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Worked solutions: Chapter 5 Measurement and data processing
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Section 5.1 Exercises
1 These results are not very accurate. The actual value of
14.70% is not close to these results. These results, however, are
precise because they are all very close in value to each other. The
average is 14.37% and the range is only ± 0.03%.
2 Percentage difference = 1
100 valueaccepted
valueaccepted valuealexperiment×
−
= 1
10070.14
37.1470.14×
−
= 2.24 %
These results are quite accurate, after all, as the percentage
difference is only 2.24%.
3 Only random uncertainties can be reduced by repeating an
experiment many times. Systematic errors, such as an incorrectly
zeroed set of scales, will occur in all the repetitions. Systematic
errors can only be reduced by more careful use and maintenance of
equipment.
4 Some solid may be spilt on the pan and would create a
systematic error if not cleaned off; the balance may not be
completely level, which would affect its accurate operation; the
balance may not return to zero after a measurement.
5 a Answers in bold.
Experiment number 1 2 3 4 5 6 7 Final volume (±0.08 cm3) 21.43
42.80 27.27 20.52 42.90 22.42 42.80
Initial volume (±0.08 cm3) 0.00 21.43 0.20 0.50 20.52 1.00
21.42
Volume of titre (±0.2 cm3) 21.43 21.37 27.07 20.02 22.38 21.42
21.38 b Not all of the results should be averaged. The results of
experiment 3, 4 and 5 are
obviously wrong (especially experiment 3) and should not be
used.
c The average is 4
38.2142.2137.2143.21 +++ = 21.40 cm3
The range is ±0.03, which is less than the random uncertainty of
the volume of the titre (±0.2 cm3) so the uncertainty in the
average titre remains at ±0.2 cm3.
6 a 4
b 4
c 5
d 2
e 5
f 1
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Worked solutions: Chapter 5 Measurement and data processing
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g 4
h 3
7 a 0.96
b 34.7
c 22.22
d 0.00306
e 0.0031
f 1.00
g 15.28
h 0.000585
i 2.44
j 0.83
8 a 5 × 108
b 4.8 × 108
c 4.78 × 108
d 4.780 × 108
9 a 485.7
b 33.79
c 1.7 × 102
d 0.18
Section 5.2 Exercises
1 a 25.82 to 25.86 cm3
b 14.1 to 14.7 cm3
c 249.5 to 250.5 cm3
2 a 06.1104.0 ×
1100 = 0.4%
b 38.21
04.0 × 1
100 = 0.2%
c 855 ×
1100 = 6%
d 325.0005.0 ×
1100 = 2%
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Worked solutions: Chapter 5 Measurement and data processing
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e 50.2405.0 ×
1100 = 0.2%
3 a 100
2 × 25.0 = 0.5
25.0 ± 0.5ºC
b 100
5 × 10.00 = 0.5
10.00 ± 0.5 cm3
c 100
1 × 2.507 = 0.025
2.507 ± 0.025 g
d 100
5 × 25.50 = 1.28
25.50 ± 1.28 cm3
4 Answers in bold. Experiment Initial
temperature (ºC)
Final temperature
(ºC)
Change in temperature
(ºC)
Absolute uncertainty in temperature change
(ºC) a 25.0 ± 0.5 48.0 ± 0.5 23 ±1 b 15 ± 1 20 ± 1 5 ±2 c 20.00
± 0.05 45.00 ± 0.05 25.0 ±0.1
5 Answers in bold. Experiment Volume of
solution 1 (cm3)
Volume of solution 2
(cm3)
Total volume (cm3)
Absolute uncertainty in volume
(cm3)
Percentage uncertainty in volume
a 25.0 ± 0.5 15 ± 1 40 ±2 4% b 2.02 ± 0.02 12.00 ± 0.02 14.02
±0.04 0.3% c 150 ± 5 25.0 ± 0.5 175 ±6 3%
6 Student A measured 400 ± 5 cm3
Student B measured 200 ± 2 cm3 + 200 ± 2 cm3 = 400 ± 4 cm3
Student B has measured a volume with less uncertainty than
student A.
