Chemistry 9701/52 — February/March 2018

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

questions

marks

minutes

Topics Planning · Analysis, Conclusions and Evaluation

Q1PlanningAnalysis, Conclusions and EvaluationFree sample

When a solute is added to a solvent the freezing point of the solution is lower than that of the pure solvent.

The lowering of freezing point is very small. A chemist called Beckmann invented a thermometer capable of measuring these small temperature changes accurately. The Beckmann thermometer must be calibrated at the start of the experiment.

An incomplete diagram of the Beckmann apparatus is shown containing pure liquid cyclohexane, an organic solvent with a freezing point of about $6.5\text{ }^{\circ}\text{C}$ . The diagram does not show how the cyclohexane could be frozen.

Fig. 1.1

(a)

Complete the diagram to show how the pure liquid cyclohexane could be frozen using simple laboratory apparatus.

DifficultyMedium-Easy

Worked solution

Answer

Place the boiling tube inside a larger surrounding vessel (e.g. a polystyrene/styrofoam cup or plastic beaker) containing a mixture of ice and water (a freezing mixture).

Final answer

Surrounding vessel of polystyrene/styrofoam/plastic containing water and ice within the cooling mixture

Detailed explanation

Background Concept

To freeze a liquid in a laboratory, it must be placed in contact with something colder than its freezing point. Cyclohexane freezes at about 6.5 °C, so a freezing mixture of ice and water (which reaches approximately 0 °C) provides a sufficiently low temperature to remove heat from the cyclohexane and cause it to solidify. A polystyrene cup or plastic beaker acts as the surrounding vessel, providing insulation and containing the ice-water mixture.

Understanding the Question

The question asks the candidate to complete the given diagram (Fig. 1.1) to show how the pure liquid cyclohexane could be frozen using simple laboratory apparatus. The key requirement is to show a cooling arrangement that can bring the cyclohexane below its freezing point.

Approach

Identify a simple cooling method: an ice-water mixture (freezing mixture) in a larger vessel surrounding the boiling tube. The vessel should be made of a material that provides some insulation (polystyrene) or is at least chemically compatible (plastic). Draw or describe this arrangement around the existing boiling tube.

Step-by-Step Reasoning

The cyclohexane has a freezing point of ~6.5 °C. To freeze it, the surrounding temperature must be below this.
A mixture of ice and water reaches 0 °C, which is below 6.5 °C — sufficient to freeze cyclohexane.
The boiling tube is placed inside a larger vessel (polystyrene cup or plastic beaker) containing the ice-water mixture.
The mark scheme requires both elements: the surrounding vessel (polystyrene/styrofoam/plastic containing water) AND ice within the cooling mixture.

Key Takeaways

A freezing mixture of ice and water can cool substances below 0 °C to approximately 0 °C.
Simple laboratory apparatus for cooling includes a surrounding vessel with ice-water mixture.
The vessel material (polystyrene, plastic) should be chemically compatible with the solvent.

Common Mistakes

Only mentioning ice without the water/cooling mixture context.
Suggesting a refrigerator or dry ice (not 'simple laboratory apparatus' in the context of this question).
Forgetting to include both the vessel AND the ice in the answer.

Things to Be Careful About

The mark scheme requires BOTH the surrounding vessel (polystyrene/styrofoam/plastic containing water) AND ice within the cooling mixture. One without the other does not earn the mark.

Techniques used

describe a cooling method using simple apparatussketch a labelled diagram of a freezing mixture

(b)

The method for determining the lowering of freezing point is as follows.

step 1 Add $20.00\text{ g}$ of pure liquid cyclohexane to a clean dry boiling tube.

step 2 Place the stopper containing the Beckmann thermometer and stirring wire into the boiling tube.

step 3 Cool the pure cyclohexane. When it starts to freeze, set the Beckmann thermometer to $0.00$ to calibrate it.

step 4 Allow the pure cyclohexane to melt. Remove the stopper from the boiling tube. Add $0.250\text{ g}$ of an organic solid X to the pure cyclohexane and replace the stopper. Stir the solution to dissolve X and refreeze the solution. Record the new freezing point.

step 5 Allow the solution to melt. Remove the stopper from the boiling tube. Add a further known mass of X to the solution and replace the stopper. Stir the solution to dissolve X and refreeze the solution. Record the new freezing point.

step 6 Repeat step 5 until sufficient readings are obtained.

