Chemistry 9701/53 — October/November 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 concentration of aqueous chloride ions can be found by titration with aqueous silver nitrate, $\text{AgNO}_3(\text{aq})$ .

\text{Ag}^+(\text{aq}) + \text{Cl}^-(\text{aq}) \rightarrow \text{AgCl}(\text{s})

The indicator used is aqueous potassium chromate(VI), $\text{K}_2\text{CrO}_4(\text{aq})$ .

As $\text{AgNO}_3(\text{aq})$ is added to aqueous chloride ions, a white precipitate of $\text{AgCl}(\text{s})$ is formed.

When all the chloride ions have reacted, further addition of $\text{AgNO}_3(\text{aq})$ leads to the formation of a red precipitate of silver chromate(VI), $\text{Ag}_2\text{CrO}_4(\text{s})$ . The first appearance of the red precipitate shows the end-point of the titration.

A student carries out an experiment to determine the number of molecules of water of crystallisation, $x$ , in hydrated barium chloride, $\text{BaCl}_2 \cdot x\text{H}_2\text{O}(\text{s})$ .

(a)

The student makes $250.0\text{ cm}^3$ of $0.0500\text{ mol dm}^{-3}\text{ AgNO}_3(\text{aq})$ to use for the titration.

(i)

Calculate the mass of solid silver nitrate, $\text{AgNO}_3(\text{s})$ , needed to make $250.0\text{ cm}^3$ of $0.0500\text{ mol dm}^{-3}\text{ AgNO}_3(\text{aq})$ .

Give your answer to two decimal places.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

$M_r(\text{AgNO}_3) = 107.9 + 14.0 + 3(16.0) = 169.9$

Amount of $\text{AgNO}_3$ :

n = cV = 0.0500 \times \frac{250.0}{1000} = 0.0125\text{ mol}

Mass of $\text{AgNO}_3$ :

m = n \times M_r = 0.0125 \times 169.9 = 2.12\text{ g}

Answer

2.12 g

Final answer

2.12 g

Detailed explanation

Background Concept

Concentration is amount of solute per unit volume: $c = \frac{n}{V}$ , so $n = cV$ . The mass of a solid is related to its amount by $m = n \times M_r$ . Volumes in cm $^3$ must be converted to dm $^3$ by dividing by 1000.

Understanding the Question

We need the mass of solid silver nitrate required to prepare 250.0 cm $^3$ of 0.0500 mol dm $^{-3}$ AgNO $_3$ (aq). This is a standard solution-preparation calculation.

Approach

First find the amount of AgNO $_3$ needed using $n = cV$ . Then convert this amount to a mass using the molar mass of AgNO $_3$ .

Step-by-Step Reasoning

Volume = 250.0 cm $^3$ = 0.2500 dm $^3$ .
$n = 0.0500 \times 0.2500 = 0.0125$ mol.
$M_r(\text{AgNO}_3) = 107.9 + 14.0 + 48.0 = 169.9$ g mol $^{-1}$ .
$m = 0.0125 \times 169.9 = 2.12375$ g, which is 2.12 g to two decimal places.

Key Takeaways

Use $n = cV$ and $m = nM_r$ . Always convert cm $^3$ to dm $^3$ before using concentration in mol dm $^{-3}$ .

Common Mistakes

Forgetting to divide the volume by 1000, giving 12.5 mol.
Using an incorrect molar mass for AgNO $_3$ .
Rounding to 2.1 g instead of 2.12 g.

Things to Be Careful About

Use the periodic-table values for atomic masses.
Give the final answer to two decimal places as requested.
Include the unit g.

Techniques used

calculate amount in moles from concentration and volumecalculate mass from amount and molar mass

(ii)

Describe how the student should make $250.0\text{ cm}^3$ of $0.0500\text{ mol dm}^{-3}\text{ AgNO}_3(\text{aq})$ starting from the mass of $\text{AgNO}_3(\text{s})$ calculated in (a)(i) in a $50\text{ cm}^3$ beaker.

