Chemistry 9701/54 — May/June 2025

Cambridge A-Level · Planning, Analysis and Evaluation · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Analysis, Conclusions and Evaluation · Planning

Q1MediumAnalysis, Conclusions and EvaluationPlanning

The relative molecular mass, $M_r$ , of an unknown volatile liquid, Y, can be determined experimentally.

A student uses the following method to determine the volume of a conical flask.

step 1 Weigh a dry conical flask.

step 2 Fill the conical flask completely with distilled water.

step 3 Weigh the conical flask when filled with distilled water.

step 4 Use a thermometer to measure the temperature of the distilled water in the conical flask.

The student collects the results shown in Table 1.1.

Table 1.1


mass of dry flask / g	31.022
mass of flask filled with water / g	161.175
temperature of water in flask / °C	23.0
laboratory atmospheric pressure / kPa	99.0

(a)

Fig. 1.1 shows the variation of the density of water with temperature.

Fig. 1.1

(i)

Use Fig. 1.1 to determine the density of water at 23.0 °C.

Give your answer to four decimal places.

1M Similar

DifficultyEasy

Worked solution

Answer

$0.9976 \text{ g cm}^{-3}$

Final answer

0.9976 g cm^-3

Detailed explanation

Background Concept

The density of water varies slightly with temperature. Between 16 °C and 30 °C, the density decreases approximately linearly from about 0.9991 g cm⁻³ to 0.9956 g cm⁻³. This is because thermal expansion increases the volume of a given mass of water as temperature rises, reducing the density. In practical work, the density of water at the working temperature must be read from a calibration graph or table to convert a mass of water into an accurate volume.

Understanding the Question

The question asks you to read the density of water at 23.0 °C from Fig. 1.1, which plots density (g cm⁻³) against temperature (°C). The answer must be given to four decimal places. The graph shows a straight line with a negative gradient, so you can interpolate directly.

Approach

Locate 23.0 °C on the x-axis (between 22 and 24), move vertically to the line, then horizontally to the y-axis to read the density. The graph has fine gridlines allowing four-decimal-place precision.

Step-by-Step Reasoning

On the x-axis, find 23.0 °C — this is the midpoint between the 22 and 24 gridlines.
Move vertically upward to the plotted line.
Move horizontally to the y-axis. The line at 23 °C sits just below the 0.9977 gridline and just above the 0.9976 gridline, at approximately 0.9976 g cm⁻³.
The mark scheme accepts 0.9976 g cm⁻³ to four decimal places.

Key Takeaways

Graph reading in Paper 5 requires precision to the number of decimal places specified.
The density of water is not exactly 1.000 g cm⁻³ at room temperature; using 1.000 would introduce error.
Linear interpolation between gridlines is sufficient for a straight-line graph.

Common Mistakes

Reading 0.9975 or 0.9977 instead of 0.9976 due to imprecise interpolation.
Giving only three decimal places (0.998) when four are required.
Confusing the axes and reading the temperature from the y-axis.

Things to Be Careful About

The question specifies four decimal places — the answer 0.9976 has exactly four.
Ensure you read at 23.0 °C, not 22 or 24 °C.
Include the unit g cm⁻³.

Techniques used

read a value from a graph to four decimal placesinterpolate between gridlines on a linear graph

(ii)

Calculate the volume of the flask, using the density of water determined in (a)(i).

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Mass of water $= 161.175 - 31.022 = 130.153 \text{ g}$

V = \frac{\text{mass}}{\text{density}} = \frac{130.153}{0.9976} = 130.466 \text{ cm}^3

Answer

$130.466 \text{ cm}^3$

Final answer

130.466 cm^3

Detailed explanation

Background Concept

The volume of a container can be determined by filling it with water, measuring the mass of water added, and dividing by the density of water at the measured temperature. This relies on the relationship $\rho = m/V$ , rearranged to $V = m/\rho$ . The mass of water is found by difference: (mass of flask + water) − (mass of dry flask). Using the correct density at the experimental temperature (rather than assuming 1.000 g cm⁻³) is essential for accuracy.

Understanding the Question

You are given: mass of dry flask = 31.022 g, mass of flask + water = 161.175 g, and the density of water at 23.0 °C = 0.9976 g cm⁻³ (from part a(i)). You must calculate the volume of the flask. The command word is "calculate", so working must be shown.

Approach

Step 1: Find the mass of water by subtracting the dry flask mass from the filled flask mass. Step 2: Divide this mass by the density to obtain the volume. Both steps are straightforward arithmetic.

Step-by-Step Reasoning

Mass of water = 161.175 − 31.022 = 130.153 g. (M1 — correct subtraction.)
Volume = mass / density = 130.153 / 0.9976 = 130.466 cm³. (M2 — correct division using the density from a(i).)
The volume is 130.466 cm³, which represents the internal volume of the conical flask.

Note: If a different density were used in a(i) (e.g. 0.9975), the volume would differ slightly — error carried forward is allowed.

Key Takeaways

The water-displacement method for flask volume requires the density at the actual water temperature.
Mass by difference: always subtract the tare (empty) mass from the filled mass.
Units: mass in g divided by density in g cm⁻³ gives volume in cm³.

Common Mistakes

Using density = 1.000 g cm⁻³ instead of 0.9976, giving 130.153 cm³ instead of 130.466 cm³.
Forgetting to subtract the dry flask mass and using 161.175 g as the water mass.
Arithmetic errors in the subtraction (31.022, not 31.023 or 31.020).

Things to Be Careful About

The subtraction must use the exact values given: 161.175 − 31.022, not rounded values.
The density used must be the one determined in a(i); ecf applies if a different (but consistently used) density is carried forward.
Report the volume to an appropriate number of significant figures (at least 5 to match the precision of the data).

Techniques used

calculate mass by subtraction of two weighingsconvert mass to volume using density

(b)

Another student determines the volume of a different conical flask to be $129.56\text{ cm}^3$ . This conical flask is used to determine the $M_r$ of unknown volatile liquid Y.

The following method is used.

step 1 Cover the opening of the dry conical flask with aluminium foil and secure using a rubber band.

step 2 Weigh the conical flask, aluminium foil and rubber band. Record the mass.

step 3 Use a syringe with a needle to make a small hole in the aluminium foil and inject approximately $5\text{ cm}^3$ of Y into the conical flask. Remove the syringe.

step 4 Place the conical flask in a water-bath containing boiling water.

step 5 Allow all of Y to evaporate. Keep the conical flask containing vaporised Y in the boiling water for a further 3 minutes.

step 6 Carefully remove the conical flask from the boiling water and dry the outside surface thoroughly.

step 7 Weigh the conical flask, aluminium foil, rubber band and contents. Record this mass.

(i)

Suggest why the conical flask containing vaporised Y is kept in boiling water for 3 minutes in step 5.

