CAMBRIDGE IGCSE™ CHEMISTRY: MATHS SKILLS WORKBOOK
All exam-style questions and sample answers in this title were written by the author. In examinations, the way marks are
awarded may be different.
aths Skills
M
Workbook answers
Chapter 1
1
2
3
4
5
6
7
1
balance ➞ grams (g)
measuring cylinder ➞ cubic centimetres (cm3)
thermometer ➞ degrees Celsius (°C)
ruler ➞ centimetres (cm)
gas syringe ➞ cubic centimetres (cm3)
Example answers:
cm3 measuring cylinder, burette, gas syringe
g
balance
cm ruler
°C
thermometer, temperature probe
a i
g
ii °C
iii s
iv cm
v m2
vi cm3
b Peer assessment
Example answer:
Index notation is needed for writing units
because cm (or m) are a measure of length,
whereas area is calculated by multiplying two
lengths. Index notation shows this as the index
‘2’ (cm2 or m2). Similarly, volume is calculated
by multiplying three lengths. This is shown by
index notation ‘3’ (cm3 or m3).
a g/s
b kg / m3
The (gaseous) product is collected and its
volume is measured.
a 692 381
b 1 949 789
c 0.17
d 0.09
e 0.000 008 5
8
a 115 362
b 0.123 451
c 0.000 002
d 232.013 4
e 104 895.82
9 a helium (smallest), neon, krypton (largest)
b Discussion
10 a 3000
b 45 000 000
c 40
d 1 230 000 000 000
11 a 0.02
b 0.000 034
c 0.000 000 009
d 0.000 43
12 diameter of a smoke particle 0.000 002;
length of edges of a crystal of table salt 0.0001
13 a i
0.003 g
ii 0.000 004 g
iii 3000 g
b i
0.004 m
ii 0.02 m
iii 0.000 000 007 m
c i
0.04 m
ii 0.2 m
d i
5000 J
ii 9000 Pa
14 a i
0.042 g
ii 0.000 402 g
iii 345 000 g
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Chapter 1 continued
b
c
15 a
b
16 a
b
17 a
b
c
d
18 a
b
c
19 a
b
c
d
20 a
b
c
d
21 a
2
i
0.000 000 074 m
ii 0.000 000 007 4 m
iii 0.000 000 704 m
i
500 000 J
ii 10 000 Pa
iii 134 000 J
i
cm
ii mm
iii kg
iiii µm
Discussion
i
1 kg
ii Divide by 1000
i
2 kg
ii 5 kJ
iii 3 kPa
iv 0.5 kg
v 2.5 kJ
vi 0.25 kPa
1.34 × 105
1.03 × 105
1.2 × 108
1.4 × 102
3.4 × 1013 carbon atoms
1.42 × 105 g
1.45 × 102 m3
3.4 × 10–3
5.4 × 10–6
5.07 × 10–4
9.754 × 10–9
1.5 × 10–10 m
3 × 10–3 g
2.3 × 10–8 g
9 × 10–4 m3
i
9 × 10–9 m
ii 9.2 × 10–8 m
iii 6 × 10–6 m
iv 7.3 × 10–5 m
b
22 a
b
23 a
b
The numbers with two digits are more
difficult to convert to standard form
because standard form is expressed using
a number that is greater or equal to 1 and
less than 10. This means that the power of
ten needed is not the same as the power
that matches the unit prefix.
i
400 000 atoms
ii 424 000 atoms
i
1 100 000 atoms
ii 1 060 000 atoms
i
0.3 g
ii 0.32 g
i
0.4 g
ii 0.42 g
Exam-style questions
1
2
a
b
a
b
c
3
a
b
As you go down Group I, the atomic
radius increases. [2]
3.44 × 10–10m [1]
[Total: 3]
a = 0.02 m [1], b = 0.02 m [1], c = 0.05 m [1]
i
0.02 × 0.02 × 0.05 = 0.000 02 m3 [2]
ii 2 × 10–5 m3 [1]
i
0.054 kg/0.000 02 m3 = 2700 kg / m3 [2]
ii 3000 kg / m3 [1]
[Total: 9]
6.02 × 1023 [1]
i
24 g [1]
ii
iii
6.02 × 1023
= 3.01 × 1023 [2]
2
3.01 × 1023
= 3.01 × 1022 [2]
10
[Total: 6]
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CAMBRIDGE IGCSE™ CHEMISTRY: MATHS SKILLS WORKBOOK
Chapter 2
1
2
3
4
5
6
7
a
b
c
a
b
c
a
42 °C
96 °C
12 °C
24 cm3
1.8 cm3
7.2 cm3
i
Discussion. The scale on the burette
is in the opposite direction to the
scale on the measuring cylinder.
The numbers on the scale on the
measuring cylinder increase going up
the cylinder. The numbers on the scale
on the burette increase going down
the burette.
Measuring cylinders are designed to
measure the volume contained in the
cylinder, whereas burettes are designed
to measure the volume dispensed from
the burette.
ii Read from top to bottom.
b i
21.3 cm3
ii 15.8 cm3
c Peer assessment
a 8.5 °C
b 36.0 °C
c 98.5 °C
Depends upon reading on actual
thermometer scale.
a 11.70 cm3
b 27.75 cm3
c 31.45 cm3
d Self-assessment
a Learner’s diagram
b Volume of acid
c pH
8
3
a
b
10 a
b
c
d
e
11 a
The independent variable is size of marble
chips. This is described in words.
Categorical
Time is continuous
Temperature is continuous
Type of solution is categorical
Atomic number is discrete
Self-assessment
i
Experiment A
Volume of acid / cm3
pH
0
1
2
3
4
5
ii
Experiment B
Time / s
Mass / g
0
30
60
90
120
b
12
A table with the column headed
‘Time / minutes’ would also be
acceptable with rows 0.5, 1.0, 1.5, 2.0,
2.5 and 3.0.