Give the name and size of any key 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 50 cm $^3$ beaker containing the 2.12 g of AgNO $_3$ and dissolve the solid.
Transfer the solution and washings through a funnel into a 250 cm $^3$ volumetric flask.
Make up to the calibration mark with distilled water.
Stopper the flask and invert it several times to mix thoroughly.

Final answer

Dissolve in water, transfer quantitatively to a 250 cm3 volumetric flask, make up to the mark, and mix.

Detailed explanation

Background Concept

Preparing a standard solution of exact concentration requires quantitative transfer: every particle of solute must end up in the volumetric flask, and the final volume must be made up exactly to the calibration mark. A 50 cm $^3$ beaker is too small to hold 250 cm $^3$ , so the solution must be transferred to a larger vessel.

Understanding the Question

Describe, in numbered steps, how to make 250 cm $^3$ of 0.0500 mol dm $^{-3}$ AgNO $_3$ starting from the solid in a 50 cm $^3$ beaker. Name and give the size of key apparatus.

Approach

Dissolve the solid, transfer it quantitatively to a 250 cm $^3$ volumetric flask, make up to the mark with distilled water, and mix thoroughly.

Step-by-Step Reasoning

Dissolve the AgNO $_3$ in a small volume of distilled water in the beaker.
Pour the solution into a 250 cm $^3$ volumetric flask through a funnel.
Rinse the beaker and funnel with distilled water and add the washings to the flask; this ensures all solute is transferred.
Add distilled water until the bottom of the meniscus sits exactly on the calibration mark.
Stopper and invert the flask several times to ensure a uniform solution.

Key Takeaways

Quantitative transfer and making up to the mark are essential for an accurate concentration. The volumetric flask is the key piece of apparatus.

Common Mistakes

Using a measuring cylinder instead of a volumetric flask.
Not washing the beaker and funnel, leaving solute behind.
Filling above the calibration mark.

Things to Be Careful About

Use distilled or deionised water.
Read the meniscus at eye level.
Name the size: 250 cm $^3$ volumetric flask.

Techniques used

dissolve a solid in watertransfer quantitatively to a volumetric flaskmake up to the mark and mix

(b)

The student uses the following method.

step 1: Dissolve $1.58\text{ g}$ of $\text{BaCl}_2 \cdot x\text{H}_2\text{O}(\text{s})$ to form $250\text{ cm}^3$ of aqueous solution. Label this solution A.
step 2: Transfer $20.0\text{ cm}^3$ of solution A into a conical flask.
step 3: Add aqueous sodium sulfate, $\text{Na}_2\text{SO}_4(\text{aq})$ , to the flask and swirl the mixture to remove barium ions from the solution.
step 4: Add 2–3 drops of $\text{K}_2\text{CrO}_4(\text{aq})$ indicator to the flask.
step 5: Titrate the contents of the flask against $0.0500\text{ mol dm}^{-3}\text{ AgNO}_3(\text{aq})$ .
step 6: Repeat steps 2 to 5 to collect sufficient data for analysis.

(i)

Suggest a suitable piece of apparatus for transferring $20.0\text{ cm}^3$ of solution A in step 2.

1M Similar

(ii)

Suggest why barium ions are removed in step 3 before performing the titration.

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(iii)

Suggest why chemically resistant gloves should be worn to carry out step 4.

1M Similar

(c)

The student’s results are shown in Table 1.1.

Table 1.1

	rough titration	titration 1	titration 2	titration 3
burette reading (final) / $\text{cm}^3$	20.10	40.55	20.75	20.90
burette reading (initial) / $\text{cm}^3$	0.00	20.25	0.05	0.30
titre / $\text{cm}^3$	20.10	20.30	20.70	20.60

The student uses the titres from titrations 2 and 3 shown in Table 1.1 to calculate a mean titre value of $20.65\text{ cm}^3$ .

(i)

Explain why only these two values are used.