1M Similar

DifficultyEasy

Worked solution

Answer

To ensure the vapour of Y in the conical flask has reached the temperature of the boiling water ( $100\text{ }^{\circ}\text{C}$ ) / thermal equilibrium has been established.

Final answer

To ensure the vapour has reached the temperature of the boiling water (100 °C)

Detailed explanation

Background Concept

In the Dumas method for determining $M_r$ of a volatile liquid, the liquid is vaporised in a sealed flask immersed in a boiling water bath. The ideal gas equation $pV = nRT$ is then applied using the temperature of the water bath. For this to be valid, the vapour inside the flask must actually be at the bath temperature. Thermal equilibration takes time, especially through the glass walls of the flask and the aluminium foil seal.

Understanding the Question

The command word is "suggest" — you must give a chemically sensible reason. The question asks specifically why the flask is kept in boiling water for an additional 3 minutes after all the liquid has evaporated. What is happening during those 3 minutes?

Approach

Consider what happens immediately after the last droplet evaporates: the vapour may still be cooler than 100 °C because heat transfer through glass takes time. The 3-minute wait ensures the vapour temperature equals the bath temperature before the flask is removed and weighed.

Step-by-Step Reasoning

Once all liquid Y has evaporated, the flask contains only vapour of Y at approximately atmospheric pressure.
However, the vapour may not yet be at 100 °C — the glass walls and the vapour itself need time to reach thermal equilibrium with the boiling water.
If the flask were removed immediately, the vapour temperature would be below 100 °C, and using $T = 373$ K in $pV = nRT$ would introduce error.
The 3-minute hold ensures thermal equilibrium: the vapour is genuinely at 100 °C when the flask is removed.

Key Takeaways

In any experiment using $pV = nRT$ , the temperature used must be the actual temperature of the gas.
Waiting periods in methods exist to allow thermal (or chemical) equilibration.
"Suggest" questions require a chemically logical reason, not a vague statement.

Common Mistakes

Saying "to make sure all the liquid has evaporated" — the question states evaporation is already complete; the 3 minutes is after that point.
Saying "to cool the flask" — the flask is kept in boiling water, not removed.
Vague answers like "to be safe" or "to make the experiment accurate" without specifying what is being equilibrated.

Things to Be Careful About

The mark scheme specifically requires the idea that the vapour reaches the temperature of the boiling water. Mentioning "equilibrium" or "100 °C" is essential.

Techniques used

explain the purpose of a timed equilibration step in an experimental procedure

(ii)

Suggest why it is not necessary to determine the mass of liquid Y injected into the conical flask by the syringe in step 3.

1M Similar

DifficultyEasy

Worked solution

Answer

The mass of vapour occupying the flask (measured at step 7) is the value used in the calculation, not the mass initially injected.

Final answer

The mass of vapour occupying the flask at step 7 is used in the calculation

Detailed explanation

Background Concept

In the Dumas method, an excess of volatile liquid is injected into the flask. When heated, the liquid vaporises and the excess escapes through the pinhole in the foil. At the end of heating, the flask contains exactly enough vapour to fill its volume at atmospheric pressure and the bath temperature. The mass of this vapour is found by weighing the flask before and after the experiment (step 7 mass − step 2 mass). The initial injected mass is irrelevant because any excess has escaped.

Understanding the Question

The command word is "suggest". The question asks why the student does not need to know how much liquid Y was injected via the syringe. What quantity actually appears in the $M_r$ calculation?

Approach

Identify what enters the calculation: $M_r = \text{mass of vapour} / n$ , where $n$ comes from $pV = nRT$ and the mass of vapour is (step 7 mass − step 2 mass). The syringe volume/mass never appears.

Step-by-Step Reasoning

An excess of liquid Y is injected (~5 cm³) — more than can fit as vapour in the flask at 100 °C.
On heating, Y vaporises; excess vapour escapes through the pinhole.
At the end, the flask contains vapour of Y at 100 °C and 99.0 kPa filling 129.56 cm³.
The mass of this vapour = (mass at step 7) − (mass at step 2) = 31.429 − 31.123 = 0.306 g.
This 0.306 g is used in $M_r = m/n$ . The 5 cm³ injected is irrelevant — it was never all retained.

Key Takeaways

In the Dumas method, the mass used is the mass of vapour remaining in the flask, not the mass injected.
The pinhole ensures only the equilibrium amount of vapour remains.
Understanding which quantity enters the calculation is key to justifying experimental design choices.

Common Mistakes

Saying "the mass cancels out" — it does not cancel; it is simply not needed because the retained mass is measured directly.
Saying "the syringe is not accurate enough" — the reason is conceptual, not instrumental.
Confusing the mass injected with the mass of vapour in the flask.

Things to Be Careful About

The answer must reference that the mass used in the calculation is the mass of vapour in the flask (step 7 minus step 2), not the injected mass.

Techniques used

explain why an excess reagent quantity need not be measured in a gravimetric method

(iii)

The student thinks that Y is toxic. Other than wearing safety glasses and a lab coat, state one safety precaution that must be taken when carrying out this experiment.

1M Similar

DifficultyEasy

Worked solution

Answer

Carry out the experiment in a fume hood / fume cupboard.

Final answer

Use a fume hood

Detailed explanation

Background Concept

When handling toxic volatile liquids, the primary risk is inhalation of vapour. A fume hood (fume cupboard) extracts vapour away from the operator's breathing zone, preventing inhalation. Other controls (gloves, lab coat, safety glasses) protect against skin and eye contact but do not address the inhalation risk from volatile toxic substances.

Understanding the Question

The command word is "state" — a brief, specific answer is required. The question excludes safety glasses and lab coat (already mentioned), so you must give a different precaution. The substance is toxic and volatile (it vaporises at 100 °C), so the main hazard is inhalation.

Approach

Identify the hazard (inhalation of toxic vapour) and the appropriate control measure (fume hood to remove vapour from the breathing zone).

Step-by-Step Reasoning

Y is toxic and volatile — heating it produces toxic vapour that can escape through the pinhole.
The primary route of exposure is inhalation.
A fume hood provides local exhaust ventilation, drawing vapour away from the student.
This is the most appropriate and specific precaution beyond PPE.

Key Takeaways

For volatile toxic substances, the inhalation route is the dominant hazard.
A fume hood is the standard control for volatile toxic reagents.
"Wear gloves" or "wash hands" are valid but less specific to the volatility aspect; a fume hood directly addresses the vapour hazard.

Common Mistakes

Repeating "wear a lab coat" or "wear safety glasses" — these are excluded by the question.
Saying "wear gloves" — while sensible, it does not address the inhalation risk from a volatile toxic substance as directly as a fume hood.
Saying "do the experiment outside" — not a standard laboratory control.