Peer assessment
Substance
pH
A
B
Dependent
variable
Independent
variable
a
i
mass
ii
time
b
i
time
ii
temperature
c
i
pH
ii
type of solution
d
i
temperature
ii
atomic number
e
9
C
Learner’s description of thinking
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CAMBRIDGE IGCSE™ CHEMISTRY: MATHS SKILLS WORKBOOK
Chapter 2 continued
13 Discussion
The tables are similar because they both have
two columns, one heading row and three rows
for the data.
The dependent variables recorded in the righthand column of the table will be different
because pH is numerical (continuous data)
but flame colour is recorded using words
(categorical data).
14 a i
Experiment A
19 Discussion. The data in each question have
a different number of significant figures.
Therefore, the calculated means also have a
different number of significant figures.
Exam-style questions
1
Time / s
Temperature
Mean
of acid / °C Test Test Test / s
1
2
3
20
Type of
metal
Mass New
Original
/g
volume volume
/ cm3
/ cm3
iron
31.4
9.0
aluminium 11.9
copper
30
a
40
ii
Experiment B
Size of
marble chip
Volume of
carbon dioxide
/ cm3
Mean
/ cm3
Test Test Test
1
2
3
b
small
medium
c
large
b
15 a
The mean is the sum of a set of values
divided by the number of values. The
table helps you calculate the mean because
the values that need to be added together
are clearly arranged next to each other
in the same row. The number of values is
clear from the column headings. There are
three tests, so there are three values.
Metal
Mass
/g
2
Volume Density
/ g per cm3
/ cm3
copper
iron
aluminium
tin
b Self-assessment
16 a 16.666 666 66 s (or 16.666 666 67)
b 17 cm3 (2 sf)
17 a 1.246 666 666 g (or 1.256 666 667)
b 1.25 g (3 sf)
18 20.03 cm3 (4 sf)
4
33.9
Volume
of
metal
/ cm3
Density
/ g per
cm3
5.0
4.0
7.9
9.4
5.0
4.4
2.7
8.8
5.0
3.8
8.9
Correct number of columns (the column
for original volume may be included to
help with the calculation) [1]
Correct number of rows [1]
Correct column headings [1]
Correct units [1]
Correct volume readings [3]
Measurements recorded in correct
location on table [1]
Correct density values [3]
The measurement with the least number
of significant figures is the volume (2 sf),
so density should be recorded to 2 sf [1]
[Total: 12]
Test
Maximum
temperature of
solution / °C
Temperature
change / °C
1
34.0
11.0
2
32.0
9.0
3
33.5
10.5
a
b
Correct temperature readings [3]
Correct temperature changes calculated [3]
c
11.0 + 9.0 + 10.5
= 10.2 °C (3 sf)
3
Answer [1]
Correct sf [1]
d
−2.5 + (−3.0) + (−4.0)
= −3.2 °C (2 sf)
3
Answer [1]
Correct sf [1]
[Total: 10]
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CAMBRIDGE IGCSE™ CHEMISTRY: MATHS SKILLS WORKBOOK
Chapter 3
a
2
b
a
C fits all the data and takes up over half
the grid.
One large square represents 200 °C.
Scale (one
Largest Do
the
large square value
data
represents:) that
can be values
plotted fit?
Does the
scale
occupy
more than
half of the
grid?
2 °C
6 °C
No
Not
applicable
20 °C
60 °C
No
Not
applicable
40 °C
120 °C
Yes
Yes
b and d
0
One large square represents 40 °C is the
best scale because the other scales do not
reach a large enough value to include the
melting points of potassium and sodium.
a and c
Element
Kr
Xe
Ra
–100
–150
–250
c
5
1.0
Ar
–200
b
3
Ne
–50
Boiling point / ºC
1
e
a
0.8
b
i
One large square represents 50 ºC.
ii One small square represents 5 ºC.
Peer assessment
Calculations could include:
90 × 3.6 = 324°, 10 × 3.6 = 36°
10% of 360° is 360/10 = 36°
Remaining angle = 360° – 36° = 324°
Or 90% = 9 × 10% = 9 × 36° = 324°
Copper 324°, tin 36°
Density / g per cm³
6
0.6
tin
silver
copper
0.4
0.2
0.0
4
5
b
a
7
Li
Na
Element
K
a
i
ii
b
i
ii
Independent variable is time
Dependent variable is volume
of hydrogen
x-axis: time
y-axis: volume of hydrogen
One small square represents 0.02 g / cm3.
Axis C
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Chapter 3 continued
8
a
i
b
ii
i
ii
9
Independent variable is concentration
of hydrochloric acid
Dependent variable is time
x-axis: concentration of
hydrochloric acid
y-axis: time
10
3.0
2.5
2.0
Mass / g
8
Mass / g
7
1.5
6
1.0
5
0.5
4
0
3
11 Peer assessment
8
12
0
1
2
3
4
Volume / cm³
5
2
7
1
6
0
1
2
3
Volume / cm³
4
5
Mass / g
0
4
3
2
1
0
6
0
1
2
3
Volume / cm³
4
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Chapter 3 continued
16 Peer assessment
60
17
3.0
13
2.5
50
1.5
Volume of oxygen / cm³
Mass / g
2.0
1.0
0.5
0
0
1
2
3
4
Volume / cm³
40
30
20
5
10
14 Peer assessment
60
15 a
0
50
The best-fit line
has an equal
number of points
above and below.
30
20
10
0
b
0
1
2
3
Volume / cm³
4
5
18
20
30
40
50
Time / s
60
70
80
90
25.5
Mass of beaker + contents / g
Mass / g
40
10
0
25.4
25.3
25.2
25.1
25.0
0
40
1
2
Time / minutes
3
4
Anomalous point.
This point can
be ignored.
30
Mass / g
The best-fit line has
two points above
and two points below.