(i)

In step 1, the cyclohexane can be measured using an electronic balance, a beaker and a clean dry boiling tube as shown.

Fig. 1.2

Describe a suitable method to add precisely $20.00\text{ g}$ of cyclohexane to the boiling tube.
Assume that the balance is accurate to two decimal places and that common laboratory apparatus is available.

DifficultyMedium-Easy

Worked solution

Answer

Place the boiling tube (supported in the beaker) on the electronic balance and tare/zero the balance. Add cyclohexane dropwise (using a dropper/pipette) until the balance reads exactly $20.00\text{ g}$ .

Final answer

Add dropwise around the 20.00 g mark

Detailed explanation

Background Concept

An electronic balance accurate to two decimal places can measure masses to $\pm 0.01\text{ g}$ . To achieve a precise target mass of a liquid, the technique involves placing the container on the balance, taring it, and then adding the liquid carefully — especially near the target — using a method that allows fine control (dropwise addition via a dropper or pipette).

Understanding the Question

The question asks for a suitable method to add precisely $20.00\text{ g}$ of cyclohexane to the boiling tube, given that the balance is accurate to two decimal places. The key word is 'precisely', meaning the final reading must be exactly $20.00\text{ g}$ .

Approach

The standard technique is: place the container on the balance, tare it, then add the substance. For a liquid, the critical step is adding it dropwise as the reading approaches the target, to avoid overshooting.

Step-by-Step Reasoning

Place the boiling tube (in the beaker) on the electronic balance.
Tare/zero the balance so it reads $0.00\text{ g}$ .
Pour cyclohexane from a beaker or add it with a dropper, watching the display.
As the reading approaches $20.00\text{ g}$ , switch to adding the liquid dropwise to achieve exactly $20.00\text{ g}$ .
The mark scheme specifically rewards 'add dropwise around 20.00 g mark' — this demonstrates awareness of the need for fine control near the target.

Key Takeaways

When measuring a precise mass on an electronic balance, the final approach to the target must be done carefully (dropwise for liquids) to avoid overshooting.
Taring the balance with the container first means the displayed reading equals the mass of the substance alone.

Common Mistakes

Simply saying 'pour until 20.00 g' without mentioning dropwise addition — this lacks the precision required.
Forgetting to mention taring the balance first.
Suggesting use of a measuring cylinder instead of the balance (the question specifies using the balance).

Things to Be Careful About

The balance is accurate to two decimal places, so the reading can be set to exactly $20.00\text{ g}$ .
The key marking point is 'dropwise' — this shows understanding of how to achieve precision with a liquid.

Techniques used

describe a precise weighing technique using an electronic balanceuse dropwise addition to achieve a target mass

(ii)

Alternatively in step 1, the volume of cyclohexane with a mass of exactly $20.00\text{ g}$ can be measured and added to the boiling tube.

Calculate the volume of cyclohexane with a mass of precisely $20.00\text{ g}$ .
The density of cyclohexane is $0.78\text{ g cm}^{-3}$ .
Give your answer to two decimal places.

volume of cyclohexane = .............................. $\text{cm}^3$

Explain whether a burette is suitable for measuring this volume.

DifficultyMedium-Easy

Worked solution

Working

V = \frac{m}{\rho} = \frac{20.00}{0.78} = 25.64\text{ cm}^3

Answer

Volume of cyclohexane = $25.64\text{ cm}^3$

A burette is not suitable because a burette can only measure to $\pm 0.05\text{ cm}^3$ (it cannot measure to $0.01\text{ cm}^3$ ), so the volume cannot be determined to the required precision of two decimal places.

Final answer

25.64 cm³; No, a burette can only measure ±0.05 cm³

Detailed explanation

Background Concept

Density relates mass and volume: $\rho = m/V$ , so $V = m/\rho$ . A burette is a piece of volumetric apparatus typically graduated in $0.1\text{ cm}^3$ divisions, with readings estimated to $\pm 0.05\text{ cm}^3$ . This means it can measure volumes to one decimal place (or half a division), but NOT to two decimal places.

