Chemistry 9701/35 — May/June 2019

Cambridge AS Level · Advanced Practical Skills · worked solutions for every part, with the mark scheme

questions

marks

120

minutes

Topics Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation · Qualitative Analysis

Q1Manipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and EvaluationFree sample

The reaction between acids and alkalis is exothermic. You will find the concentration of a monoprotic acid, HZ, by a thermometric method using a solution of sodium hydroxide of known concentration.

\text{HZ(aq)} + \text{NaOH(aq)} \rightarrow \text{NaZ(aq)} + \text{H}_2\text{O(l)}

FA 1 is a solution of acid HZ.
FA 2 is $2.00 \text{ mol dm}^{-3}$ sodium hydroxide, NaOH.

(a)

Method

Place the thermometer into FA 1. Record the temperature of FA 1 in the table. This is the temperature when the volume of FA 2 is 0.0.
Rinse and dry the thermometer.
Place the thermometer into FA 2. Record the temperature of FA 2 in the table. This is the temperature when the volume of FA 1 is 0.0.
Fill a burette with FA 1.
Support the plastic cup in the $250\text{ cm}^3$ beaker.
From the burette transfer $35.0\text{ cm}^3$ of FA 1 into the plastic cup.
Use the $50\text{ cm}^3$ measuring cylinder to measure $5.0\text{ cm}^3$ of FA 2.
Transfer the $5.0\text{ cm}^3$ of FA 2 into the plastic cup. Stir the mixture and record the highest temperature.
Tip out the solution, rinse the plastic cup with water, shake it to remove excess water and replace the cup in the beaker.
Rinse and dry the thermometer.
Use the burette to transfer $30.0\text{ cm}^3$ of FA 1 into the plastic cup.
Use the measuring cylinder to transfer $10.0\text{ cm}^3$ of FA 2 into the plastic cup.
Stir the mixture and record the highest temperature.
Tip out the solution, rinse the plastic cup with water, shake it to remove excess water and replace the cup in the beaker.
Rinse and dry the thermometer.
Continue the experiment using the volumes of FA 1 and FA 2 given in the table and record the maximum temperature of each mixture.

volume FA 1 / $\text{cm}^3$	40.0	35.0	30.0	25.0	20.0	15.0	10.0	5.0	0.0
volume FA 2 / $\text{cm}^3$	0.0	5.0	10.0	15.0	20.0	25.0	30.0	35.0	40.0
temperature / $^\circ\text{C}$

DifficultyMedium-Easy

Worked solution

Answer

All thermometer readings recorded to $.0$ or $.5\text{ }^\circ\text{C}$ (e.g. $22.0$ , $22.5$ , $23.0$ etc.).

The temperature rise for each mixture is calculated as:

\Delta T = T_{\text{max}} - \frac{T_{\text{acid}} + T_{\text{alkali}}}{2}

The candidate's $\Delta T$ values should be within the tolerance of the supervisor's maximum $\Delta T$ (tolerance depends on the magnitude of the supervisor's $\Delta T$ ; e.g. $\delta = 1.5\text{ }^\circ\text{C}$ if supervisor's $\Delta T$ is between $8.5$ and $16.0\text{ }^\circ\text{C}$ ).

Final answer

Readings recorded to 0.5°C precision; ΔT within tolerance of supervisor's value

Detailed explanation

Background Concept

In a thermometric titration, the reaction between an acid and an alkali is exothermic, so the temperature of the mixture rises as the reaction proceeds. The maximum temperature reached for each mixture is recorded. The initial temperature is taken as the average of the separate temperatures of the acid and alkali before mixing, since both solutions are at approximately room temperature but may differ slightly.

The temperature change ( $\Delta T$ ) for each mixture is calculated as:

\Delta T = T_{\text{max}} - \frac{T_{\text{acid}} + T_{\text{alkali}}}{2}

Understanding the Question

This part requires the candidate to carry out the experiment and record all temperature readings in the table. The mark scheme awards three independent marks (B1, B1, B1) for: (I) recording readings to the correct precision (to $.0$ or $.5\text{ }^\circ\text{C}$ ), (II) the candidate's $\Delta T$ being within a specified tolerance of the supervisor's $\Delta T$ , and (III) a similar tolerance check (likely for a different volume or the maximum $\Delta T$ ).