20
10
0
c
7
0
1
2
3
Volume / cm³
4
5
Peer assessment
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CAMBRIDGE IGCSE™ CHEMISTRY: MATHS SKILLS WORKBOOK
Chapter 3 continued
Exam-style questions
1
a
i
0
H2O
b
Compound
H 2S
H2Se
i
ii
copper 288° [1], sulfur 72° [1]
H2Te
–10
copper
Melting point / °C
–20
sulfur
–30
–40
–50
Correct angles [1]
–60
–70
Largest sector drawn starting at
the top [1]
–80
Labels or key [1]
c
–90
Axes [1]
Bars appropriate width and spacing [1]
Correct height of bars [1]
ii
100
0
Compound
H 2S
H2Se
H2Te
H 2O
–50
–100
2
Axes [1]
Bars appropriate width and spacing [1]
Correct height of bars [1]
H2O does not fit the pattern. [1]
gas state [1]
[Total: 8]
b
i
ii
a
Covellite 33% [1], chalcocite 20% [1],
digenite 22% [1]
50
40
30
20
10
0
8
60
3
Solubility in 100 g water / g
Boiling point / °C
50
A bar chart is best [1] to compare the
percentage of copper in different minerals
because the percentages do not form
part of a whole [1] and so this cannot
be shown on one pie chart. In order to
compare the percentages, several pie
charts would need to be drawn, whereas a
bar chart makes it easier to compare the
percentages all in one chart. [1]
[Total: 11]
0
10
20
30
40
50
Temperature / °C
60
70
80
a
Axes drawn correctly and labelled [1]
Appropriate scales used on axes [1]
Points plotted correctly [1]
b
c
Anomalous point correctly circled [1]
Smooth best-fit curve [1] with
approximately equal number of points
above and below [1]
[Total: 6]
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CAMBRIDGE IGCSE™ CHEMISTRY: MATHS SKILLS WORKBOOK
Chapter 4
1
2
3
4
Verbal descriptions
a The boiling point of the alkanes increases
as the number of carbons increases.
i
ii
–160 °C
70 °C
Fertiliser
Fraction N Percentage N Fraction P Percentage P Fraction K
A
1
3
33.3%
B
3
4
75%
C
2
3
66.6%
D
1
4
25%
a
b
a
b
Percentage K
33.3%
1
3
33.3%
About 1%
Just
1
less than
Just less than
25%
1
6
16.6%
1
6
16.6%
1
4
25%
1
2
50%
1
3
About
1
100
potassium
potassium 50%, phosphorus 25%
5
25 cm3
Find 0.025 g on the x-axis. Read up to the
graph line. Then read across to the y-axis.
c 0.035 g
d Find 35 cm3 on the y-axis. Read across
to the graph line. Then read down to
the x-axis.
6 a 6 cm3
b 19 cm3
c 2.5 minutes
d 3.5 minutes
7 These values are found by extrapolation of
the graph.
a 0.055 g plus description by learner and
peer assessment
b 45 cm3 plus description by learner and
peer assessment
8 a 46 g
b 55 g
c 5.4 cm3
9 28 g in 100 g of water.
10 a i
The graph is a curve and not a
straight line.
ii The graph cannot be extended using a
ruler. The shape of the curve must be
extended by hand.
b i
The value will depend on the extended
curve drawn by the learner. Actual
value is 13.3 g in 100 g of water.
9
b
4
ii
A straight-line graph as in question
9 can be more accurately extended
than the curved graph. It is difficult to
tell from the graph exactly where the
graph should extend.
11 a Statements A and D
b Statements A and C
c Statements B and E
12 Discussion
13 a The gradient up the hill gets greater and
greater until the top of the hill is reached.
The top of the hill is flat so the gradient is
zero. The gradient down the hill starts
very high but gets gradually less.
b The gradient up the hill starts high but
gets gradually less as the top of the hill is
reached. The top of the hill is flat so the
gradient is zero. The gradient down the
hill starts low but gets larger.
14 a The graph has the steepest gradient at the
start. The gradient gradually decreases
until about 3 minutes when it becomes
zero (horizontal).
This means that the reaction has the
fastest rate at the start. The rate of
reaction gradually decreases until the
reaction stops.
b
The reactions stops at about 3 minutes.
change in vertical (y) values
change in mass
=
change in horizontal (x) values change in volume
37 − 10 27
=
=
4−1
3
15 gradient =
= 9.0 g / cm3 (2 sf)
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Chapter 4 continued
18 Answer should be correct for the appropriate
tangent line drawn by the learner.
40
58 − 23
35
For example,
= = 0.51 cm3 / s
70 − 2
68
for tangent drawn as shown
60
37 – 10 = 27 g
10
0
16
4 – 1 = 3 cm³
0
2
3
4
Volume / cm³
change in vertical (y) values
change in mass
gradient =
=
change in horizontal (x) values change in volume
39 − 8 31
=
=
5−1 4
e
58–23=35cm 3
50
ng
en
tl
in
40
Ta
20
Volume of oxygen / cm³
Mass / g
30
30
70 – 2 = 68 s
20
1
10
= 7.75
0
= 7.8 g / cm3 (2 sf)
0
10
20
30
40
50
Time / s
60
70
80
90
40
19 a and b
Answer should be correct for the
appropriate tangent line drawn by
the learner.
36 − 11 25
For example,
= = 6.4 cm3 / minute
10
5 – 1 = 4 cm³
40
30
nt
e
ng
e
lin
a
aT
36 – 11 = 25 cm³
20
4 – 0.1 = 3.9 minutes
10
25
3.9
= 6.4 cm³ / minute
b gradient =
0
1
2
3
Volume / cm³
4
5
0
17 Learner comment on how close calculated
densities are to the official densities.
10
3.9
for tangent drawn as shown
Volume of hydrogen / cm3
20
0
4 − 0.1
39 – 8 = 31 g
Mass / g
30
1
2
3
Time / minutes
4
5
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Chapter 4 continued
c
b
The rate of reaction is 0 cm3 / minute.