1M Similar

(ii)

Calculate the percentage error in the titre volume for titration 3.

Show your working.

1M Similar

(d)

The equation for the reaction of silver nitrate with barium chloride is shown.

2\text{AgNO}_3(\text{aq}) + \text{BaCl}_2(\text{aq}) \rightarrow \text{Ba}(\text{NO}_3)_2(\text{aq}) + 2\text{AgCl}(\text{s})

(i)

Calculate the amount, in mol, of $\text{AgNO}_3(\text{aq})$ in the mean titre of $20.65\text{ cm}^3$ .

1M Similar

(ii)

Calculate the amount, in mol, of $\text{BaCl}_2(\text{aq})$ in $250\text{ cm}^3$ of solution A.

1M Similar

(iii)

Calculate the value of $x$ in the formula $\text{BaCl}_2 \cdot x\text{H}_2\text{O}$ .

2M Similar

(e)

Another student uses a different experimental method to check the value of $x$ obtained by the method described in (b).

Give a brief description of another method, not involving titration, that could be used to determine the value of $x$ in the formula $\text{BaCl}_2 \cdot x\text{H}_2\text{O}(\text{s})$ . Write your answer using a series of numbered steps.

Your plan should include details of the following:

the apparatus and method you would use
the measurements you would make.

You are provided with standard laboratory apparatus.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

Weigh a clean, dry crucible (and lid) using a balance.
Add a known mass of $\text{BaCl}_2 \cdot x\text{H}_2\text{O}$ (s) to the crucible and weigh again.
Heat the crucible strongly with a Bunsen burner, using a pipeclay triangle, until the water of crystallisation is driven off.
Allow to cool in a desiccator, then weigh.
Repeat heating, cooling and weighing until constant mass is obtained.
Calculate the mass of water lost = initial mass – final mass.
Use the moles of water lost and moles of anhydrous $\text{BaCl}_2$ to find $x$ .

Final answer

Heat a known mass of hydrated BaCl2 to constant mass, measure mass loss, and calculate x from moles of water lost.

Detailed explanation

Background Concept

Water of crystallisation can be removed by heating. The mass lost is the mass of water. From the mass of anhydrous BaCl $_2$ remaining and the mass of water lost, the mole ratio $n(\text{H}_2\text{O}) : n(\text{BaCl}_2)$ gives $x$ .

Understanding the Question

Plan another method, not involving titration, to determine $x$ in $\text{BaCl}_2 \cdot x\text{H}_2\text{O}$ . Include apparatus, method and measurements.

Approach

Use heating to drive off water, measuring masses before and after. The key is heating to constant mass to ensure all water is removed.

Step-by-Step Reasoning

Weigh an empty crucible and lid.
Add a known mass of hydrated salt and weigh again.
Heat strongly with a Bunsen burner; the pipeclay triangle supports the crucible.
Cool in a desiccator and weigh.
Repeat until constant mass, proving complete dehydration.
Mass of water lost = initial hydrated mass – final anhydrous mass.
Moles of anhydrous BaCl $_2$ = final mass / 208.3; moles of water = mass of water / 18.0; $x = n(\text{H}_2\text{O}) / n(\text{BaCl}_2)$ .

Key Takeaways

Heating to constant mass is essential for accurate dehydration experiments.

Common Mistakes

Not heating to constant mass.
Using the initial hydrated mass to calculate moles of BaCl $_2$ .
Forgetting to cool before weighing.

Things to Be Careful About

Use the anhydrous mass to calculate moles of BaCl $_2$ .
Ensure all water is removed before the final weighing.

Techniques used

heat to constant massmeasure masses before and after heatingcalculate moles of water lost

Q2MediumPlanningAnalysis, Conclusions and Evaluation

Effusion is the process in which a gas escapes through a small hole.

A student investigates the relationship between rate of effusion and relative molar mass of a gas using the apparatus shown in Fig. 2.1.

