.
[2]
+
H is needed for the reaction with manganite. If acid is used in excess, then the
volume does not matter.
[2]
Pole pink colour.
[1]
 0.02 
(vi)
25.5
1000
MnO4  5.1 104 mole
HCOO– = 5.1 × 10–4 × 2.5
HCOO– in 25 cm3 = 1.28 × 10–3 mole
Concentration of solution of HCOO– =
1.28  103
 1000
25
= 0.051 mol dm–3 [1]
3
(vii)
1.28  10  250
25
 1.28  102 
1
2
Mg(HCOO)2 in 5 cm = 6.4 × 10–3
Concentration of Mg(HCOO)2 =
6.4  103  1000
5
= 1.28 mol dm–3 [1]
(b)
(c)
(d)
2.
Independent variable: temperature
Dependent variable: concentration
[1]
H will be positive as reaction is endothermic. With increase in temperature position of
equilibrium will shift forwards.
[2]
Barium sulfate precipitate will be formed.
[1]
M/J 15/P51/Q2
At high temperatures a mixture of the iodine and hydrogen gases reacts to form an equilibrium
with gaseous hydrogen iodide.
H2(g) + I2(g)  2HI(g)
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
85
Chemistry A-Level P-5
(a)
(b)
June 2015/P51
(i)
Write an expression for the equilibrium constant, Kc, based on concentration,
for this reaction.
[1]
(ii)
If the starting concentration of both iodine and hydrogen was a mol dm–3 and
it was found that 2y mol dm–3 of hydrogen iodide had formed once equilibrium
had been established, write Kc in terms of a and y.
[1]
The experiment for the equilibrium constant from (a)(ii) can be re-written as shown
below.
y=
a Kc
2+ K c
[1]
dm 3
In an experiment, air was removed from a 1
flask and amounts of hydrogen and
iodine gases were mixed together such that their initial concentrations were both a mol
dm–3. This mixture was allowed to come to equilibrium at 760 K in the flask. The
equilibrium concentration of iodine, (a – y) mol dm–3, was then measured.
The experiment was repeated for various initial concentrations, a mol dm–3, and the
results were recorded in the table below.
(i)
Complete the table to give the values of y mol dm–3 to three decimal places.
a mol dm–3
0.200
0.500
0.800
1.000
1.500
2.100
2.800
3.400
3.800
4.200
4.900
(a – y) mol dm–3
0.022
0.050
0.252
0.200
0.365
0.570
0.652
0.700
0.867
0.868
1.150
y mol dm–3
0.178
[2]
(ii)
Plot a graph to show how y mol
hydrogen and iodine, a mol dm–3.
P-5 (Planning, Analysis and Evaluation)
dm–3
varies with initial concentrations of
Yearly Worked Solutions
86
Chemistry A-Level P-5
June 2015/P51
[1]
(c)
(iii)
Use your points to draw a line of best fit.
(i)
Determine the slope of your graph. State the co-ordinate of both points you
used for your calculation. Record the value of the slope to three significant
figures.
co-ordinates of both points used
[1]
………..….
……………
slope = ………………
(ii)
[2]
Use the value of your slope and the equation in (b) to calculate the value of Kc.
Your working must be shown.
[2]
(d)
Explain why, for safety reasons, it is necessary to remove air from the 1 dm 3 flask. [1]
(e)
One of the experiments in (b) was repeated in a 500 cm 3 flask instead of the 1 dm 3
flask.
What effect, if any, would this have on the rate of reaction and the value of Kc
measured?
[2]
(f)
The reaction of hydrogen and iodine to form hydrogen iodide is exothermic.
H2(g) + I2(g)  2HI(g)
H = –9.6 kJ mol–1
(i)
On your graph, draw and label the line you would expect if the experiment was
performed at 1000 K instead of 760 K.
[1]
(ii)
What effect, if any, would be higher temperature have on the value of Kc? [1]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
87
Chemistry A-Level P-5
June 2015/P51
Solution:
HI
H2 I2 
2
(a)
(b)
(i)
Kc =
(ii)
Kc =
(i)
[1]
4y 2
a - y 
[1]
2
a mol dm–3
0.200
0.500
0.800
1.000
1.500
2.100
2.800
3.400
3.800
4.200
4.900
(a–y) mol dm–3
0.022
0.050
0.252
0.200
0.365
0.570
0.652
0.700
0.867
0.868
1.150
y mol dm–3
0.178
0.450
0.548
0.800
1.135
1.530
2.148
2.700
2.933
3.332
3.750
[2]
(ii)
3.5
4.6
[1]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
88
Chemistry A-Level P-5
June 2015/P51
(iii)
[1]
(c)
(i)
Co-ordinates off both points used (0,0)

(4.6, 3.51)
y 2  y1
x2  x1
3.5  0
4.6  0
y = 0.761
slope = 0.761
(ii)
0.761 
[2]
Kc
2  Kc
1.522  0.761 Kc  Kc
1.522  0.239 Kc
Kc  6.368
K c = 40.6
[2]
(d)
(e)
Hydrogen is explosive with oxygen at high temperature.
The reaction rate will increase whereas the value of K c will remain the same.
(f)
(i)
(ii)
It will decrease.
P-5 (Planning, Analysis and Evaluation)
[1]
[2]
[1]
[1]
Yearly Worked Solutions
Chemistry A-Level P-5
89
November 2014/P52
October/November 2014 Paper 52
1.
ON 14/P52/Q1
A solder is an alloy of metals which is used to join other metal pieces together.
A specialist solder that can be used to join together pieces of aluminium is made from a mixture
by mass of 65% zinc, 20% aluminium and 15% copper.
You are to plan an experimental procedure to confirm the composition of a powdered sample
of this solder, by adding reagents and then extracting from the mixture each of the following in
sequence;
(i)
the copper metal,
(ii)
the aluminium as aluminium hydroxide,
(iii)
the zinc as zinc hydroxide.
You are provided with
●
a sample of this solder, with approximate mass 4 g,
●
1.00 mol dm–3 sulfuric acid,
●
1.00 mol dm–3 ammonia.
No other reagents should be used. Standard laboratory equipment is available including a
balance, accurate to two decimal places.
(a)
Complete the flowchart below to show the order in which the reagents would be added
to the solder to allow you to extract and separate the components as copper metal,
(Step 1),
aluminium hydroxide, (Step 2), and zinc hydroxide, (Step 3).
You are reminded that aqueous ammonia contains both the base OH – and the
complex-forming
molecule NH3.
(b)
(c)
[5]
For some of the steps in the procedure you would need to be careful to add an
appropriate quantity of a reagent.
For each step of your procedure explain why particular quantities of reagent should be
chosen
Step 1:
Step2:
Step3:
[4]
The aluminium hydroxide and zinc hydroxide that have been extracted are difficult to
dry so it is better to convert them to their oxides.
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
90
Chemistry A-Level P-5
(d)
(e)
(f)
November 2014/P52
Describe how this could be done and how you would make sure that each hydroxide
had been completely converted into its oxide.
[2]
The purpose of the experiment is to con firm the composition of the solder.
When the experiment is carried out state
●the measurements that would be taken,
●what you would do to the copper to make sure that the correct value is obtained.
[2]
If the mass of aluminium oxide obtained was 1.50 g, calculate the mass of aluminium
that was present in the solder.
( Ar: Al, 27.0; O, 16.0)
[1]
Even if the experimental difficulties of extracting all of the copper from the mixture were
overcome, it would be difficult to obtain an accurate mass of copper from this
experiment.
Suggest why
[1]
Solution:
(a)
Ammonia form hydrogen bond with water: while trichloromethane only form dipole-dipole
forces of attraction.
(b)
Step 1: Excess sulfuric acid should be used to dissolve the zinc and aluminium ,
Step 2: Excess aqueous ammonia should be used to precipitate hydroxide and to
completely dissolve the zinc hydroxide.
Step 3: There should be enough sulfuric acid to neutralize ammonia and
precipitate zinc hydroxide but not so much that the zinc hydroxide dissolve.
[5]
(c)
(d)
Heat both the hydroxides to constant mass.
Measure mass of copper, zinc hydroxide and aluminium hydroxide.
The copper should be washed with water and
dried.
[4]
[2]
[2]
54  1.5
(e)
Mass of Al =
(f)
102
=0.794 g
The mass of copper is small.
P-5 (Planning, Analysis and Evaluation)
[1]
[1]
Yearly Worked Solutions
Chemistry A-Level P-5
2.
91
November 2014/P52
ON 14/P52/Q2
The acid dissociation constant, Ka, of a weak monoprotic acid, HA, is to be determined from the
measurement of the pH change that occurs when it is titrated with an aqueous solution of sodium
hydroxide.
2.70 g of HA was dissolved in distilled water to make exactly 250.0 cm 3 of solution.
25.00 cm3 of the solution was pipetted into a beaker.
The pH of the acid in the beaker was measured and recorded in the table below.
A burette was then filled with aqueous sodium hydroxide and the 25.00 cm 3 of HA was titrated
by adding volumes of the aqueous sodium hydroxide to the beaker as indicated in the table
below.
After each addition the pH was measured and the value recorded.
(a)
Plot a graph to show how the pH of the mixture changes with the volume of added
aqueous sodium hydroxide as shown in the table.
Draw a smooth curve, using the plotted points on your graph, to produce a titration
curve for the addition of aqueous sodium hydroxide to the acid HA.
[2]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
Chemistry A-Level P-5
92
November 2014/P52
(b)
Circle any points on the graph that are anomalous and suggest a reason why this
might occur.
(c)
What would be a suitable range of pH values in which an indicator would change
colour to identify the end point of this neutralisation?
[1]
(d)
30.00 cm3 of aqueous sodium hydroxide is required to neutralise 25.00 cm 3 of HA
and the equation for the neutralisation is shown.
NaOH + HA → NaA + H2O
(i)
Excluding water, state the three ions or molecules that will be present in the
highest concentration when 15.00 cm 3 of aqueous sodium hydroxide has
been added to 25.00 cm 3 of HA.
[1]
(ii)
State and explain how the concentrations of these ions or molecules
compare.
[2]
(e)
Use your graph to determine the pH obtained when 15.00 cm 3 of aqueous sodium
hydroxide is added to 25.00 cm 3 of HA. Use this pH to determine the value of Ka for
HA.
[3]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
93
Chemistry A-Level P-5
(f)
(i)
(ii)
(g)
November 2014/P52
Use your answer to (e) and the initial pH of HA from the table to calculate the
concentration of HA in mol dm–3.
[2]
Calculate the initial concentration of HA, in g dm–3, and use this together with
your answer to (f)(i) to calculate the relative molecular mass, Mr, of HA.
(Remember that 2.70 g of HA was dissolved in distilled water to make exactly
250.0 cm3 of solution.)
[1]
Even if the experiment is done very carefully with very accurate apparatus, the
answer obtained for the molecular mass of HA is likely to be subject to error. Suggest
why.
[1]
Solution:
(a)
[2]
(b)
(c)
Point at 12.00 cm 3.
Insufficient sodium hydroxide was added.
Between 6.5 and 11
P-5 (Planning, Analysis and Evaluation)
[2]
[1]
Yearly Worked Solutions
Chemistry A-Level P-5
(d)
(e)
(i)
[2]
(1.259  104 )2
1.259  104
= 1.259 × 10–4
pH = -lg[H+]
[H+] = 3.89 × 10–3 mol dm3
[H+] = [A-]
(3.89 103 ) 2
=0.120 mol dm-3
4
1.259 10
1000
Concentration of HA = 2.7 ×
= 10.8 g dm-3
250
10.8
Mr of HA =
2(0.12)
[HA] =
(ii)
[1]
[ H  ][ A ]
[ HA]
Ka =
(g)
November 2014/P52
Na+ , A and HA .
They will all have the same concentration because half of the HA has been
neutralized by NaOH
pH = 3.9
(i)
(ii)
Ka =
(f)
94
[3]
[2]
=45
[1]
The graph is drawn by hand and is not very accurate Also, the experiment has not been
repeated.
[1]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
95
Chemistry A-Level P-5
June 2014/P52
May/June 2014 Paper 52
1.
MJ 14/P52/Q1
The liquids trichloromethane and water separate into two immiscible layers when shaken
together and allowed to stand.
Ammonia can dissolve in both of these layers. The distribution of ammonia between these two
solvents is called partition, where the concentration of ammonia in each solvent will be different.
The partition coefficient represents the ratio of the distribution.
You are to plan an experiment, using a titration with sulfuric acid, to determine the value of the
partition coefficient of ammonia between water and trichloromethane at room temperature.
The following information gives some of the hazards associated with trichloromethane and
ammonia.
Trichloromethane:
Anaesthetic if inhaled. Dangerously irritating to the respiratory system.
Ammonia:
An aqueous solution with a concentration of less than 3 mol dm –3 may cause harm to eyes
or in a cut. At greater concentrations aqueous ammonia should not be inhaled and it cases
irritation to the eyes and skin.
You are provided with the following:

Trichloromethane

Aqueous ammonia of concentration 5.00 mol dm –3

Sulfuric acid, of concentration 0.500 mol dm –3

Distilled water for dilution of aqueous ammonia.
(a)
Explain why ammonia is likely to be more soluble in water than in trichloromethane.
[2]
(b)
Define the partition coefficient, Kpartition, for ammonia between water and
trichloromethane. State whether the partition coefficient you have defined will be
greater or less than 1.
[2]
(c)
In the experiment, explain whether it is important that the volumes of water and
trichloromethane are the same.
[1]
(d)
Write an equation for the reaction of aqueous ammonia and sulfuric acid.
2 NH3 +H2SO4  (NH4 )2SO4
(e)
(f)
(g)
(h)
(i)
(i)
The partition coefficient for ammonia distributed between water and
trichloromethane is approximately 25.
Calculate the concentration of aqueous ammonia that should be used so that
a 25.0 cm3 sample of the aqueous ammonia layer would require approximately
24.0 – 26.0 cm3 of 0.500 mol dm–3 sulfuric acid for complete neutralization.
Then state the factor by which the 5.00 mol dm -3 aqueous ammonia should be
diluted to give that concentration.
[2]
(ii)
Describe, in details, how you would dilute 5.00 mol dm –3 aqueous ammonia to
make 250 cm3 of aqueous ammonia ready for use in the experiment.
[2]
Other than the use of eye protection and gloves, state one safety precaution you would
take while setting up the experiment.
[1]
State a suitable indicator for use in the titration of a sample taken from the experiment.
Explain your answer.
[2]
It is unnecessary to titrate both layers of the partition. Explain why it would be better to
titrate a simple of the aqueous layer rather than the trichloromethane layer.
[1]
Once a mean titre has been calculated, outline the steps you would take to calculate
the partition coefficient.
[2]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
96
Chemistry A-Level P-5
June 2014/P52
Solution:
(a)
Water can form hydrogen bonds with ammonia whereas trichlomethane has dipole-dipole
interactions only.
K partition 
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
2.
NH3 water 
NH3 trichloromethane 
Kpartition will be greater than 1.
[1]
It is not important since the concentration is independent of the volumes of the solutions. [1]
2NH3 + H2SO4 → (NH4)2 SO4
M1V1 M2V2
(i)

n1
n2
0.5  25 M2  25

1
2
M2 = 1 mol dm–3
5.00 mol dm–3 of aqueous ammonia should be diluted by a factor of 5.
[2]
3
–3
3
(ii)
Pipette 50 cm of the 500 mol dm aqueous ammonia into a 250 cm volumetric
flask. Make up to the mark with distilled water. Then shake the flask.
Place a bung on top of the flasks containing aqueous ammonia to avoid the inhalation of
ammonia.
[1]
Methyl orange with a pH range from 0 to 4 should be used as the indicator since this is a
weak base against strong acid titration reaction.
[1]
The concentration of ammonia in the aqueous larger, so there is less error in the
determination of its concentration.
[1]
Calculate the moles of ammonia in the aqueous layer, subtract from the original to obtain
moles in the trichlormethane.
[2]
MJ 14/P52/Q2
Nitrogen(II) oxide, NO, can be oxidized by ozone, O 3, in the atmosphere to form nitrogen(IV)
oxide, NO2.
It is possible to simulate this process in the laboratory to measure the rate at which this reaction
takes place.
In this experiment, nitrogen(II) oxide is reacted with ozone. Since the concentration of
nitrogen(II) oxide, [NO], in the air is very low, specialist equipment is required and [NO] oxide
is measured as the number of molecules present in a volume of 1 cm 3.
The results obtained from the experiment are shown below.
(a)
Time/s
Concentration of NO/108
molecules cm–3
0
27.0
30
21.9
60
17.7
90
14.4
120
13.2
150
9.45
180
7.66
210
6.21
240
5.03
270
4.08
300
3.31
Use the results obtained to plot a graph to show the relationship between [NO] and
more.
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
Chemistry A-Level P-5
(b)
(c)
(d)
(e)
97
June 2014/P52
[2]
On your graph circle the single result that you consider to be the most anomalous.
Suggest a reason why anomalous results may occur during the experiment to measure
the rate of a reaction.
[2]
Use construction lines at concentrations of nitrogen(II) oxide equal to 13.5×108
molecules cm –3 and 6.75 × 108 molecules cm–3 to determine the order of reaction with
respect to nitrogen(II) oxide.
Show the construction lines on your graph and your working.
[3]
(i)
Use your graph to calculate the initial rate of the reaction.
(ii)
In the reaction between ozone and nitrogen(II) oxide the order of reaction with
respect to ozone is 1.
In the experiment the initial concentration of ozone used was 4 × 1011
molecules cm–3.
Calculate the value of the rate constant for the reaction and give its units. [2]
1 mole contains 6.02 × 1023 particles.
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
Chemistry A-Level P-5
(f)
98
June 2014/P52
Convert the initial rate in (d)(i) to a value with units of mol dm–3 s–1.
(If you have no answer to (d)(i) you may use 3.0 × 106 as the value of the initial rate).
[2]
The concentration of ozone used in the experiment is considerably greater than the
concentration of nitrogen(II) oxide.
Explain why this is necessary for the experiment, in order to determine the order with
respect to nitrogen(II) oxide.
[2]
Solution:
(a)
[2]
(b)
Point at 120s.
Concentration has been measured before the recorded time.
P-5 (Planning, Analysis and Evaluation)
[2]
Yearly Worked Solutions
99
Chemistry A-Level P-5
(c)
(d)
The half-life is approximately 98s.
Because the half-lines are equal, the reaction is first order.
(i)
Gradient =
(e)
(f)
k
[3]
27  0
 0.216
0  125
Initial rate = 0.216 ×108 molecules cm–3 s–1
(ii)
June 2014/P52
0.216  10
 2  1014 cm3 molecules–1 s–1
27  108  4  1011
[2]
8
0.216×108 ×103
=3.59×10-14 mold m–3 s–1
6.02×1023
[2]
[2]
The ozone concentration should remain constant during the experiment so that the rate
depends only on the concentration of NO.
[2]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
100
Chemistry A-Level P-5
November 2013/P51 & 52
October/November 2013 Paper 51 & 52
1.
O/N 13/P51, O/N 13/P52/Q1
Ammonium nitrate, NH4NO3, is soluble in water (approximately 2.5 mol/100 g at 25°C). The
molar enthalpy of solution of a solid is defined as the enthalpy change when one mole of the
solid is dissolved in water.
NH4NO3 (s)
(a)
(b)
(c)
NH+4(aq)  NO3- (aq)
Hsoln = +26.5 kJ mol-1
(i)
Predict how the temperature of water, initially at 25°C would change as
ammonium nitrate is dissolved. Explain this prediction in terms of lattice energy
and the enthalpy of hydration of ions.
Prediction of the temperature change:
Explanation:
(ii)
In the space below, sketch a graph to show your prediction of temperature
change with concentration. Use two labelled axes and include an origin. [4]
If you were to carry out an experiment to investigate how the temperature change of
the solution varies as the concentration changes name:
(i)
the independent variable
(ii)
the dependent variable
[1]
You are to plan an experiment to determine as accurately as possible how the
temperature change varies when different solutions are made, each with different
concentrations of ammonium nitrate. You are reminded that the approximate solubility
of ammonium nitrate is 2.5 mol/100 g at 25°C.
The following information gives some of the hazards associated with ammonium
nitrate.
Ammonium nitrate NH4NO3. Contact with combustible material may cause fire.
Explosive when mixed with combustible material.
Do not allow the salt to become contaminated with organic matter and do not
grind it.
Solutions should be diluted to less than 0.5 mol dm –3 for disposal.
(d)
You should use only standard apparatus found in a school or college laboratory. Draw
a diagram of the apparatus and experimental set up you would showing clearly the
following:
(i)
The apparatus used, such as the reaction vessel, and how the thermometer
will be positioned in order to measure the temperature of the solution as
accurately as possible.
(ii)
How the apparatus will be insulated.
Label each piece of apparatus used, indicating its size or capacity and both
the temperature range and the precision of the thermometer.
[3]
Using the apparatus shown in (c) design an experiment to test your prediction in (a)(ii)
of how the temperature change of the solution varies with solutions of different
concentration.
In addition to the apparatus normally found in a laboratory you are provided with the
following materials:
a supply of solid ammonium nitrate.
distilled (deionized) water.
Give a step-by-step description of how you would carry out the experiment to include:
(i)
The number of experiments you would do,
(ii)
The temperature measurements you would take,
(iii)
The volume of water you would use,
(iv)
A calculation of show the maximum mass of ammonium nitrate you could use
for your volume of water in (iii) and a range of masses for the other
experiments.
[Ar: H, 1.0; N, 14.0; O, 16.0]
[4]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solution
101
Chemistry A-Level P-5
(e)
(f)
November 2013/P51 & 52
State one hazard that must be considered when planning the experiment and describe
a precaution that should be taken to keep risks from this hazard to a minimum. You
may use the information (c) if you wish.
[1]
In order to test your prediction in (a)(ii), you would need to plot a graph. In the space
below, draw a table with appropriate headings, in which you would record all your
experimental data and calculated values necessary for the construction of the graph.
The headings must include the appropriate units.
[2]
Solution:
(a)
(i)
Prediction of the temperature change:
The
temperature
would
decrease.
Explanation: The lattice energy is more exothermic than the sum of the enthalpies
of hydration.
(ii)
25
Temperature /o
C
Concentration / mol /100g of water
[4]
(b)
(c)
(i)
the independent variable: concentration
(ii)
the dependent variable: temperature change
(i) & (ii)
[1]
Thermometer
Range -10- 110oC
o
precise upto 0.1 C
Lid
250 cm
plastic
cup
3
Lagging
Water
(to which NH4 NO3
is to be added)
[3]
(d)
(i)
(ii)
(iii)
(iv)
Measure out five different masses of solid ammonium nitrate between 20100 grams.
To the plastic cup, add 50 cm 3 of distilled water, using a measuring cylinder.
Measure the initial temperature of water using the thermometer and then add the
measured mass of ammonium nitrate. Record the final temperature.
Repeat experiment for other masses.
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solution
102
Chemistry A-Level P-5
November 2013/P51 & 52
2.5 mol of NH4NO3 dissolves in 100 cm 3 water.
x mol dissolve in  50 cm3 water
x
2.5
 1.25 mol
2
Mr of NH4NO3 = 28 + 4 + 48
Maximum mass that dissolves
(e)
(f)
= 80
= moles × Mr
= 1.25 × 80
= 100 g.
[4]
Ammonium nitrate may cause a fire or explosion so it must not be ground up.
Mass of
NH4NO3
Mass of
water /g
Initial
Temperature /°C
Final
Temperature /°C
[1]
Concentration of
NH4NO3 /mol/100g
[2]
2.
O/N 13/P51, O/N 13/P52/Q2
The solubility of hydrated sodium sulfate, Na2SO4·10H2O, in water increases with temperature.
At a temperature between 25°C and 70°C there is a transition and the solubility becomes that
of Na2SO4. The units of solubility are grams per one hundred grams of water, g/100 g water.
An experiment was carried out to investigate this solubility and determine the transition
temperature between the two forms of sodium sulfate.

An empty boiling tube was weighed and the mass recorded.

Some distilled water was added to the boiling tube and the new mass recorded.

A small sample of hydrated sodium sulfate was added and this new mass recorded.

The boiling tube was carefully heated with stirring until all the solid had dissolved.

The apparatus was cooled slowly while constantly stirring and the temperature
recorded when the first crystals appeared in the tube.
(a)
The results of several of these experiments are recorded below.
Process the results in the table to calculate the solubility, in g/100 g water, of the sodium
sulfate for each of the temperatures listed.
Record these values to two decimal places in the additional columns of the table. You
may use some or all of the columns.
Label the columns you use.
For each column you use include units where appropriate and an expression to show
how your values are calculated.
Use the column headings A to H for these expression (e.g. A–B).
A
B
C
D
E
Experiment
number
Mass
of
boiling
tube
/g
Mass
of
boiling
of tube
+ water
/g
Mass
of
boiling
tube +
water +
solid /g
Crystallising
temperature
/°C
1
10.20
35.20
36.45
0.0
2
10.35
30.35
31.60
10.0
3
10.10
35.10
40.10
20.0
P-5 (Planning, Analysis and Evaluation)
F
G
H
Yearly Worked Solution
103
Chemistry A-Level P-5
November 2013/P51 & 52
4
9.80
29.20
36.96
30.0
5
9.95
32.95
44.06
40.0
6
9.90
34.90
46.35
50.0
7
9.70
30.70
40.32
60.0
8
10.45
30.45
39.55
70.0
9
10.05
35.05
46.30
80.0
10
10.10
40.10
53.45
90.0
[3]
(b)
(c)
(d)
Plot a graph to show the variation of solubility (y-axis) with temperature (x-axis).
Draw two curves of best fit and extrapolate to locate their intersection at the transition
temperature.
[4]
From your graph, state the transition temperature and the solubility at which it occurs.
[2]
(i)
In an attempt to repeat the 4th experiment using the same masses of water
and solid, the temperature was mistakenly read and recorded before crystals
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solution
104
Chemistry A-Level P-5
(e)
(f)
November 2013/P51 & 52
appeared. Place a cross on your graph to represent the point that would have
been obtained.
(ii)
If this was a valid point, which effect would this have on your transition
temperature? Explain your answer.
[2]
It was found that all the mass recordings in columns C and D had been made with a
balance that had been zeroed incorrectly and they should all have been 0.3 g smaller.
The mass recorded in column B can be considered to be accurate. Using the corrected
masses from experiment 6 calculate the new value of the solubility. By comparing this
with the original solubility value for experiment 6 calculate the percentage error
difference.
[2]
From the pattern of solubility demonstrated by your graph, predict and explain whether
the dissolving of the two forms of sodium sulfate in water are exothermic or
endothermic reactions.
Prediction for Na2SO4 · 10H2O
Explanation
Prediction for Na2SO4
Explanation
[2]
Solution:
(a)
Use the column headings A to H for these expression (e.g. A–B).
A
B
Experiment Mass of
number
boiling
tube /g
1
2
3
4
5
6
7
8
9
10
10.20
10.35
10.10
9.80
9.95
9.90
9.70
10.45
10.05
10.10
C
D
E
F
G
H
Mass of
Mass of
Crystallising Mass
Mass
F
×100 /
Solubility
boiling of boiling tube temperature of solid
of
G
tube +
+ water +
°C
(D–C)/ water
g/100 g water
water /g
solid /g
g
(C–B)/g
35.20
30.35
35.10
29.20
32.95
34.90
30.70
30.45
35.05
40.10
36.45
31.60
40.10
36.96
44.06
46.65
40.32
39.55
46.30
53.45
0.0
10.0
20.0
30.0
40.0
50.0
60.0
70.0
80.0
90.0
1.25
1.25
5.00
7.76
11.11
11.75
9.62
9.10
11.25
13.35
25.00
20.00
25.00
19.40
23.00
25.00
21.00
20.00
25.00
30.00
5.00
6.25
20.00
40.00
48.30
47.00
45.81
45.50
45.00
44.50
[3]
(b)
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solution
105
Chemistry A-Level P-5
November 2013/P51 & 52
55.00
50.00
49.50
45.00
40.00
35.00
30.00
/
sobility
g/100g
water
25.00
20.00
15.00
10.00
5.00
0.00
0.0
10.0
20.0
30.0
40.0
42.0
50.0
60.0
70.0
80.0
90.0
o
Temperature / C
(c)
(d)
(e)
[4]
Transition temperature =
42.0°C
Solubility at 42.0°C
=
49.5 g/100 g water
[2]
(i)
(ii)
Transition temperature would be higher as intersection of curves would be
at higher temperature.
[2]
Mass of water =
25.00 – 0.30 = 24.70 g.
Mass of solid =
(46.65 – 0.3) – (39.00 – 0.3) = 11.75 g
Solubility
% error
P-5 (Planning, Analysis and Evaluation)
=
11.75
 100
24.70
=
47.6 g / 100 g water
=
(47.6  47.0)
 100
47.0
=
1.28%
[2]
Yearly Worked Solution
Chemistry A-Level P-5
(f)
106
Prediction for Na2SO4·10H2O: Dissolving is endothermic
Explanation: Solubility increases with increase in temperature.
Prediction for Na2SO4: Dissolving is exothermic
Explanation: Solubility is decreased by increasing temperature.
P-5 (Planning, Analysis and Evaluation)
November 2013/P51 & 52
[2]
Yearly Worked Solution
107
Chemistry A-Level P-5
November 2013/P53
October/November 2013 Paper 53
1.
O/N 13/P53/Q1
Air, which is 99% nitrogen and oxygen, is slightly soluble in water. At 25°C a saturated solution
of air in water has a concentration of 19 cm 3 dm–3. When water is boiled all the dissolved air is
boiled out of solution.
(a)
(i)
The molar enthalpies of solution for nitrogen and oxygen are:
Hsoln N2 = –1.04 kJ mol–1 and Hsoln O2 = –1.20 kJ mol–1
Predict how the solubility of air in water will change as the temperature is
increased. Explain this prediction using Le Chatelier’s principle in terms of the
equilibrium between air and the aqueous solution as the temperature is
increased.
Prediction:
Explanation:
(ii)
Display your prediction in the form of a sketch graph for the solubility of air
between 0°C and 100°C, labelling clearly the axes. Include labelled points to
indicate the solubility of air at 25°C and 100°C.
0
0
(b)
[4]
If you were to carry out an experiment to investigate ow the solubility of air varies as
the temperature increases name.
(i)
the independent variable:
(ii)
the dependent variable:
[1]
Solution:
(a)
(i)
Prediction: Solubility of air in water will decrease with increase in temperature.
Explanation: As shown by the negative Hsol values above, dissolving of air is an
exothermic process, therefore an increase in temperature will shift the equilibrium in
reverse direction, decreasing the solubility at air.
(ii)
19
Solubility / cm 3 dm -3
0
0
25
o
Temperature / C
100oC
[4]
(b)
(i)
(ii)
the independent variable:
temperature
the dependent variable: solubility of air
P-5 (Planning, Analysis and Evaluation)
[1]
Yearly Worked Solutions
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Chemistry A-Level P-5
2.
November 2013/P53
O/N 13/P53/Q2
When heated, aqueous hydrogen peroxide, H 2O2, decomposes to form oxygen and water.
2H2 O2(aq)  2H2O(I) + O2(g)
The decomposition can also occur at room temperature if a suitable catalyst is added. Both of
the solids, manganese(IV) oxide and lead(IV) oxide, will catalyse the decomposition.
The following information gives some of the hazards associated with manganese(IV) oxide and
lead(IV) oxide.
Manganese(IV) oxide: Poisoning can occur by inhalation or swallowing the powder.
Lead(IV) oxide: Poisoning can occur by inhalation or swallowing the powder. The powder
can also cause skin irritation.
You are provided with a 0.300 mol dm –3 solution of hydrogen peroxide and a syringe with a
capacity of 100 cm 3.
(a)
Provide the following information about experiments you would carry out to collect
oxygen from the decomposition of hydrogen peroxide and to determine, using identical
masses, which of the two catalysts was the most efficient at promoting this
decomposition.