You do not need to draw a tangent line
because you can see that the curve is
horizontal at this point. The gradient (and
therefore the rate of reaction) must be
zero. The reaction has finished.
i
ii
Exam-style questions
a
i
ii
b
i
180 °C [1]
The melting point of potassium
is 65 ºC, so the difference is
180 – 65 = 115 °C [1]
The melting point gradually decreases
(going down Group I) [1]
A sensible estimate for the melting
point of rubidium is from 30 °C
to 40 °C [1]
Lithium oxide contains just under
50% lithium (by mass) [1]
Sodium oxide contains about 75%
sodium (by mass) [1]
[Total: 6]
ii
c
i
ii
2
Solubility in 100 g of water / g
a
b
3
11
c
i
ii
iii
Volume of oxygen
at 30 s is 26 cm³.
Reaction finishes when
maximum reached.
Final / maximum
volume is 46 cm³.
50
46
cii
40
Volume of O2 / cm³
1
The rate of reaction is equal to
the gradient which is calculated by
dividing volume by time, so the units
are cm3 / s [1]
The rate of reaction is fastest at the
start of the reaction. The rate of
reaction decreases until it is 0 cm3 / s
[1]
26 cm3 [1]
46 cm3 [1]
80 seconds [1]
60
30
ci
26
20
a ii 55
50
10
a ii 45
40
0
10
15
20
30
b 37 40
Temperature / °C
50
60
0
c iii
0
10
20
30
40
50
Time / s
60
70
80
90
Time taken for reaction
to finish is 80 seconds.
i
45 g [1]
ii 55 g [2]
37 °C [1]
[Total: 6]
[Total: 4]
a
The graph shows that as time increases
the volume of oxygen produced
also increases, until it reaches a
maximum value. [1]
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CAMBRIDGE IGCSE™ CHEMISTRY: MATHS SKILLS WORKBOOK
Chapter 5
1
2
3
4
5
a
b
a
b
c
d
e
a
b
a
b
c
d
e
a
b
6
7
a
b
c
d
a
i
40 + 32 + 16 × 4
ii 40 + 12 + 16 × 3
Peer assessment
1 + 14 + 16 × 3
24 + 32 + 16 × 4
39 + 55 + 16 × 4
24 + (14 + 16 × 3) × 2
(14 + 1 × 4) × 2 + 32 + 16 × 4
i
40 + 32 + 16 × 4 = 136
ii 40 + 12 + 16 × 3 = 100
Peer assessment
1 + 14 + 16 × 3 = 63
24 + 32 + 16 × 4 = 120
39 + 55 + 16 × 4 = 155
24 + (14 + 16 × 3) × 2 = 148
(14 + 1 × 4) × 2 + 32 + 16 × 4 = 132
i
30.1 × 1023
ii 3.01 × 1024
The calculator answer may give the
answer in standard form.
6.02 × 1024
6.02 × 1022
6.02 × 1025
1.204 × 1023
i and ii
8
9
Energy / kJ
H H Cl Cl
Overall
energy
change
10 a
b
c
12
2
× 100 = 11% (2 sf)
18
40
× 100 = 40%
40 + 12 + 16 × 3
12
× 100 = 12%
40 + 12 + 16 × 3
16 × 3
× 100 = 48%
40 + 12 + 16 × 3
11 Answers given to two significant figures.
a
b
c
14
× 100 = 22%
1 + 14 + 16 × 3
24
× 100 = 16%
24 + (14 + 3 × 16) × 2
There are two N atoms in the chemical
formula, so percentage
mass =
12 a
b
13
b
H Cl H Cl
15
Discussion
i
The diagram shows that the energy
change for bond breaking is positive
by use of an up arrow.
ii The diagram shows that the energy
change for bond making is negative by
use of a down arrow.
iii The energy levels of the reactants and
products mean that an arrow showing
the overall enthalpy change must point
down, meaning that the overall energy
change is negative (exothermic).
14 × 2
× 100 = 21%
2 × (14 + 4) + 32 + (16 × 4)
0.22
× 100 = 67% (2 sf)
0.33
The percentage yield is less than in the
worked example because less product has
been made. This may be because more
magnesium oxide was lost as smoke and
not included in the final weighed amount.
1.2
× 100 = 75%
1.6
14 a
Bonds
made
Progress of reaction
b
1 × [H–H] + 1 × [Br–Br] = 436 + 193
= +629 kJ / mol
b 2 × [H–Br] = 2 × 362 = 724 kJ / mol
c Overall energy change = 629 – 724
= –95 kJ / mol. The reaction is exothermic.
Discussion. The student’s answer is incorrect.
H2O contains two H atoms, so the top number
of the calculation should be 2. The correct
calculation is:
H H Cl Cl
Bonds
broken
a
The student has used the mass of other
metals and has not used the mass of gold.
mass of gold = 85 – 22.1 = 62.9 g
62.9
× 100 = 74%
85
70
× 100 = 93%
75
16 a
Discussion. Naturally occurring bromine
is a mixture of two isotopes with the
abundances shown in Table 5.3. The
relative atomic mass shown in the Periodic
Table is the average mass of a naturally
occurring sample of bromine.
50.69 × 79 + 49.21 × 81
= 80 (2 sf)
100
37.40 × 185 + 62.60 × 187
= 186 (3 sf)
100
b
17
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Chapter 5 continued
b
i
B
ii C
iii A
iv C
v A
vi B
Peer assessment
19 a
Number of moles =
18 a
b
c
d
20 a
b
c
d
21 a
b
c
d
22 a
23 a
b
c
mass
10
=
= 0.25 moles
Mr
24 + 16
Discussion. The correct mathematical
formula should include the two physical
quantities in the question and the
unknown value.