The command word is implicit: "record the temperature" — this is about technique and data quality.

Approach

Ensure all thermometer readings are recorded to the nearest $0.5\text{ }^\circ\text{C}$ (i.e. ending in $.0$ or $.5$ ).
Calculate $\Delta T$ for the mixture that gives the maximum temperature rise using the formula above.
Compare with the supervisor's value — the tolerance depends on the magnitude of the supervisor's $\Delta T$ .

Step-by-Step Reasoning

Mark I (Recording precision): Thermometers used in this experiment are typically graduated in $0.5\text{ }^\circ\text{C}$ divisions. All readings must be recorded to the nearest $0.5\text{ }^\circ\text{C}$ , meaning they end in $.0$ or $.5$ . For example, $22.0$ , $22.5$ , $23.5$ are acceptable; $22.3$ or $22.7$ are not.
Mark II (ΔT within tolerance): The examiner calculates the supervisor's maximum $\Delta T$ from the table. The candidate's $\Delta T$ at the same volume is compared. The tolerance $\delta$ depends on the supervisor's $\Delta T$ :
- If supervisor's $\Delta T > 16.0\text{ }^\circ\text{C}$ : 1 mark for within $2.0\text{ }^\circ\text{C}$ , 2 marks for within $1.0\text{ }^\circ\text{C}$
- If supervisor's $\Delta T$ is $8.5\text{--}16.0\text{ }^\circ\text{C}$ : 1 mark for within $1.5\text{ }^\circ\text{C}$ , 2 marks for within $1.0\text{ }^\circ\text{C}$
- If supervisor's $\Delta T$ is $4.5\text{--}8.0\text{ }^\circ\text{C}$ : 1 mark for within $1.0\text{ }^\circ\text{C}$ , 2 marks for within $0.5\text{ }^\circ\text{C}$
- If supervisor's $\Delta T \leq 4.0\text{ }^\circ\text{C}$ : 1 mark for within $0.5\text{ }^\circ\text{C}$
Mark III: Similar tolerance check, likely for a second data point or the overall quality.

Key Takeaways

Always record thermometer readings to the precision of the instrument (here, $0.5\text{ }^\circ\text{C}$ ).
The initial temperature is the average of the two separate solutions' temperatures.
$\Delta T$ is calculated relative to this average initial temperature.

Common Mistakes

Recording readings to $0.1\text{ }^\circ\text{C}$ when the thermometer only reads to $0.5\text{ }^\circ\text{C}$ .
Using only the acid temperature (or only the alkali temperature) as the initial temperature instead of the average.
Forgetting to subtract the initial temperature from $T_{\text{max}}$ .

Things to Be Careful About

Ensure all readings in the table end in $.0$ or $.5$ .
The tolerance is generous for large $\Delta T$ values but tight for small ones.
Stirring must be thorough to reach the true maximum temperature.

Techniques used

record temperature readings to correct precisioncalculate maximum temperature risecompare with supervisor's value

(b)

(i)

Plot a graph of temperature of solution ( $y$ -axis) against volume of FA 2 added ( $x$ -axis) on the grid. Select a scale on the $y$ -axis to include a temperature of $2\text{ }^\circ\text{C}$ above your maximum thermometer reading. Label any points you consider anomalous.

Draw two lines of best fit through the points on your graph, the first for the increase in temperature and the second for the decrease in temperature of the mixtures. Extrapolate the two lines so they intersect.

Fig. 1.1

DifficultyMedium

Worked solution

Answer

Choose linear scales so the graph occupies more than half the available length on both axes. The $x$ -axis covers $0$ to $40\text{ cm}^3$ of FA 2, and the $y$ -axis covers from the lowest temperature to $2\text{ }^\circ\text{C}$ above the maximum reading.
Label both axes with quantity and unit (e.g. "temperature / °C" and "volume of FA 2 / cm³").
Plot all recorded points accurately (minimum 7 points).
Draw two lines of best fit: one through the points showing increasing temperature, and one through the points showing decreasing temperature.
Extrapolate both lines so they intersect.
Label any anomalous points.

Final answer

Graph with two extrapolated lines of best fit intersecting at the maximum temperature point

Detailed explanation

Background Concept

In a thermometric titration, as the volume of alkali added increases, the temperature of the mixture rises because the neutralisation reaction is exothermic. Once the equivalence point is reached, no more heat is produced, and the temperature begins to fall because the excess alkali (at room temperature) cools the mixture. This produces a characteristic "inverted V" or peak-shaped graph.