a fully labelled diagram of the apparatus to be used that would ensure that no
oxygen would be lost when the experiment was carried out.

a calculation of the maximum volume in cm 3 of the aqueous hydrogen peroxide
that could be used such that the oxygen produced would not exceed the
volume of the syringe.

a statement of the measurements you would take that would allow you to say
which of the catalysts was most efficient.
The molar volume of a gas at 25°C is 24.0 dm 3.
Please continue into the space provided on the next page if necessary.
[6]
(b)
What other feature of the catalyst should be controlled?
[1]
(c)
If one of the experiments takes 2 minutes to complete, draw a sketch graph with
labelled axes showing how the volume of oxygen produced will vary with time between
0 and 3 minutes.
[2]
(d)
State the hazards that might be encountered when using the solids required in this
experiment and give the one essential precaution you would take to make sure these
chemicals were handled safely during the experiments.
[1]
Solution:
(a)
Catalyst
Thistle
Funnel
Tightly fitted
bung
Gas Syringe
Conical flask
H2O2
(aq)
Start stopwatch immediately after mixing catalyst and H2O2(aq) and note time taken for 30.0
cm3 of oxygen to be evolved.
Maximum volume of oxygen = 100 cm3
100
Moles of oxygen
=
 0.00416 moles
24000
Moles of H2O2
= 2 × 0.00416
P-5 (Planning, Analysis and Evaluation)
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Chemistry A-Level P-5
Volume of H2O2
(b)
(c)
November 2013/P53
= 0.00833 moles
moles
=
conc.
0.00833
=
0.300
= 0.02777 dm3
= 27.8 cm3
[6]
[1]
Surface area of catalyst.
Volume of
Oxygen/cm 3
0
1
2
3
Time / mins
(d)
3.
[2]
[1]
Lead(IV) oxide causes skin irritation. Wear safety gloves.
O/N 13/P53/Q3
In the fractional distillation of two liquids which are miscible (dissolve in each other) in all
proportions the more volatile of the two will distil first. At any temperature the composition of
the vapour in equilibrium with the liquid has a higher proportion of the more volatile component
which has a lower boiling point.
An experiment was carried out to investigate the boiling points of mixture of tetrachloromethane,
CCl4, and tetrachloroethane, C2H2Cl4.
A convenient method for representing the composition of the mixtures, both liquid and vapour,
is to use the concept of mole fraction. For example, if the liquid mixture consists of 0.15 mole
of liquid A and 0.35 mole of liquid B, the mole fraction A is
(a)
(b)
(c)
0.15
i.e. 0.30.
(0.15  0.35)
The results of several of these experiments are recorded below.
Temperature/
°C
120.0
108.5
99.3
93.0
89.3
83.3
79.9
76.0
Mole fraction
CCl4 liquid
0.000
0.100
0.200
0.300
0.400
0.600
0.800
1.000
Mole fraction
CCl4 vapour
0.000
0.469
0.552
0.800
0.861
0.918
0.958
1.000
Calculate the relative molecular masses (M,s) of CCl4 and C2H2Cl4.
[Ar: H, 1.0; C, 12.0; Cl, 3.5]
[1]
(i)
The vapour from the equilibrium at 108.5°C was analysed and found to
consist of 7.22 g of CCl4 and 8.92 g of C2H2Cl4. Show clearly by calculation
that this gives a mole fraction of 0.469 for CCl4 vapour.
(ii)
The vapour from the equilibrium at 83.3°C was analysed and found to consist
of 14.14 g of CCl4 and 1.38 g of C2H2Cl4. Show clearly by calculation that this
gives a mole fraction of 0.918 for CCl4 vapour.
[2]
On the same axes, plot two graphs, one for the liquid and one for the vapour, to show
the variation in temperature (y-axis) with the mole fraction compositions (x-axis) of both
the liquid and the vapour.
P-5 (Planning, Analysis and Evaluation)
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110
November 2013/P53
Draw two lines of best fit. Each line could be either a curved line or a straight line.
[4]
(d)
(e)
(f)
(g)
Circle and label on the graph the point you consider to be the most anomalous.
Do not circle or label any other point.
If it is assumed that the analysis was carried out accurately, suggest a reason why the
point might be anomalous.
[2]
By drawing an appropriate construction line on your graphs, determine the mole
fraction of CCl4 in the vapour which is in equilibrium with a liquid with a mole fraction
of 0.500 CCl4.
[2]
The temperatures were measured using a thermometer calibrated in 0.1°C
graduations. If the thermometer had only been calibrated in 1.0°C graduations,
calculate the percentage error which would result from the determination of the boiling
points of each of the two pure liquids.
[2]
(i)
Use your graphs to state whether CCl4 CCl4 or C2H2Cl4 distills first from a
mixture of the two liquids.
(ii)
Explain your answer (i)
[2]
P-5 (Planning, Analysis and Evaluation)
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November 2013/P53
Solution:
(a)
(b)
Mr of CCl4 = 12 + 4 (35.5) = 154
Mr of C2H2Cl4 = (12)2 + 2(1) + 4(35.5) = 168
(i)
8.92
 0.0531 mol
168
moles of C2H2Cl4 =
mole fraction CCl4 =
(ii)
[1]
7.22
moles of CCl4 =
 0.0469 mol
154
moles of CCl4 =
0.0469
 0.469 mol
0.0469  0.0531
14.14
 0.0918 mol
154
moles of C2H2Cl4 =
1.38
 0.00821 mol
168
mole fraction CCl4 =
0.0918
 0.9178  0.918 mol
0.0918  0.00821
[2]
(c)
P-5 (Planning, Analysis and Evaluation)
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November 2013/P53
130.0
120.0
110.0
o
Temperature / C
100.0
90.0
80.0
×
70.0
0.000
0.200
0.400
0.600
0.800
mole fraction
1.000
[4]
[4]
(d)
(e)
Anomalous point is at T = 99.3°C (vapour curve).a
It may be due to analysis being made at a temperature that was too low.
0 : 892
(f)
For
C2H2Cl4 :
For
CCl4 :
(i)
(ii)
CCl4.
The vapour produced when the mixture is heated, contains a greater proportion of
CCl4 than the liquid.
[2]
(g)
0.5
 100  0.417%
12.0
0.5
 100  0.658%
76
P-5 (Planning, Analysis and Evaluation)
[2]
[2]
[2]
[2]
Yearly Worked Solutions
113
Chemistry A-Level P-5
June 2013/P52
May/June 2013 Paper 52
1.
M/J 13/P52/Q1
Calcium hydroxide, Ca(OH)2, is slightly soluble in water, approximately 1 g dm –3 at 25°C. The
molar enthalpy of solution of a solid is defined as the enthalpy change when one mole of the
solid dissolves in water.
(a)
(i)
predict how the solubility of calcium hydroxide in water changes as the
temperature is increased. Explain this prediction using i.e. Chaltelier’s Principle
in terms of the equilibrium between the solid calcium hydroxide and the
aqueous solution, as show in the equation below.
Ca(OH)2(s)  Ca2+ (as) + 2OH–(aq)
(ii)
Hsoln = –16.7 kJ mol–1
Predict how the solubility will change as the temperature is increased.
Explanation:
Display your prediction in the form of a sketch graph between 0°C and 100°C.
Label the axes with units and give numerical values to ensure that the line
clearly shows the solubility at 25°C.
0
0
(b)
(c)
[4]
If you were to carry out an experiment to investigate how the solubility of calcium
hydroxide varies as the temperature increases name,
(i)
the independent variable,
(ii)
the dependent variable
[1]
You are to plan an experiment to determine as accurately as possible the concentration
of a saturated aqueous solution of calcium hydroxide by titration with hydrochloric acid.
You are reminded that the approximate solubility of calcium hydroxide is 1 g dm –3 at
25°C.
The following information gives some of the hazards associated with calcium hydroxide
and hydrochloric acid.
The following information gives some of the hazards associated with calcium hydroxide
and hydrochloric acid.
Hydrochloric acid, HCl (aq)
Corrosive; Causes burns: Irritating to respiratory system.
Solutions equal to or more concentrated than 6.5 mol dm –3 are corrosive.
Solutions equal to or more concentrated than 2 mol dm –3 but more dilute than 6.5
mol dm–3 are said to be irritant.
Calcium hydroxide, Ca(OH)2(s)
Irritant; risk of serious damage to eyes.
You are provided with the following materials:
250 cm3 of saturated calcium hydroxide,
50 cm3 of 2.00 mol dm–3 hydrochloric acid
Give a step-by-step description of how you would carry out the experiment by including:
(i)
a balanced equation for the reaction between aqueous calcium hydroxide and
hydrochloric acid,
P-5 (Planning, Analysis and Evaluation)
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Chemistry A-Level P-5
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(d)
June 2013/P52
a list of apparatus with volumes where appropriate.
a suitable indicator with relevant colours.
a calculation of the approximate concentration of a saturated aqueous solution
of calcium hydroxide in mol dm –3 at 25°C,
[Ar : H, 1.0; O, 16.0; Ca, 40.1]
a detailed method for the dilution of the hydrochloric acid such that when a
titration is carried out the two reacting volumes are approximately equal at the
end-point. The relevant calculations and reasoning must be shown in full.
a detailed method for carrying out sufficient titrations to allow an accurate endpoint to be obtained.
an outline calculation to show how the results are to be used to determine the
accurate concentration of the aqueous calcium hydroxide.
[Ar: Cl, 35.5].
State one hazard that must be considered when planning the experiment and describe
a precaution that should be taken to keep risks from this hazard to a minimum. You
should use the information in (c).
[2]
Solution:
(a)
(i)
Solubility will decrease as the temperature is increased.
Explanation: The forward reaction being exothermic is not favoured by high
temperatures, therefore increase in temperature will shift the above equilibrium in
backward direction, decreasing the solubility of calcium hydroxide.
(ii)
Solubility/gdm
-3
1
0
0
10
20 25 30
40
50
60
70
80
90
100
Temperature /oC
[4]
(b)
(c)
(i)
(ii)
the independent variable: Temperature
the dependent variable: Solubility of calcium hydroxide.
(i)
Ca(OH)2 +2HCl  CaCl2 +2H2O
(ii)
250 cm3 titration flask
50.0 cm3 burette
25.0 cm3 pipette
Use methyl orange as indicator which shows red colour in acidic medium and yellow
colour in alkaline medium.
Concentration of Ca(OH)2 at 25°C = 1 g dm–3.
Mr at Ca(OH)2 = 40.1 + 2(16) + 2(1) = 74.1
(iii)
(iv)
Moles =
Mass 1.0
= 0.0135 moles

Mr
74.1
Concentration =
(v)
[1]
moles
0.0135
= 0.0135 mol dm–3

volume
1
Moles of HCl: moles of Ca(OH)2 = 2 : 1
For the reacting volumes to be equal, ratio of their concentrations must also be 2:1
Formula for dilution: M1  v1  M2  v 2
2.0  v1  (2  0.0135)  250
P-5 (Planning, Analysis and Evaluation)
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June 2013/P52
v1  3.375 cm3  3.38 cm3
(vi)
(vii)
Pipette out 3.38 cm 3 of 2.0 mol dm–3 hydrochloric acid and transfer it to a dilution
flask of 250 cm 3 capacity. Fill the flask upto the mark with distilled water and shake
well. The solution obtained has a concentration of 0.027 mol dm –3.
Use a pipette to transfer 25.0 cm 3 of the Ca(OH)2 solution to a titration flask and add
a few drops of indicator to it. Now fill a burette with the HCl solution prepared and
gradually. Add it to the Ca(OH)2 solution until end point is reached. Note the volume
of HCl added. Repeat the titration three times to find the average titre.
Let average titre
= x cm3
 x

 0.027  moles.
 1000

Moles of HCl = 
1
x

 0.027  moles
Moles of Ca(OH)2 =  
 2 1000

Concentration of Ca(OH)2
(d)
2.
=
=
1
x
1000

 0.027 
2 1000
25
=
0.027
x mol dm–3
50
moles
volume
[8]
2.00 mol dm–3 hydrochloric acid is an irritant. Eye protection, for example goggles,
be worn, Wear gloves.
should
[2]
M/J 13/P52/Q2
Hydrated copper (II) sulfate can be represented as CuSO 4 × H2O where × is the number of
molecules of H2O for each CuSO4. When the compound is heated, it loses the molecules of
water leaving anhydrous copper (II) sulfate.
A suggested equation is:
CuSO4 H2O(s)  CuSO4(s) + ×H2O(g)
An experiment is carried out to attempt to determine the value of x.

An open crucible is weighed and the mass recorded.

A sample of hydrated cooper (II) sulfate is added to the crucible and the new mass
recorded.

The crucible with hydrated copper (II) sulfate is heated strongly five minutes and
allowed to cool back to room temperature.