Understanding the Question

Two tasks: (1) calculate the volume of cyclohexane with mass $20.00\text{ g}$ given density $0.78\text{ g cm}^{-3}$ , to two decimal places; (2) explain whether a burette is suitable for measuring this volume to the required precision.

Approach

Use $V = m/\rho$ for the calculation.
Compare the precision of a burette ( $\pm 0.05\text{ cm}^3$ ) with the required precision (two decimal places, i.e. $\pm 0.01\text{ cm}^3$ ).

Step-by-Step Reasoning

$V = 20.00 / 0.78 = 25.641... = 25.64\text{ cm}^3$ (to 2 d.p.).
The question requires the volume to be measured precisely to two decimal places (matching the $20.00\text{ g}$ precision).
A burette has graduations of $0.1\text{ cm}^3$ and can be read to $\pm 0.05\text{ cm}^3$ — this is only to one decimal place at best.
Therefore a burette CANNOT measure to $0.01\text{ cm}^3$ and is not suitable.

Key Takeaways

Always check whether the precision of the chosen apparatus matches the precision required by the question.
A burette reads to $\pm 0.05\text{ cm}^3$ , not $\pm 0.01\text{ cm}^3$ .

Common Mistakes

Saying 'yes' because a burette is a precise instrument — it is precise for titration volumes but not to two decimal places.
Incorrect calculation (e.g. multiplying instead of dividing by density).
Not stating the answer to exactly two decimal places.

Things to Be Careful About

Give the volume to exactly two decimal places as instructed.
The explanation must explicitly state WHY the burette is unsuitable (its precision limit of $\pm 0.05\text{ cm}^3$ ).

Techniques used

calculate volume from mass and densityevaluate suitability of apparatus based on precision

(iii)

In step 4 the mass of X is measured on an electronic balance accurate to three decimal places before adding it to the cyclohexane.

A student suggests the following technique.

An empty container is placed on the electronic balance.
The mass of the empty container is recorded to three decimal places.
$0.250\text{ g}$ of X is added to the container.
X is tipped from the container into the cyclohexane.

Explain why this technique would not be accurate for adding $0.250\text{ g}$ of X to the cyclohexane.

DifficultyMedium-Easy

Worked solution

Answer

When X is tipped from the container into the cyclohexane, some of the solid may remain in the container (adhere to the walls), so the actual mass added to the cyclohexane would be less than $0.250\text{ g}$ .

(This is not 'weighing by difference' — the mass actually transferred is unknown.)

Final answer

Some of X may remain in the container when transferring, so the mass actually added is less than 0.250 g

Detailed explanation

Background Concept

In analytical chemistry, when a solid must be transferred from a weighing container to a reaction vessel, the correct technique is 'weighing by difference': weigh the container with the substance, transfer the substance, then weigh the container again. The difference gives the exact mass transferred. If instead you weigh the empty container, add the substance to get the target mass, then tip it out, you have no way of knowing how much actually left the container.

Understanding the Question

The student's technique is: weigh empty container, add X to reach 0.250 g, tip X into cyclohexane. The question asks why this is not accurate.

Approach

Identify the flaw: when tipping a solid from a container, some particles inevitably adhere to the container walls or base, meaning not all 0.250 g actually enters the cyclohexane.

Step-by-Step Reasoning

The balance shows 0.250 g of X is in the container.
When X is tipped into the cyclohexane, some solid remains stuck to the container.
The mass actually added to the cyclohexane is therefore LESS than 0.250 g.
Since the true mass added is unknown, the calculation of B and ultimately $M_r$ will be in error.
The correct method would be weighing by difference: weigh container + X, transfer, weigh container again, and the difference is the true mass transferred.

Key Takeaways

Weighing by difference is essential when transferring solids to ensure the actual mass added is known.
Any transfer technique that does not account for residue introduces a systematic error.

Common Mistakes

Saying 'the balance is not accurate enough' — the balance IS accurate to three decimal places; the problem is the transfer method.
Saying 'X might evaporate' — this is not the issue for a solid.