Things to Be Careful About

The question asks for ONE precaution. Do not list several.
The answer must be specific to the toxicity and volatility of Y.

Techniques used

identify an appropriate laboratory safety control for a toxic volatile substance

(c)

Table 1.2 shows the student’s results.

Table 1.2


temperature of water-bath / °C	100.0
laboratory atmospheric pressure / kPa	99.0
mass of conical flask, aluminium foil and rubber band measured in step 2 / g	31.123
mass of conical flask, aluminium foil, rubber band and Y measured in step 7 / g	31.429
mass of Y in flask in step 7 / g	0.306
volume of conical flask used / $\text{cm}^3$	129.56

(i)

Calculate the percentage error in the mass measured in step 2.

Show your working.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

The balance reads to $0.001\text{ g}$ , so the uncertainty in a single reading is $\pm 0.0005\text{ g}$ .

\text{percentage error} = \frac{0.0005 \times 2}{31.123} \times 100 = 3.21 \times 10^{-3}\% \;(= 0.00321\%)

Answer

$3.21 \times 10^{-3}\%$

Final answer

3.21 × 10^-3 %

Detailed explanation

Background Concept

Percentage error (or percentage uncertainty) is calculated as $\frac{\text{absolute uncertainty}}{\text{measured value}} \times 100$ . For a digital balance reading to 0.001 g, the uncertainty in a single reading is taken as ±0.0005 g (half the smallest division). Some mark schemes express this as $0.0005 \times 2 = 0.001$ g to represent the full range of uncertainty (from −0.0005 to +0.0005, a span of 0.001 g). The percentage error quantifies how significant the instrument's precision is relative to the quantity being measured.

Understanding the Question

The mass measured in step 2 is 31.123 g (a single balance reading). The balance reads to 0.001 g. You must calculate the percentage error in this single measurement. The command word is "calculate" and you must "show your working".

Approach

State the uncertainty of the balance: ±0.0005 g per reading.
Apply the percentage error formula: (absolute uncertainty / measured value) × 100.
The mark scheme uses $0.0005 \times 2$ in the numerator, representing the total uncertainty range of 0.001 g.

Step-by-Step Reasoning

Balance precision: reads to 0.001 g → uncertainty = ±0.0005 g.
Total absolute uncertainty = $0.0005 \times 2 = 0.001$ g (the full ± range).
Percentage error = $\frac{0.001}{31.123} \times 100 = 0.003213\% = 3.21 \times 10^{-3}\%$ .
This very small percentage error confirms the balance is highly precise for this mass.

Key Takeaways

Percentage error = (absolute uncertainty / measured value) × 100.
For a balance reading to 0.001 g, uncertainty = ±0.0005 g (or expressed as a range of 0.001 g).
Small percentage errors indicate the instrument is appropriate for the quantity measured.

Common Mistakes

Using 0.001 g as the uncertainty without the ×2 convention, giving a different (but arguably valid) answer.
Dividing by the wrong mass (e.g. 0.306 g instead of 31.123 g).
Forgetting to multiply by 100 to get a percentage.
Reporting as a decimal (0.0000321) instead of a percentage.

Things to Be Careful About

The question asks for the error in the mass measured in step 2 (31.123 g), not the mass of Y (0.306 g).
The mark scheme's convention of $0.0005 \times 2$ should be followed.
Express the answer in scientific notation or as a percentage to 3 significant figures.

Techniques used

determine the absolute uncertainty of a balance readingcalculate percentage error from absolute uncertainty and measured value

(ii)

The student assumes that the vapour formed from liquid Y is an ideal gas. The ideal gas equation is shown.

pV = nRT

\begin{aligned} p &= \text{pressure of gas measured in Pa} \\ V &= \text{volume of gas in m}^3 \\ n &= \text{number of moles of gas} \\ R &= \text{molar gas constant} \\ T &= \text{temperature in K} \end{aligned}

Use the ideal gas equation and the results in Table 1.2 to calculate the amount, in mol, of Y in the conical flask in step 7.

Assume that the temperature of the vapour formed from liquid Y is $100\text{ }^{\circ}\text{C}$ .

2M Similar

DifficultyMedium

Worked solution

Working

Convert to SI units:

$p = 99.0\text{ kPa} = 99\,000\text{ Pa}$
$V = 129.56\text{ cm}^3 = 129.56 \times 10^{-6}\text{ m}^3$
$T = 100\text{ }^{\circ}\text{C} = 373\text{ K}$

n = \frac{pV}{RT} = \frac{99\,000 \times 129.56 \times 10^{-6}}{8.31 \times 373}

n = \frac{12.826}{3100.6} = 4.14 \times 10^{-3}\text{ mol}

Answer

$n = 4.14 \times 10^{-3}\text{ mol}$

Final answer

4.14 × 10^-3 mol

Detailed explanation

Background Concept

The ideal gas equation $pV = nRT$ relates the pressure, volume, temperature, and amount of an ideal gas. In the Dumas method, the vapour of the volatile liquid fills the flask at atmospheric pressure and the bath temperature, so the number of moles of vapour can be calculated directly. All quantities must be in SI units: pressure in Pa, volume in m³, temperature in K, and $R = 8.31$ J mol⁻¹ K⁻¹.

Understanding the Question

You must calculate the number of moles of vapour Y in the flask using the data in Table 1.2. The command word is "calculate" and you must use the ideal gas equation. The temperature is assumed to be 100 °C (373 K). The pressure is 99.0 kPa and the volume is 129.56 cm³.

Approach

Convert all quantities to SI units.
Rearrange $pV = nRT$ to $n = pV/(RT)$ .
Substitute and evaluate.

Step-by-Step Reasoning

Pressure: $99.0\text{ kPa} = 99\,000\text{ Pa}$ (multiply by 1000).
Volume: $129.56\text{ cm}^3 = 129.56 \times 10^{-6}\text{ m}^3$ (since $1\text{ cm}^3 = 10^{-6}\text{ m}^3$ ). Equivalently, $1.2956 \times 10^{-4}\text{ m}^3$ .
Temperature: $100\text{ }^{\circ}\text{C} = 100 + 273 = 373\text{ K}$ .
Rearrange: $n = \frac{pV}{RT}$ .
Substitute: $n = \frac{99\,000 \times 129.56 \times 10^{-6}}{8.31 \times 373}$ .
Numerator: $99\,000 \times 129.56 \times 10^{-6} = 12.826\text{ J}$ (or Pa·m³).
Denominator: $8.31 \times 373 = 3100.6\text{ J mol}^{-1}$ .
Result: $n = 12.826 / 3100.6 = 4.138 \times 10^{-3}\text{ mol}$ .

Key Takeaways

Always convert to SI units before substituting into $pV = nRT$ .
$1\text{ cm}^3 = 10^{-6}\text{ m}^3$ and $1\text{ kPa} = 1000\text{ Pa}$ .
Temperature must always be in Kelvin for gas law calculations.
The mark scheme awards M1 for correct unit conversions and M2 for correct rearrangement and evaluation.