The crucible with the contents is then reweighed and the mass recorded.
(a)
Calculate the relative formula mases, Mr, of CuSO4 and H2O.
[Ar: H, 1.0; O, 16.0; S, 32.1; Cu, 63.5]
[1]
(b)
The results of several of these experiments are recorded below.
Process the results in the table to calculate both the number of moles of anhydrous
copper (II) sulfate and the number of moles of water.
Record these values in the additional columns of the table.
You may use some or all of the columns.
Masses should be recorded to two decimal places, while the numbers of moles
should be recorded to three significant figures.
Label the columns you use. For each column you use include units where appropriate
and an expression to show how your values are calculated.
You may use the column headings A to G for these expressions (e.g. A–B)
P-5 (Planning, Analysis and Evaluation)
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Chemistry A-Level P-5
A
Mass of
crucible
/g
15.20
15.10
14.95
15.15
15.05
14.90
14.92
15.30
15.07
15.01
(c)
B
Mass of
crucible +
CuSO4
×H2O
/g
16.76
16.90
16.95
17.25
17.32
17.24
17.42
17.99
17.96
18.09
C
Mass of
crucible +
CuSO4
/g
June 2013/P52
D
E
F
G
16.20
16.25
6.23
16.49
16.47
16.43
16.52
17.02
16.92
16.98
[2]
Plot a graph to show the relationship between the number of moles of anhydrous
copper (II) sulfate, CuSO4 (x-axis), and the number of moles of water (y-axis).
Draw the line of best fit. It is recommended that you do not include the origin in your
choice of scaling.
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June 2013/P52
[3]
Circle and label on the graph any point(s) you consider to be anomalous.
For each anomalous point gave a different reason why it is anomalous clearly indicating
which point you are describing.
[3]
Determine the slope of the graph. You must mark clearly on the graph any construction
lines and show clearly in your calculation how the intercepts were used in the
calculation of the slope.
[3]
Comment on the reliability of the data provided in (b).
[1]
(i)
Use the value of the slope of your graph calculated in (e) to suggest the correct
formula for hydrated copper (II) sulfate.
(ii)
Explain your answer to (i)
[2]
(d)
(e)
(f)
(g)
Solution:
(a)
Mr of H2O
Mr of CuSO4
=
=
18.0
159.6
[1]
(b)
A
B
C
D
E
F
G
Mass of
crucible
/g
Mass of
crucible +
CuSO4
×H2O
/g
Mass of
crucible
+
CuSO4
/g
Mass
of
CuSO4
/g
(C–A)
Mass of
H2O /g
(B–C)
Moles of
CuSO4
Moles of
H2O
D
/mol
159.6
E
/mol
18.0
15.20
16.76
16.20
1.00
0.56
0.00627
0.0311
15.10
16.90
16.25
1.15
0.65
0.00721
0.0361
14.95
16.95
6.23
1.28
0.72
0.00802
0.0400
15.15
17.25
16.49
1.34
0.76
0.00840
0.0422
15.05
17.32
16.47
1.42
0.85
0.00890
0.0472
14.90
17.24
16.43
1.53
0.81
0.00959
0.0450
14.92
17.42
16.52
1.60
0.90
0.0100
0.0500
15.30
17.99
17.02
1.72
0.97
0.0108
0.0539
15.07
17.96
16.92
1.85
1.04
0.0116
0.0578
15.01
18.09
16.98
1.97
1.11
0.0123
0.0617
[2]
(c)
P-5 (Planning, Analysis and Evaluation)
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moles of H2O/mol
0.0800
0.0700
0.0600
0.0584
0.0500
0.0400
0.0140
0.0130
0.0120
0.0116
0.0110
0.0100
0.0090
0.0080
0.0075
0.0070
0.0300
0.0060
0.0375
moles of CuSo4/mol
[3]
(d)
(e)
Point 5: The anhydrous CuSO4 had decomposed/prior to heating the crucible was wet.
Point 6: All of the water was not driven off the copper sulphate crystals.
=
(f)
(g)
[3]
0.0584  0.0300
=
0.0116  0.006
0.0284
0.0056
= 5.07
[3]
The data obtained is reliable because most of the points lie on the line and there are only a
few anomalies.
[1]
(i)
CuSO4 · 5H2O
(ii)
The slope of the graph is the ratio H2O : CuSO4 and it is 5 : 1 therefore
x = 5.
[2]
P-5 (Planning, Analysis and Evaluation)
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Chemistry A-Level P-5
119
November 2012/P52
October/November 2012 Paper 52
1.
O/N 12/P52/Q1
There are three oxides of lead, PbO, PbO2 and Pb3O4, all of which can be reduced to metallic
lead by hydrogen.
The following information gives some of the hazards associated with these compounds.
Lead oxides
Lead(II) oxide (PbO) Lead(IV) oxide (PbO2) Dilead(II) lead(IV) oxide (Pb3O4)
Toxic Dangerous for the environment
Harmful by inhalation and if swallowed. Danger of cumulative effects.
Hydrogen Extremely flammable. Readily forms an explosive mixture with air. Mixtures
between 4 and 74% by volume are explosive.
An unknown sample of an oxide of lead can be identified by investigating the molar ratio of
oxygen atoms to lead atoms.
You are to plan an experiment to investigate the molar ratio of oxygen atoms to lead atoms
in the oxide sample. Your plan should result in a correct identification of the oxide.
(a)
(b)
Calculate the number of moles of oxygen atoms that combine with one mole of lead
atoms in each of the three oxides.
[2]
Draw a sketch graph to show how the number of moles of oxygen atoms varies with
the number of moles of lead for lead(II) oxide, PbO. Draw two more sketch graphs to
show this relationship for the other two oxides.
Label clearly each axis and each graph.
c
(c)
(d)
(e)
[2]
In the experiment you are about to plan, identify the following.
(i)
the independent variable
(ii)
the dependent variable
[1]
Draw a diagram of the apparatus and experimental set up you would use to determine
the chemical formula of the oxide. Your apparatus should use only standard items
found in a school or college laboratory and should show clearly
(i)
how the hydrogen gas needed for the reduction is prepared, naming the
chemicals (reagents) to be used,
(ii)
how the oxide of lead will be heated,
(iii)
how any excess hydrogen is dealt with safely.
Label each piece of apparatus used.
[3]
Using the apparatus shown in (d), design a laboratory experiment which will enable
you to determine the chemical formula of the oxide.
Give a step-by-step description of how you would carry out the experiment by
(i)
stating a suitable mass of the oxide of lead,
(ii)
stating how you would ensure that the decomposition is complete,
P-5 (Planning, Analysis and Evaluation)
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November 2012/P52
(iii)
showing by calculation the minimum volume of hydrogen, measured at 25°C,
that would be needed to reduce the mass of oxide of lead stated in (i) above.
For calculation purposes, you may assume that the oxide of lead is PbO,
(iv)
stating how you would use your results to reach a conclusion.
[Ar: H, 1.0; O, 16.0; Pb, 207.0; the molar volume of a gas at 25 °C is 24.0 dm 3]
[4]
State one hazard that must be considered when planning the experiment and describe
a precaution that should be taken to minimise the risk from this hazard.
[1]
Draw up a table with appropriate headings to show the data you would record when
carrying out your experiments and the values you would calculate in order to determine
graphically the formula of the oxide. The headings should include appropriate units.
[2]
(f)
(g)
Solution:
(a)
PbO  1:1
PbO2  1:2
3
Pb3O4  1:1.33  
4
[2]
(b)
PbO 2
Pb 3O2
Moles of O2
PbO
0
0
Moles of Pb
[2]
(c)
(i)
(ii)
moles of lead
moles of oxygen
[1]
(d)
HCL
[3]
Zn
(e)
(i)
(ii)
(iii)
Take 2g lead oxide
Keep re-heating until there is a constant volume of gas in the syringe and mass
become constant.
PbO + H2  Pb + H2O
2g
x
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
121
Chemistry A-Level P-5
November 2012/P52
4
= 0.09
223
1:1
= 0.09 × 24000 cm
= 204 cm3
1 mole of H2 reduces one of lead oxide by above stoichiometric equation. By
calculating the number of moles of lead remaining and subtracting it from the initial
mass, we obtain oxygen mass. The plot a graph.
[4]
Moles =
(iv)
(f)
Lead oxide is toxic if inhaled so wear a mask when performing the experiment.
[1]
(g)
Mass of oxide
(g)
Mass of lead
(g)
Mass of oxygen
(g)
Moles of
oxygen
Moles of
lead
[2]
2.
O/N 12/P52/Q2
In aqueous solution, glucose can be slowly hydrolysed. The reaction appears to be first-order
with respect to the glucose. As the hydrolysis proceeds, samples of the glucose solution can
be analysed at regular intervals and the concentrations recorded.
If the reaction is first-order, the following equation can be used to verify this.
log10 a – log10 (a-x) = kt
where a is the initial concentration of glucose, x is the decrease in the concentration of the
glucose, a-x is the glucose concentration at any time t and k is a constant.
A plot of log10 (a-x) against time will be linear for a first-order reaction and the slope will be
equal to –k.
(a)
The experimentally determined values of such a hydrolysis experiment carried out at
298 K are recorded below.
You should use a value of 1.000 mol dm –3 for a.
Process the results in the table to enable you to plot a graph of log 10 (a-x) against time
t.
Record these values to three significant figures in the additional columns of the table.
Label the columns you use. For each column you use, include units where appropriate
and an expression to show how your values are calculated.
You may use the column headings A to D in these expressions (e.g. A-B).
A
B
0
decrease in
the glucose
concentration
/ mol dm–3
0.000
30
0.101
60
0.193
100
0.259
130
0.370
time/min
P-5 (Planning, Analysis and Evaluation)
C
D
Yearly Worked Solutions
122
Chemistry A-Level P-5
180
0.469
210
0.551
240
0.573
270
0.617
300
0.655
November 2012/P52
[2]
(b)
(c)
(d)
Present the data calculated in (a) in graphical form. Draw the line of best fit.
In plotting this graph, it is necessary to show an origin for both axes.
Remember that the values of log10 (a-x) are negative.
[3]
Circle and label on the graph in (b) any point(s) you consider to be anomalous.
For each anomalous point give a different reason why it is anomalous clearly indicating
which point(s) you are describing.
[3]
Comment on the reliability of the data provided in (a).
P-5 (Planning, Analysis and Evaluation)
[1]
Yearly Worked Solutions
123
Chemistry A-Level P-5
(e)
(f)
(g)
November 2012/P52
Determine the slope of the graph. Mark clearly on the graph any construction lines and
show clearly in your calculation how the intercepts were used in the calculation of the
slope.
Record the value of the slope to three significant figures with appropriate units. [3]
Do the results and your graph confirm the relationship log10 a – log10 (a-x) = kt?
Explain your answer.
[1]
On your graph, draw another line to show how an increase in temperature would affect
your results.
[2]
Solution:
(a)
A
B
C
D
time/min
decrease in the
glucose concentration
/ mol dm–3
(a–x)
(1–B)
(mol/dm3)
Log c
0
0.000
1.00
0.00
30
0.101
0.899
–0.0462
60
0.193
0.807
–0.0931
100
0.259
0.741
–0.130
130
0.370
0.630
–0.201
180
0.469
0.531
–0.275
210
0.551
0.449
–0.348
240
0.573
0.427
–0.370
270
0.617
0.383
–0.417
300
0.655
0.345
–0.462
(b)
P-5 (Planning, Analysis and Evaluation)
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Chemistry A-Level P-5
50
0
×
100
150
200
November 2012/P52
250
300
×
-0.1
×
×
-0.2
×
×
log C
-0.3
×
×
(2.60,-0.4)
-0.4
×
×
-0.5
(c)
(d)
(e)
decreased
Temperature
At t = 100 min  recorded time is later than sample withdrawn.
At t = 210 min  recorded time is earlier than sample withdrawn.
The data is reliable as straight line through the origin is formed and there are very few
anomalous points.
(50, – 0.075) (260, – 0.4)
y2 – y
x2 – x
–0.4 – (–0.075)
=
= –0.00155 min–1
260 – 50
The relationship is confirmed as a straight line through the origin is produced.
m=
(f)
P-5 (Planning, Analysis and Evaluation)
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125
Chemistry A-Level P-5
June 2012/P52
May/June 2012 Paper 52
1.
M/J 12/P52/Q1
When ammonium nitrate(V), NH4NO3, is heated it decomposes completely into nitrogen(I)
oxide, N2O, and water vapour, H2O, which if allowed to cool will condense to liquid water.
The stoichiometric equation for this decomposition is
NH4NO(s) → N2O(g) + 2H2O(l).
The following information gives some of the hazards associated with ammonium nitrate(V).
Ammonium nitrate(V) NH4NO3
Oxidising: Contact with combustible material may cause fire. Explosive when mixed with
combustible material.
Do not allow the salt to become contaminated with organic matter and do not grind it.
You are to plan an experiment to investigate the molar ratio of nitrogen(I) oxide and
ammonium
nitrate(V) at 25°C, and confirm that it remains unchanged as the mass of ammonium
nitrate(V) changes.
(a)
(i)
(ii)
(iii)
(b)
(c)
(d)
Predict quantitatively how the number of moles of nitrogen(I) oxide varies as
the number of moles of ammonium nitrate(V) increases, if the products are
measured at room temperature 25 °C.
Predict quantitatively how the sum of the number of moles of water vapour and
nitrogen(I) oxide varies as the number of moles of ammonium nitrate(V)
increases, if the products are measured at 110 °C.
Display both your predictions in the form of sketch graphs on the axes below.
Label clearly each axis and each graph line.
[3]
In the experiment you are about to plan to test your prediction in (a)(i) at 25 °C, identify
the following.
(i)
the independent variable
(ii)
the dependent variable
[2]
Draw a diagram of the apparatus and the experimental set up you would use to carry
out this experiment. Your apparatus should use only standard items found in a school
or college laboratory and show clearly
(i)
how the solid will be heated,
(ii)
how the water vapour will be condensed into a liquid and collected. Ice is
available,
(iii)
how the nitrogen(I) oxide will be collected.
Label each piece of apparatus used, indicating its size or capacity.
[3]
Using the apparatus shown in (c) design a laboratory experiment to test your prediction
in (a)(i) for an experiment at 25 °C.
In addition to the standard apparatus present in a laboratory you are provided with the
following materials.
a sample of solid ammonium nitrate(V)
P-5 (Planning, Analysis and Evaluation)
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126
Chemistry A-Level P-5
(e)
(f)
June 2012/P52
crushed ice
Give a step-by-step description of how you would carry out the experiment,
(i)
to produce enough results to give sufficient data to plot a graph as in (a)(iii),
(ii)
by stating the volumes of nitrogen(I) oxide you would collect,
(iii)
by calculating the mass of ammonium nitrate(V) needed to produce one of the
volumes of nitrogen(I) oxide suggested in (ii),
(iv)
by stating how you would ensure that decomposition was complete.
[Ar : H, 1.0; N, 14.0; O, 16.0; the molar volume of a gas at 25 °C, 24.0 dm3]
[4]
State one hazard that must be considered when planning the experiment and
describe a precaution that should be taken to minimise the risk from this hazard. [1]
Draw a table with appropriate headings to show the data you would record when
carrying out your experiments and the values you would calculate in order to construct
a graph to support or reject your prediction in (a)(i). The headings should include the
appropriate units.
[2]
Solution:
(a)
(i)
(ii)
Increasing the number of moles of NH4NO3, causes the moles of N2O to increase
as well.
As the moles of NH4NO3 increase, the moles of both N2O and H2O also increase.
Hence the sum of number of moles of products also increases.
(iii)
O
moles of products
110 C
25OC
moles of NH 4NO 3
(b)
(i)
(ii)
moles of NH4NO3
moles of N2O
[2]
(c)
P-5 (Planning, Analysis and Evaluation)
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127
Chemistry A-Level P-5
June 2012/P52
thermometer
50cm 3
Ice
NH4NO 3
[3]
(d)
(i)
(ii)
(iii)
Use of least 6 different masses of NH4NO3
The volume of N2O collected using these masses should be 20, 25, 30, 35, 40 and
45 cm3
To calculate the mass of NH4NO3 used when 20cm3 of N2O is evolved:
moles =
(e)
20
= 0.00083
24000
moles ratio =1:1
mass = 0.00083  80
= 0.066g
(iv)
To ensure complete decomposition, heat each mass of NH4NO3 till the volume of
N2O collected in gas syringe becomes constant
[4]
NH4NO3 is an oxidant and causes explosive reactions when mixed with combustible
materials. Keep it away from any source of combustible material.
[1]
Mass of
NH4NO3 /g
Vol of N2O/
cm3
Moles of
NH4NO3/ mol
Moles of
N2O/ mol
[2]
2
M/J 12/P52/Q2
The variation of the volume with pressure of a fixed mass of any ideal gas at constant
temperature may be represented by a relationship known as Boyle’s law,
PV = constant
where P is the pressure of the gas, V is the volume of the gas.
A gas such as carbon dioxide, under certain conditions of temperature and pressure, does not
always behave as an ideal gas.
An experiment was carried out on carbon dioxide to investigate its behaviour.