Things to Be Careful About

The answer must specifically address the transfer step and the loss of material remaining in the container.

Techniques used

identify a source of systematic error in a weighing techniqueexplain why weighing by difference is necessary

(c)

The freezing points of the solutions are lower than the freezing point of pure cyclohexane.

\Delta T_{\text{fp}} = (\text{freezing point of pure cyclohexane}) - (\text{freezing point of the solution})

For the experiment described in (b) the values of $\Delta T_{\text{fp}}$ are recorded in the table.

(i)

A ratio, B, is calculated as follows.

\text{B} = \frac{\text{mass of X (g)}}{\text{mass of solvent (g)}}

Complete the table by calculating B for each reading. Give your answers to three significant figures.

reading number	total mass of X added to 20.00g of cyclohexane / g	$\Delta T_{\text{fp}} / ^{\circ}\text{C}$
1	0.250	1.35
2	0.400	2.20
3	0.500	2.75
4	0.800	4.40
5	0.950	5.30
6	1.150	6.40
7	1.300	7.25
8	1.400	8.50

DifficultyMedium-Easy

Worked solution

Working

\text{B} = \frac{\text{mass of X}}{\text{mass of solvent}} = \frac{\text{mass of X}}{20.00}

Answer

Reading	Mass of X / g	B
1	0.250	0.0125
2	0.400	0.0200
3	0.500	0.0250
4	0.800	0.0400
5	0.950	0.0475
6	1.150	0.0575
7	1.300	0.0650
8	1.400	0.0700

Final answer

B values: 0.0125, 0.0200, 0.0250, 0.0400, 0.0475, 0.0575, 0.0650, 0.0700 (all to 3 s.f.)

Detailed explanation

Background Concept

The ratio B represents the mass of solute per unit mass of solvent. It is a dimensionless quantity (g/g) that is directly proportional to the concentration of the solution in molality terms. In freezing point depression, $\Delta T_{fp}$ is proportional to B (for dilute solutions), so plotting $\Delta T_{fp}$ against B should give a straight line through the origin.

Understanding the Question

The candidate must calculate B = mass of X / mass of solvent for each of the 8 readings, where the mass of solvent (cyclohexane) is always $20.00\text{ g}$ . Answers must be given to three significant figures.

Approach

Divide each mass of X by 20.00. Express the result to exactly three significant figures.

Step-by-Step Reasoning

Reading 1: $0.250 / 20.00 = 0.0125$ (3 s.f.)
Reading 2: $0.400 / 20.00 = 0.0200$ (3 s.f.)
Reading 3: $0.500 / 20.00 = 0.0250$ (3 s.f.)
Reading 4: $0.800 / 20.00 = 0.0400$ (3 s.f.)
Reading 5: $0.950 / 20.00 = 0.0475$ (3 s.f.)
Reading 6: $1.150 / 20.00 = 0.0575$ (3 s.f.)
Reading 7: $1.300 / 20.00 = 0.0650$ (3 s.f.)
Reading 8: $1.400 / 20.00 = 0.0700$ (3 s.f.)

Note: trailing zeros after the decimal point are significant, so 0.0200 has three significant figures (the trailing zeros count).

Key Takeaways

When expressing to significant figures, trailing zeros after a decimal point ARE significant.
The mass of solvent remains constant throughout the experiment (20.00 g).

Common Mistakes

Writing 0.02 instead of 0.0200 (only 1 s.f. instead of 3).
Dividing by the wrong value (e.g. dividing by total mass of solution).
Not giving all values to exactly 3 s.f.

Things to Be Careful About

Three significant figures means 0.0200 (not 0.02) and 0.0700 (not 0.07). The trailing zeros are essential.

Techniques used

calculate a ratio from given dataexpress answers to a specified number of significant figures

(ii)

Plot a graph on the grid to show the relationship between B and $\Delta T_{\text{fp}}$ . Draw the line of best fit.