Common Mistakes

Using 99.0 Pa instead of 99 000 Pa.
Using 129.56 m³ instead of $129.56 \times 10^{-6}$ m³.
Using 100 K instead of 373 K.
Forgetting to convert °C to K (using 100 instead of 373).
Using $R = 8.314$ vs $8.31$ — both are acceptable.

Things to Be Careful About

The volume conversion: $129.56\text{ cm}^3$ to m³ is $\times 10^{-6}$ , NOT $\times 10^{-3}$ .
The pressure is 99.0 kPa (not 101.3 kPa standard) — use the given value.
Report to 3 significant figures: $4.14 \times 10^{-3}$ mol.

Techniques used

convert units to SI (kPa to Pa, cm^3 to m^3, °C to K)rearrange the ideal gas equation to solve for nsubstitute values into pV = nRT

(iii)

Calculate the $M_r$ of Y.

1M Similar

DifficultyEasy

Worked solution

Working

M_r = \frac{\text{mass of Y}}{n} = \frac{0.306}{4.138 \times 10^{-3}} = 73.9 \;(\approx 74)

Answer

$M_r = 73.9$

Final answer

73.9

Detailed explanation

Background Concept

The relative molecular mass ( $M_r$ ) is numerically equal to the molar mass in g mol⁻¹. It is calculated as $M_r = m/n$ , where $m$ is the mass of the substance and $n$ is the number of moles. In the Dumas method, both $m$ (from the balance) and $n$ (from the ideal gas equation) are determined experimentally.

Understanding the Question

You must calculate $M_r$ of Y using the mass of Y from Table 1.2 (0.306 g) and the moles calculated in part (c)(ii). The command word is "calculate".

Approach

Simply divide the mass of Y by the number of moles: $M_r = m/n$ .

Step-by-Step Reasoning

Mass of Y in flask (step 7) = 0.306 g (given in Table 1.2).
Moles of Y (from c(ii)) = $4.138 \times 10^{-3}$ mol.
$M_r = 0.306 / (4.138 \times 10^{-3}) = 73.95$ .
To appropriate significant figures: $M_r \approx 73.9$ (or 74).

Note: ecf applies — if a different value of $n$ was obtained in (c)(ii), using that value here still earns the mark provided the division is correct.

Key Takeaways

$M_r$ is dimensionless (relative), but numerically equals the molar mass in g mol⁻¹.
The calculation is a simple division; the challenge is in obtaining $n$ correctly in (c)(ii).
The result (~74) is consistent with a small organic molecule (e.g. $\text{C}_4\text{H}_{10}$ has $M_r = 58$ ; $\text{C}_5\text{H}_{12}$ has $M_r = 72$ ; a close match).

Common Mistakes

Inverting the division: $n/m$ instead of $m/n$ .
Using the wrong mass (e.g. 31.429 g instead of 0.306 g).
Using moles from (c)(ii) incorrectly (e.g. $4.14 \times 10^{3}$ instead of $4.14 \times 10^{-3}$ ).

Things to Be Careful About

The mass of Y is 0.306 g (the difference between step 7 and step 2 masses), already given in the table.
ecf from (c)(ii) is allowed.
Report to 3 significant figures.

Techniques used

calculate molar mass from mass and moles

(d)

The actual temperature of the vapour formed from liquid Y is less than $100\text{ }^{\circ}\text{C}$ . Describe and explain the effect this has on the calculated value of the $M_r$ .

1M Similar

DifficultyMedium

Worked solution

Answer

The calculated $M_r$ would be greater than the true value. At a lower actual temperature, more moles of vapour occupy the flask (at the same pressure and volume), so the mass of vapour measured in step 7 is greater than it would be at $100\text{ }^{\circ}\text{C}$ . Since the student uses $T = 373\text{ K}$ (too high) in $n = pV/RT$ , the calculated $n$ is too small, and $M_r = m/n$ is too large.

Final answer

The calculated Mr would be greater than the true value because the mass of vapour measured is greater (more moles fit at lower T) and the calculated n is too small (using T = 373 K).

Detailed explanation

Background Concept

In the Dumas method, the flask is assumed to contain vapour at the bath temperature (100 °C). If the actual vapour temperature is lower, the physical situation changes: at the same pressure and volume, a lower temperature means more moles of gas are present (from $n = pV/RT$ ). More moles means more mass of vapour in the flask. The student, however, calculates $n$ using $T = 373$ K, which underestimates the true number of moles. The mass measured is the actual (larger) mass. Thus $M_r = m_{\text{actual}} / n_{\text{calculated}}$ is too large.

Understanding the Question

The command words are "describe and explain". You must state the effect on $M_r$ (greater) AND give the reasoning. The key insight is that the mass measured changes (it is larger than it would be at true 100 °C) because more vapour fits in the flask at a lower temperature.

Approach

Consider what happens physically if the vapour is at, say, 95 °C instead of 100 °C:

At lower T, more moles fit in the same V at the same p → more mass of vapour in the flask → step 7 mass is greater.
The student still uses T = 373 K → calculated n is smaller than true n.
$M_r = m/n$ : numerator too large, denominator too small → $M_r$ too large.

Step-by-Step Reasoning

If actual vapour temperature < 100 °C, then at the same $p$ and $V$ , the actual number of moles $n_{\text{actual}} = pV/(RT_{\text{actual}})$ is greater than $n_{\text{calc}} = pV/(R \times 373)$ .
More moles of vapour means more mass of Y in the flask when weighed at step 7. The mass measured (0.306 g) is therefore greater than the mass that would be present if the vapour were truly at 100 °C.
The student calculates $n$ using $T = 373$ K, obtaining a value that is too small (since the true T is lower, the true $n$ is larger).
$M_r = m/n$ : the measured $m$ is too large AND the calculated $n$ is too small → both effects make $M_r$ too large (greater than true value).

The mark scheme's key point: "mass of vapour measured is greater (all other measurements are the same)" — this is the primary explanation.

Key Takeaways

Systematic errors from wrong temperature assumptions affect both the measured mass AND the calculated moles.
At lower temperature, more gas molecules fit in a fixed volume at fixed pressure.
When evaluating errors, trace the effect through each quantity in the calculation chain.

Common Mistakes

Saying $M_r$ would be smaller (wrong direction).
Only stating the effect without explaining why the mass is greater.
Focusing only on the $n$ calculation error without mentioning that the mass itself is also affected.
Saying "the temperature is wrong so the answer is wrong" without specifying direction or mechanism.

Things to Be Careful About

The mark scheme requires BOTH the direction (greater) AND the reason (mass of vapour is greater). One without the other scores zero.
The reasoning is about the actual mass being larger because more moles fit at lower T, not about the calculation of $n$ alone.