A calibrated glass tube was filled with a sample of carbon dioxide.
P-5 (Planning, Analysis and Evaluation)
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Chemistry A-Level P-5



(a)
June 2012/P52
The tube was attached to a calibrated pressure pump.
The pressure and the volume of the gas sample were recorded.
The measured pressure on the gas was increased and the new volume recorded.
The results of the experiment are recorded in the table below.
A graph of V against P has been plotted for you.
Process the results in the table to enable you to plot two further graphs:

PV against P

1/V against P
Record these values to three significant figures in the additional columns of the table.
Label the columns you use. For each column you use include units where appropriate
and an expression to show how your values are calculated.
You may use the column headings A to D for these expressions (e.g. A–B).
A
pressure of
the gas / kPa
B
volume of the
gas /cm3
335
9.09
298
10.9
243
13.9
205
17.0
170
20.8
145
24.4
110
27.0
101
35.0
80.0
45.5
60.0
60.0
C
B
[2]
This graph shows the relationship between the volume of the gas and the pressure of the gas.
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solution
Chemistry A-Level P-5
(b)
(c)
129
June 2012/P52
Plot a graph to show the relationship between the product of the pressure and volume,
PV, of the gas and the pressure, P, of the gas. Draw the line of best fit.
[1]
Plot a graph to show the relationship between the reciprocal of the volume of the gas
and the pressure (1/V against P) of the gas. Begin the scales on both axes at 0. Draw
the line or curve of best fit.
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solution
Chemistry A-Level P-5
(d)
(e)
(f)
(g)
130
June 2012/P52
[3]
Circle and label on the graph in (c) any point(s) you consider to be anomalous.
For each anomalous point give a different reason why it is anomalous, clearly indicating
which point(s) you are describing.
[2]
Determine the initial slope of the graph in (c). Mark clearly on the graph any
construction lines and show in your calculation how the intercepts were used in the
calculation of the slope.
[2]
(i)
Does the initial shape of your graph in (c) confirm the equation PV = constant?
(ii)
Explain your answer in (i) above.
(iii)
Why is the graph of volume against pressure provided inappropriate for the
verification of Boyle’s law?
[3]
(i)
Explain why it was important to measure the initial slope of the graph in (e).
(ii)
What is the significance of the value of the initial slope?
[2]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solution
131
Chemistry A-Level P-5
June 2012/P52
Solution:
(a)
A
pressure of
the gas / kPa
B
C
B
volume of the
gas /cm3
PV (kPa/cm3)
(A  B)
1 (cm–3)
V
( 1B )
335
9.09
3050
0.110
298
10.9
3250
0.0917
243
13.9
3380
0.0719
205
17.0
3490
0.0588
170
20.8
3540
0.0481
145
24.4
3540
0.0410
110
27.0
2970
0.0370
101
35.0
3540
0.0286
80.0
45.5
3600
0.0220
60.0
60.0
3600
0.0167
[1]
(b)
×
×
×
×
×
×
×
×
×
×
(c)
P-5 (Planning, Analysis and Evaluation)
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Chemistry A-Level P-5
June 2012/P52
×
0.11
0.10
×
0.09
0.08
×
0.07
Cm
-3
(2.25, 0.07)
0.06
×
0.05
×
×
0.04
×
0.03
×
×
0.02
× (1.15, 0.02)
0.01
0
0
100
200
300
400
P/kpa
(d)
(e)
At point7, volume is measured at5 a higher temperature
At point4, volume is measured at a lower temperature
(225, 0.067); (75,0.02)
2
y –y
1
M=
2
x – 225
=
(f)
(g)
0.02 - 0.067
= 3.13  10–4
75 - 225
(i)
Yes
(ii)
The initial section of the graph is a straight line
(iii)
The graph curves at the ending region
(i)
The is the region where the graph is a straight line
(ii)
It is the value of 1K
P-5 (Planning, Analysis and Evaluation)
[2]
[2]
[3]
Yearly Worked Solution
133
Chemistry A-Level P-5
November 2011/P52
October/November 2011 Paper 52
1.
O/N 11/P52/Q1
When potassium nitrate dissolves in water, the temperature of the solution goes down because the
enthalpy of solution is endothermic.
You are to plan an experiment to investigate how the solubility of potassium nitrate varies with
temperature. The units of solubility are grams per one hundred grams of water (g / 100g water).
(a)
(i)
(ii)
Predict how the solubility of potassium nitrate will change if the solution temperature is
increased.
Explain your prediction using the fact that dissolving potassium nitrate is endothermic.
prediction
explanation
Display your prediction in the form of a sketch graph, labelling clearly the axes.
[3]
(b)
In the experiment you are about to plan, identify the following.
(i)
the independent variable
(ii)
the dependent variable
[2]
Design a laboratory experiment to test your prediction in (a).
In addition to the standard apparatus present in a laboratory you are provided with the following
materials,
a boiling tube,
a looped wire stirrer,
a thermometer covering the temperature range 0 °C to 100 °C.
Describe how you would carry out the experiment. You should

ensure a wide range of results suitable for analysis by graph,

decide on the amounts of water and potassium nitrate to use,

measure the amounts of the two reagents,

heat the apparatus,

decide at what point the temperature of the solution is to be taken.
[7]
(d)
State a hazard that must be considered when planning the experiment and describe precautions
that should be taken to keep risks to a minimum.
[1]
(c)
(e)
Draw a table with appropriate headings to show the data you would record when carrying out
your experiments and the values you would calculate in order to construct a graph to support or
reject your prediction in (a). The headings must include the appropriate units.
[2]
Solution:
(a)
(i)
Prediction: The solubility increases with rise in temperature.
Explanation: Since the forward reaction is endothermic then a rise in temperature will favour
the forward reaction, increasing solubility.
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
134
November 2011/P52
Solubility/g 100g
Chemistry A-Level P-5
0
O
temp / C
(ii)
[3]
(b)
(i)
(ii)
(c)






(d)
(e)
the independent variable temperature
the dependent variable solubility
[2]
Take 5 beaker
Add 25 cm3 of distilled water in each beaker. Which is equal to 25 gram.
Now take KNO3 a petri dish and weigh the KNO3.
Now dissolve the KNO3 in the beaker at the 25Co temperature to make saturated
Subtract the amount of KNO3 remained from the original KNO3 to amount of KNO3 disolved.
Take second beaker, take other petri dish, weigh KNO3, and dissolved it in the beaker by
keeping temperature at 35 C˚ and get dissolved amount as above.

Repeat this experiment for other three observation at 45 C˚, 55 C˚, 65 C˚.

Table the values.
[7]
Hop apparatus handle with tongs or heat resistant gloves.
[1]
Temperature / C˚
Man
dissolved/ g
Mass of water
dissolved / g
Solubility (g/100g)
35 C˚
45 C˚
55 C˚
65 C˚
[2]
2.
O/N 11/P52/Q2
Chemical reactions occur more rapidly as the temperature of the reaction mixture increases. The
mathematical relationship that summarises this is
log10 (rate of reaction)=
–E A
19T
where EA is the activation energy of the reaction and T is the absolute temperature in Kelvin and the
rate of reaction can be taken as the reciprocal of the time taken in seconds (1/time).
An experiment was carried out to investigate this relationship using dilute hydrochloric acid and aqueous
sodium thiosulfate.

20 cm3 of dilute hydrochloric acid was placed in a boiling tube contained in a water bath.

20 cm3 of aqueous sodium thiosulfate was added to the dilute hydrochloric acid, while stirring
and a stopwatch started.

The temperature of the water bath was recorded.

After a period of time the liquid became cloudy (opaque) due to the formation of a precipitate of
sulfur.

As soon as this cloudiness (opacity) appeared the time was recorded.
P-5 (Planning, Analysis and Evaluation)
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135
Chemistry A-Level P-5

(a)
(b)
November 2011/P52
The temperature of the water bath was raised and the whole experiment repeated.
The results of several such experiments are recorded below.
Process the results in the table to calculate log10 (rate of reaction), the reciprocal of the absolute
temperature (1/T) and the ‘rate of reaction’ (1/time). You should expect the values of log10 (rate
of reaction) to be negative.
Record these values to three significant figures in the additional columns of the table.
Label the columns you use. For each column you use include units where appropriate and an
expression to show how your values are calculated.
You may use the column headings A to F for these expressions (e.g. A–B).
[3]
A
B
C
Temperature
/˚C
Absolute
temperature
/K
Time /s
20.0
293
60.3
30.0
303
46.8
40.0
313
41.6
45.0
318
31.6
50.0
323
28.8
55.0
328
25.1
60.0
333
21.0
65.0
338
20.4
70.0
343
18.1
80.0
353
15.1
D
E
F
Plot a graph to show the relationship between log10 (rate of reaction) and the reciprocal of the
absolute temperature. You are reminded that the values for log10 (rate of reaction) are negative.
Draw the line of best fit.
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
Chemistry A-Level P-5
136
November 2011/P52
[3]
(c)
Circle and label on the graph any point(s) you consider to be anomalous. For each anomalous
point give a different reason why it is anomalous, clearly stating which point you are describing.
(d)
Comment on whether the results obtained can be considered as reliable.
(e)
Determine the slope of the graph. Mark clearly on the graph any construction lines and show
clearly in your calculation how the values from the intercepts were used in the calculation of the
slope.
(f)
Using the value of the slope of your graph calculated in (f) calculate a value for the activation
energy, EA. Correct use of the equation will produce an answer in kJ mol–1.
(g)
By considering the movement of particles in the reaction explain why the rate of reaction
increases with increasing temperature.
[3]
[1]
[2]
[1]
[2]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
137
Chemistry A-Level P-5
November 2011/P52
Solution:
(a)
A
B
C
D
E
F
Temperature
/˚C
Absolute
temperature
/K
Time /s
(1/T)/K–1
1÷B
(1/time)/s–1
1÷C
Log10 (Rate of
reaction)
Log10 E
20.0
293
60.3
0.00340
0.0166
–1.78
30.0
303
46.8
0.00330
0.0213
–1.67
40.0
313
41.6
0.00319
0.0240
–1.62
45.0
318
31.6
0.00314
0.0316
–1.50
50.0
323
28.8
0.00310
0.0347
–1.46
55.0
328
25.1
0.00305
0.0398
–1.40
60.0
333
21.0
0.00300
0.0476
–1.32
65.0
338
20.4
0.00295
0.0490
–1.31
70.0
343
18.1
0.00291
0.0552
–1.26
80.0
353
15.1
0.00283
0.0662
–1.18
[3]
(b)
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
138
Chemistry A-Level P-5
November 2011/P52
-200
-10
-101
-1.00
-1.30
-1.40
log10 (Rate)
-1.20
-1.50
-1.60
-1.70
×
2.50
2.60
2.70
2.80
2.90
3.00
-3
3.10
3.20
3.30
-1.80
3.40
-1
(1/T) x10 /K
[3]
(c)
Points 3 and 7 are off the line.
Point 3 – Time is started too early or Time to passed opacity point.
Point Time is started late or Time to prior opacity.
(d)
Most of the points lies on the best fit line, so results reliable.
(e)
Co-ordinate 2 –
Co-ordinate 1 –
(–1.26, –1.67)
(0.00291, 0.00330)
Gradient
=
[3]
[1]
–1.67 – (1.26)
0.00330 – 0.00291
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
139
Chemistry A-Level P-5
=
– 0.41
3.9  10
–4
November 2011/P52
Gradient = – 1.05
[2]
(f)
– EA
EA
= slope × 19
= – 1.05 × 19
= 19.95
[1]
(g)
Increase in temperature, increase the kinetic energy of particles, which increase the frequency of
collision of molecules and number of molecules exceeding activation energy increase. So rate of
reaction increases.
[2]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
140
Chemistry A-Level P-5
June 2011/P52,51,53
May/June 2011 Paper 52, 51 & 53
1.
MJ 11/P52,51,53/Q1
The carbonates of group II in the periodic table decompose on heating forming an oxide and
carbon dioxide.
X is any group II cation (e.g. Mg2+)
XCO3  XO + CO2
This decomposition occurs because the positively charged cations polarise (distort) the C—O
bond in the carbonate ion causing the ion to break up. The charge density of the group II. This
affects the decomposition rate.
You are to plan an experiment to investigate how the rate of decomposition of a group II
carbonate varies as the group is descended. The rate can be conveniently measured by
finding the time taken to produce the same volume of carbon dioxide from each carbonate.
(a)
(i)
(ii)
Predict how the rate of decomposition of the group II carbonates will change
as the group is descended.
Explain this prediction in terms of the charge density of the cation as the group
is descended.
prediction
explanation
Display your prediction in the form of a sketch graph, clearly labelling the axes.
0
(b)
(c)
(d)
[3]
In the experiment you are about to plan, identify the following.
(i)
the independent variable
(ii)
the dependent variable
[2]
Draw a diagram of the apparatus and experimental set up you would use to carry out
this experiment. Your apparatus should use only standard items found in a school or
college laboratory and show clearly the following.
(i)
the apparatus used to heat the carbonate
(ii)
how the carbon dioxide will be collected
Label each piece of apparatus used, indicating its size or capacity.
[2]
Using the apparatus shown in (c) design a laboratory experiment to test your prediction
in (a).
In addition to the standard apparatus present in a laboratory you are provided with the
following materials,
samples of the carbonates of magnesium, calcium, strontium and barium, a stopwatch/clock with second hand.
Give a step-by-step description of how you would carry out the experiment by stating
(i)
the gas volume you would collect from each carbonate,
P-5 (Planning, Analysis and Evaluation)
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June 2011/P52,51,53
(ii)
(e)
(f)
(g)
how you would calculate the mass of each carbonate to ensure that this volume
of carbon dioxide is produced,
(iii)
how you would control the factors in the heating so that different carbonates
can be compared.
[4]
State a hazard that must be considered when planning the experiment and describe
precautions that should be taken to keep risks to a minimum.
[2]
Draw a table with appropriate headings to show the data you would record when
carrying out your experiments and the values you would calculate in order to construct
a graph to support or reject your prediction in (a). The headings must include the
appropriate units.
[2]
This simple experiment is likely to produce only approximate results.
Suggest an improvement to your apparatus or an alternative apparatus that may
improve the reliability of the results.
[1]
Solution
(a)
(i)
Prediction: Rate of decomposition increases as we go down the group.
Explanation: As we go down the group atomic radii increases effective nuclear
charge remains the same so charge density decreases and polarizing power
decreases.
Rate of Decomposition
(ii)
0
Group Two Carbonates
(down the group)
[3]
(b)
(i)
(ii)
Group two carbonate.
Time taken
[2]
(c)
CaCO3
(fixed
mass)
Bunsen
Burner
[2]
(d)
(1)
(2)
(3)
Take 5g mass of magnesium, calcium, strontium and barium carbonate.
Use an electronic balance to weigh out the mass.
Place each of the weighed samples in a crucible and heat.
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(4)
(e)
(f)
The crucible should be placed inside an enclosed tube which is air tight and has a
syringe attached.
(5)
As soon as the heating starts, start the stopwatch and stop when the volume gas in
the syringe becomes constant.
(6)
Bunsen burner should be placed at same distance from the crucible and same
strength of the burner used.
[4]
The apparatus can become hot, so use heat proof gloves.
[2]
Type of carbonate
Time / s
Rate / s-1
MgCO3
CaCO3
SrCO3
BrCO3
[2]
(g)
Electrical Heating Coil
[1]
2.
MJ 11/P52 51 53/Q2
When sodium nitrate, NaNO3, is heated, it decomposes into sodium nitrite, NaNO2, and oxygen.
A suggested equation is:2NaNO3  2NaNO2 + O2
An experiment was carried out to attempt to confirm this.