By drawing two lines of best fit — one for the rising portion and one for the falling portion — and extrapolating them to their intersection, we can determine the volume at which the maximum temperature would have occurred (the equivalence point). This corrects for the fact that the actual maximum may not have been recorded exactly at the equivalence volume due to the discrete volume increments used.

Understanding the Question

The candidate must plot temperature against volume of FA 2 added, using the data from part (a). The graph must have:

Appropriate linear scales occupying more than half the grid on both axes
The $y$ -axis extended $2\text{ }^\circ\text{C}$ above the maximum reading
All points plotted accurately
Two separate lines of best fit (not a single smooth curve)
Extrapolation to find the intersection

The command word is "plot" and "draw" — this is a practical graphing task.

Approach

Examine the data range and choose scales that make good use of the grid.
Plot all points to within half a small square.
Identify the rising and falling portions of the graph.
Draw a straight line of best fit through the rising points and another through the falling points.
Extend both lines until they intersect.
Mark any anomalous points (points that don't fit either line) with a circle or annotation.

Step-by-Step Reasoning

Mark I (Scales and axes): The $x$ -axis must span $0$ to $40\text{ cm}^3$ and occupy at least 5 large squares. The $y$ -axis must include the full temperature range plus $2\text{ }^\circ\text{C}$ above the maximum and occupy at least 6 large squares. Both axes must be labelled with the quantity name and/or unit.
Mark II (Plotting): At least 7 points must be plotted accurately. If a point should lie on a line, it must be on the line; if it should not, it must be within half a small square of its true position.
Mark III (Lines of best fit): Two separate lines must be drawn — one for the increasing temperature region and one for the decreasing temperature region. These can be straight or smoothly curved. A single continuous curve through all points does NOT earn this mark. Anomalous points (if labelled) are ignored when drawing the lines.

Key Takeaways

The thermometric titration graph has a characteristic peak shape.
Two lines of best fit (not one curve) are required to find the equivalence point by extrapolation.
Anomalous points should be identified and labelled but excluded from the lines.
A single smooth curve through all points cannot earn the mark for two lines, nor can the intersection be read off.

Common Mistakes

Drawing a single smooth curve through all points instead of two separate lines.
Not extrapolating the lines to find the intersection.
Choosing scales that make the graph too small or that don't include the extra $2\text{ }^\circ\text{C}$ on the $y$ -axis.
Forgetting to label axes with units.
Not identifying anomalous points.

Things to Be Careful About

The mark scheme explicitly states that a continuous curve cannot score either the "two lines" mark (b(i)III) or the intersection mark (b(ii)).
Points marked as anomalous are ignored when drawing the lines of best fit.
The $y$ -axis scale must include $2\text{ }^\circ\text{C}$ above the maximum thermometer reading.

Techniques used

plot points on a graph with correct scalesdraw two lines of best fitextrapolate lines to find intersection

(ii)

The intersection on your graph occurs at the volume of FA 2 that reacted to form a neutral solution.

Determine the volumes of FA 1 and FA 2 required to form a neutral solution.

.............................. $\text{cm}^3$ of FA 1 neutralises .............................. $\text{cm}^3$ of FA 2.

DifficultyMedium-Easy

Worked solution

Answer

Read the volume of FA 2 at the intersection of the two extrapolated lines (to within $0.25\text{ cm}^3$ of the examiner's value).

Volume of FA 1 = $40.0 -$ volume of FA 2.

Both volumes must be given to 1 decimal place.

For example, if the intersection occurs at $20.0\text{ cm}^3$ of FA 2:

$20.0\text{ cm}^3$ of FA 1 neutralises $20.0\text{ cm}^3$ of FA 2.

Final answer

Volumes read from intersection to 1 dp; FA 1 = 40.0 − FA 2

Detailed explanation

Background Concept

The intersection of the two extrapolated lines on the thermometric titration graph represents the equivalence point — the volume of alkali that exactly neutralises the acid present. Since the total volume of each mixture is kept constant at $40.0\text{ cm}^3$ (volume of FA 1 + volume of FA 2 = 40.0), once the volume of FA 2 is read from the graph, the volume of FA 1 is simply $40.0 - V_{\text{FA 2}}$ .