ii
iii
iv
volume (dm3)
molar volume (24 dm3)
mass (g)
Number of moles =
Mr
moles
Concentration =
volume (dm3)
moles
Concentration =
volume (dm3)
Number of moles =
b i
volume = number of moles × molar volume
= 2 × 24 = 48 dm3
ii mass = number of moles × Mr
= 0.5 × (40 + 12 + 16 × 3) = 50 g
iii number of moles = concentration × volume
= 0.1 × 0.25
= 0.025 moles
iv
13
volume =
mass
36
= = 18
number of moles 2
Description of how to rearrange:
mass (g)
number of moles =
Mr
volume
48
molar gas volume =
= = 24 dm3
number of moles 2
24 a
285
= 3 moles
24 + 35.5 × 2
10
= 0.1 moles
40 + 12 + 16 × 3
1800
1.8 kg = 1800 g,
= 100 moles
2 × 1 + 16
volume
36
Number of moles =
= = 1.5 moles
molar volume 24
3
= 0.125 moles
24
12
= 0.5 moles
24
6
6000 cm3 = 6 dm3, = 0.25 moles
24
number of moles 0.5
Concentration =
=
= 0.5 mol / dm3
volume
1
0.5
= 1 mol / dm3
0.5
1
= 0.5 mol / dm3
2
0.5
500 cm3 = 0.5 dm3, = 1 mol / dm3
0.5
i
Mr =
number of moles 0.04
=
= 0.08 dm3
concentration
0.5
i
Sodium hydroxide
ii Flask
iii 25.0 cm3
b i
Sulfuric acid
ii Burette
iii 22.0 cm3
c Discussion. The answers should be
different because in the first experiment
the solution with known concentration is
in the flask and so has a volume of
25.0 cm3, but in the second experiment the
solution with known concentration is in
the burette. In this experiment, the volume
of the solution with known concentration
is equal to the titration volume (22.0 cm3).
25 a 1 mole of NaOH reacts with 1 mole
of HCl.
b i
0.04 moles of NaOH react with
0.04 moles of HCl.
ii 0.16 moles of NaOH react with
0.16 moles of HCl.
iii 0.2 moles of NaOH react with
0.2 moles of HCl.
26 a 2 moles of NaOH react with 1 mole
of H2SO4.
b i
0.04 moles of NaOH react with
0.02 moles of H2SO4.
ii 0.16 moles of NaOH react with
0.08 moles of H2SO4.
iii 0.4 moles of NaOH react with
0.2 moles of H2SO4.
27 a Number of moles (NaOH) =
concentration (NaOH) × volume
25
(NaOH) = 0.2 ×
= 0.005 moles.
1000
b The ratio of NaOH : HCl from the
chemical equation is 1 : 1, so the number
of moles (HCl) = 0.005 moles.
number of moles
c i
Concentration =
3
volume (dm )
ii
iii
iv
Answer to part b
Burette
21.5 cm3
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Chapter 5 continued
d
28 a
b
c
29 a
b
30 a
b
31 a
b
number of moles (HCl)
volume (dm3)(HCl)
0.005
=
= 0.233 mol / dm3 (3 sf)
21.5/1000
Number of moles (H2SO4)
= concentration (H2SO4) × volume (H2SO4)
= 0.1 × 22/1000 = 0.0022 moles.
The ratio NaOH : H2SO4 from the
chemical equation is 2 : 1.
Number of moles (NaOH)
= 2 × number of moles (H2SO4)
= 2 × 0.0022 = 0.0044 moles
When concentration multiplies by two,
the volume of carbon dioxide produced
in a minute is also multiplied by two.
The variables are directly proportional.
i
0.2 × 2 = 0.4 so 26 × 2 = 52 cm3
ii The relationship between the variables
is true for any multiplication factor.
0.1 × 4 = 0.4 so 13 × 4 = 52 cm3 can also
be used as the calculation.
iii The graph should be a linear graph
that passes through the origin (0, 0).
When the concentration of sodium
thiosulfate is multiplied by two, the time
for the cross to disappear divides by two.
The variables are inversely proportional.
i
0.05 × 2 = 0.1 so 132/2 = 66 s
ii The relationship between the variables
is true for any multiplication factor.
0.025 × 4 = 0.1 so 264/4 = 66 s can also
be used as the calculation.
Mr of CuCO3 = 124, Mr of CuO = 80,
Mr of CO2 = 44
i
124 g of CuCO3 produces 80 g of CuO
124
So
= 1 g of CuCO3
80
g of CuO
124
80
× 31 g
So 31 g of CuCO3 produces
124
of CuO = 20 g
ii Description of method above.
24 g of magnesium produces 40 g of MgO
(a ratio of 2 : 2 is the same as 1 : 1).
24
40
So = 1 g of Mg produces g of MgO.
24
So 0.96 g of Mg produces 0.96 × =
24
1.6 g of MgO.
24 g of Mg produces 40 g of MgO.
24
g of Mg produces 1 g of MgO.
40
24
So × 0.2 g of Mg produces 0.2 g of MgO.
40
So
33 a
So 0.12 g of Mg is needed to produce 0.2 g
of MgO.
28 g of N2 produces 2 × (14 + 3 × 1) =
34 g of NH3.
So
number of moles (NaOH)
volume (dm3)(NaOH)
0.0044
=
= 0.176 mol / dm3
25.0/1000
124
14
b
Concentration (NaOH) =
produces
32 a
40
Concentration =
28
34
= 1 g of N2 produces g of NH3.
28
28
So 0.7 g of N2 produces
34
× 0.7 = 0.85 g of NH3.
28
b
3 × (2 × 1) = 6 g of H2 produces
2 × (14 + 3 × 1) = 34 g of NH3.
6
34
g of H2 produces = 1 g of NH3.
34
34
6
So × 0.17 = 0.03 g of H2 produces
34
So
0.17 g of NH3.
34 a
b
Mr of CuCO3 = 124, Mr of CuO = 80,
Mr of CO2 = 44
i
1 mole of CuCO3
produces 1 mole of CuO.
Number of moles of CuCO3
=
c
mass of CuCO3 372
= = 3 moles
Mr of CuCO3
124
This means that 3 moles of CuO
are produced.
Rearranging gives:
Mass of CuO
= number of moles of CuO × Mr of CuO
= 3 × 80 = 240 g
ii Description of method above.
1 mole of CuCO3
produces 1 mole of CO2.