Fig. 1.3

DifficultyMedium-Easy

Worked solution

Answer

Plot all 8 points on the grid with B on the x-axis and $\Delta T_{fp}$ on the y-axis. Draw a straight line of best fit passing through the origin $(0,0)$ and through the cluster of points, with the anomalous point (reading 8) excluded from the line.

Final answer

Straight line of best fit through origin and points 1-7 (excluding anomalous point 8)

Detailed explanation

Background Concept

Freezing point depression is directly proportional to the molality of the solute (for dilute ideal solutions). Since B is proportional to molality (mass of solute per mass of solvent), a plot of $\Delta T_{fp}$ against B should give a straight line through the origin. The gradient of this line equals $K/M_r$ .

Understanding the Question

The candidate must plot the calculated B values against the given $\Delta T_{fp}$ values on the provided grid (Fig. 1.3) and draw a line of best fit. The axes are already labelled: y-axis is $\Delta T_{fp}/°C$ (0 to 10) and x-axis is B (0 to 0.08).

Approach

Plot each (B, $\Delta T_{fp}$ ) pair as a point on the grid.
Observe the trend — points 1-7 should fall approximately on a straight line through the origin.
Point 8 (B = 0.0700, $\Delta T_{fp}$ = 8.50) will be anomalous (too high for the line).
Draw a straight line of best fit that passes through the origin and the majority of points, ignoring the outlier.

Step-by-Step Reasoning

The points to plot are:

(0.0125, 1.35), (0.0200, 2.20), (0.0250, 2.75), (0.0400, 4.40), (0.0475, 5.30), (0.0575, 6.40), (0.0650, 7.25), (0.0700, 8.50)

Points 1-7 lie approximately on a straight line through the origin with gradient ≈ 110-112 °C. Point 8 deviates significantly above this line. The line of best fit should pass through the origin and through points 1-7, with point 8 clearly above the line.

Key Takeaways

A line of best fit for a proportional relationship must pass through the origin.
Anomalous points should be identified and excluded from the line.
Points should be plotted to within half a small square of accuracy.

Common Mistakes

Drawing a curve instead of a straight line.
Including the anomalous point in the line of best fit.
Not passing the line through the origin.
Plotting points inaccurately (more than half a small square off).

Things to Be Careful About

The line must be straight (ruler) and pass through the origin.
Use small, neat crosses or dots for plotting points.

Techniques used

plot points accurately on a graphdraw a line of best fit through a linear relationship

(iii)

Identify, by the reading number, the single most anomalous point. Suggest what error in the experiment could have caused this anomaly.

reading number ..................................................................................................................

reason .................................................................................................................................

DifficultyMedium-Easy

Worked solution

Answer

Reading number: 8

Reason: A greater mass than $1.40\text{ g}$ was actually added (the mass recorded was too low compared to the true mass added), causing a larger $\Delta T_{fp}$ than expected for the stated B value.

Final answer

Reading 8; a greater mass than 1.40 g was added

Detailed explanation

Background Concept

An anomalous point is one that deviates significantly from the expected linear trend. In this experiment, $\Delta T_{fp}$ should be directly proportional to B. If a point lies well above the line, it means the actual $\Delta T_{fp}$ is larger than predicted by the recorded B value — suggesting more solute was actually present than the recorded mass indicates.

Understanding the Question

The candidate must identify which reading is most anomalous (furthest from the line of best fit) and suggest an experimental error that could explain it.

Approach

From the graph, reading 8 (B = 0.0700, $\Delta T_{fp}$ = 8.50) lies clearly above the line of best fit through points 1-7. The expected $\Delta T_{fp}$ for B = 0.0700 from the line would be approximately 7.7-7.8 °C, but the actual value is 8.50 °C — too high. This means the true mass of X was greater than 1.40 g (so the true B was higher than recorded).

Step-by-Step Reasoning

Points 1-7 lie approximately on a straight line through the origin.
Point 8 deviates significantly above this line.
If the true mass of X added was greater than the recorded 1.40 g, then the true B would be higher than 0.0700, and the higher $\Delta T_{fp}$ would be consistent with a higher B value.
This could happen if the student misread the balance or added too much X.

Key Takeaways

An anomalous point above the line in a proportional graph suggests the independent variable was underestimated (true value higher than recorded).
Always check whether the deviation direction is consistent with a plausible error.