An empty boiling tube was weighed and the mass recorded.

A sample of sodium nitrate was added to the boiling tube and the new mass recorded.

The boiling tube and sodium nitrate was heated strongly for five minutes and then
allowed to cool back to room temperature.

The boiling tube and contents was then reweighed and the mass recorded.
(a)
Calculate the relative molecular masses (Mr) of NaNO3 and NaNO2.
[Ar: N, 14.0; O, 16.0; Na, 23.0]
[1]
(b)
The results of several such experiments are recorded below.
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(c)
June 2011/P52,51,53
A
B
C
D
E
F
G
Mass of
boiling
tube /g
Mass of
boiling
tube +
NaNO3
/g
Mass of
NaNO3
/g
(B – A)
Mass of
NaNO3
/g
(C – A)
Mass of
NaNO3
(D ÷ 85)
Mass off
NaNO3
(D ÷ 85)
Mass off
NaNO2
(E ÷ 69)
9.90
13.10
12.50
10.05
14.73
13.91
10.25
14.20
13.46
9.80
12.67
12.65
9.60
14.56
13.63
10.30
15.80
14.76
11.05
17.18
15.50
10.00
17.00
15.68
9.75
17.65
16.16
10.15
18.48
16.84
Process the results in the table to calculate the number of moles of sodium nitrate and
the number of moles of sodium nitrite.
Record these values in the additional columns of the table. You may use some or all of
the columns.
Masses should be recorded to two decimal places. Numbers of moles should be
recorded to two significant figures.
Label the columns you use. For each column you use include units where appropriate
and an expression to show how your values are calculated. You may use the column
headings A to G for these expressions (e.g. A–B).
[2]
Plot a graph to show the relationship between the number of moles of sodium nitrate
and the number of moles of sodium nitrite.
Draw the line of best fit.
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[3]
(d)
(e)
(f)
Circle and label on the graph any point(s) you consider to be anomalous.
For each anomalous point give a different reason why it is anomalous clearly indicating
which point you are describing.
[3]
Determine the slope of the graph. Mark clearly on the graph any construction lines and
show clearly in your calculation how the intercepts were used in the calculation of the
slope.
[3]
(i)
Does the value of the slope of your graph calculated in (e) confirm the equation
given in (a) or not?
(ii)
Explain your answer in (f)(i) above.
[2]
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Solution:
(a)
NaNO3
= 85
NaNO2
= 69
[1]
(b)
B
C
D
E
F
G
Mass of
boiling
tube /g
Mass of
boiling
tube +
NaNO3
/g
Mass of
boiling
tube
NaNO2
/g
Mass of
NaNO3
/g
(B – A)
Mass of
NaNO3
(C – A)
Mass off
NaNO3
(D ÷ 85)
Mass off
NaNO2
(E ÷ 69)
9.90
13.10
12.50
3.20
2.60
0.038
0.038
10.05
14.73
13.91
4.68
3.86
0.055
0.056
10.25
14.20
13.46
3.95
3.21
0.046
0.047
9.80
12.67
12.65
2.87
2.87
0.034
0.042
9.60
14.56
13.63
4.96
4.03
0.058
0.058
10.30
15.80
14.76
5.50
4.46
0.065
0.080
11.05
17.18
15.50
6.13
4.45
0.072
0.064
10.00
17.00
15.68
7.00
5.68
0.082
0.082
9.75
17.65
16.16
7.90
6.41
0.093
0.093
10.15
18.48
16.84
8.33
6.69
0.098
0.097
[2]
(c)
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1.0
10
0.093
9
0.9
Po
0.8
in t
7
8
7
moles NaNO2
0.7
6
0.6
4
5
0.5
3
in
Po
t4
×
0.4
0.038
2
0.3
0.2
0.2
0.093
0.038
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
moles NaNO 3
(d)
Point 4:
Point 7:
(e)
(f)
(i)
(ii)
[3]
Incomplete decomposition or heated for less time.
Sample may be damp so more weight is taken or due to escape of gasses
few solid may be lost.
[3]
Co-ordinate 1 - (0.038, 0.093)
Co-ordinate 2 - (0.038, 0.093
0.093  0.038
Slope

0.093  0.038
=1
[3]
Slope confirm the equation.
From the equation the mole ratio is 1:1. The slope of the graph 1, it confirms the
1:1 of the equation.
[2]
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Chemistry A-Level P-5
November 2010/P52&53
October/November 2010 Paper 52 & 53
1.
O/N 10/P52/Q1, O/N 10/P53/Q1
Aqueous hydrogen peroxide decomposes into oxygen gas and water. The reaction is normally
very slow but is catalyzed by solid manganese(IV) oxide.
2H2O2 (aq)  2H2O(l) + O2 (g)
You are to plan an experiment to investigate how the rate of the catalyzed decomposition of
aqueous hydrogen peroxide depends on its concentration.
(a)
The rate of decomposition depends on the number of hydrogen peroxide molecules
present in a given volume of solution.
(i)
Use this information to predict and explain how the rate of decomposition of
the hydrogen peroxide depends on the concentration.

Prediction:

Explanation:
(ii)
Display your prediction in the form of a sketch graph below:
rate /s –1
0
(b)
(c)
(d)
0
[H2 O2 ]/ mol dm–3
[3]
An approximate method for the determination of the rate of decomposition of the
hydrogen peroxide is to measure the time taken to collect a fixed volume of oxygen.
The volume is kept the same throughout a series of experiments. The rate of the
reaction can be represented by the reciprocal of the time taken.
Rate of reaction  1/time taken
In the experiment you are about to plan identify the following.
(i)
the independent variable:
(ii)
the dependent variable:
[2]
Draw a diagram of the apparatus you would use in the experiment. Your apparatus
should use only standard items found in a school or college laboratory and show clearly
the following:
(i)
how the volume of the oxygen will be collected and measured.
(ii)
how you will make sure that none of the oxygen is lost.
[3]
Label each piece of apparatus used, indicating its size or capacity.
Using the apparatus shown in (c) design a laboratory experiment to test your prediction
in (a)
In addition to the standard apparatus present in a laboratory you are provided with the
following materials.

2.00 mol dm–3 aqueous hydrogen peroxide.

a supply of manganese(IV) oxide
(i)
Complete the table below to show how you would prepare five solutions of
aqueous hydrogen peroxide. Make sure that the correct units are recorded.
Expt. No.
Volume of
H2O2
Volume of
H2O
Concentration of
H2O2
1
2
3
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4
5
(ii)
(e)
(f)
Give a step-by-step description of how you would use your apparatus shown
in (c) to carry out one complete experiment.
[4]
State a problem which might be experienced by someone having to carry out these
experiments alone.
[1]
Draw a table with appropriate headings to show the data you would record when
carrying out your experiments and the values you would calculate in order to construct
a graph to support or reject your prediction in (a). The headings should include
appropriate units.
[2]
Solution:
(a)
(i)
Prediction:
Explanation:
Role of decomposition increases as the concentration increase.
Some increase in concentration. More molecule of H2O2 are present
in given volume of solution.
(ii)
rate/s-1
0
0
[H2O2] / mol dm
-3
[3]
(b)
(i)
(ii)
the independent variable:
concentration of H2O2.
the dependent variable: time taken.
[2]
(c)
10cm 2
syringe
250ml
[3]
(d)
(i)
Expt. No.
Volume of
H2O2
Volume of
H2O
Concentration of H2O2
1
25 cm3
0 cm3
2.0
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Chemistry A-Level P-5
2
20 cm3
5 cm3
2.0
3
15
cm3
10
cm3
2.0
10
cm3
15
cm3
2.0
20
cm3
2.0
4
5
(ii)
(e)
(f)
November 2010/P52&53




5
cm3
cm3
cm3
Add 20
of H2O2 and 5
of H2O in flask.
Add a certain amount of manganese(IV) oxide.
Record the volume change of syringe.
Note down the time.
[4]
Starting the reaction and the stopwatch simultaneously is difficult.
H2O2
concentration in
dm3
[1]
Rate / s–1
Time
[2]
2.
O/N 10/P52/Q2, O/N 10/P53/Q2
When a solute is added to two solvents, A and B, which do not mix, some of the solute dissolves
in each of the solvents and an equilibrium is set up between the two solvents. At equilibrium
the ratio of the two concentrations is a constant known as the Partition Coefficient, K.
concentration in solvent A
=K
concentration in solvent B
An experiment was carried out to determine K for succinic acid, HO2CCH2CH2CO2H, between
water (boiling point 100°C) and diethyl ether, (C2H5)2O, (boiling point 35°C).

100cm3 of distilled water and 100 cm 3 of diethyl ether were transferred to a conical
flask.

A sample of succinic acid was added, the flask was stoppered and the mixture
thoroughly shaken until all of the solid and dissolved.

A 10.0cm3 sample of the water layer was removed and titrated with 0.10 mol dm –3
aqueous sodium hydroxide using phenolphthalein as an indicator.

A 25.0 cm3 sample of the diethyl ether layer was removed and a small amount of water
added. This was then titrated with 0.020 mol dm–3 aqueous sodium hydroxide using
phenolphthalein as an indicator.

The experiment was repeated using the same volumes of water and diethyl ether but
decreasing masses of succinic acid.
(a)
The results of the series of titrations are recorded below.
A
B
C
Expt.
No.
Volume of 0.10 mol
dm–3 NaOH reacting
with 10.0 cm3 of the
water layer /cm 3
Volume of 0.020
mol dm–3 NaOH
reacting with
25.0 cm3 of the
ether layer /cm3
1
24.3
18.6
2
22.5
17.3
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D
E
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Chemistry A-Level P-5
(b)
November 2010/P52&53
3
20.3
15.6
4
18.8
13.1
5
16.3
12.5
6
13.8
10.6
7
10.3
7.9
8
6.8
6.9
9
5.0
3.8
10
2.5
1.9
Process the results in the table to calculate the concentration of the succinic acid in
each layer.
Record these values of three significant figures in the additional columns of the table.
Label each column, including units and an expression to show how your values are
calculated.
You may use the column headings A to E your expression.
[3]
Present the concentration of the succinic acid in each layer in graphical form. Draw the
line of best fit.
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Chemistry A-Level P-5
(c)
(d)
(e)
(f)
November 2010/P52&53
[3]
Circle on the graph any point(s) you consider to be anomalous.
For any point circled on the graph suggest an error in the conduct of the experiment
that might have led to this anomalous result.
[3]
(i)
Determine the value of K from your graph. Mark clearly on the graph any
construction lines and show clearly in your calculation how the intercepts were
used in the calculation of the slope.
(ii)
By considering the data you have processed and the graph you have drawn,
decide if the experimental procedure described is suitable for the determination
of the Partition Coefficient, K. Explain your reasoning.
[3]
In the experimental procedure a small volume of water was added to the diethyl ether
prior to the titration with aqueous sodium hydroxide. The flask was constantly shaken
during the titrations. What was the purpose of this technique?
[1]
Using a burette, the error associated with a titration depends on the value of the titre.
Comment on the magnitude of the titres recorded in the table in (a) and indicate, with
reasons, which have the highest error.
[2]
Solution:
(a)
A
Expt.
No.
B
C
Volume of
Volume of
0.10 mol dm–3
0.020 mol
NaOH
dm–3 NaOH
reacting with reacting with
10.0 cm3 of
25.0 cm3 of
the water
the ether layer
layer /cm3
/cm3
D
E
Concentration of
Concentration of
succinic acid in H2O / succinic acid in ether
mol dm–3
/ mol dm–3
B
10 

 0.1 1000  1000 


C
25 

 0.2  1000  1000 


1
24.3
18.6
0.243
0.015
2
22.5
17.3
0.225
0.014
3
20.3
15.6
0.203
0.0125
4
18.8
13.1
0.188
0.0105
5
16.3
12.5
0.163
0.010
6
13.8
10.6
0.138
0.00848
7
10.3
7.9
0.103
0.00632
8
6.8
6.9
0.068
0.00552
9
5.0
3.8
0.050
0.00304
10
2.5
1.9
0.025
0.00152
[3]
(b)
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November 2010/P52&53
260
243
240
225
220
188
180
-3
concentration of succinic acid in H2O 1 mol dm / x 10
3
203
200
163
160
140
139
120
103
100
80
68
60
50
40
25
20
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
concentration of succinic acid in ether
(c)
For point x:
For point y:
(d)
(i)
g=
=
(ii)
[3]
More concentration in ether is taken or less concentration in water is taken.
Less concentration of in ether is taken or more in water is taken.
[3]
y 2 - y1
x 2 - x1
228 - 64
14 - 4
= 16.4
Yes, since the graph produces a straight line.
P-5 (Planning, Analysis and Evaluation)
[3]
Yearly Worked Solutions
Chemistry A-Level P-5
153
November 2010/P52&53
(e)
To ensure that all the reacted of acid and Base are completely mixed with each
other.
[1]
(f)
Low titra values have a higher percentage error. Since p.e. =
P-5 (Planning, Analysis and Evaluation)
x
100 .
x
[2]
Yearly Worked Solutions
154
Chemistry A-Level P-5
November 2010/P51
October/November 2010 Paper 51
1.
O/N 10/P51/Q1
When aqueous sodium chloride, NaCl, is added to aqueous lead nitrate, Pb(NO3)2, a white
precipitate of lead chloride, PbCl2, is produced. A suggested stoichiometric equation is:
Pb(NO3 )2 (aq) + 2NaCl(aq)  PbCl2 (s) + 2NaNO3 (aq)
In separate experiments, different volumes of 0.20 mol dm –3 aqueous sodium chloride are
added to a fixed volume of 0.10 mol dm –3 aqueous lead nitrate. In each case, the precipitate is
filtered, washed with distilled water and thoroughly dried. The mass of the precipitate is
recorded.
You are to plan an experiment to investigate this reaction in order to confirm or reject the
stoichiometry of the equation.
(a)
(b)
By considering the suggested stoichiometric equation, predict and explain how the
number of moles of the precipitate, PbCl2, will change as the number of moles of NaCl
added increases.