As worked out in part b, 372 g CuCO3
contains 3 moles, so 3 moles of CO2
are produced.
Mass CO2
= number of moles CO2 × Mr of CO2
= 3 × 44 = 132 g
24
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Chapter 5 continued
35 a
Mr of SO2 = 64 g
1 mole of S produces 1 mole of SO2.
Number of moles S =
2
a
i
mass of S 3.2
=
Mr of S
32
N=
= 0.1 moles
Mass SO2
= number of moles of SO2 × Mr of SO2
= 0.1 × 64 = 6.4 g
b
Moles =
ii
volume
molar volume
So volume = moles × molar volume
= 0.1 × 24 = 2.4 dm3
c
36 a
b
Number of moles of SO2 =
= 0.02 moles
This means that 0.02 moles × 1.5 = 0.03
moles of H2 are required.
Mass of H2 = number of moles H2 × Mr
= 0.03 × 2 = 0.06 g of H2
Exam-style questions
1
a
b
i
ii
b
Mr of CO(NH2)2
= 12 + 16 + (14 + 2 × 1) × 2 = 60
Percentage by mass of
28
N = × 100 = 47% (to 2 sf) [2]
60
1 mole NH3 produces 1 mole NH4NO3
mass of NH4NO3
Mr of NH4NO3
50000
= 625 moles
=
80
Number of moles =
c
3
a
Note the mass in kilograms (kg) should be
converted to grams (g).
625 moles of NH3 are required.
mass NH3 required
= number of moles of NH3 × Mr of NH3
= 625 × 17 = 10 625 g or 10.6 kg (3 sf) [3]
This mass may also be calculated without
moles using the reacting masses method.
volume = number of moles × molar volume
= 625 × 24 = 15 000 dm3 [1]
[Total: 10]
Mr of Fe2O3 = 56 × 2 + 16 × 3 = 160
Number of moles of Fe2O3
=
mass of Fe2O3 16 000
=
Mr of Fe2O3
160
= 100 moles [2]
b
64 + 32 + 4 × 16 = 160 [2]
160 + 5 × (2 × 1 + 16) = 250 [2]
i
ii
5 × 18
× 100 = 36% [2]
250
iii
[Total: 6]
c
15
28
× 100 = 21% (to 2 sf) [2]
132
iii
mass of SO2 3.2
=
Mr of SO2
64
mass of NH3 0.34
=
Mr of NH3
17
28
× 100 = 35%
80
Note there are two lots of N in the
formula (14 + 14 = 28) [2]
Mr of (NH4)2SO4
= (14 + 4 × 1) × 2 + 32 + 16 × 4 = 132
Percentage by mass of
N=
= 0.05 moles
Mass S
= number of moles of S × Mr of S
= 0.05 × 32 = 1.6 g
Mr of NH3 = 14 + 3 × 1 = 17,
Mr of N2 = 2 × 14 = 28
According to the chemical equation,
1 mole of N2 produces 2 moles of NH3.
mass of N2 21
Number of moles N2 =
=
Mr of N2
28
= 0.75 moles
This means that 2 × 0.75 = 1.5 moles of NH3
are produced.
mass of NH3 = number of moles NH3 × Mr
= 1.5 × 17 = 25.5 g
3 moles of H2 produce 2 moles of NH3.
So 1.5 moles H2 produce 1 mole of NH3.
Number of moles NH3 =
Mr of NH4NO3
= 14 + 4 × 1 + 14 + 3 × 16 = 80
Percentage by mass of
i
The ratio Fe2O3 : Fe is 1 : 2, so 200
moles of Fe are produced. [1]
Mass of Fe = Ar of Fe × moles of
Fe = 56 × 200 = 11 200 g [1]
actual yield
× 100
theoretical yield
10 000
=
× 100 = 89%
11 200
Percentage yield =
[2]
The ratio Fe2O3 : CO2 is 1 : 3, so 300
moles of CO2 are produced. [1]
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Chapter 5 continued
ii
Number of moles of CO2
=
volume of CO2
molar volume
So volume of CO2
= number of moles of CO2
× molar volume
= 300 × 24 = 7200 dm3. [2]
4
[Total: 9]
a
Concentration of NaOH =
number of moles of NaOH
volume of NaOH in dm3
So number of moles of NaOH
= concentration of NaOH
× volume of NaOH
25
= 0.1 ×
= 0.0025 moles. [3]
1000
b
c
The ratio HCl : NaOH is 1 : 1 so 0.0025
moles of HCl react. [1]
Concentration of HCl
number of moles of HCl
volume of HCl in dm3
0.0025
=
= 0.16 mol / dm3 [3]
15.6/1000
=
16
[Total: 7]
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Chapter 6
1
2
3
4
5
6
7
17
a
Correctly labelled drawing of a cube
with units.
b Surface area of one face calculated as
length of edge × length of edge or (length
of edge)2
Total surface area = 6 × calculated value
for surface area of one face
c Self-assessment
Surface area of cube = 6 × area of one square
face = 6 × 3 × 3 = 54 mm2
Surface area of octahedron = 8 × area of one
triangular face = 8 × 4 = 32 mm2
a Surface area of cube = 6 × area of one
square face = 6 × 1 = 6 cm2
b Surface area of one
cube = 6 × 0.52 = 1.5 cm2
Total surface area of eight
cubes = 8 × 1.5 = 12 cm2
a Discussion. If you imagine that the
side is unrolled then its width has to
equal the distance around the lid
(the circumference) because there is
no overlap.