Fig. 2.1

The following method is used:

step 1: Turn the tap and remove any gas from the syringe through the side-arm tube, by pushing in the plunger.
step 2: Add $50\text{ cm}^3$ of the gas being tested to the syringe through the side-arm tube.
step 3: Remove the gas from the syringe, through the side-arm tube, by pushing in the plunger.
step 4: Add $70\text{ cm}^3$ of the gas being tested to the syringe through the side-arm tube.
step 5: Turn the tap to connect the syringe to the tube with the aluminium foil and small hole.
step 6: Allow the syringe plunger to fall and start a timer when the volume of gas in the syringe reaches $60\text{ cm}^3$ .
step 7: Stop the timer when the volume of gas in the syringe reaches $10\text{ cm}^3$ . Record the time taken.
step 8: Repeat steps 1 to 7 with different gases.

(a)

Suggest why the student adds $50\text{ cm}^3$ of the gas being tested to the syringe in step 2 and then removes this gas in step 3.

1M Similar

DifficultyEasy

Worked solution

Answer

To ensure no air or other gas remains in the syringe.

Final answer

To ensure no air or other gas remains in the syringe.

Detailed explanation

Background Concept

In gas experiments, the apparatus must be completely filled with the gas being tested to avoid contamination from atmospheric air. Air is a mixture of gases (primarily nitrogen and oxygen) with different molar masses. If air remains in the syringe, the effusion rate measured will be an average of the test gas and air, leading to inaccurate results.

Understanding the Question

The student performs steps 2 and 3: adding 50 cm³ of gas and then pushing it out before adding the final 70 cm³. The question asks for the purpose of this 'rinsing' or 'purging' step.

Approach

Think about what is in the syringe before step 2. It contains air. If the student just added 70 cm³ of test gas directly, the syringe would contain a mixture of air and test gas. By adding and removing 50 cm³ first, the air is displaced. Adding another 70 cm³ afterwards ensures the syringe contains only the pure test gas.

Step-by-Step Reasoning

Before step 2, the syringe and connecting tubes contain air.
Adding 50 cm³ of test gas in step 2 mixes with the air.
Removing this mixture in step 3 pushes the contaminated gas out through the side-arm tube.
This process is repeated (implicitly or by the subsequent addition) to ensure the final 70 cm³ added in step 4 is pure test gas.
The mark scheme reward is simply stating that this ensures no air or other gas remains.

Key Takeaways

Always purge gas syringes and reaction vessels with the gas being tested to displace atmospheric air. This is a fundamental technique in quantitative gas experiments.

Common Mistakes

Students often write 'to clean the syringe' or 'to remove moisture'. While purging can remove some moisture, the primary chemical reason is to remove air (nitrogen, oxygen, etc.) which would affect the molar mass and effusion rate of the gas mixture.

Things to Be Careful About

Be precise with terminology. Say 'air' or 'atmospheric gas', not just 'impurities'. The mark scheme specifically looks for 'no air / other gas remains'.

Techniques used

purge the apparatusensure gas purity

(b)

The student’s results are shown in Table 2.1.

Table 2.1

gas	hydrogen, $\text{H}_2$	helium, $\text{He}$	neon, $\text{Ne}$	argon, $\text{Ar}$	krypton, $\text{Kr}$
relative molar mass, $M$	2.0	4.0	20.2	39.9	83.8
$\sqrt{\frac{1}{M}}$
time taken / s	10.8	15.3	34.5	39.7	70.4
rate of effusion / $\text{cm}^3\text{ s}^{-1}$

\text{rate of effusion} = \frac{\text{volume of gas}}{\text{time taken}}

(i)

Complete Table 2.1.

Give the values for $\sqrt{\frac{1}{M}}$ to three significant figures.