7 c = Vn =
250.0500.0 = 2.00 mol dm–3
Uncertainty = 0.2% + 1.5 = 1.7%
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Worked solutions: Chapter 5 Measurement and data processing
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8 n(NaCl) = 45.3599.22
325.1+
= 0.02267 mol
The concentration of the solution was 2.00 mol dm–3 ± 2%.
c = Vn =
100.067022.0 = 0.2267 mol dm–3
Percentage uncertainty in m(NaCl) = 325.1005.0
× 1
100 = 0.4%
Percentage uncertainty in n(NaCl) = 0.4%
Percentage uncertainty in V = 100
5.0 × 1
100 = 0.5%
Percentage uncertainty in c(NaCl) = 0.4 + 0.5 = 0.9%
Absolute uncertainty in c(NaCl) = 100
9.0× 0.2267 = 0.002 mol dm–3
The concentration of the solution is 0.227 ± 0.002 mol dm–3.
Section 5.3 Exercises
1 Volume is inversely proportional to pressure.
2
3 a C b A c B
4
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Worked solutions: Chapter 5 Measurement and data processing
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5
Chapter 5 Review questions
1 Yes, it is possible to obtain a set of experimental results
which are precise, but not accurate. The results may all be very
close to one another, with a very small range (therefore precise)
but, perhaps due to a systematic error, they are far from the
theoretical result, and so are not accurate.
2 The average of the five results is 10.51 cm3, with a range of
±0.04 cm3. The percentage difference between the expected volume
(10 cm3) and the actual volume (10.51 cm3) is
1100
101051.10
×− = 5%.
The graduated cylinder is not very accurate.
3 Random uncertainties are minor uncertainties inherent in any
measurement. In a measuring cylinder, the random uncertainty will
vary depending on the size of the measuring cylinder. The value is
usually recorded on the glassware. For example, a 100 cm3 measuring
cylinder will typically have a random uncertainty of ±0.1 cm3. A
systematic error will occur when a piece of equipment is not used
correctly, consistently. In the case of the measuring cylinder, a
systematic error will occur if the top, rather than the bottom, of
the meniscus is read to find a volume. The value will consistently
be too high.
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Worked solutions: Chapter 5 Measurement and data processing
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4 To minimize the random uncertainty in a burette, a number of
different experiments should be performed and the concordant
results should be averaged.
5 a 0.266
b 15.13
c 1348
6 a 36.1408.0
×1
100 = 0.6%
b 014.0002.0
×1
100 = 14%
c 0.215.0×
1100 = 2%
7 a 100
5.0 × 25.00 = 0.1250 cm3
25.00 cm3 ± 0.13 cm3
b 100
2 × 124.3 = 2.486 g
124.3 g ± 2.5 g
c 100
5 × 0.257 = 0.0129 mol
0.257 mol ± 0.013 mol
8 a m(copper) = 3.694 ± 0.004 g
b ∆T(solution) = 21 ± 1ºC
c m(beaker + contents) = 115.27 ± 0.06 g
9 a 694.3004.0 ×
1100 = 0.1%
b 0.21
1 × 1
100 = 5%
c 27.115
06.0 × 1
100 = 0.05%
10 a Percentage error in mass = 324.0002.0 ×
1100 = 0.6%
n(HCl) = 0.00889 mol ± 0.6%
b Percentage error in concentration = 997.0005.0 ×
1100 = 0.5%
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Worked solutions: Chapter 5 Measurement and data processing
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Percentage error in volume = 0.50
75.3 × 1
100 = 7.5%
Total percentage error = 0.5 + 7.5 = 8%
n(NaOH) = (0.997 ± 0.005) × 1000
75.30.50 ± = 0.0499 ± 8%
c Percentage error in mol = 99009.003000.0
× 1
100 = 0.3%
Percentage error in volume = 250.00015.0 ×
1100 = 0.6%
Total percentage error = 0.3 + 0.6 = 0.9%
c(NaCl) = 250.0
99009.0 = 0.0400 mol dm–3 ± 0.9%
11 a 100
6.0 × 0.008 89 = ±5 × 10–5 mol
b 100
8 × 0.0499 = ±4 × 10–3 mol
c 100
9.0 × 0.0400 = ±4 × 10–4 mol dm–3
12 Volume is proportional to temperature at constant
pressure.
