Chemistry 9701/51 — 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.

mass of $\text{AgNO}_3(\text{s}) =$ .............................. g

1M Similar

DifficultyEasy

Worked solution

Working

n(\text{AgNO}_3) = cV = 0.0500 \times \frac{250.0}{1000} = 0.0125 \text{ mol}

M_r(\text{AgNO}_3) = 107.9 + 14.0 + 3(16.0) = 169.9 \text{ g mol}^{-1}

m = n \times M_r = 0.0125 \times 169.9 = 2.12375 \text{ g}

Answer

mass of $\text{AgNO}_3(\text{s})$ = 2.12 g

Final answer

2.12 g

Detailed explanation

Background Concept

The amount of a substance in moles is related to its concentration and the volume of solution by $n = cV$ , where $n$ is the amount in mol, $c$ is the concentration in mol dm $^{-3}$ , and $V$ is the volume in dm $^3$ . To convert an amount into a mass of solid, use $m = n \times M_r$ , where $m$ is the mass in g and $M_r$ is the relative molecular mass (molar mass) in g mol $^{-1}$ . These two relationships underpin nearly every volumetric calculation in A-Level chemistry.

Understanding the Question

The student needs 250.0 cm $^3$ of a 0.0500 mol dm $^{-3}$ silver nitrate solution. The task is to find the mass of solid AgNO $_3$ that must be weighed out to prepare this solution. The volume is given in cm $^3$ , so it must first be converted to dm $^3$ before substitution into $n = cV$ .

Approach

Convert 250.0 cm $^3$ to dm $^3$ by dividing by 1000.
Calculate the amount of AgNO $_3$ needed using $n = cV$ .
Calculate the relative molecular mass of AgNO $_3$ from its formula.
Convert the amount to a mass using $m = n \times M_r$ .
Round the answer to two decimal places.

Step-by-Step Reasoning

Convert the volume: $250.0 \text{ cm}^3 = 0.2500 \text{ dm}^3$ because $1 \text{ dm}^3 = 1000 \text{ cm}^3$ .

Calculate the amount of AgNO $_3$ needed:

n = cV = 0.0500 \times 0.2500 = 0.0125 \text{ mol}

Calculate the relative molecular mass of AgNO $_3$ :

M_r = 107.9 + 14.0 + 3(16.0) = 169.9 \text{ g mol}^{-1}

(Ag = 107.9, N = 14.0, O = 16.0).

Convert to mass:

m = n \times M_r = 0.0125 \times 169.9 = 2.12375 \text{ g}

Rounded to two decimal places, the mass is 2.12 g.

Key Takeaways

Always convert cm $^3$ to dm $^3$ before using $n = cV$ .
The mass of solid required is found by multiplying the amount in moles by the molar mass.
Round only at the final stage to the precision requested.

Common Mistakes

Forgetting to convert cm $^3$ to dm $^3$ , which makes the answer 1000 times too large.
Using an incorrect relative molecular mass (e.g. missing the three oxygen atoms in AgNO $_3$ ).
Rounding intermediate values, which can shift the final answer.

Things to Be Careful About

The answer must be given to two decimal places: 2.12 g, not 2.1 g or 2.123 g.
$M_r(\text{AgNO}_3) = 169.9$ g mol $^{-1}$ — check each element's atomic mass before adding.

Techniques used

convert volume from cm³ to dm³calculate amount of solute from concentration and volumeconvert amount to mass using relative molecular 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 $\text{AgNO}_3(\text{s})$ and swirl or stir to dissolve the solid completely.
Pour the solution through a funnel into a 250 cm $^3$ volumetric flask.
Rinse the beaker, stirring rod and funnel with distilled water and add the washings to the flask.
Make up to the calibration mark with distilled water, adding the final drops with a dropping pipette.
Stopper the flask and invert it several times to mix thoroughly.

Final answer

Dissolve in a small volume of water, transfer with washings to a 250 cm³ volumetric flask, make up to the mark, stopper and invert to mix.

Detailed explanation

Background Concept

To prepare a solution of accurately known concentration, the solid must be dissolved and transferred quantitatively into a volumetric flask of the stated volume, then made up to the calibration mark. "Quantitatively" means that no solute is lost — every crystal and every drop of washings must reach the flask. The volumetric flask is the key piece of apparatus because it is calibrated to contain an exact volume (here 250.0 cm $^3$ ).

Understanding the Question

The student has weighed the calculated mass of AgNO $_3$ into a 50 cm $^3$ beaker. They must now describe, in numbered steps, how to turn this into exactly 250.0 cm $^3$ of solution, naming the key apparatus and its size. The question explicitly requires a series of numbered steps.

Approach

The standard procedure has three credited stages: (1) dissolve the solid in a small volume of water, (2) transfer the solution and washings into the volumetric flask, (3) make up to the mark and mix. Each stage must be described with the correct apparatus named.