Give the values for rate of effusion to two decimal places.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

gas	hydrogen, $\text{H}_2$	helium, $\text{He}$	neon, $\text{Ne}$	argon, $\text{Ar}$	krypton, $\text{Kr}$
relative molar mass, $M$	2.0	4.0	20.2	39.9	83.8
$\sqrt{\frac{1}{M}}$	0.707	0.500	0.222	0.158	0.109
time taken / s	10.8	15.3	34.5	39.7	70.4
rate of effusion / $\text{cm}^3\text{ s}^{-1}$	4.63	3.27	1.45	1.26	0.71

Final answer

sqrt(1/M): 0.707, 0.500, 0.222, 0.158, 0.109; Rate: 4.63, 3.27, 1.45, 1.26, 0.71

Detailed explanation

Background Concept

Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass: $\text{rate} \propto \frac{1}{\sqrt{M}}$ . To test this, we calculate $\sqrt{\frac{1}{M}}$ (which is $\frac{1}{\sqrt{M}}$ ) and the rate of effusion ( $\frac{\text{volume}}{\text{time}}$ ) for each gas.

Understanding the Question

We are given a table with molar masses ( $M$ ) and times. We need to calculate two new rows: $\sqrt{\frac{1}{M}}$ to 3 significant figures, and rate of effusion to 2 decimal places. Note: The volume of gas effusing is $60 - 10 = 50 \text{ cm}^3$ (from steps 6 and 7).

Approach

For each gas, calculate $\frac{1}{M}$ , then take the square root. Round to 3 s.f.
Calculate rate = $\frac{50}{\text{time}}$ . Round to 2 d.p.

Step-by-Step Reasoning

Hydrogen ( $M=2.0$ ):
$\sqrt{\frac{1}{2.0}} = \sqrt{0.5} = 0.7071... \rightarrow 0.707$ (3 s.f.)
Rate = $\frac{50}{10.8} = 4.629... \rightarrow 4.63$ (2 d.p.)

Helium ( $M=4.0$ ):
$\sqrt{\frac{1}{4.0}} = \sqrt{0.25} = 0.500$ (3 s.f.)
Rate = $\frac{50}{15.3} = 3.267... \rightarrow 3.27$ (2 d.p.)

Neon ( $M=20.2$ ):
$\sqrt{\frac{1}{20.2}} = \sqrt{0.04950...} = 0.2224... \rightarrow 0.222$ (3 s.f.)
Rate = $\frac{50}{34.5} = 1.449... \rightarrow 1.45$ (2 d.p.)

Argon ( $M=39.9$ ):
$\sqrt{\frac{1}{39.9}} = \sqrt{0.02506...} = 0.1583... \rightarrow 0.158$ (3 s.f.)
Rate = $\frac{50}{39.7} = 1.259... \rightarrow 1.26$ (2 d.p.)

Krypton ( $M=83.8$ ):
$\sqrt{\frac{1}{83.8}} = \sqrt{0.01193...} = 0.1092... \rightarrow 0.109$ (3 s.f.)
Rate = $\frac{50}{70.4} = 0.7102... \rightarrow 0.71$ (2 d.p.)

Key Takeaways

Always pay attention to significant figure instructions. $\sqrt{\frac{1}{M}}$ is calculated from $M$ , so the number of s.f. in $M$ doesn't strictly limit it, but 3 s.f. is standard. Rate is volume/time; volume is exactly 50 cm³ (from 60 to 10), so use the time's precision.

Common Mistakes

Calculating $\sqrt{M}$ instead of $\sqrt{\frac{1}{M}}$ . Forgetting to use the volume change (50 cm³) instead of the total volume (70 cm³) or initial volume (60 cm³). The timer starts at 60 cm³ and stops at 10 cm³, so 50 cm³ has effused.

Things to Be Careful About

The mark scheme requires 3 s.f. for $\sqrt{\frac{1}{M}}$ and 2 d.p. for rate. 0.5 must be written as 0.500 to show 3 s.f. 0.710 must be written as 0.71 to show 2 d.p. (trailing zeros after decimal are significant, but here 0.71 is sufficient for 2 d.p.).

