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IB Chemistry Measuring, modelling and reasoning worked solutions

Worked solutions to every practice question in Measuring, modelling and reasoning of Photon Chemistry: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question Q01

[1 mark]
How many significant figures are present in the recorded length 0.004070\,\text{m}?
A. 2 B. 3 C. 4 D. 6
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Answer: C
C counts the two non-zero digits, the interior zero and the final recorded zero.
A counts only non-zero digits; B discards the final zero; D also counts the two leading decimal zeros.

Final answer

C: four.

Question Q02

[1 mark]
A test gauge reads 9.71\,\mathrm{mm} for an exact 10.0\,\mathrm{mm} reference; reading uncertainty is \pm0.02\,\mathrm{mm}. Give the percentage error to two significant figures. A. 2.9\%B. 3.0\%C. 29\%D. 97\%
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Answer: A
The reference belongs in the denominator: 100|9.71-10.0|/10.0=2.9\%.
B divides by the measured length; C uses a factor of 1000; D gives the reading as a percentage of the reference without subtracting. The uncertainty is not the error.

Final answer

A: 2.9\%.

Question Q03

[1 mark]
Subtract two ruler positions, each uncertain by \pm0.06\,\mathrm{cm}. Under maximum-bound addition, what absolute uncertainty belongs to the travel distance? A. \pm0.00\,\mathrm{cm}B. \pm0.03\,\mathrm{cm}C. \pm0.06\,\mathrm{cm}D. \pm0.12\,\mathrm{cm}
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Answer: D
Two readings contribute: \Delta d=0.06+0.06=0.12\,\text{cm}.
A subtracts uncertainties as if they cancel; B halves the uncertainty of one reading, giving 0.06/2=0.03\,\text{cm}; C includes just one reading.

Final answer

D: \pm0.12\,\mathrm{cm}.

Question Q04

[1 mark]
A spreadsheet's straight-line fit to two measured variables has R^2=0.997. Which conclusion follows? A. The gradient is positive.B. The line accounts for most dependent-variable variation.C. The independent variable is proved causal.D. Each reading has 0.3\% uncertainty.
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Answer: B
A high coefficient of determination indicates a close fit to the variation in the data.
A mistakes fit quality for the sign of the gradient; C confuses association with proof of causation; D treats the unexplained fraction as measurement uncertainty.

Final answer

B: most dependent-variable variation is accounted for.

Question Q05

[1 mark]
A gauge calibration is supported only from 4.0 to 14.0\,\mathrm{mm}. An unknown corresponds to 21.0\,\mathrm{mm}. What best supports reporting it? A. Report: a line is valid everywhere.B. Repeat only at 9.0\,\mathrm{mm}.C. Extend standards near 21.0\,\mathrm{mm}.D. Delete the largest calibration point.
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Answer: C
Standards near the unknown test whether the fitted relationship remains valid there.
A assumes untested extrapolation is valid; B improves precision within the old range but does not test the new range; D discards relevant data without evidence.

Final answer

C: extend and validate calibration near the reading.

Question Q06

[1 mark]
One repeated drop time is much larger than the rest. The record documents that the timer continued after landing because it failed to stop. Which treatment is justified? A. Exclude and document the timing failure.B. Keep: all readings are automatically valid.C. Exclude solely for being furthest from the mean.D. Present the mean as its measured replacement.
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Answer: A
The recorded instrument failure gives independent evidence that the reading does not measure the intended interval.
B ignores known invalid data; C uses disagreement alone as grounds for deletion; D invents a measurement and conceals the failure.

Final answer

A: exclude with the recorded timing-failure reason.

Tutorial practice

Question Q07

[4 marks]
Use exact SI prefix conversions.
(a) Convert 7.36\,\text{cm}^{3} to cubic decimetres.   [1]
(b) Convert 0.0824\,\text{m} to millimetres.   [1]
(c) Write 0.000583\,\text{g} in scientific notation in grams.   [1]
(d) Convert the mass in (c) to micrograms.   [1]
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(a) 0.00736\,\text{dm}^{3} [1]
(b) 82.4\,\text{mm} [1]
(c) 5.83\times10^{-4}\,\text{g} [1]
(d) 583\,\mu\text{g} [1]
Note: Allow ECF from (c) in (d) for a correct conversion to micrograms; the unit is still required.

Final answer

0.00736\,\mathrm{dm^3}; 82.4\,\mathrm{mm}; 5.83\times10^{-4}\,\mathrm g; 583\,\mu\mathrm g.

Question Q08

[4 marks]
A balance reads 15.12,15.11,15.13,15.12\,\mathrm g for a certified 15.00\,\mathrm g mass, treated as exact. (a) Find the mean and range.   [2] (b) Explain the evidence for precision and systematic error.   [2]
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(a) 15.12\,\text{g} for the mean.
0.02\,\text{g} for the range. [2]
(b) The small range shows the repeated measurements are closely grouped / precise.
All readings exceed the certified mass; the mean is high by 0.12\,\text{g}, indicating a common positive bias. [2]
Note: Award [2] for both correct answers in (a). Repeating measurements alone does not remove the calibration bias. A small range does not establish accuracy.