Prediction:

Explanation:
[2]
State a limiting factor that must be taken into account when increasing the volume of
the aqueous sodium chloride added.
Sketch the graph which would result if, after some of the experiments, the NaCl is in
excess. Start your graph with no NaCl added.
PbCl 2 / mol
0
0
NaCl /mol
[3]
(c)
In the experiment you are about to plan, identify the following:
(i)
the independent variable:
(ii)
the dependent variable:
(iii)
another variable to be controlled.
[2]
(d)
(e)
Design a laboratory experiment to test your prediction in (a).
You are provided with 250 cm 3 of 0.20 mol dm–3 aqueous sodium chloride.
(i)
Outline how you would prepare 250 cm 3 of 10 mol dm–3 aqueous lead nitrate,
[Ar : N, 14; O, 16; Pb, 207]
(ii)
Give a step by step description of how you would carry out one experiment.
You should state

the volumes of each solution to be used,

how the volumes will be measured.

how you would dry the precipitate.
[6]
In the table below:

enter appropriate headings to show additional data you would record when
carrying out your experiments and the values you would calculate in order to
construct a graph to support or reject your prediction in (a). The headings
should include the appropriate units.

enter the volumes from your plan in (d).
P-5 (Planning, Analysis and Evaluation)
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
(f)
November 2010/P51
enter suitable volumes for four further experiments.
[2]
How would you ensure that at the end of each experiment the precipitate was
thoroughly dried?
[1]
Solution:
(a)
(b)
Prediction: Increases the number of moles of NaCl by 1, increases PbCl2 by 0.5.
Explanation: By the equation 2 moles of NaCl produce 2 mole of PbCl2.
No. of moles of Pb(NO3)2.
[2]
PbC l2/ mol
0
0
NaCl / mol
[3]
(c)
(d)
(i)
(ii)
(iii)
(i)
no. of moles of NaCl
no. of moles of PbCl2 produced.
temperature.
(ii)

C=
mass
0.2 =
1000
×
Mr
mass
[2]
3
Solution in cm
×
331
1000
250
mass = 16.55




Add 16.27 of NaCl and dissolve it into 10 cm 3 of water in a beaker.
Add the solution in volumetric flask and add water upto the mark.
Take a fixed volume of 25 cm 3 of Pb(NO3)2 solution using a pipette into 4
different beaker, add 10 cm 3, 20 cm3, 30 cm3 and 40 cm 3 of NaCl solution
to each using a beaker.
Add the solutions and mix in evaporating dish. After mixing the solutions are
heated to evaporate water (heat will constant mass) then note down the
mass of the precipitates.
[6]
(e)
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
156
Chemistry A-Level P-5
Volume of
Pb(NO3)2
/cm3
Volume of
NaCl (aq)
/ cm3
Mass of
precipitate
/g
November 2010/P51
No. of
moles of
NaCl
M
V
1000
No. of
moles of
NaCl
n
mass
Mr
25
25
25
25
25
(f)
2.
[2]
[1]
Reheat to a constant mass.
O/N 10/P51/Q2
The melting point of solid water is 0°C. This is the same as the freezing point of water. This
freezing point can be lowered (depressed) by the addition of a solute, such as glucose. The
extent of the freezing point depression depends on the number of particles of solute
dissolved in the solution.
The freezing point depression, Tf, is proportional to the molal concentration, cm, of the solution.
Tf, = Kf, cm
where Kf is the freezing point depression constant.
The molal concentration (molality) of a solution is defined as the number of moles of
a solute dissolved in one kilogram of water e.g. a one molal solution has one mole of
solute dissolved in one kilogram of water.
An experiment was carried out to investigate the relationship between Tf and cm.

A weighed sample of distilled water was placed in a boiling tube.

A weighed sample of glucose was added.

The mixture was stirred until a solution was obtained.

The tube was placed in a freezing apparatus to lower the temperature.

The freezing point of the solution was measured precisely and the freezing point
depression calculated.
(a)
Calculate the Mr of glucose C6H12O6.
[Ar : H, 1.0; C, 12.0; O, 16.0]
[1]
(b)
The results of the experiment are recorded below.
A
B
C
mass of
water /g
mass of
glucose /g
freezing point
depression
Tf /°C
100
10.0
1.03
100
12.2
1.26
100
18.0
2.09
100
23.3
2.40
100
27.7
2.86
100
30.9
3.22
100
33.1
3.31
100
38.6
3.98
100
42.3
4.37
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D
E
F
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157
November 2010/P51
(c)
Process the results in the table to calculate the molality of the glucose solution. This
will enable you to plot a graph to show how the freezing point depression, Tf, varies
with the molality of the solution.
Record these values to three significant figures in the additional columns of the table.
You may use some or all of the columns.
Label the columns you use.
For each column you use include units where appropriate and an expression to show
how your values are calculated. You may use the column headings A to F for this
purpose.
[2]
Present the data calculated in (b) in graphical form. Draw the line of best fit.
(d)
Circle on the graph any point(s) you consider to be anomalous.
[3]
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For any point circled on the graph suggest an error in the conduct of the experiment
that might have led to this anomalous result.
[3]
(i)
Determine the value of Tf /cm from your graph. This is the freezing point
depression constant Kf. Mark clearly on the graph any construction lines and
show clearly in your calculation how the intercepts were used in the calculation
of the slope.
(ii)
By considering the data you have processed and the graph you have drawn,
decide if the experimental procedure described is suitable for the determination
of the freezing point depression constant Kf. Explain your reasoning.
[3]
When the experiment was repeated using sodium chloride instead of glucose as the
solute, the freezing point depressions were found to be twice the value obtained in the
glucose experiment for each molality.
Using the information given at the start of the question suggest a reason for this. [1]
Using your suggestion from (f) predict the effect on the freezing point depression if a
weak acid such as ethanoic acid was used instead of glucose or sodium chloride as
the soluble.
[1]
(e)
(f)
(g)
Solution:
(a)
180
[1]
(b)
A
B
C
D
E
mass of
water /g
mass of
glucose /g
freezing
point
depression
Tf /°C
no. of
moles of
glucose
(B/180)
Molal conc.
cm/mol
kgt ×10
(Molality)
100
10.0
1.03
0.060
0.6
100
12.2
1.26
0.070
0.7
100
18.0
2.09
0.10
1.0
100
23.3
2.40
0.13
1.29
100
27.7
2.86
0.15
1.54
100
30.9
3.22
0.17
1.72
100
33.1
3.31
0.18
1.84
100
38.6
3.98
0.21
2.14
100
42.3
4.37
0.24
2.35
F
[2]
(c)
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5.00
4.00
7
∆T/c°
3.00
2.00
1.00
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
0.00
2.4
molality mol / Kg
(d)
(e)
[3]
For point 3, more glucose, or less water or temperature is measured after freezing point.
For point 7, too much water is added, or less glucose is added or temperature is measured
before freezing point.
[3]
(i)
(ii)

4.35  1.25
2.31  0.7

3.1
1.61
= 1.92°C kg mol–1
Most of the points lies on the line, so suitable for determining the freezing point
depression.
[3]
P-5 (Planning, Analysis and Evaluation)
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Chemistry A-Level P-5
(f)
(g)
160
November 2010/P51
NaCl is more soluble in water than glucose. NaCl produces twice the number of
particles.
[1]
Ethanoic acid is less ionize, so less number of particles, so less depression in boiling
point.
[1]
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May/June 2010 Paper 52
1.
M/J 10/P52/Q1
At any temperature, equilibrium can be established between a liquid X and the vapour given off
by that liquid.
X (I)  X (g)
As the temperature of the system is raised, so the pressure exerted by the vapour increases.
When the vapour pressure becomes equal to the surrounding (ambient) air pressure, the liquid
boils.
When the liquid contains a dissolved solid (solute), the vapour above the liquid is reduced.
You are to plan an experiment to investigate how the boiling point of an aqueous solution of
potassium chloride depends on the concentration of the solution.
(a)
(i)
(ii)
By considering how the vapour pressure changes as the concentration of the
aqueous potassium chloride increases, predict and explain how the boiling
point of the solution will be affected by the concentration of the solution.
Predict how the boiling point will change
Explanation
Display your prediction in the form of the sketch graph, labelling clearly the
point representing the boiling point of pure water and its value.
boiling point
0
concentr ation of the potassium chlor ide
(b)
(c)
(d)
[3]
In the experiment you are about to plan, identify the following:
(i)
the independent variable
(ii)
the dependent variable
[2]
Draw a diagram of the apparatus you would use in the experiment. Your apparatus
should use only standard items found in a school or college laboratory and should show
clearly.
(i)
how the solution will be heated and over-heating of the solution prevented.
(ii)
how the thermometer will be positioned. Remember you are investigating an
equilibrium.
Label each piece of apparatus used, indicating its size or capacity and the temperature
range that the thermometer should cover.
[3]
When investigating how the boiling point of a solution changes with concentration, it is
convenient to represent the concentrations of the solute as a molality.
The molality of a solution is defined as the number of moles of a solute
dissolved in one kilogram of water e.g. a one molal solution has one mole of
solute dissolved in one kilogram of water.
In addition to the standard apparatus present in a laboratory you are provided with the
following materials.
100 g of distilled/deionized water (you should take particular note of this limited
supply of water)
solid potassium chloride, KCl
Give a step-by-step description of how you would:
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(i)
(e)
(f)
(g)
prepare a series of solutions of potassium chloride that can be used in the
apparatus you have shown in (c) to give sufficient data to plot a graph as in
(a)(ii).
(ii)
show how you would calculate the molality of one of these solutions.
[Ar : K, 39.1; Cl, 35.5]
[3]
State a hazard that must be considered when planning the experiment.
[1]
State a limiting factor that must be taken into account when increasing the
concentration of the aqueous potassium chloride.
[1]
Draw up a table with appropriate headings to show the data you would record when
carrying out your experiments and the values you would calculate in order to construct
a graph to support or reject your prediction in (a). The headings should include the
appropriate units.
[2]
Solution:
(a)
(i)
Predict how the boiling point will change: The boiling point increase with the
increase in concentration of solution.
Explanation: Boiling point depends upon vapour pressure, this decrease with
increasing concentration as a result boiling point increase.
(ii)
boiling point
0
concentr ation of the potassium chlor ide
[3]
(b)
(c)
(i)
(ii)
(i)
(ii)
the independent variable: concentration KCl.
the dependent variable: temperature or boiling point.
P-5 (Planning, Analysis and Evaluation)
[2]
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Thermometer Co
100
50
100 cm 3 beaker
0
[3]
(d)
(i)
(ii)


Divide the 100 cm3 of water in equal 0.5 para of 20 cm 3.
Make 0.5 molal solution first by adding 0.766 gram in beaker
containing 20 cm3 of water.
molality =
mass
Mr
0.5 =
1000
×
solvent in grams
Mass of KCl
1000
×
76.6
20
Mass of KCl = 0.766 g
Similarly dissolved 1.53 g of KCl to make 1 molal 2.30 g to make 1.5 molal,
3.06 to make 2 moles and 3.89 g to make 2.5 molal solution.

Determine the boiling point one by one of all solution.
[3]
Wear gloves when touching the heated beaker.
[1]
The solubility of KCl
[1]

(e)
(f)
(g)
Sr.No.
Water (g)
KCl (g)
Molality
(mol/kg–1)
1.
20
0.766
0.5
2.
20
1.53
1.0
3.
20
2.30
1.5
4.
20
3.06
2.0
5.
20
3.83
2.5
Boiling point
or
Temperature
°C
[2]
2.
M/J 10/P52/Q2
When aqueous sodium hydroxide, NaOH, is added to aqueous copper(II) sulfate, CuSO4, a
precipitate of copper(II) hydroxide, Cu(OH)2, is produced.
The stoichiometric equation for this reaction is:
P-5 (Planning, Analysis and Evaluation)
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CuSO4 (aq) + 2NaOH (aq)  Cu(OH)2 (s) + Na2SO4 (aq)
An experiment was carried out to investigate this stoichiometry.

A weighed sample of copper(II) sulfate-5-water, CuSO4.5H2O, was completely
dissolved in distilled water.

An excess of aqueous sodium hydroxide was added.

The resultant precipitate was filtered off.

The precipitate was washed thoroughly and completely dried.