Techniques used

calculate square root of inverse molar masscalculate rate from volume and time

(ii)

Identify the dependent variable in this experiment.

1M Similar

(iii)

Identify a variable, other than temperature, that is controlled when carrying out this experiment.

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(c)

Plot a graph on the grid in Fig. 2.2 to show the relationship between rate of effusion and $\sqrt{\frac{1}{M}}$ .

Use a cross ( $\times$ ) to plot each data point. Draw a suitable line of best fit.

Fig. 2.2

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Plot the following points on the grid in Fig. 2.2 using crosses ( $\times$ ):

$(0.707, 4.63)$
$(0.500, 3.27)$
$(0.222, 1.45)$
$(0.158, 1.26)$
$(0.109, 0.71)$

Draw a straight line of best fit that passes through the origin $(0,0)$ and as close to all points as possible.

Final answer

Graph with 5 points plotted and a straight line of best fit through the origin.

Detailed explanation

Background Concept

Graham's law predicts a linear relationship: $\text{rate} = k \sqrt{\frac{1}{M}}$ . A graph of rate (y-axis) against $\sqrt{\frac{1}{M}}$ (x-axis) should be a straight line passing through the origin. The gradient of this line is the constant $k$ .

Understanding the Question

Plot the calculated values from part (b)(i) on the given grid. The x-axis is $\sqrt{\frac{1}{M}}$ (0.0 to 0.8) and the y-axis is rate of effusion (0.0 to 5.0). Draw a line of best fit.

Approach

Read the coordinates from the table: $(x, y) = (\sqrt{\frac{1}{M}}, \text{rate})$ .
Plot each point carefully.
Draw a straight line through the origin that minimizes the distance to all points.

Step-by-Step Reasoning

Point 1 (Hydrogen): $x = 0.707$ , $y = 4.63$ . This is near the top right.
Point 2 (Helium): $x = 0.500$ , $y = 3.27$ . Mid-right.
Point 3 (Neon): $x = 0.222$ , $y = 1.45$ . Lower left.
Point 4 (Argon): $x = 0.158$ , $y = 1.26$ . This point is slightly below the line (anomalous).
Point 5 (Krypton): $x = 0.109$ , $y = 0.71$ . Bottom left.

The line should start at $(0,0)$ and go through $(0.707, 4.63)$ . The gradient is roughly $\frac{4.63}{0.707} \approx 6.55$ .

Key Takeaways

When plotting, ensure axes are read correctly. The line of best fit for a proportional relationship must pass through the origin. Don't force the line through an anomalous point; let it represent the general trend.

Common Mistakes

Plotting points incorrectly (swapping x and y). Forcing the line through all points including the anomalous one. Not drawing the line through the origin when the theory predicts it should.

Things to Be Careful About

Use a sharp pencil. Plot crosses ( $\times$ ), not dots. The line of best fit should have roughly equal numbers of points on either side (excluding the anomalous point if it's a clear outlier).

Techniques used

plot data points on a graphdraw line of best fit

(d)

Circle one point on the graph in Fig. 2.2 which you consider to be most anomalous.

Suggest one reason for this anomaly. Assume there is no error in $\sqrt{\frac{1}{M}}$ .

1M Similar

DifficultyMedium

Worked solution

Answer

Circle the point for argon ( $0.158, 1.26$ ) as it is the most anomalous (furthest below the line of best fit).

Reason: The measured (recorded) time for argon was greater than the true time taken (or the measured rate was lower than the true rate).

Final answer

Argon point circled; measured time was greater than the true time.

Detailed explanation

Background Concept

An anomalous result is a data point that does not fit the general trend or line of best fit. In this experiment, the trend is linear through the origin. If a point is below the line, its y-value (rate) is lower than expected for its x-value ( $\sqrt{\frac{1}{M}}$ ). Since rate = volume / time, a lower rate means a larger time was recorded.

Understanding the Question

Identify the most anomalous point on the graph and explain why, assuming no error in $\sqrt{\frac{1}{M}}$ (i.e., molar masses are correct).