Final answer

Mean 15.12\,\mathrm g; range 0.02\,\mathrm g; precise with positive 0.12\,\mathrm g bias.

Question Q09

[4 marks]
A food-colour assay tests liquid batches with a pH probe. P has pH 4.32; Q has [\mathrm H^+]=7.9\times10^{-3}\,\mathrm{mol\,dm^{-3}}. (a) Find [\mathrm H^+] of P.   [1] (b) Find the pH of Q.   [1] (c) Why prefer a calibrated probe to universal indicator for reporting these values to two decimal places?   [2]
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(a) [\mathrm{H^+}]=10^{-4.32}=4.8\times10^{-5}\,\mathrm{mol\,dm^{-3}} [1].
(b) \mathrm{pH}=-\log_{10}(7.9\times10^{-3})=2.10 [1].
(c) A probe provides a numerical reading with much finer resolution [1]. Universal indicator gives a colour over a range of \mathrm{pH} values, so colour matching only estimates \mathrm{pH} [1].

Final answer

4.8\times10^{-5}\,\mathrm{mol\,dm^{-3}}; pH 2.10; calibrated numerical resolution exceeds coarse colour matching.

Question Q10

[8 marks]
A spectrometer quantifies dissolved reporter forms \ce{U(aq) <=> V(aq)}. Independent equilibrium readings give [\mathrm U]=(0.0400\pm0.0010) and [\mathrm V]=(0.160\pm0.004)\,\mathrm{mol\,dm^{-3}}, with maximum absolute uncertainties. A reference method gives K=4.30 at the same temperature. (a) Calculate experimental K.   [1] (b) Add percentage uncertainties for the quotient to estimate percentage and absolute uncertainty in K.   [3] (c) Calculate reference percentage error. Does the reference lie inside the estimated interval?   [2] (d) Interpret a possible systematic effect and identify a calibration check.   [2]
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(a) K=0.160/0.0400=4.00 [1].
(b) \mathrm{U} percentage uncertainty =100(0.0010/0.0400)=2.50\% [1]; \mathrm{V} is also 100(0.004/0.160)=2.50\%, so the total is 5.00\% [1]. Absolute uncertainty =4.00(0.0500)=0.20, giving K=4.00\pm0.20 [1].
(c) Percentage error =100|4.00-4.30|/4.30=6.98\% [1]. The reference is outside 3.80 to 4.20 [1].
(d) If the reference is reliable, the discrepancy exceeds the estimated uncertainty and suggests an unaccounted effect; it does not identify its cause [1]. Check the instrument response against standards of known concentration [1].

Final answer

K=4.00\pm0.20; 5.00\% bound; 6.98\% reference error; 4.30 outside 3.80–4.20; check against standards.

Question Q11

[5 marks]
A bromine-uptake study uses different known amounts of a pure, open-chain hydrocarbon. Only ordinary alkene C=C groups react. The apparatus consumes a fixed amount of bromine even when no hydrocarbon is added. All bromine-reactive sites react fully.
\begin{array}{c|rrrr}n(\text{hydrocarbon})/\mathrm{mmol}&0.000&0.100&0.200&0.300\\\hline n(\mathrm{Br_2})\text{ consumed}/\mathrm{mmol}&0.050&0.350&0.650&0.950\end{array}
(a) Use the blank to calculate the bromine consumed by the hydrocarbon in the 0.200 mmol run.   [1]
(b) Deduce the number of double bonds per molecule and verify it using a different non-zero run.   [2]
(c) Explain why using the uncorrected result of the 0.100 mmol run would give a misleading answer.   [1]
(d) The blank varies between runs. Suggest one change to the procedure that would improve the correction.   [1]
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(a) The blank consumes 0.050\,\mathrm{mmol}, so corrected uptake is 0.650-0.050=0.600\,\mathrm{mmol} [1].
(b) 0.600/0.200=3.00, so three C=C bonds are present because each consumes one bromine molecule [1]. The 0.100 mmol run gives (0.350-0.050)/0.100=3.00, confirming the count [1].
(c) 0.350/0.100=3.50 includes apparatus consumption; it does not represent a fractional double bond in a pure molecule [1].
(d) Run a matched blank alongside each sample under the same timing and conditions, rather than using a single blank for all runs [1].

Final answer

0.600\,\mathrm{mmol} corrected; three C=C bonds; another corrected run gives three; uncorrected gives 3.50; use matched blanks.