Techniques used

identify the direction of systematic error from a temperature assumptionexplain how a lower actual vapour temperature increases the measured mass of vapour

(e)

The boiling point of methylbenzene is $111\text{ }^{\circ}\text{C}$ .

Suggest why the $M_r$ of methylbenzene cannot be determined using this method.

1M Similar

DifficultyEasy

Worked solution

Answer

The boiling water bath is at $100\text{ }^{\circ}\text{C}$ , which is below the boiling point of methylbenzene ( $111\text{ }^{\circ}\text{C}$ ), so methylbenzene would not vaporise (would not boil) in the water-bath.

Final answer

The temperature of the water bath (100 °C) is not high enough to boil methylbenzene (bp 111 °C)

Detailed explanation

Background Concept

The Dumas method requires the liquid to be fully vaporised in the flask. This is achieved by heating the flask in a boiling water bath, which provides a maximum temperature of 100 °C (at atmospheric pressure). For the method to work, the liquid must have a boiling point below 100 °C so that it vaporises completely when immersed in the boiling water. If the boiling point exceeds 100 °C, the liquid cannot fully vaporise at the available temperature.

Understanding the Question

The command word is "suggest". You must explain why methylbenzene (bp = 111 °C) cannot be analysed by this method. The water bath provides a maximum temperature of 100 °C. Compare this with 111 °C.

Approach

The Dumas method requires complete vaporisation. The maximum temperature available is 100 °C (boiling water). Methylbenzene boils at 111 °C. Since 100 < 111, the liquid will not boil/vaporise fully.

Step-by-Step Reasoning

The water bath is at 100 °C (the temperature of boiling water at ~99 kPa, approximately 100 °C).
Methylbenzene has a boiling point of 111 °C.
At 100 °C, methylbenzene is below its boiling point — it remains a liquid and does not fully vaporise.
Without complete vaporisation, the flask does not contain vapour of methylbenzene at 100 °C and atmospheric pressure, so the ideal gas calculation is invalid.
Therefore, this method cannot determine the $M_r$ of methylbenzene.

Key Takeaways

The Dumas method is limited to liquids with boiling points below the temperature of the heating bath.
A boiling water bath provides a maximum of ~100 °C; for higher-boiling liquids, an oil bath or other heating method at higher temperature would be needed.
Understanding the physical requirement (complete vaporisation) is essential for evaluating method limitations.

Common Mistakes

Saying "methylbenzene is not volatile" — it is volatile, just not volatile enough at 100 °C.
Saying "methylbenzene decomposes at 100 °C" — it does not; it simply does not boil.
Saying "the method requires the liquid to be a gas at room temperature" — it must be a gas at the bath temperature.

Things to Be Careful About

The key comparison is: bath temperature (100 °C) < boiling point (111 °C). State both values.
The answer must reference the temperature of the water bath being insufficient, not just "it doesn't work".

Techniques used

compare the boiling point of a substance with the maximum temperature available in the apparatus

Q2MediumAnalysis, Conclusions and EvaluationPlanning

A student carries out an experiment to determine the charge of an aqueous ion, $\text{M}^{n+}(\text{aq})$ , of metal M.

The student prepares $100.0\text{ cm}^3$ of $1.00\text{ mol dm}^{-3}$ aqueous copper(II) nitrate, $\text{Cu}(\text{NO}_3)_2(\text{aq})$ , to use in the experiment.

(a)

Calculate the mass of $\text{Cu}(\text{NO}_3)_2\cdot3\text{H}_2\text{O}(\text{s})$ required to prepare $100.0\text{ cm}^3$ of $1.00\text{ mol dm}^{-3}$ $\text{Cu}(\text{NO}_3)_2(\text{aq})$ .

1M Similar

DifficultyEasy

Worked solution

Working

n = c \times V = 1.00 \times 0.100 = 0.100 \text{ mol}

M_r(\text{Cu(NO}_3)_2 \cdot 3\text{H}_2\text{O}) = 63.5 + 2(14.0 + 48.0) + 3(18.0) = 241.5

\text{mass} = n \times M_r = 0.100 \times 241.5 = 24.15 \text{ g}

Answer

24.15 g

Final answer

24.15 g

Detailed explanation

Background Concept

When preparing a standard solution from a hydrated salt, the water of crystallisation must be included in the relative formula mass ( $M_r$ ). The number of moles of the hydrated salt equals the number of moles of the anhydrous salt dissolved, because each formula unit of the hydrate releases one formula unit of the salt into solution.

Understanding the Question

We need the mass of $\text{Cu(NO}_3)_2 \cdot 3\text{H}_2\text{O}$ that, when dissolved and made up to $100.0 \text{ cm}^3$ , gives a solution of $1.00 \text{ mol dm}^{-3}$ in $\text{Cu(NO}_3)_2$ . The command word is "calculate".

Approach

Find moles needed from $n = cV$ .
Compute $M_r$ of the hydrated salt (include the three waters).
Multiply moles by $M_r$ to get mass.

Step-by-Step Reasoning

Volume in dm³: $100.0 \text{ cm}^3 = 0.1000 \text{ dm}^3$ .
Moles of $\text{Cu(NO}_3)_2$ : $1.00 \times 0.1000 = 0.100 \text{ mol}$ .
$M_r$ : Cu = 63.5; $2 \times$ N = 28.0; $6 \times$ O = 96.0; $3 \times$ H₂O = $3 \times 18.0 = 54.0$ . Total = 241.5.
Mass = $0.100 \times 241.5 = 24.15 \text{ g}$ .

Key Takeaways

Always use the $M_r$ of the hydrated form when weighing out a hydrate to prepare a solution of known concentration of the anhydrous salt.

Common Mistakes

Using $M_r = 187.5$ (anhydrous) instead of 241.5 (tri-hydrate), giving 18.75 g.
Forgetting to convert cm³ to dm³.

Things to Be Careful About

The question specifies the trihydrate $\cdot 3\text{H}_2\text{O}$ ; do not omit the water of crystallisation.
Report to 4 significant figures (or 2 decimal places in grams) as the mark scheme expects 24.15 g.

Techniques used

calculate moles from concentration and volumedetermine relative formula mass of hydrated saltconvert moles to mass

(b)

Describe the steps the student should take to prepare $100.0\text{ cm}^3$ of $1.00\text{ mol dm}^{-3}$ $\text{Cu}(\text{NO}_3)_2(\text{aq})$ starting from the mass calculated in (a) supplied in a small beaker.

Give the name and capacity of any apparatus used.