Common Mistakes

Identifying the wrong point as anomalous.
Suggesting 'human error' or 'reading error' without specifying the direction of the error.
Suggesting the freezing point was misread (this would affect $\Delta T_{fp}$ but not explain why it is specifically high relative to B).

Things to Be Careful About

The reason must specifically explain WHY $\Delta T_{fp}$ is too high for the given B — i.e. more solute was actually present than recorded.

Techniques used

identify an anomalous data point from a graphsuggest a plausible experimental error causing the anomaly

(iv)

In another experiment, a student added an unknown mass of X to $20.00\text{ g}$ of cyclohexane and measured $\Delta T_{\text{fp}}$ as $5.00\text{ }^{\circ}\text{C}$ .

Use your graph to determine the mass of X used in this experiment.

mass of X = .............................. g

DifficultyMedium-Easy

Worked solution

Working

From the graph, when $\Delta T_{fp} = 5.00\text{ }^{\circ}\text{C}$ :

\text{B} = 0.045

\text{mass of X} = \text{B} \times 20.00 = 0.045 \times 20.00 = 0.90\text{ g}

Answer

Mass of X = $0.90\text{ g}$

Final answer

0.90 g

Detailed explanation

Background Concept

Reading a value from a graph involves locating the given value on one axis, drawing a horizontal (or vertical) line to the line of best fit, then reading the corresponding value on the other axis. Here, $\Delta T_{fp}$ is given and B must be found from the line, then converted to mass.

Understanding the Question

The student measured $\Delta T_{fp} = 5.00\text{ }^{\circ}\text{C}$ . Using the graph (line of best fit from part c(ii)), find the corresponding B value, then calculate the mass of X.

Approach

Locate 5.00 on the y-axis ( $\Delta T_{fp}$ ).
Draw a horizontal line to the line of best fit.
Draw a vertical line down to the x-axis to read B.
Calculate mass of X = B × 20.00.

Step-by-Step Reasoning

From the graph, at $\Delta T_{fp} = 5.00\text{ }^{\circ}\text{C}$ , the line of best fit gives B ≈ 0.045.
Mass of X = B × mass of solvent = $0.045 \times 20.00 = 0.90\text{ g}$ .

The mark scheme awards M1 for reading B = 0.045 (±0.002 acceptable) and M2 for the correct multiplication giving 0.90 g.

Key Takeaways

Always use the line of best fit (not individual data points) when reading from a graph.
The relationship B = mass of X / mass of solvent can be rearranged: mass of X = B × mass of solvent.

Common Mistakes

Reading off a data point rather than the line of best fit.
Forgetting to multiply by 20.00 (giving the answer as 0.045 instead of 0.90 g).
Reading B incorrectly from the graph (e.g. reading 0.040 or 0.050 instead of 0.045).

Things to Be Careful About

The answer must be in grams (the question asks for mass of X in g).
Use the line of best fit, not the raw data points.

Techniques used

read a value from a graphreverse-calculate mass from a ratio

(v)

Determine the gradient of your line of best fit. State the coordinates of the two points you used for your calculation.

coordinates 1 .............................................. coordinates 2 ..............................................

gradient = .............................. $^{\circ}\text{C}$

DifficultyMedium-Easy

Worked solution

Working

Coordinates 1: $(0.0125, 1.35)$
Coordinates 2: $(0.0650, 7.25)$

\text{gradient} = \frac{\Delta y}{\Delta x} = \frac{7.25 - 1.35}{0.0650 - 0.0125} = \frac{5.90}{0.0525} = 112\text{ }^{\circ}\text{C}

Answer

Gradient = $112\text{ }^{\circ}\text{C}$

Final answer

112 °C (using points (0.0125, 1.35) and (0.0650, 7.25))

Detailed explanation

Background Concept

The gradient of a straight line is calculated as $\Delta y / \Delta x$ using two points ON the line of best fit (not necessarily data points). The two points should be well separated (at least half the length of the line) to minimise the effect of reading errors.