Question Q12

[6 marks]
A model dry gas has molar mass 40.0\,\mathrm{g\,mol^{-1}}. A 0.400\,\mathrm g sample is measured at constant pressure 100\,\mathrm{kPa}. At 300\,\mathrm K and 350\,\mathrm K its measured volumes are 249\,\mathrm{cm^3} and 291\,\mathrm{cm^3}. Treat the gas as ideal over this interval.
(a) Calculate the amount and both values of PV/n in \mathrm{J\,mol^{-1}}.   [3]
(b) Use the two points on a plot of PV/n against T to estimate R, including its unit.   [2]
(c) Name one assumption or instrument check needed before interpreting a disagreement with an accepted R value as nonideal behaviour.   [1]
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Use n=m/M=0.400/40.0=0.0100\,\mathrm{mol} [1]. Convert pressure to 1.00\times10^5\,\mathrm{Pa} and volumes to 2.49\times10^{-4},2.91\times10^{-4}\,\mathrm{m^3}. The two values of PV/n are 2490,2910\,\mathrm{J\,mol^{-1}} [2]. The gradient is (2910-2490)/(350-300)=8.4\,\mathrm{J\,mol^{-1}\,K^{-1}} [2], representing R under the ideal model. Check calibration, gas dryness/purity, leaks, constant pressure or constant amount before interpreting disagreement [1]. Rounded volume inputs do not justify extra precision in this two-point gradient.

Final answer

0.0100\,\mathrm{mol}; 2.49\times10^3,2.91\times10^3\,\mathrm{J\,mol^{-1}}; 8.4\,\mathrm{J\,mol^{-1}\,K^{-1}}; check calibration, leaks or gas state.

Advanced practice

Question Q13

HL[1 mark]
A dimensionless index is calculated as Z=a^{3}b^{2}. The percentage uncertainties in a and b are 2.0\% and 0.50\%. What is the percentage uncertainty in Z using the small-change maximum-bound power rule?
A. 1.0\% B. 2.5\% C. 6.5\% D. 7.0\%
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Answer: D
\Delta Z/Z=3\Delta a/a+2\Delta b/b, giving 3(2.0)+2(0.50)=7.0\%.
A includes only the weighted contribution from b; B ignores both exponents; C applies the exponent of a but omits the exponent of b.

Final answer

D: 7.0\%.

Question Q14

HL[5 marks]
After a temperature step, a logger reading approaches the new temperature. The magnitude of its remaining error is modelled by E(t)=E_{0}\mathrm{e}^{-t/\tau}, where E_{0}=6.40\,{}^{\circ}\text{C} and \tau=18.0\,\text{s}. The symbol \mathrm{e} is the base of natural logarithms.
(a) Calculate E at t=27.0\,\text{s} to three significant figures.   [2]
(b) Determine the time when E first reaches 0.500\,{}^{\circ}\text{C}. Give the answer to three significant figures.   [3]
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(a) E=6.40\mathrm{e}^{-27.0/18.0}.
E=1.43\,{}^{\circ}\text{C} (unrounded 1.42803302). [2]
(b) 0.500/6.40=\mathrm{e}^{-t/18.0}.
\ln(0.500/6.40)=-t/18.0.
t=-18.0\ln(0.500/6.40)=45.9\,\text{s} (unrounded 45.89001308). [3]
Note: Accept 1.42 to 1.44 degrees C and 45.8 to 46.0 s. Award [2] for a bald correct answer in (a) and [3] in (b). Use natural logarithms or a consistent equivalent method.

Final answer

1.43\,{}^\circ\mathrm C; 45.9\,\mathrm s.

Question Q15

[8 marks]
A colour assay uses a linear calibration only from 0.000 to 0.800\,\mathrm{mmol\,dm^{-3}}. Standards at c=0.000,0.400,0.800\,\mathrm{mmol\,dm^{-3}} give absorbances 0.030,0.350,0.670. A technician dilutes 10.00\,\mathrm{cm^3} of an unknown to 50.00\,\mathrm{cm^3} and measures absorbance 0.430.
(a) Use the standards to obtain the slope and intercept, including any units.   [2]
(b) Calculate the original concentration and decide whether the measured diluted sample lies inside the calibration range.   [3]
(c) Explain the direction of error if c is calculated by dividing sample absorbance by the slope without correcting for the intercept.   [1]
(d) State one absorbance control and one action making the volumetric dilution reliable.   [2]
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The slope is (0.670-0.030)/0.800=0.800\,\mathrm{dm^3\,mmol^{-1}} [1]; intercept is 0.030, dimensionless [1]. Diluted concentration is (0.430-0.030)/0.800=0.500\,\mathrm{mmol\,dm^{-3}} [1]. The factor is 50.00/10.00=5.000, so original concentration is 2.50\,\mathrm{mmol\,dm^{-3}} [1]. The measured diluted sample is within the calibration interval [1]. Omitting the intercept gives 0.5375 diluted and 2.6875\,\mathrm{mmol\,dm^{-3}} original, so the result is high [1]. Hold wavelength/path/cuvette conditions fixed [1]. Condition the pipette with stock, deliver correctly, adjust the final mark at eye level or mix thoroughly [1; one appropriate action].

Final answer

Slope 0.800\,\mathrm{dm^3\,mmol^{-1}}; intercept 0.030; diluted 0.500, original 2.50\,\mathrm{mmol\,dm^{-3}}; in range; omitted correction gives high result; optical and dilution controls.

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Photon Chemistry: Study Guide for the IB Diploma (SL & HL)

Measuring, modelling and reasoning from the core ideas to full exam questions, with question types, worked examples and graded practice.

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