Write your answer using a series of numbered steps.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

Add a small volume of distilled water to the small beaker and dissolve the solid $\text{Cu(NO}_3)_2 \cdot 3\text{H}_2\text{O}$ completely, stirring with a glass rod.
Transfer the solution quantitatively into a $100 \text{ cm}^3$ volumetric flask, rinsing the beaker and glass rod with distilled water and adding the washings to the flask.
Add distilled water up to the calibration mark on the neck of the volumetric flask, then stopper and invert the flask several times to mix thoroughly.

Final answer

See numbered steps: dissolve in small volume of distilled water; transfer to 100 cm³ volumetric flask with washings; make up to calibration mark and invert to mix.

Detailed explanation

Background Concept

A standard solution of precise concentration is prepared using a volumetric flask, which is calibrated to contain an exact volume at a specific temperature. The key principle is quantitative transfer — every drop of solute must reach the flask, hence the use of washings.

Understanding the Question

The command word is "describe" and the question asks for numbered steps starting from the weighed solid. Three marks are available (M1, M2, M3), each corresponding to a critical stage: dissolving, transferring with washings, and making up to the mark with mixing.

Approach

Structure the answer chronologically: dissolve → transfer → make up → mix. Name the apparatus (volumetric flask with capacity) and mention washings explicitly.

Step-by-Step Reasoning

M1: The solid must first be dissolved in a small volume of distilled water in the beaker. Using only a small volume ensures the final volume does not exceed the flask's capacity before topping up.
M2: The solution is transferred to a 100 cm³ volumetric flask. The beaker and glass rod must be rinsed (washed) with distilled water and the washings added to the flask to ensure all solute is transferred quantitatively.
M3: Distilled water is added carefully until the bottom of the meniscus reaches the calibration mark. The flask is then stoppered and inverted repeatedly to ensure uniform concentration.

Key Takeaways

Always specify the type and capacity of the volumetric apparatus.
"With washings" is a mark-bearing phrase — it demonstrates understanding of quantitative transfer.
Inverting the stoppered flask is the correct mixing technique (not stirring with a rod inside the flask).

Common Mistakes

Saying "add water to 100 cm³" without mentioning the calibration mark or volumetric flask.
Omitting the washings step.
Suggesting stirring in the volumetric flask rather than inverting.

Things to Be Careful About

The question requires the name AND capacity of apparatus — "volumetric flask" alone without "100 cm³" loses the mark.
Do not say "measuring cylinder" — a volumetric flask is required for accurate standard solution preparation.

Techniques used

describe a standard solution preparation procedurename appropriate volumetric apparatus

(c)

The student sets up the electrochemical cell shown in Fig. 2.1 to investigate the effect of changing the concentration of $\text{M}^{n+}(\text{aq})$ on the measured cell potential, $E_{\text{cell}}$ .

Fig. 2.1

Suggest the function of the item labelled A in Fig. 2.1.

1M Similar

(d)

The student uses the apparatus in Fig. 2.1 to measure the cell potentials, using six solutions each with a different concentration of $\text{M}^{n+}(\text{aq})$ .

Table 2.1 shows the results obtained by the student.

Table 2.1

concentration of $\text{M}^{n+}(\text{aq})$ / $\text{mol dm}^{-3}$	cell potential, $E_{\text{cell}}$ / V	$\log [\text{M}^{n+}]$
$1.00 \times 10^{-2}$	3.295	$-2.00$
$5.00 \times 10^{-3}$	3.304	$-2.30$
$1.00 \times 10^{-3}$	3.325	$-3.00$
$5.00 \times 10^{-4}$	3.343	$-3.30$
$1.00 \times 10^{-4}$	3.354	$-4.00$
$1.00 \times 10^{-5}$	3.384	$-5.00$

\text{electrode potential of } \text{M}^{n+}(\text{aq})/\text{M}(\text{s}) = E_{\text{Cu}} - E_{\text{cell}}

\text{electrode potential of } \text{Cu}^{2+}(\text{aq})/\text{Cu}(\text{s}), E_{\text{Cu}} = 0.337\text{ V}

Complete Table 2.1. Record your values to three decimal places.

1M Similar

DifficultyEasy

Worked solution

Working

E_{\text{M}^{n+}/\text{M}} = E_{\text{Cu}} - E_{\text{cell}} = 0.337 - E_{\text{cell}}

$\log[\text{M}^{n+}]$	$E_{\text{cell}}$ / V	Electrode potential / V
$-2.00$	3.295	$-2.958$
$-2.30$	3.304	$-2.967$
$-3.00$	3.325	$-2.988$
$-3.30$	3.343	$-3.006$
$-4.00$	3.354	$-3.017$
$-5.00$	3.384	$-3.047$

Answer

See completed table above.

Final answer

Electrode potentials: -2.958, -2.967, -2.988, -3.006, -3.017, -3.047 V

Detailed explanation

Background Concept

In an electrochemical cell, $E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}$ . Here, copper is the cathode (higher $E^\ominus$ ) and M is the anode. So $E_{\text{cell}} = E_{\text{Cu}} - E_{\text{M}}$ , which rearranges to $E_{\text{M}} = E_{\text{Cu}} - E_{\text{cell}}$ .

Understanding the Question

The formula is given: electrode potential of $\text{M}^{n+}/\text{M} = E_{\text{Cu}} - E_{\text{cell}}$ , with $E_{\text{Cu}} = 0.337 \text{ V}$ . We must compute this for each row and record to three decimal places.

Approach

For each row, subtract $E_{\text{cell}}$ from 0.337.

Step-by-Step Reasoning

$0.337 - 3.295 = -2.958$
$0.337 - 3.304 = -2.967$
$0.337 - 3.325 = -2.988$
$0.337 - 3.343 = -3.006$
$0.337 - 3.354 = -3.017$
$0.337 - 3.384 = -3.047$

All values are negative, consistent with M being a very reactive metal (strong reducing agent, very negative $E^\ominus$ ).

Key Takeaways

When a cell potential is larger than the reference electrode potential, the unknown electrode potential will be negative. The subtraction order matters: $E_{\text{unknown}} = E_{\text{reference}} - E_{\text{cell}}$ .

Common Mistakes

Reversing the subtraction: $E_{\text{cell}} - E_{\text{Cu}}$ gives positive values (wrong sign).
Recording to only two decimal places.

Things to Be Careful About

The question specifies three decimal places. Each answer must be recorded to exactly three decimal places (e.g., -2.958, not -2.96).

Techniques used

subtract cell potential from copper electrode potentialrecord values to three decimal places

(e)

Identify the independent variable in this experiment.

1M Similar

(f)

(i)

Plot a graph on the grid in Fig. 2.2 to show the relationship between the electrode potential of $\text{M}^{n+}(\text{aq})/\text{M}(\text{s})$ and $\log [\text{M}^{n+}]$ . Use a cross (×) to plot each data point. Draw a straight line of best fit.