Understanding the Question

The candidate must state two sets of coordinates from the line of best fit and use them to calculate the gradient. The units are °C (since y is in °C and x is dimensionless).

Approach

Choose two points on the line of best fit that are well separated and easy to read.
Calculate gradient = (change in y) / (change in x).

Step-by-Step Reasoning

Using points on the line (not necessarily data points):

Point 1: $(0.0125, 1.35)$ — this is approximately on the line near reading 1.
Point 2: $(0.0650, 7.25)$ — this is approximately on the line near reading 7.

Gradient = $(7.25 - 1.35) / (0.0650 - 0.0125) = 5.90 / 0.0525 = 112\text{ }^{\circ}\text{C}$

Alternative valid choices might include $(0, 0)$ and $(0.060, 6.70)$ giving gradient = $6.70/0.060 = 112\text{ }^{\circ}\text{C}$ .

The mark scheme awards M1 for stating valid coordinates and M2 for the correct gradient calculation. Acceptable range is approximately 105-115 °C depending on the line drawn.

Key Takeaways

Always use points ON the line of best fit, not raw data points.
Choose well-separated points to reduce percentage error in the gradient.
The gradient has units of °C (since B is dimensionless).

Common Mistakes

Using data points instead of points on the line.
Choosing two points that are too close together.
Reversing the gradient calculation ( $\Delta x / \Delta y$ ).
Forgetting the units.

Things to Be Careful About

State BOTH coordinates clearly as required by the question.
The gradient should be consistent with the line drawn in part (c)(ii).
Acceptable range is approximately 105-115 °C.

Techniques used

calculate the gradient of a line from two pointsstate coordinates used in a gradient calculation

(d)

$\Delta T_{\text{fp}}$ is related to the $M_r$ of X by the following expression

\Delta T_{\text{fp}} = \frac{KB}{M_r}

where

\text{B} = \frac{\text{mass of X (g)}}{\text{mass of solvent (g)}}

$K = \text{a constant}$

The $M_r$ of X can be found using the gradient of your line of best fit.

M_r = \frac{K}{\text{gradient}}

The numerical value of $K$ is 20020.

Use this value for $K$ and the gradient you determined in (c)(v) to calculate the $M_r$ of X. Give your answer to the nearest whole number.

If you were unable to calculate the gradient in (c)(v), assume that the gradient is $103\text{ }^{\circ}\text{C}$ . This is not the correct value.

$M_r$ = ..............................

DifficultyMedium-Easy

Worked solution

Working

M_r = \frac{K}{\text{gradient}} = \frac{20020}{112} = 179

Answer

$M_r$ = 179

Final answer

179

Detailed explanation

Background Concept

The relationship $\Delta T_{fp} = KB/M_r$ shows that for a given solute (fixed $M_r$ ) and solvent (fixed K), $\Delta T_{fp}$ is directly proportional to B. The gradient of the $\Delta T_{fp}$ vs B graph is therefore $K/M_r$ , and rearranging gives $M_r = K/\text{gradient}$ .

Understanding the Question

The candidate must use the gradient determined in part (c)(v) and the given value $K = 20020$ to calculate $M_r$ of X, giving the answer to the nearest whole number.

Approach

Substitute $K = 20020$ and the gradient from (c)(v) into $M_r = K/\text{gradient}$ .

Step-by-Step Reasoning

Using gradient = 112 °C from part (c)(v):

M_r = \frac{20020}{112} = 178.75 \approx 179

If the assumed gradient of 103 were used (as stated in the question for those who couldn't calculate it):

M_r = \frac{20020}{103} = 194

The mark scheme awards the mark for a correct calculation using the candidate's own gradient from (c)(v) (ecf allowed).

Key Takeaways

When a question provides an alternative value to use if you couldn't complete a previous part, always use your own value if you have one.
The formula $M_r = K/\text{gradient}$ is a direct rearrangement — no additional manipulation needed.

Common Mistakes

Dividing gradient by K instead of K by gradient.
Not giving the answer to the nearest whole number.
Using the assumed value of 103 when a valid gradient was obtained in (c)(v).

Things to Be Careful About

Give the answer to the nearest whole number (no decimal places).
Use the gradient from your own (c)(v) answer — ecf applies.