Understanding the Question

The candidate must read the volume of FA 2 at the intersection point from their graph, then calculate the corresponding volume of FA 1. The mark scheme requires the volume of FA 2 to be within $0.25\text{ cm}^3$ of the examiner's value, and both volumes must be stated to 1 decimal place.

Approach

Locate the intersection of the two extrapolated lines on the graph.
Read the $x$ -coordinate (volume of FA 2) to 1 decimal place.
Subtract from $40.0$ to get the volume of FA 1.
State both values.

Step-by-Step Reasoning

The intersection point gives the volume of FA 2 at the equivalence point. This is read directly from the $x$ -axis.
Since $V_{\text{FA 1}} + V_{\text{FA 2}} = 40.0\text{ cm}^3$ for every mixture, the volume of FA 1 is $40.0 - V_{\text{FA 2}}$ .
The mark scheme allows a "discontinuity" at the intersection (i.e. the lines can meet at a sharp peak rather than a smooth curve), but a continuous curve cannot earn this mark.
Both volumes must be to 1 dp (e.g. $20.0$ , not $20$ ).

Key Takeaways

The intersection of the two lines gives the equivalence volume.
The total volume is constant at $40.0\text{ cm}^3$ , so the two volumes are complementary.
Precision in reading the graph (to $0.1\text{ cm}^3$ ) is required.

Common Mistakes

Reading the volume to the nearest whole number instead of 1 dp.
Forgetting that the total volume is $40.0\text{ cm}^3$ and not calculating FA 1 correctly.
Using a continuous curve (which cannot earn this mark).

Things to Be Careful About

The volume must be read to within $0.25\text{ cm}^3$ of the examiner's value.
Both answers must be to 1 decimal place.
If there is no maximum temperature (i.e. the graph doesn't peak), neither this mark nor b(i)III can be awarded.

Techniques used

read volume from graph intersectioncalculate complementary volume

(c)

(i)

Calculate the number of moles of sodium hydroxide, FA 2, required to obtain a neutral solution in this experiment.

moles of $\text{NaOH}$ = .............................. mol

DifficultyEasy

Worked solution

Working

\text{moles of NaOH} = \frac{2.00 \times V_{\text{FA 2}}}{1000}

where $V_{\text{FA 2}}$ is the volume of FA 2 read from the graph in part (b)(ii).

For example, if $V_{\text{FA 2}} = 20.0\text{ cm}^3$ :

\text{moles of NaOH} = \frac{2.00 \times 20.0}{1000} = 0.0400\text{ mol}

Answer

$0.0400\text{ mol}$ (using the example volume of $20.0\text{ cm}^3$ )

Final answer

0.0400 mol (example; depends on candidate's graph reading)

Detailed explanation

Background Concept

The number of moles of a solute in solution is calculated from:

\text{moles} = \frac{\text{concentration (mol dm}^{-3}\text{)} \times \text{volume (cm}^3\text{)}}{1000}

This is because concentration is defined as moles per cubic decimetre, and $1\text{ dm}^3 = 1000\text{ cm}^3$ .

Understanding the Question

The candidate must use the volume of FA 2 (NaOH, concentration $2.00\text{ mol dm}^{-3}$ ) determined in part (b)(ii) to calculate the number of moles of NaOH that reacted to reach the equivalence point.

Approach

Take the volume of FA 2 from part (b)(ii).
Substitute into the formula: moles = (2.00 × volume) / 1000.
Give the answer to 3 or 4 significant figures.

Step-by-Step Reasoning

The concentration of NaOH is given as $2.00\text{ mol dm}^{-3}$ .
The volume of FA 2 at the equivalence point is read from the graph (e.g. $20.0\text{ cm}^3$ ).
Moles of NaOH = $\frac{2.00 \times 20.0}{1000} = 0.0400\text{ mol}$ .
The answer must be to 3 or 4 significant figures (e.g. $0.0400$ or $0.040$ ).

Key Takeaways

The formula moles = concentration × volume / 1000 is fundamental to all solution stoichiometry.
The volume used is the one read from the graph, not from the table.

Common Mistakes

Using the wrong volume (e.g. from the table rather than the graph intersection).
Forgetting to divide by 1000 (giving an answer 1000 times too large).
Incorrect significant figures.