Fig. 2.2

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Plot all six points as crosses (×) on the grid in Fig. 2.2 using the completed values from part (d):

$\log[\text{M}^{n+}]$	$E_{\text{M}^{n+}/\text{M}}$ / V
$-2.00$	$-2.958$
$-2.30$	$-2.967$
$-3.00$	$-2.988$
$-3.30$	$-3.006$
$-4.00$	$-3.017$
$-5.00$	$-3.047$

Draw a single straight line of best fit through the points (the point at $\log = -3.30$ lies slightly above the line and should be excluded from the fit).

Final answer

Six points plotted as crosses with a straight line of best fit showing a positive gradient of approximately 0.030 V per unit log

Detailed explanation

Background Concept

When plotting experimental data, each point should be marked with a clear cross (×) at the correct coordinates. A line of best fit for linear data should pass as close as possible to all points, with roughly equal scatter above and below. Anomalous points that deviate significantly may be excluded from the line.

Understanding the Question

The grid in Fig. 2.2 has axes: $\log[\text{M}^{n+}]$ on the x-axis (range -5 to -2) and electrode potential on the y-axis (range -3.06 to -2.94). Two marks are available: M1 for correct plotting, M2 for a straight line of best fit.

Approach

Read off each (x, y) pair from the completed table.
Plot each as a cross on the grid.
Draw a straight line that best represents the trend, noting that the point at (-3.30, -3.006) is slightly above the line.

Step-by-Step Reasoning

Point 1: (-2.00, -2.958) — near the top right of the grid.
Point 2: (-2.30, -2.967) — slightly left and below point 1.
Point 3: (-3.00, -2.988) — middle of the grid.
Point 4: (-3.30, -3.006) — this point is noticeably above where a straight line through the others would predict (expected ≈ -2.997), making it the anomalous point.
Point 5: (-4.00, -3.017) — lower left region.
Point 6: (-5.00, -3.047) — bottom left corner.

The line of best fit should pass close to points 1, 2, 3, 5, and 6, with point 4 lying above it.

Key Takeaways

Use crosses, not dots or circles, for data points.
The line of best fit is a straight line for linear relationships; do not join points.
Anomalous points are identified by their deviation from the expected trend.

Common Mistakes

Plotting points inaccurately (e.g., misreading the small grid squares).
Drawing a curve through the points instead of a straight line.
Including the anomalous point in the line of best fit, distorting the gradient.

Things to Be Careful About

The y-axis has small increments (0.02 V per major division, 0.002 V per small square). Precision in plotting matters for M1.
The line must be straight (ruler), not a freehand curve.

Techniques used

plot data points on a graph griddraw a straight line of best fit

(ii)

Circle the one point on the graph that you consider to be most anomalous.

Suggest one reason why this anomaly may have occurred during the experimental procedure. Assume no error was made in the measurement of the cell potential.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The point at $\log[\text{M}^{n+}] = -3.30$ (concentration $5.00 \times 10^{-4} \text{ mol dm}^{-3}$ ) is circled as most anomalous — it lies above the line of best fit.

Reason: the actual concentration of $\text{M}^{n+}$ in this solution was lower than the recorded value of $5.00 \times 10^{-4} \text{ mol dm}^{-3}$ (e.g., due to incomplete dilution or pipette error), resulting in a higher measured electrode potential than expected.

Final answer

Point at log[Mⁿ⁺] = -3.30 is anomalous; the actual concentration was lower than the recorded 5.00 × 10⁻⁴ mol dm⁻³

Detailed explanation

Background Concept

An anomalous point deviates from the expected linear trend. In this experiment, a lower-than-recorded concentration of $\text{M}^{n+}$ would give a more negative electrode potential (by the Nernst equation), which corresponds to a larger $E_{\text{cell}}$ and hence a less negative calculated electrode potential when using the recorded (higher) concentration value — placing the point above the line.

Understanding the Question

We must identify which point is furthest from the line of best fit and suggest a procedural reason (not a measurement error in $E_{\text{cell}}$ ) for the anomaly.

Approach

Check the residuals from the line. The point at (-3.30, -3.006) sits about 0.009 V above where the line predicts (~-2.997), making it the most anomalous.

Step-by-Step Reasoning

If the true concentration were lower than $5.00 \times 10^{-4}$ , the true $\log[\text{M}^{n+}]$ would be more negative (further left on the x-axis), but the student plotted it at -3.30. The measured $E_{\text{cell}}$ reflects the true (lower) concentration, giving a larger $E_{\text{cell}}$ and thus a less negative electrode potential — the point appears higher (less negative y) than expected at x = -3.30.
Possible cause: the dilution to make $5.00 \times 10^{-4} \text{ mol dm}^{-3}$ was not performed accurately (e.g., the pipette delivered slightly less than the intended volume, or the volumetric flask was over-filled).

Key Takeaways

Anomalous points in concentration-dependent measurements often arise from dilution or pipetting errors rather than voltmeter reading errors.

Common Mistakes

Choosing the wrong point as anomalous.
Suggesting "the voltmeter reading was wrong" — the question explicitly excludes measurement error in $E_{\text{cell}}$ .
Giving a vague answer like "human error" without specifying what went wrong.

Things to Be Careful About

The question says "assume no error was made in the measurement of the cell potential" — the reason must relate to the concentration preparation, not the voltmeter.

Techniques used

identify an anomalous data point from a graphsuggest a procedural cause for an anomalous result

(iii)

Suggest how the reliability of the data shown in Table 2.1 could be improved.

1M Similar

(g)

(i)

Use Fig. 2.2 to determine the gradient of the line of best fit.

State the coordinates of both points you used in your calculation. These must be selected from your line of best fit. Give the gradient to three significant figures.

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Points on the line of best fit (read from the graph):

(x_1, y_1) = (-5.00, -3.047), \quad (x_2, y_2) = (-2.00, -2.958)

\text{gradient} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2.958 - (-3.047)}{-2.00 - (-5.00)} = \frac{0.089}{3.00} = 0.0297

Answer

Gradient = $0.0297$ V (to three significant figures)

Final answer

0.0297 V

Detailed explanation

Background Concept

The gradient of a line of best fit is calculated by selecting two well-separated points on the line (not data points) and computing $\Delta y / \Delta x$ . Using points far apart minimises the effect of reading errors.

Understanding the Question

We must use the line drawn in (f)(i), state two coordinates from that line in $(x, y)$ form, and calculate the gradient to three significant figures. Two marks: M1 for valid coordinates, M2 for correct gradient.

Approach

Choose the two most widely separated points on the line that are easy to read from the grid.

Step-by-Step Reasoning

The line passes through approximately (-5.00, -3.047) at the left end and (-2.00, -2.958) at the right end.
$\Delta y = -2.958 - (-3.047) = +0.089$
$\Delta x = -2.00 - (-5.00) = +3.00$
Gradient = $0.089 / 3.00 = 0.0297$ (3 s.f.)