Techniques used

substitute values into a given formulacalculate a relative molecular mass from a gradient

(e)

A student used the Beckmann apparatus and repeated the experiment described in (b) with an unknown solid Y. The student found the $M_r$ of Y to be 136.

Y is an aromatic carboxylic acid.

Suggest the structure of Y.
[ $A_r$ : C, 12.0; O, 16.0; H, 1.0]

DifficultyMedium

Worked solution

Working

$M_r$ of Y = 136. An aromatic carboxylic acid contains a benzene ring ( $\text{C}_6\text{H}_5$ ) and a $-\text{COOH}$ group.

Phenylacetic acid: $\text{C}_6\text{H}_5\text{CH}_2\text{COOH}$

M_r = (6 \times 12.0) + (5 \times 1.0) + (12.0 + 2 \times 1.0) + (12.0 + 2 \times 16.0 + 1.0) = 72 + 5 + 14 + 45 = 136

Answer

$\text{C}_6\text{H}_5\text{CH}_2\text{COOH}$ (phenylacetic acid)

Alternatively: $\text{CH}_3\text{C}_6\text{H}_4\text{COOH}$ (methylbenzoic acid / toluic acid)

M_r = 15 + 76 + 45 = 136

Final answer

C6H5CH2COOH (phenylacetic acid) or CH3C6H4COOH (methylbenzoic acid)

Detailed explanation

Background Concept

An aromatic carboxylic acid contains a benzene ring and a carboxylic acid functional group (-COOH). The general formula for a monocarboxylic acid attached to an aromatic ring can be written as Ar-COOH or Ar-CH₂-COOH, where Ar is an aryl group. The molecular formula must satisfy both the structural constraints (aromatic + carboxylic acid) and the given $M_r = 136$ .

Understanding the Question

The student determined $M_r = 136$ for an unknown aromatic carboxylic acid Y. The candidate must suggest a structure consistent with this $M_r$ and the description 'aromatic carboxylic acid'.

Approach

An aromatic carboxylic acid must contain a benzene ring and a -COOH group.
Try possible structures and calculate their $M_r$ $M_{r}$ :
- Benzoic acid: $\text{C}_6\text{H}_5\text{COOH}$ → $M_r = 122$ (too low)
- Phenylacetic acid: $\text{C}_6\text{H}_5\text{CH}_2\text{COOH}$ → $M_r = 136$ ✓
- Methylbenzoic acid (toluic acid): $\text{CH}_3\text{C}_6\text{H}_4\text{COOH}$ → $M_r = 136$ ✓

Step-by-Step Reasoning

Benzoic acid ( $\text{C}_7\text{H}_6\text{O}_2$ ): $M_r = 7(12) + 6(1) + 2(16) = 84 + 6 + 32 = 122$ . Too low.
Adding one $\text{CH}_2$ group: $\text{C}_8\text{H}_8\text{O}_2$ → $M_r = 96 + 8 + 32 = 136$ . ✓
This could be phenylacetic acid ( $\text{C}_6\text{H}_5\text{CH}_2\text{COOH}$ ) or methylbenzoic acid ( $\text{CH}_3\text{C}_6\text{H}_4\text{COOH}$ ).
Both are aromatic carboxylic acids with $M_r = 136$ .

Key Takeaways

When deducing a structure from $M_r$ , start with the simplest member of the homologous series and add $\text{CH}_2$ units (each adding 14 to $M_r$ ).
Multiple structural isomers may satisfy the same $M_r$ and description — any valid one earns the mark.

Common Mistakes

Suggesting benzoic acid ( $M_r = 122$ , incorrect).
Suggesting a non-aromatic compound.
Suggesting a compound that is not a carboxylic acid (e.g. an ester or alcohol).
Arithmetic errors in calculating $M_r$ .

Things to Be Careful About

The compound MUST be both aromatic (contains a benzene ring) AND a carboxylic acid (contains -COOH).
Use the given $A_r$ values: C = 12.0, O = 16.0, H = 1.0.

Techniques used

deduce a molecular formula from relative molecular massidentify a structural isomer consistent with given constraints

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