13 a
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Worked solutions: Chapter 5 Measurement and data processing
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b
c
d
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Worked solutions: Chapter 5 Measurement and data processing
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e
14 a Ba(OH)2(aq) + H2SO4(aq) → BaSO4(s) + 2H2O(l)
b The electrical conductivity decreased until 15 cm3 of sulfuric
acid had been added, then it increased.
c As sulfuric acid is added, it reacts with the barium ions and
hydroxide ions in solution, forming solid barium sulfate and water,
neither of which conduct electricity. When all of the barium
hydroxide has reacted, the conductivity starts to increase again
due to the excess of H+ ions and SO42– ions in solution.
15
16 a At t = 0 min, gradient = 5.08.1 = 3.6 g min–1
At t = 1.5 min, gradient = 0.1
8.29.3 − = 1.1 g min–1
b The gradient has the greater value at t = 0.
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Worked solutions: Chapter 5 Measurement and data processing
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Chapter 5 Test
Part A: Multiple-choice questions
Question Answer Explanation 1 B The four masses are very close
to each other (are precise); however, the average
mass is affected by the white powder, making it too high (not
accurate). 2 C Answers A and B have leading zeros, which do not
count as significant figures.
The trailing zero in answer C takes the total of significant
figures to 3. 3 C The answer should be given to the minimum number
of significant figures in the
data used. This is 3 (2.50 dm3). 4 B 2.050 + 0.002 = 2.052 and
2.050 – 0.002 = 2.048
so the range is 2.048 to 2.052 g. 5 B
n = RTPV
; V must be in dm3 (divide by 1000) and the least accurate data
is given
to 3 significant figures; therefore, the answer must also be to
3 significant figures. 6 D
81
× 1
100 = 12.5%
7 C c2 =
2
1
VVc
When dividing or multiplying the % errors are added, so the
answer
has an error of 5%. 8 D % error in mass = 0.1%, % error in
volume = 2%
Percentage uncertainties are given to 1 significant figure only,
so answer is 2%. 9 D A horizontal line indicates that the variable
being represented on the y-axis does
not have any dependence on the x-axis variable. 10 A The size of
the marble chips was being varied—the independent variable (on
the
x-axis). Time was being measured, so it is the dependent
variable and should be on the y-axis.
Part B: Short-answer questions
1 The student should ensure that equipment has been zeroed
correctly. Parallax error should be avoided by ensuring that
readings are taken at eye level with eyes directly in front of the
scale.
(2 marks)
2 Results in bold.
Experiment 1 2 3 4 5 6 Final volume (cm3 ± 0.08 cm3)
18.45 36.90 19.10 37.35 19.40 37.77
Initial volume (cm3 ± 0.08 cm3)
0.10 18.45 0.05 19.10 10.00 19.40
Volume of titre (cm3) 18.35 18.45 19.05 18.25 9.40 18.37
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Worked solutions: Chapter 5 Measurement and data processing
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a Random uncertainty is ±0.16 or ±0.2 cm3. (1 mark)
b The result of experiments 3 and 5 should be omitted. (3
marks)
c 4
37.1825.1845.1835.18 +++ = 18.36 cm3 ± 0.2 cm3 (range ± 0.1 is
less than random
uncertainty) (2 marks)
3 a The rate of formation of nitrogen decreases with time. (2
marks)
b The volume of N2 eventually remains constant because one (or
both) of the reactants has been used up and no more nitrogen gas is
being produced.
(1 mark)
4 a As NaOH is added to the HCl, the pH increases very gradually
at first, but then increases rapidly when the equivalence point is
approaching (equal numbers of mole of NaOH and HCl). The pH then
continues to increase gradually as NaOH is in excess.
(3 marks) b The change in pH is 7.2 when the volume of NaOH
added is equal to 25 cm3.
(1 mark)
Part C: Data-based questions
1 a
(4 marks)
b Independent variable: concentration of potassium iodide
Dependent variable: mass of lead iodide precipitated (2
marks)
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Worked solutions: Chapter 5 Measurement and data processing
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c The mass of lead iodide precipitated is proportional to the
concentration of potassium iodide used.
(1 mark)
2 a
(6 marks)
b Slope (gradient) = runrise = 3100.1
)2.3(15.3−×−
−− = –6320
∴–6320 = –Ea ÷ 8.314
∴ Ea = 52.5 kJ mol–1 (3 marks)
c Percentage difference = 53
5.5253− × 1
100 = 0.9%
(2 marks)