b Area of side
= height × width = 20 × 2πr
= 20 × 2 × π × 2 = 251 cm2
c Total area of lid and base
= 2 × π × r2 = 2 × π × 22 = 25 cm2
d Total surface area of container
= area of side + total area of lid and base
= 251 + 25 = 276 cm2
Surface area of cylinder
= 2 × area of one circular face + area of curved
surface (side)
The diameter = 0.5 cm so the radius = 0.25 cm
Area of circular face = π × 0.252
Area of curved surface = π × 0.5 × 10
So surface area of cylinder
= 2 × π × 0.252 + π × 0.5 × 10 = 16 cm2 (2 sf)
a Discussion
b i
Surface area of 3 cm × 3 cm × 3 cm cube
= 6 × 9 = 54 cm2
Volume of 3 cm × 3 cm × 3 cm cube
= 33 = 27 cm3
Surface area : volume ratio is 54 : 27
or 2/cm
ii
c
8
a
b
Surface area of separate
1 cm × 1 cm × 1 cm cubes
= 27 × 6 × 1 = 162 cm2
Volume of separate cubes
= 27 × 1 = 27 cm3
Surface area : volume ratio is 162 : 27
or 6/cm
Discussion. Thinking about the answer
before doing the calculation helps to
spot if the answer to the calculation is
not expected. This can alert you that a
mistake has been made in the calculation.
i
Surface area = 6 × 4 × 4 = 96 cm2
Volume = 43 = 64 cm3
Surface area : volume ratio is 96 : 64
or 1.5/cm
ii Surface area = 8 × 6 × 2 × 2 = 192 cm2
Volume = 8 × 23 = 64 cm3
Surface area : volume ratio is 192 : 64
or 3/cm
iii Surface area = 64 × 6 × 1 = 384 cm2
Volume = 64 × 1 = 64 cm3
Surface area : volume ratio is 384 : 64
or 6/cm
As the larger cube is broken into smaller
cubes, the area of the exposed surface
increases. The volume remains the same,
so the surface area : volume ratio increases.
Exam-style questions
1
a
b
c
B surface area [1]
The radius of the cylinder is 0.25 cm [1]
Surface area of single stick of chalk
= 2 × π × 0.252 + 2 × π × 0.25 × 8 = 12.96 cm2
or 13 cm3 (2 sf)
[2]
The length of the broken pieces of chalk
is
d
8
= 2 cm [1]
4
Surface area of broken chalk
= 4 × (2 × π × 0.252 + 2 × π × 0.25 × 2)
= 14.14 cm2 or 14 cm3 (2 sf) [3]
The surface area of the broken chalk is
slightly greater [1], meaning that more of
the chalk is exposed to the acid [1] and
the reaction happens more quickly. [1]
[Total: 11]
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Chapter 6 continued
2
a
b
c
d
e
f
18
Surface area of hexagonal tube
= 6 × 0.5 × 25 = 75 cm2 [2]
Total surface area of 100 hexagonal tubes
= 100 × 75 = 7500 cm2 [1]
Volume of block = 10 × 10 × 25 = 2500 cm3
[1]
Surface area : volume ratio is 7500 : 2500
or 3/cm [1]
i
Surface area of empty block
= 4 × 10 × 25 = 1000 cm2 [1]
ii Surface area : volume ratio is
1000 : 2500 or 0.4 / cm [1]
The surface area : volume ratio of the
block containing hexagonal tubes
is greater. [1]
[Total: 8]
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Applying more than one skill
1
a
b
i
ii
i
ii
iii
c
d
2
Mr (CaCO3) = 40 + 12 + (3 × 16)
= 100 [1]
Mr (CO2) = 12 + (2 × 16) = 44 [1]
44 g [1]
4.4 g [1]
2.2 g [1]
Time
Total mass
/g
Loss of
mass / g
0 (Start)
79.4
0.0
1
78.9
0.5
2
78.3
1.1
3
78.2
1.2
4
77.7
1.7
5
77.5
1.9
6
77.5
1.9
a
Volume of sodium hydroxide added (from
burette readings):
Experiment 1: 22.00 cm3 [1]
Experiment 2: 22.20 cm3 [1]
Experiment 3: 22.10 cm3 [1]
b
c
22.0 + 22.2 + 22.1
= 22.10 cm3 [2]
3
i
ii
moles = concentration × volume [1]
number of moles of sodium hydroxide
= concentration of sodium hydroxide
× volume of sodium hydroxide
added (dm3)
22.1
=1×
[1]
1000
d
e
[7]
See graph.
All points should be correctly plotted. [1]
The best-fit line should be a curve that is
horizontal between 5 and 6 minutes. [1]
The line should pass somewhere between
the points at 2 and 3 minutes. [1]
= 0.0221 moles [1]
i
1 mole [1]
ii 0.0221 moles of ethanoic acid [1]
concentration of ethanoic acid
number of moles of ethanoic acid
[1]
volume of ethanoic acid in flask (dm3)
0.0221
=
= 0.884 mol / dm3 [1]
0.025
=
[Total: 12]
3
a
i
[1]
1200
2.0
800
1.0
Temperature / °C
Loss of mass / g
1000
1.5
0.5
0.0
600
400
0
1
2
3
4
Time / minutes
5
6
anomalous point
200
e
i
ii
iii
19
The rate of reaction is shown by the
gradient of the graph. [1]
The rate of reaction is greatest at the
start. It then gets gradually less until
it is zero. [1]
The reaction has stopped when the
rate of reaction (gradient) is zero,
which occurs at about 5 minutes. This
answer will depend upon the line of
best fit drawn. [1]
[Total: 18]
0
0
1
2
ii
Smooth curve of appropriate shape
[1]
Roughly equal number of points
above and below curve [1]
Correctly identifies horizontal section
of graph as showing melting point [1]
Answer within range 1060–1080 °C [1]
iii
3
4
5
7
6
Time / minutes
8
9
10
11
12
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Applying more than one skill continued
i
ii
To find the melting point, the ring
would have to be melted, which would
destroy the ring. [1]
The purity of the gold may be
checked by finding the actual density
of the ring. [1]
density = mass /volume
mass = 85.0 g [1]
volume = 9.5 cm3 – 5.0 cm3 = 4.5 cm3 [1]
density =
85.0
= 18.9 g/cm3 [1]
4.5
This is less than the density of pure
gold, so the ring is not made of pure
gold. [1]
c
i
gold
palladium
nickel
zinc
ii
4
a
b
c
5
20
a
time [1]
volume of hydrogen [1]
time on horizontal axis [1]
suitable scale on each axis [2]
0
1
80
60
40
0
B [1]
65 g [1]
2
= 0.03 g [1]
65
40 + 12 + 16 × 3 [1]
The mass of carbon dioxide produced
= relative molecular mass of carbon
dioxide = 12 + 16 × 2 = 44 g [1]
44
iii = 0.44 g [1]
100
Loss of mass is not a suitable method to
investigate the rate of reaction of zinc
with hydrochloric acid because the mass
lost is so small it is difficult to measure.