Step-by-Step Reasoning

Stage 1 — dissolving (M1): Add a small volume of distilled water to the beaker and swirl or stir until all the solid has dissolved. A small volume is used so that the beaker can be rinsed effectively afterwards.

Stage 2 — transfer (M2): Pour the solution through a funnel into a 250 cm $^3$ volumetric flask. Rinse the beaker, the stirring rod and the funnel with distilled water and add these washings to the flask. This ensures that no AgNO $_3$ is left behind — quantitative transfer.

Stage 3 — making up and mixing (M3): Add distilled water until the liquid level is near the calibration mark, then add the final drops with a dropping pipette so the bottom of the meniscus sits exactly on the mark. Stopper the flask and invert it several times to mix the solution thoroughly.

Key Takeaways

The volumetric flask is the essential apparatus for preparing an exact volume of solution.
Washing the beaker, rod and funnel is essential for quantitative transfer.
The solution must be mixed by inverting after making up to the mark.

Common Mistakes

Omitting the washings — this loses solute and makes the concentration too low.
Overshooting the calibration mark — the solution must be discarded and restarted.
Mixing before making up to the mark, or not mixing at all.

Things to Be Careful About

Name the 250 cm $^3$ volumetric flask specifically; the size is part of the mark.
The final drops must be added with a dropping pipette to avoid overshooting.

Techniques used

dissolve a solid in a small volume of watertransfer the solution quantitatively to a volumetric flaskmake up to the calibration mark and mix thoroughly

(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.

1M Similar

(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.

percentage error = ..............................

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Each burette reading has an uncertainty of $\pm 0.05$ cm $^3$ , and a titre is the difference of two readings, so the total uncertainty is $2 \times 0.05 = 0.10$ cm $^3$ .

\text{percentage error} = \frac{2 \times 0.05}{20.60} \times 100 = \frac{0.10}{20.60} \times 100 = 0.4854\%

Answer

percentage error = 0.49%

Final answer

0.49%

Detailed explanation

Background Concept

The burette has an uncertainty of $\pm 0.05$ cm $^3$ on each reading. A titre is calculated as the difference between the final and initial readings, so the total uncertainty is the sum of the two reading uncertainties: $2 \times 0.05 = 0.10$ cm $^3$ . Percentage error is the absolute uncertainty divided by the measured value, multiplied by 100.

Understanding the Question

Calculate the percentage error in the titre volume for titration 3, whose titre is 20.60 cm $^3$ . Working must be shown.

Approach

Determine the absolute uncertainty: $2 \times 0.05 = 0.10$ cm $^3$ .
Divide by the titre (20.60 cm $^3$ ) and multiply by 100.

Step-by-Step Reasoning

The titre for titration 3 is 20.60 cm $^3$ . The burette readings are recorded to the nearest 0.05 cm $^3$ , so each reading has an uncertainty of $\pm 0.05$ cm $^3$ . Since the titre is the difference between two readings, the total uncertainty is $0.05 + 0.05 = 0.10$ cm $^3$ .

Calculate the percentage error:

\text{percentage error} = \frac{0.10}{20.60} \times 100 = 0.4854\% \approx 0.49\%

The mark scheme accepts 0.48(54)% — both 0.48% and 0.49% are reasonable roundings.

Key Takeaways

Burette error is doubled because two readings are taken (final and initial).
Percentage error = (absolute uncertainty / measured value) × 100.

Common Mistakes

Using only 0.05 cm $^3$ instead of 0.10 cm $^3$ — forgetting that two readings are involved.
Using the wrong titre (e.g. the mean 20.65 instead of 20.60 for titration 3).

Things to Be Careful About

The titre for titration 3 is 20.60 cm $^3$ , not the mean 20.65 cm $^3$ .
Show the working clearly, as the mark is for the calculation as well as the value.

Techniques used

calculate percentage errorpropagate the burette reading uncertainty

(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 $20.65\text{ cm}^3$ .

amount of $\text{AgNO}_3 =$ .............................. mol

1M Similar

(ii)

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

amount of $\text{BaCl}_2 =$ .............................. mol

1M Similar

DifficultyMedium-Easy

Worked solution

Working

From the equation, 2 mol $\text{AgNO}_3$ reacts with 1 mol $\text{BaCl}_2$ .

n(\text{BaCl}_2) \text{ in } 20.0 \text{ cm}^3 = \frac{0.0010325}{2} = 0.00051625 \text{ mol}

n(\text{BaCl}_2) \text{ in } 250 \text{ cm}^3 = 0.00051625 \times \frac{250}{20} = 0.006453125 \text{ mol}

Answer

amount of $\text{BaCl}_2$ = 6.45 × 10⁻³ mol

Final answer

6.45 × 10⁻³ mol

Detailed explanation

Background Concept

The balanced equation shows that 2 mol AgNO $_3$ reacts with 1 mol BaCl $_2$ . The amount of BaCl $_2$ in the 20.0 cm $^3$ aliquot is therefore half the amount of AgNO $_3$ used. The aliquot is only a fraction of the total 250 cm $^3$ solution, so the amount must be scaled up by the factor $250/20 = 12.5$ .