Things to Be Careful About

The answer must be to 3 or 4 significant figures.
This value is carried forward into part (c)(ii), so an error here will affect the next calculation (ecf applies).

Techniques used

calculate moles from concentration and volume

(ii)

Hence calculate the concentration of HZ in FA 1.

concentration of HZ = .............................. $\text{mol dm}^{-3}$

DifficultyMedium-Easy

Worked solution

Working

The reaction is 1:1, so moles of HZ = moles of NaOH from (c)(i).

\text{concentration of HZ} = \frac{\text{moles of HZ} \times 1000}{V_{\text{FA 1}}}

where $V_{\text{FA 1}}$ is the volume of FA 1 from part (b)(ii).

For example, if moles of HZ = $0.0400$ and $V_{\text{FA 1}} = 20.0\text{ cm}^3$ :

\text{concentration of HZ} = \frac{0.0400 \times 1000}{20.0} = 2.00\text{ mol dm}^{-3}

Answer

$2.00\text{ mol dm}^{-3}$ (example; depends on candidate's values)

Final answer

2.00 mol dm⁻³ (example; depends on candidate's graph reading)

Detailed explanation

Background Concept

Since HZ is a monoprotic acid and the reaction with NaOH is 1:1, the moles of HZ that reacted equal the moles of NaOH at the equivalence point. The concentration of HZ is then found from:

\text{concentration} = \frac{\text{moles} \times 1000}{\text{volume in cm}^3}

Understanding the Question

The candidate must use the moles of NaOH calculated in (c)(i) (which equals the moles of HZ due to the 1:1 stoichiometry) and divide by the volume of FA 1 used (from part (b)(ii)) to find the concentration of HZ.

Approach

Moles of HZ = moles of NaOH (from c(i)) because the ratio is 1:1.
Use the volume of FA 1 from part (b)(ii).
Apply: concentration = (moles × 1000) / volume.
Give answer to 3 or 4 significant figures.

Step-by-Step Reasoning

From the equation: $\text{HZ} + \text{NaOH} \rightarrow \text{NaZ} + \text{H}_2\text{O}$ , the mole ratio is 1:1.
Therefore, moles of HZ = moles of NaOH = value from (c)(i).
Volume of FA 1 = $40.0 - V_{\text{FA 2}}$ (from part b(ii)).
Concentration of HZ = $\frac{\text{moles from (c)(i)} \times 1000}{V_{\text{FA 1}}}$ .
Example: if (c)(i) gives $0.0400\text{ mol}$ and $V_{\text{FA 1}} = 20.0\text{ cm}^3$ , then concentration = $\frac{0.0400 \times 1000}{20.0} = 2.00\text{ mol dm}^{-3}$ .

Key Takeaways

The 1:1 stoichiometry means moles of acid = moles of base at equivalence.
Error carried forward (ecf) is allowed — if (c)(i) is wrong but used correctly here, the method mark is still available.

Common Mistakes

Using the volume of FA 2 instead of FA 1 in the denominator.
Forgetting to multiply by 1000 (giving concentration in mol cm⁻³).
Not recognising the 1:1 ratio and trying to apply a different stoichiometry.

Things to Be Careful About

The volume in the denominator is the volume of FA 1 (the acid), not FA 2.
Answer must be to 3 or 4 significant figures.
ECF applies: if (c)(i) is incorrect but this calculation uses that value correctly, credit is given for the method.

Techniques used

calculate concentration from moles and volumeapply 1:1 stoichiometry

(d)

Explain how you would use the data obtained to calculate the enthalpy change of neutralisation of HZ. You do not need to carry out the calculation.

DifficultyMedium

Worked solution

Answer

Determine $\Delta T$ from the graph (or table): $\Delta T = T_{\text{max}} - \text{initial temperature}$ , where the initial temperature is the average of the temperatures of FA 1 and FA 2 before mixing.
Calculate the heat energy released using $Q = mc\Delta T$ , where $m$ is the total mass of the solution ( $40.0\text{ g}$ , assuming density = $1\text{ g cm}^{-3}$ ) and $c = 4.18\text{ J g}^{-1}\text{K}^{-1}$ .
Divide the heat energy produced by the moles of NaOH neutralised (from part (c)(i)) to obtain the enthalpy change of neutralisation:

\Delta H_{\text{neut}} = \frac{-Q}{\text{moles of NaOH}}

(the negative sign indicates an exothermic reaction).