Note: the gradient is positive because as $\log[\text{M}^{n+}]$ increases (becomes less negative), the electrode potential also increases (becomes less negative).

Key Takeaways

Always use points on the line, not data points, for gradient calculations.
Choose widely separated points for accuracy.
Express coordinates in $(x, y)$ form as required.

Common Mistakes

Using data points instead of points on the line of best fit.
Choosing points too close together, amplifying reading error.
Reporting the gradient to fewer than three significant figures.
Getting the sign wrong (the gradient is positive).

Things to Be Careful About

The mark scheme requires coordinates in the form $(x, y)$ — writing them as separate x and y values without parentheses may not score M1.

Techniques used

select two points on a line of best fitcalculate gradient from two coordinates

(ii)

For the electrode equilibrium,

\text{M}(\text{s}) \rightleftharpoons \text{M}^{n+}(\text{aq}) + n\text{e}^-

the Nernst equation can be written as shown.

E = E^{\ominus} + \frac{2.303 RT}{nF} \log [\text{M}^{n+}]

\begin{aligned} E &= \text{electrode potential} \\ E^{\ominus} &= \text{standard electrode potential} \\ R &= \text{molar gas constant} \\ T &= \text{temperature in K} \\ F &= \text{Faraday constant} \end{aligned}

The equation for a straight line is $y = mx + c$ .

State which parts in the Nernst equation correspond to $y$ , $m$ and $c$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

Comparing $E = E^{\ominus} + \dfrac{2.303RT}{nF}\log[\text{M}^{n+}]$ with $y = mx + c$ :

$y = E$
$m = \dfrac{2.303RT}{nF}$
$c = E^{\ominus}$

Final answer

y = E, m = 2.303RT/nF, c = E

Detailed explanation

Background Concept

Many physical equations can be rearranged into the form $y = mx + c$ to extract information graphically. The gradient gives a physically meaningful quantity (here, related to the number of electrons transferred), and the intercept gives the standard value (here, the standard electrode potential).

Understanding the Question

The Nernst equation is already in the correct form: $E$ (the y-axis variable) equals a constant ( $E^\ominus$ , the intercept) plus a coefficient ( $2.303RT/nF$ , the gradient) multiplied by $\log[\text{M}^{n+}]$ (the x-axis variable).

Approach

Match each term in the Nernst equation to its counterpart in $y = mx + c$ by identifying which quantity is plotted on each axis.

Step-by-Step Reasoning

The y-axis is electrode potential $E$ , so $y = E$ .
The x-axis is $\log[\text{M}^{n+}]$ , so $x = \log[\text{M}^{n+}]$ .
The coefficient multiplying $x$ is $2.303RT/nF$ , so $m = 2.303RT/nF$ .
The constant term (value of $E$ when $\log[\text{M}^{n+}] = 0$ , i.e., at 1 mol dm⁻³) is $E^\ominus$ , so $c = E^\ominus$ .

Key Takeaways

Recognising the linear form of an equation allows graphical determination of physical constants (gradient → $n$ ; intercept → $E^\ominus$ ).

Common Mistakes

Writing $m = 2.303RT$ (omitting $nF$ in the denominator).
Confusing $y$ with $\log[\text{M}^{n+}]$ or $x$ with $E$ .

Things to Be Careful About

The question asks specifically for $y$ , $m$ , and $c$ — three separate marks. Each must be clearly identified.

Techniques used

compare a physical equation to the straight-line equation y = mx + cidentify corresponding terms

(iii)

Use the value that you calculated for the gradient of your line of best fit in (g)(i) and the Nernst equation to calculate the value of $n$ in $\text{M}^{n+}(\text{aq})$ .

The experiment is carried out at $25.0\text{ }^{\circ}\text{C}$ .

(If you were unable to determine an answer to (g)(i), then use the value 0.0285 for the gradient. This is not the correct value.)

2M Similar

DifficultyMedium

Worked solution

Working

From (g)(ii): $\text{gradient} = \dfrac{2.303RT}{nF}$

Rearranging for $n$ :

n = \frac{2.303RT}{\text{gradient} \times F}

Substituting $R = 8.31 \text{ J mol}^{-1} \text{K}^{-1}$ , $T = 298 \text{ K}$ , $F = 96500 \text{ C mol}^{-1}$ , gradient $= 0.0297$ :

n = \frac{2.303 \times 8.31 \times 298}{0.0297 \times 96500} = \frac{5707}{2866} = 1.99

Answer

$n = 2$

Therefore the ion is $\text{M}^{2+}(\text{aq})$ .

Final answer

n = 2

Detailed explanation

Background Concept

The Nernst equation relates the electrode potential to the concentration of the electroactive species. The gradient of a plot of $E$ versus $\log[\text{M}^{n+}]$ is $2.303RT/nF$ , which depends inversely on $n$ — the number of electrons transferred. Measuring the gradient therefore allows determination of the charge on the ion.

Understanding the Question

We use the gradient from (g)(i) and the expression for $m$ from (g)(ii) to calculate $n$ . The experiment is at 25.0°C = 298 K. The answer must be rounded to a whole number (since $n$ represents a count of electrons).

Approach

Set the gradient equal to $2.303RT/nF$ .
Rearrange to isolate $n$ .
Substitute known values.
Round to the nearest whole number.

Step-by-Step Reasoning

From (g)(ii): gradient $= 2.303RT/nF$
Rearranging: $n = 2.303RT/(\text{gradient} \times F)$
Numerator: $2.303 \times 8.31 \times 298 = 5707$ (to 4 s.f.)
Denominator: $0.0297 \times 96500 = 2866$ (to 4 s.f.)
$n = 5707/2866 = 1.99$
Rounded to a whole number: $n = 2$

If the fallback gradient of 0.0285 were used: $n = 5707/(0.0285 \times 96500) = 5707/2750 = 2.08$ , which still rounds to 2.

Key Takeaways

The Nernst equation's gradient provides a direct experimental route to determine the number of electrons in a half-reaction — a key application of electrochemistry in analytical chemistry.

Common Mistakes

Forgetting to convert 25.0°C to 298 K (using 25 instead of 298).
Using $F = 96500$ incorrectly (e.g., using 9650).
Not rounding to a whole number (writing $n = 1.99$ or $n = 2.0$ ).
Algebraic errors in rearranging (dividing by $n$ instead of multiplying).

Things to Be Careful About

The mark scheme requires the rearrangement AND correct substitution (M1), then the correctly calculated value rounded to a whole number (M2).
Use $T = 298 \text{ K}$ (from 25.0°C), not 25.
The final answer must be a whole number: $n = 2$ , not 1.99 or 2.0.

Techniques used

rearrange the Nernst equation to solve for nsubstitute values for R, T, F and the experimental gradientround to a whole number