Understanding the Question

Given the amount of AgNO $_3$ in the mean titre (found in (d)(i)), find the amount of BaCl $_2$ in the whole 250 cm $^3$ of solution A.

Approach

Halve the AgNO $_3$ amount to find the BaCl $_2$ in the 20.0 cm $^3$ aliquot (stoichiometry).
Multiply by $250/20 = 12.5$ to scale to the full 250 cm $^3$ .

Step-by-Step Reasoning

Amount of BaCl $_2$ in the 20.0 cm $^3$ aliquot:

0.0010325 / 2 = 0.00051625 \text{ mol}

Scale to 250 cm $^3$ :

0.00051625 \times \frac{250}{20} = 0.00051625 \times 12.5 = 0.006453125 \text{ mol}

This is approximately $6.45 \times 10^{-3}$ mol.

Key Takeaways

Use the stoichiometric ratio from the balanced equation.
The scaling factor is total volume / aliquot volume = 250/20 = 12.5.

Common Mistakes

Forgetting to halve for the 2:1 stoichiometry.
Using the wrong scale factor (e.g. 20/250 instead of 250/20).

Things to Be Careful About

The mark scheme: = (d)(i) / 2 × (250/20) = (d)(i) × 6.25.

Techniques used

use stoichiometry to find moles of BaCl₂scale from the aliquot to the total volume

(iii)

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

$x =$ ..............................

2M Similar

DifficultyMedium

Worked solution

Working

M_r(\text{BaCl}_2 \cdot x\text{H}_2\text{O}) = \frac{1.58}{0.006453125} = 244.84 \text{ g mol}^{-1}

x = \frac{244.84 - 208.3}{18} = \frac{36.54}{18} = 2.03

Answer

$x$ = 2

Final answer

x = 2

Detailed explanation

Background Concept

The molar mass of the hydrated salt is found from $M_r = m/n$ , using the mass of the sample (1.58 g) and the amount of BaCl $_2$ found in (d)(ii). The difference between this molar mass and the molar mass of anhydrous BaCl $_2$ (208.3 g mol $^{-1}$ ) is the mass of water of crystallisation per mole. Dividing by 18 (the molar mass of water) gives $x$ .

Understanding the Question

Given 1.58 g of BaCl $_2 \cdot x$ H $_2$ O contains the amount of BaCl $_2$ found in (d)(ii), find $x$ .

Approach

Calculate $M_r$ (hydrate) = mass / amount.
Subtract $M_r$ (BaCl $_2$ ) = 208.3.
Divide by 18 to find $x$ , and round to an integer.

Step-by-Step Reasoning

First, $M_r$ (BaCl $_2$ ) = 137.3 + 2(35.5) = 208.3 g mol $^{-1}$ .

From (d)(ii), the amount of BaCl $_2$ in 250 cm $^3$ is 0.006453125 mol, and this is the amount in the 1.58 g sample.

Molar mass of the hydrate:

M_r = \frac{1.58}{0.006453125} = 244.84 \text{ g mol}^{-1}

Mass of water per mole:

244.84 - 208.3 = 36.54 \text{ g mol}^{-1}

Number of water molecules:

x = \frac{36.54}{18} = 2.03 \approx 2

So $x = 2$ , and the formula is BaCl $_2 \cdot 2$ H $_2$ O.

Key Takeaways

$M_r = m/n$ .
Water of crystallisation is found from the mass difference between hydrate and anhydrous salt.
$x$ must be an integer.

Common Mistakes

Using the wrong $M_r$ for BaCl $_2$ (e.g. forgetting the two chlorines).
Not rounding $x$ to an integer.
Using the amount in the 20 cm $^3$ aliquot instead of the 250 cm $^3$ solution.

Things to Be Careful About

$x$ must be an integer (mark scheme requirement).
Some working must be shown for both marks.

Techniques used

calculate relative molecular mass from mass and amountdeduce water of crystallisation from the molar-mass difference

(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

Worked solution

Answer

Weigh a clean, dry crucible (with lid) and record its mass.
Add a known mass of $\text{BaCl}_2 \cdot x\text{H}_2\text{O}(\text{s})$ (e.g. about 1–2 g) and record the total mass.
Heat the crucible strongly with a Bunsen burner (supported on a pipeclay triangle) to drive off the water of crystallisation.
Cool the crucible in a desiccator and reweigh.
Repeat heating, cooling and reweighing until constant mass is obtained (no further mass loss).
Mass of water lost = mass before heating − mass after heating.
Amount of water = mass of water / 18; amount of BaCl $_2$ = mass of residue / 208.3.
$x$ = amount of water / amount of BaCl $_2$ .