Final answer

Find ΔT from graph, calculate Q = mcΔT, divide Q by moles of NaOH from (c)(i)

Detailed explanation

Background Concept

The enthalpy change of neutralisation is the heat energy released when one mole of water is formed from the reaction between an acid and a base under standard conditions. In a thermometric experiment, the heat released by the reaction is absorbed by the solution, causing its temperature to rise. By measuring this temperature change and knowing the mass and specific heat capacity of the solution, we can calculate the heat released using:

Q = mc\Delta T

where $m$ is the mass of the solution (in grams), $c$ is the specific heat capacity of water ( $4.18\text{ J g}^{-1}\text{K}^{-1}$ ), and $\Delta T$ is the temperature change (in °C or K).

The enthalpy change per mole is then:

\Delta H = \frac{-Q}{n}

where $n$ is the number of moles of water formed (equal to the moles of acid or base neutralised, since the reaction is 1:1). The negative sign indicates that the reaction is exothermic.

Understanding the Question

The command word is "explain" — the candidate must describe the method for calculating $\Delta H_{\text{neut}}$ from the data obtained, without actually performing the calculation. The mark scheme awards three independent marks for: (1) obtaining $\Delta T$ , (2) using $Q = mc\Delta T$ , and (3) dividing by moles of alkali neutralised.

Approach

State how to find $\Delta T$ — either from the graph (extrapolated maximum) or from the table.
State the use of $Q = mc\Delta T$ with appropriate values for $m$ and $c$ .
State the division by moles of NaOH (from c(i)).

Step-by-Step Reasoning

Mark 1 (ΔT): The temperature change can be obtained from the graph as the difference between the extrapolated maximum temperature and the initial temperature. The initial temperature is the average of the separate temperatures of FA 1 and FA 2 (since they may differ slightly). Alternatively, $\Delta T$ can be read directly from the table as $T_{\text{max}} - T_{\text{initial}}$ . The mark scheme allows either "rise in temperature" or the explicit formula.
Mark 2 (Q = mcΔT): The heat energy released is calculated using $Q = mc\Delta T$ . The mass $m$ is the total volume of the mixture ( $40.0\text{ cm}^3$ ) assuming a density of $1.0\text{ g cm}^{-3}$ , giving $m = 40.0\text{ g}$ . The specific heat capacity $c = 4.18\text{ J g}^{-1}\text{K}^{-1}$ (assumed same as water). $\Delta T$ is in °C (equivalent to K for a temperature difference).
Mark 3 (Divide by moles): The enthalpy change of neutralisation is the heat energy per mole of water formed. Since the reaction is 1:1, moles of water = moles of NaOH neutralised = the value from part (c)(i). So:

\Delta H_{\text{neut}} = \frac{-Q}{\text{moles of NaOH}}

The negative sign is needed because the reaction is exothermic (heat is released).

Key Takeaways

The three-step method for calculating enthalpy change from calorimetry data: find ΔT → calculate Q → divide by moles.
The mass of solution is assumed to be equal to the total volume in cm³ (density = 1 g cm⁻³).
The specific heat capacity is assumed to be that of water.
The enthalpy change is negative for an exothermic reaction.

Common Mistakes

Using only the acid volume or only the alkali volume as the mass, instead of the total volume (40.0 cm³).
Forgetting the negative sign for exothermic reactions.
Dividing by the wrong number of moles (e.g. using moles of acid instead of the moles from (c)(i), though these are equal in a 1:1 reaction).
Not specifying that the initial temperature is the average of the two separate solutions.
Using the wrong value for specific heat capacity or not stating it.

Things to Be Careful About

The mark scheme specifically says "allow use of rise in temperature" — so stating ΔT as the rise is acceptable.
The moles used for division must be the moles of alkali neutralised, which is the answer from (c)(i).
Units must be consistent: Q in J, moles in mol, giving ΔH in J mol⁻¹ (convert to kJ mol⁻¹ by dividing by 1000).
The question says "you do not need to carry out the calculation" — so a description of the method earns the marks, not a numerical answer.

Techniques used

determine temperature change from graphapply Q = mcΔTdivide heat by moles to find enthalpy change

The rest of this paper

2 more questions

Q2Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation14M
Q3Qualitative Analysis · Manipulation, Measurement and Observation14M

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