The mass of the precipitate was recorded.
(a)
Calculate the relative formular mass, Mr, of each of the following.
[Ar : Cu, 63.5; S, 32.1; O, 16.0; H, 1.0]
CuSO4.5H2O
Mr
Cu(OH)2
Mr
[1]
(b)
(c)
(d)
The results of the experiment are recorded below.
A
B
mass of
CuSO4.5H2O
/g
mass of
Cu(OH)2 /g
2.50
0.78
6.24
1.95
9.99
3.12
14.98
4.20
19.97
6.24
24.96
7.80
29.95
9.36
34.94
11.81
42.43
13.26
C
D
E
Process the results in the table to calculate the number of moles of CuSO 4.5H2O used
and the number of moles of Cu(OH) 2 produced, to enable you to plot a graph to show
the relative number of moles of the CuSO4.5H2O and Cu(OH)2.
Record these values to three decimal places in the additional columns of the table.
You may use some or all of the columns.
Label the columns you use.
For each column you use, include units where appropriate and an expression to show
how you values are calculated.
You may use the column headings A to E in your expressions.
[2]
Why was it not necessary to know the concentration of the aqueous sodium hydroxide?
[1]
Present the data calculated in (b) in graphical form. Draw the line of best fit.
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[3]
(e)
(f)
(g)
Circle on the graph any point(s) you consider to be anomalous.
For any point circled on the graph suggest an error in the conduct of the experiment
that might have led to an anomalous result.
[3]
Determine the slope of the graph. Mark clearly on the graph any construction lines and
show clearly in your calculation how the intercept were used in the calculation of the
slope.
[3]
State whether the result in (f) confirms or not the stoichiometry of the equation for the
reaction between aqueous copper(II) sulfate and aqueous sodium hydroxide.
Comment on your conclusion.
[2]
Solution:
(a)
CuSO4.5H2O
Cu(OH)2
Mr
Mr
249.6
97.5
[1]
(b)
A
B
C
D
Mass of
CuSO4.5H2O
/g
Mass of
Cu(OH)2
/g
Moles of
CuSO4.5H2O
A = Mr
Moles of
Cu(OH)2
P-5 (Planning, Analysis and Evaluation)
E
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Chemistry A-Level P-5
(c)
(d)
June 2010/P52
2.50
0.78
0.01
0.008
6.24
1.95
0.025
0.020
9.99
3.12
0.04
0.032
14.98
4.20
0.06
0.043
19.97
6.24
0.08
0.064
24.96
7.80
0.10
0.080
29.95
9.36
0.12
0.096
34.94
11.81
0.14
0.121
42.43
13.26
0.17
0.136
[2]
[1]
Sodium hydroxide is in excess.
0.18
0.16
0.14
Cu(OH)2
0.12
0.10
0.08
0.06
0.04
0.02
0.00
0.02
0.04
0.06
0.08
0.10
0.12
0.14
0.16
0.18
0.00
CuSO 4
[3]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
Chemistry A-Level P-5
(e)
(f)
(g)
167
June 2010/P52
Point 4:
Not an excess NaOH is used.
or sulphate is less used and recorded in greater amount.
Point 8: Recorded mass of sulphate is less than actual value or greater amount of Cu(OH)2
is recorded, than taken.
[3]
co-ordinate:
(0.025, 0.14)
co-ordinate-2:
(0.02, 0.121)
0.121 0.02
=
0.14  0.025
= 0.9
[3]
The slope of the equation proves the stoichiometry.
The slope give the number of moles of Cu(OH) 2 that is according to the stoichiometry, the
value of slope is very close to one.
[2]
P-5 (Planning, Analysis and Evaluation)
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May/June 2010 Paper 51
1.
M/J 10/P51/Q1
The neutralisation of an acid by a base is exothermic.
In this experiment the following solutions are available.
2 mol dm–3 sulfuric acid, H2SO4
3 mol dm–3 sodium hydroxide, NaOH
The equation for the reaction is:
2NaOH(aq) + H2SO4(aq)  Na2SO4(aq) + 2H2O(l)
(a)
2 mol dm–3 H2SO4 is gradually added to a fixed volume of 3 mol dm –3 NaOH in a 150cm3
plastic cup, while stirring continuously. The temperature of the solution, measured with
a thermometer, increases until the alkali is just neutralised. On further addition of the
cold acid the temperature of the solution slowly falls.
Select an appropriate volume, x cm3, of 3 mol dm–3 NaOH to use in the experiment.
…………………. cm3
–3
Calculate the volume 2 mol dm H2SO4 that will just neutralise x cm3 of 3 mol dm –3
NaOH.
Sketch the graph you would expect to obtain as the acid is added.
Label the neutralization point.
temper ature
of the
solution /°C
0
(b)
(c)
(d)
(e)
(f)
(g)
volume of acid added / cm3
[3]
This experiment can be used to determine the enthalpy change of neutralization for
the reaction. To ensure reliable results the experiment should be repeated a number
of times.
When sulfuric acid is added to the fixed volume of aqueous sodium hydroxide in this
experiment.
(i)
the independent variable is
(ii)
the dependent variable is
(iii)
the other variable that need to be controlled are
[3]
In carrying out the experiment, what apparatus would you use to accurately measure
the independent variable?
[1]
Explain how you would use this apparatus to control the independent variable.
[1]
Identity and assess:
(i)
a risk associated with the plastic cup used in this experiment,
(ii)
a risk associated with the 3mol dm –3 NaOH.
[1]
Describe how the risks in (e) can be kept to a minimum for
(i)
the plastic cup,
(ii)
the 3 mol dm–3 NaOH
[1]
In the space below draw a table to show column headings for all the measurements
you would make during the experiment.
P-5 (Planning, Analysis and Evaluation)
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(h)
(i)
June 2010/P51
Including in the table one or more columns for any calculated values needed to
determine the enthalpy change of neutralization.
[2]
Show how you would calculate the total heat energy produced in the plastic cup up to
the point when the sulfuric acid has just neutralized the sodium hydroxide.
You may use T to represent the temperature change.
[4.3 J of heat energy raise the temperature of 1 cm3 of any solution by 1°C.]
[1]
The enthalpy change of neutralization is the energy change associated with the
reaction shown by the following equation.
NaOH(aq) + 21 H2SO4(aq)  21 Na2SO4(aq) + H2O(l)
Show how you would convert the energy change expressed in (h) into a value, in kJ
mol–1, for the enthalpy change of neutralization, Hneutralizatiion.
Show clearly the sign and the expression for the enthalpy change.
Hneutralizatiion = ……….. ………………………… kJ mol–1 [2]
sign
expression
Solution:
50 cm3
(a)
c1v1 c 2v 2

n1
n2
2  v1 3  50

1
2
v1 (H2SO4) = 375 cm3
temper ature
of the
solution /°C
37cm
3
0
50cm
volume of acid added /cm 3
[3]
(b)
(c)
(d)
(e)
(f)
(i)
Volume of acid
(ii)
Rise in temperature
(iii)
Loss of heat to surrounding and initial temperature.
Burette or pipette.
Adding the calculated volume of acid or add acid dropwise.
(i)
The apparatus is unstable and liquid may spilled out or become hot.
(ii)
NaOH is very corrosive so it may damage the skin.
(i)
Place plastic cup in a beaker.
(ii)
Wear gloves and google.
[3]
[1]
[1]
[1]
[1]
(g)
Sr.
No.
Volume of
Acid in cm3
P-5 (Planning, Analysis and Evaluation)
Initial
Temperature °C
Final
Temperature °C
Rise in
Temperature °C
Yearly Worked Solutions
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[2]
(h)
q = mc∆T
q=
50+37.5 +4.2 × ΔT (kj)
1000
(i)
q=Y kj
Y (kj) = 0.15 mol of NaOH
x:1
[1]
ΔHneutralisation =
sign
2.
0.15 mol of NaOH
Y(kj)
kJ mol-1
expression
[2]
M/J 10/P51/Q2
A student reads in a text-book about ideal and non-ideal mixtures of liquids.
When two different liquids are mixed, bonds between molecules in each of the different
liquids are broken and new bonds are formed between the different molecules in the mixture.
In an ideal mixture the energy used to break bonds between molecules in the different liquids
is approximately the same as the energy released when new bonds are formed between
molecules in the mixture.
In a non-ideal mixture the energy used to break bonds in the different liquids is not the same
as that released in bond formation.
The breaking of bonds is an endothermic process.
The formation of bonds is an exothermic process.
In order to test this information, the student investigates a number of mixtures of:
(i)
propane-1-ol, CH3CH2CH2OH, and propan-2-ol, CH3CHOHCH3
(ii)
ethanol, CH3CH2OH, and cyclohexane, C6H12.
The boiling point of each pure liquid and each mixture is measured. The thermometer used has
graduations at 0.2°C.
The results for the experiment with propan-1-ol and propan-2-ol, which was carried out four
times, are shown.
Temperature of boiling
mixture / °C
Volume/cm3
% (by
volume) of
propan-1-ol
in mixture
Mean
boiling
temperature
/°C
Propane-1Propan-21
2
3
4
ol
ol
0
20.00
82.1
82.6
82.7
82.2
4.00
16.00
85.3
85.4
85.5
85.4
8.00
12.00
88.5
88.4
88.1
88.2
12.00
8.00
91.3
90.6
91.2
91.4
16.00
4.00
94.2
94.0
94.3
94.3
20.00
0
97.1
97.3
97.2
97.8
(a)
Indicate below any results that you consider to be anomalies and that should not be
included when calculating the mean boiling temperature.
[1]
(b)
Complete the table above to show the following:
(i)
the percentage (%) by volume of propane-1-ol in each of the liquids/mixtures.
(ii)
the mean boiling temperature for each of the liquids/mixtures.
[2]
The experiment is repeated using mixtures of ethanol and cyclohexane. The results of these
experiments are given in the table below.
volume/cm3
ethanol
cyclohexane
P-5 (Planning, Analysis and Evaluation)
%
(by volume) of
ethanol in
mixture
mean
boiling
temperature
/ °C
Yearly Worked Solutions
171
Chemistry A-Level P-5
0
2.00
6.00
10.00
14.00
18.00
20.00
(c)
20.00
18.00
14.00
10.00
6.00
2.00
0
June 2010/P51
0
10.0
30.0
50.0
70.0
90.0
100.0
81.4
66.5
65.0
65.3
65.5
68.0
78.5
For both graphs below, use as much of the y-axis as possible for experimental data.
Do not start your temperature scales at 0°C.
Draw a graph of mean boiling temperature (y-axis) against the % of propan-1-ol (xaxis) in the mixture of propan-1-ol and propan-2-ol.
Draw a graph of mean boiling temperature (y-axis) against the % of ethanol (x-axis) in
the mixture of ethanol and cyclohexane.
(d)
[3]
The graph you have drawn for propane-1-ol and propane-2-ol is typical of ideal
mixtures of liquids.
The graph you have drawn for ethanol and cyclohexane is typical of some non-ideal
mixtures of liquids.
From the shape of your graph explain whether the mixing of ethanol and cyclohexane
is overall in endothermic or an exothermic process.
The mixing of ethanol and cyclohexane is
P-5 (Planning, Analysis and Evaluation)
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because
What type of intermolecular forces exist between the molecules of
(i)
pure ethanol,
(ii)
pure cyclohexane
(iii)
ethanol and cyclohexane in the mixture?
[3]
Solution:
(a)
(b)
90.6 and 97.8, difference is greater as compared to other value.
Volume/cm3
Propane-1of
Propan-2-ol
0
4.00
8.00
12.00
16.00
20.00
20.00
16.00
12.00
8.00
4.00
0
Temperature of boiling
mixture / °C
1
2
3
4
82.1
85.3
88.5
91.3
94.2
97.1
82.6
85.4
88.4
90.6
94.0
97.3
82.7
85.5
88.1
91.2
94.3
97.2
82.2
85.4
88.2
91.4
94.3
97.8
[1]
% (By
Mean
volume) of
boiling
propan-1- temperature
ol in
/°C
mixture
0
82.4
20
85.4
40
88.3
60
91.3
80
94.2
100
97.2
[2]
(c)
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Boiling Temperature/C
o
100
90
80
0
20
40
60
80
100
0
20
40
60
80
100
Boiling Temperature
80
70
60
% of ethanol
(d)
[3]
The mixing of ethanol and cyclohexane is endothermic
because more energy is required to break intermolecular forces them energy released.
(i)
(ii)
(iii)
3.
Hydrogen Bonding
vander waal forces.
vander waal forces
[3]
M/J 10/P51/Q3
A group of students perform an experiment to confirm that the formula of magnesium oxide is
MgO. Each student is provided with a different length of magnesium ribbon which is coiled and
heated in a crucible fitted with a lid. The magnesium reacts with oxygen to form magnesium
oxide.
The instructions for the experiment are as follows:

Weigh the empty crucible and lid.
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



Coil the length of magnesium ribbon and place it in the bottom of the crucible.
Reweigh the crucible and lid with the magnesium.
Heat the crucible with a Bunsen burner.
Periodically lift the crucible lid for a very short period of time. This allows air to
enter the crucible.

Each time the lid of lifted, take care to minimize the loss of any white smoke
which is some of the powder formed.

When the reaction appears to have stopped, remove the crucible lid and heat
the crucible and its contents strongly for 2 minutes.

Cool and reweigh the crucible, lid and the contents.
The results of the experiment are given below:
(a)
(b)
Student
Mass of
crucible
and lid
/g
Mass of
crucible and
lid +
magnesium /
g
Mass of
crucible and
lid +
magnesium
oxide / g
1
25.37
26.62
27.50
2
25.18
27.01
3
25.44
4
Mass of
magnesium
/g
Mass of
magnesium
oxide / g
28.19
1.83
3.01
27.73
29.19
2.29
3.75
25.26
27.71
24.96
2.45
5
25.39
28.11
29.84
2.72
4.45
6
25.04
27.89
28.54
2.85
3.50
7
25.13
28.08
29.93
Complete the table above for student 1, student 4 and student 7.
Plot the data for student 1 and for student 7 on the graph on the next page.
[1]
If the formula of magnesium oxide is MgO, the straight line indicates the mass of
magnesium oxide formed from a given mass of magnesium.
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
175
Chemistry A-Level P-5
June 2010/P51
5.00
mass of
magnesium
oxide /g
4.00
3.00
2.00
1.00
0.00
0.00
1.00
2.00
3.00
mass of magnesium /g
(c)
(d)
(e)
(f)
Choose a mass of Mg and use the straight line to determine the corresponding mass
of MgO formed.
…………….. g of magnesium form
…………….. g of magnesium oxide. [1]
Show by calculation that the masses of magnesium and magnesium oxide selected in
(b) correspond to a formula of MgO.
[Ar : O, 16.0; Mg, 24.3]
[1]
The point plotted for student 6 shows a large deviation from the straight line.
Refer to the instructions for the experiment and suggest a possible explanation for this
anomalous result.
[1]
The result for student 4 could not be plotted on the graph. Suggest an error in carrying
out the experiment that could have led to the result.
[1]
Student 1 added a few drops of water to the cooled residue in the crucible.
The residue and water reacted to produce ammonia gas, NH3.
Explain why this observation reduces confidence in this experiment as a method for
determining the formula of magnesium oxide.
[1]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions
176
Chemistry A-Level P-5
June 2010/P51
Solution:
(a)
(b)
(c)
Student
Mass of
crucible
and lid /
g
Mass of
crucible
and lid +
magnesium
/g
Mass of
crucible and
lid +
magnesium
oxide / g
Mass of
magnesium
/g
Mass of
magnesium
oxide / g
1
25.37
26.62
27.50
1.25
2.13
2
25.18
27.01
28.19
1.83
3.01
3
25.44
27.73
29.19
2.29
3.75
4
25.26
27.71
24.96
2.45
5
25.39
28.11
29.84
2.72
4.45
6
25.04
27.89
28.54
2.85
3.50
7
25.13
28.08
29.93
2.95
4.8
2.0 g of magnesium form 3.25 g of magnesium oxide.
MgO
:
Mg
3.25
:
2.00
Oxygen
:
Mg
1.25
:
2.00
Mole =
1.25
16
:
2.00
2.43
0.078
:
0.082
0.078
0.078
:
0.082
0.078
:
:
1.05  1
1
1
1
MgO
(d)
(e)
(f)
[1]
[1]
[1]
Too less mass of magnesium oxide, mass is lost in the form of smoke, or some magnesium
oxide is escaped with lit.
[1]
The crucible lid has been connected when weighing the mass.
[1]
Magnesium must have related with nitrogen to form Mg3N2.
[1]
P-5 (Planning, Analysis and Evaluation)
Yearly Worked Solutions