Approach

Look at the graph. Which point is furthest from the line of best fit?
The argon point (0.158, 1.26) is below the line. The line predicts a rate of ~1.55 for $x=0.158$ , but the measured rate is 1.26.
Lower rate = longer time. So the student took too long to stop the timer, or started it late.

Step-by-Step Reasoning

Identify anomaly: The argon point is clearly below the line of best fit. Hydrogen, helium, neon, and krypton are close to or on the line. Argon is the outlier.
Analyze direction: The point is below the line, meaning the measured rate (1.26) is lower than the predicted rate (~1.55).
Relate to measurement: Rate = $\frac{50}{\text{time}}$ . A lower calculated rate means the time used in the calculation was larger than it should have been.
Conclusion: The student recorded a time that was too long (measured time > true time). This could be due to human reaction time starting or stopping the timer late.

Key Takeaways

When analyzing anomalies, determine if the measured value is higher or lower than expected, and link this to the raw measurement (time, volume, temperature, etc.).

Common Mistakes

Saying 'the student made a mistake' without explaining what kind of mistake. Not linking the position of the point (above/below line) to the direction of the error (time too long/short).

Things to Be Careful About

The mark scheme accepts 'measured time > true time' or 'measured rate < true rate'. Be specific about which variable was affected.

Techniques used

identify anomalous resultanalyze error in time measurement

(e)

Graham’s law of effusion can be expressed as:

\text{the rate of effusion of a gas is proportional to } \sqrt{\frac{1}{M}}

State whether or not the student’s results support Graham’s law of effusion.

Explain your answer, using the graph in Fig. 2.2.

1M Similar

(f)

Suggest how the position of the plotted points relative to the line of best fit in Fig. 2.2 is related to the reliability of the results.

1M Similar

(g)

The student then repeats this method to determine the value of $M$ of a sample of natural gas.

The time recorded in step 7 is $31.6\text{ s}$ .

(i)

Use the graph in Fig. 2.2 and the student’s result to calculate the value of $M$ for this sample.

2M Similar

DifficultyMedium

Worked solution

Working

Step 1: Calculate the rate of effusion for natural gas.
Volume effused = $60 - 10 = 50 \text{ cm}^3$
Time taken = $31.6 \text{ s}$
$\text{Rate} = \frac{50}{31.6} = 1.582 \text{ cm}^3\text{ s}^{-1}$

Step 2: Read $\sqrt{\frac{1}{M}}$ from the graph.
On the graph, find $y = 1.58$ on the y-axis. Read across to the line of best fit, then down to the x-axis.
$\sqrt{\frac{1}{M}} \approx 0.225$ (allow $0.22 - 0.23$ )

Step 3: Calculate $M$ .
$\frac{1}{M} = (0.225)^2 = 0.0506$
$M = \frac{1}{0.0506} = 19.8 \text{ g mol}^{-1}$
(Allow $18 - 21$ depending on reading)

Answer

$M \approx 19.8 \text{ g mol}^{-1}$ (Accept range $18 - 21$ )

Final answer

19.8 g mol^-1 (range 18-21)

Detailed explanation

Background Concept

To find the molar mass of an unknown gas using Graham's law, we calculate its rate of effusion, find the corresponding $\sqrt{\frac{1}{M}}$ value from the calibration graph (plot of rate vs $\sqrt{\frac{1}{M}}$ ), and then solve for $M$ .

Understanding the Question

The student tests natural gas. Time = 31.6 s. Calculate $M$ using the graph.

Approach

Calculate rate = $\frac{50}{31.6}$ .
Use the graph to find $\sqrt{\frac{1}{M}}$ for this rate.
Square the value to get $\frac{1}{M}$ .
Invert to get $M$ .