The mass lost in the reaction with calcium
carbonate is measurable on a balance with
two decimal places. [2]
[Total: 8]
i
ii
iii
points plotted correctly [1]
anomalous point identified [1]
smooth curve [1]
roughly equal number of points above
and below line [1]
20
Angle for gold should be 270 degrees
and other angles approximately
matching diagram above. [1]
Labels or key [1]
A pie chart is most suitable to show
percentages that make up a whole. [1]
[Total: 14]
i
ii
iii
i
ii
iv
v
vi
100
Volume of hydrogen / cm3
b
b
2
3
4
6
5
Time / minutes
7
8
9
10
i
The rate of reaction at 1 minute is
faster than at 6 minutes. [1]
ii gradient [1]
iii Between 7 minutes and 8 minutes
(where the graph becomes horizontal)
[2]
c i
number of moles
mass
0.1
=
=
relative formula mass
24
= 0.004 moles [2]
ii 1:1 [1]
iii Number of moles H2 = number of
moles Mg = 0.004 moles [1]
d i
volume = moles × molar volume [1]
ii volume of
hydrogen = 0.004 × 24 = 0.096 dm3 [1]
iii 0.096 × 1000 = 96cm3 [1]
e
i
ii
iii
iv
25
= 0.025 dm3 [1]
1000
number of moles of HCl
= concentration × volume
= 0.1 × 0.025 = 0.0025 moles [1]
2.5 × 10–3 moles [1]
Ratio Mg : HCl is 1 : 2
number of moles magnesium
=
number of moles HCl
2
= 0.00125 moles [2]
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Applying more than one skill continued
6
a
i
ii
iii
b
c
i
ii
i
ii
iii
7
6.50 g [1]
b
i
63 × 69.1 +65 × 30.9
= 63.6 [2]
100
ii
i
ii
i ii
64 [1]
64 + 12 + 16 × 3 = 124 [1]
64 + 35.5 × 2 = 135 [1]
135
= 1.09 g [1]
124
1.09 × 6.2 = 6.76 g [1]
d
e
percentage yield =
=
28
26
24
22
20
0
b
c
9
a
Time / minutes
4
6
5
Time / minutes
7
B
8
9
10
[1]
C
Volume / cm3
Mean
Titration Titration Titration titration
volume /
1
2
3
cm3
16.45
Temperature / °C
8
25.80
9
25.60
10
25.20
[3]
b
i
16.40
16.35
16.40
Correct number of columns [1]
Correct column headings (including
units and /) [1]
Correct measurements [3]
Correct mean [1]
moles = concentration × volume
= 0.1 ×
16.4
= 1.64 × 10–3 moles [2]
1000
ii
1.64 × 10–3 moles [1]
iii
concentration =
=
21
3
i–iii
6.50
× 100 = 96% [1]
6.76
i
2
A constant starting temperature [1]
B rapid temperature increase [1]
C gradual temperature decrease [1]
i
Roughly equal number of points
above and below line [1]
ii See graph in part a ii [1]
iii 28.0 °C (or correct for graph drawn)
[1]
iv Energy is transferred (lost) to the
surroundings [1]
[Total: 11]
[Total: 9]
a
1
A
actual yield
× 100
theoretical yield
8
30
ethene
C2H4 [1]
ethane
C2H6 [1]
1,2-dibromoethane C2H4Br2 [1]
ethene
1 : 2 [1]
ethane
1 : 3 [1]
1,2-dibromoethane 1 : 2 : 1 [1]
ethene
CH2 [1]
ethane
CH3 [1]
1,2-dibromoethane CH2Br [1]
The reaction is exothermic. [1]
The energy level of the products is
lower than the reactants. [1]
1 × [C=C] + 4 × [C–H] + 1 × [Br–Br]
= 612 + 4 × 412 + 193 = 2453 kJ / mol [2]
1 × [C–C] + 4 × [C–H] + 2 × [C–Br]
= 1 × 348 + 4 × 412 + 2 × 276
= –2548 kJ / mol [2]
overall enthalpy change
= 2453 – 2548 = –95 kJ / mol [1]
[Total: 16]
a
c
ii
Excess (unreacted) magnesium [1]
[Total: 26]
Temperature / °C
v
moles
volume
1.64 × 10–3
= 0.0656 mol / dm3 [1]
25/1000
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CAMBRIDGE IGCSE™ CHEMISTRY: MATHS SKILLS WORKBOOK
Applying more than one skill continued
c
i
180 [1]
ii mass = moles × relative formula mass
= 1.64 × 10–3 × 180 = 0.2952 g [1]
iii 0.310 g [1]
mass of pure product
iv percentage purity =
mass
of impure product
0.2952
=
= 95% [1]
0.310
[Total: 14]
10 a
i
ii
iii
1 : 5 [1]
5 moles [1]
64 + 32 + 4 × 16 + 5 × (16 + 2) = 250 [1]
iv
1
= 0.004 moles [1]
250
v
number of H2O molecules
= number of moles × Avogadro’s
constant
= 0.004 × 6.02 × 1023
= 2.41 × 1021 molecules [1]
64
× 100 = 26% (2 sf) [1]
250
relative formula mass of white copper
sulfate = 64 + 32 + 4 × 16 = 160 [1]
64
× 100 = 40% (2 sf)
160
[1]
[Total: 8]
b
22
i
ii
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