Step-by-Step Reasoning

Rate calculation:
$\text{Rate} = \frac{50}{31.6} = 1.5823... \text{ cm}^3\text{ s}^{-1}$
Reading the graph:
Find $1.58$ on the y-axis (rate). Move horizontally to the line of best fit. Move vertically down to the x-axis ( $\sqrt{\frac{1}{M}}$ ).
The value is approximately $0.225$ (neon was $0.222$ with rate $1.45$ , so $1.58$ should be slightly higher, around $0.225-0.23$ ).
Let's use $0.225$ .
Calculating $M$ :
$\sqrt{\frac{1}{M}} = 0.225$
$\frac{1}{M} = 0.225^2 = 0.050625$
$M = \frac{1}{0.050625} = 19.75 \approx 19.8 \text{ g mol}^{-1}$

If the student reads $0.22$ : $M = \frac{1}{0.22^2} = \frac{1}{0.0484} = 20.7$
If the student reads $0.23$ : $M = \frac{1}{0.23^2} = \frac{1}{0.0529} = 18.9$

Acceptable range: $18 - 21$ .

Key Takeaways

When using a graph to find a value, read carefully. The graph is a calibration curve. Once you have the intermediate value ( $\sqrt{\frac{1}{M}}$ ), do the algebra to find the final quantity ( $M$ ).

Common Mistakes

Forgetting to square the x-axis value to get $\frac{1}{M}$ . Forgetting to invert to get $M$ . Using the wrong volume (e.g., 60 or 70 instead of 50).

Things to Be Careful About

State symbols are not required for $M$ , but units are good practice ( $\text{g mol}^{-1}$ or just 'relative molar mass'). The graph reading will have uncertainty, so a range of answers is acceptable in the mark scheme.

Techniques used

calculate rate from volume and timeread value from graphcalculate molar mass from graph value

(ii)

Natural gas is a mixture of mainly methane, $\text{CH}_4$ , with small amounts of other gases.

Suggest what your calculated value of the $M$ of natural gas in (g)(i) tells you about the other gases in the mixture.

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DifficultyMedium-Easy

Worked solution

Answer

The calculated $M$ ( $\approx 19.8$ ) is greater than the $M$ of methane ( $\text{CH}_4 = 16$ ).

This suggests that the other gases in the mixture have a greater average molar mass than methane (or greater than 16).

Final answer

The calculated M > 16, so the other gases have a greater average molar mass than methane.

Detailed explanation

Background Concept

Natural gas is mainly methane ( $\text{CH}_4$ , $M = 12 + 4(1) = 16$ ). The experiment gives an average molar mass for the mixture. If the average is higher than 16, the impurities must be heavier gases (higher molar mass) to pull the average up.

Understanding the Question

The calculated $M$ is ~19.8. Methane is 16. What does this tell us about the other gases?

Approach

Compare calculated $M$ (19.8) with methane $M$ (16).
Since $19.8 > 16$ , the mixture is heavier than pure methane.
Therefore, the other gases must be heavier than methane.

Step-by-Step Reasoning

Pure methane has $M = 16$ .
The natural gas sample has an average $M \approx 19.8$ .
The average is a weighted mean: $M_{\text{avg}} = x_{\text{CH}_4} M_{\text{CH}_4} + x_{\text{other}} M_{\text{other}}$ .
Since $M_{\text{avg}} > M_{\text{CH}_4}$ , it must be that $M_{\text{other}} > M_{\text{CH}_4}$ .
Therefore, the other gases have a molar mass greater than 16.

Key Takeaways

The average molar mass of a mixture lies between the molar masses of its components. If the average is higher than the main component, the minor components must be heavier.

Common Mistakes

Saying 'the other gases are heavier'. Must specify 'heavier than methane' or 'molar mass > 16'.

Things to Be Careful About

Be precise. Don't just say 'they are heavier'. Say 'their average molar mass is greater than 16' or 'greater than methane'.

Techniques used

compare calculated molar mass to known values

(h)

The experiment described in (g) is repeated at a higher temperature.

Suggest how the rate of effusion for this sample of natural gas would change, if at all.

Explain your answer.

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