Final answer

See working — gravimetric method: heat to constant mass, mass loss gives moles of water and hence x.

Detailed explanation

Background Concept

Water of crystallisation can be determined gravimetrically. Heating the hydrated salt drives off the water, leaving anhydrous BaCl $_2$ . The loss in mass equals the mass of water driven off. From the masses of water and anhydrous salt, the mole ratio gives $x$ .

Understanding the Question

Another student wants to check the value of $x$ using a method that does not involve titration. The candidate must describe a complete plan in numbered steps, including apparatus, method, and measurements. Standard laboratory apparatus is available.

Approach

The classic gravimetric method: heat a known mass of the hydrate to constant mass, measure the mass loss, and calculate the mole ratio of water to BaCl $_2$ .

Step-by-Step Reasoning

Apparatus: A crucible (with lid), a pipeclay triangle, a tripod, a Bunsen burner, a desiccator, and a balance.

Method:

Weigh the empty crucible with its lid.
Add a known mass of the hydrated salt and weigh again — the difference is the mass of hydrate used.
Heat strongly to drive off the water.
Cool in a desiccator (to prevent reabsorption of moisture) and reweigh.
Repeat heating, cooling and reweighing until constant mass — this ensures all the water has been driven off.

Calculations:

Mass of water = mass of hydrate − mass of anhydrous residue.
Amount of water = mass of water / 18.
Amount of BaCl $_2$ = mass of residue / 208.3.
$x$ = amount of water / amount of BaCl $_2$ .

Key Takeaways

Gravimetric analysis measures a mass change to find composition.
Heating to constant mass ensures complete dehydration.
Cooling in a desiccator prevents reabsorption of water.

Common Mistakes

Not heating to constant mass — incomplete dehydration gives a low value of $x$ .
Weighing hot samples — convection currents affect the balance reading.
Not cooling in a desiccator — the anhydrous salt reabsorbs moisture.

Things to Be Careful About

The mark scheme also accepts an alternative precipitation method (e.g. precipitate BaSO $_4$ with sulfate, filter, wash, dry to constant mass, and use the mass of precipitate).
The plan must include apparatus, measurements, and a key step (heat to constant mass).

Techniques used

design a gravimetric methodheat to constant masscalculate water of crystallisation from the mass loss

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

(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

Working

For $\sqrt{\frac{1}{M}}$ : $\sqrt{\frac{1}{2.0}} = 0.707$ , $\sqrt{\frac{1}{4.0}} = 0.500$ , $\sqrt{\frac{1}{20.2}} = 0.222$ , $\sqrt{\frac{1}{39.9}} = 0.158$ , $\sqrt{\frac{1}{83.8}} = 0.109$ .

For rate of effusion: $\text{rate} = \frac{50}{\text{time}}$ .
$\text{H}_2: \frac{50}{10.8} = 4.63$ ; $\text{He}: \frac{50}{15.3} = 3.27$ ; $\text{Ne}: \frac{50}{34.5} = 1.45$ ; $\text{Ar}: \frac{50}{39.7} = 1.26$ ; $\text{Kr}: \frac{50}{70.4} = 0.71$ .

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 / cm $^3$ 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 of effusion 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 $\sqrt{\frac{1}{M}}$ and the rate of effusion ( $\frac{\text{volume}}{\text{time}}$ ) for each gas.

Understanding the Question

Complete the table by calculating $\sqrt{\frac{1}{M}}$ to 3 significant figures and the rate of effusion to 2 decimal places. The volume of gas effusing is 50 cm³ (from 60 cm³ down to 10 cm³).

Approach

Use a calculator to compute the values. Pay close attention to the requested significant figures and decimal places.

Step-by-Step Reasoning

Calculate $\sqrt{\frac{1}{M}}$ :
- H₂: $\sqrt{1/2.0} = 0.7071... \rightarrow 0.707$ (3 s.f.)
- He: $\sqrt{1/4.0} = 0.500$ (3 s.f.)
- Ne: $\sqrt{1/20.2} = 0.2222... \rightarrow 0.222$ (3 s.f.)
- Ar: $\sqrt{1/39.9} = 0.1581... \rightarrow 0.158$ (3 s.f.)
- Kr: $\sqrt{1/83.8} = 0.1092... \rightarrow 0.109$ (3 s.f.)
Calculate rate of effusion ( $\frac{50}{\text{time}}$ ):
- H₂: $50 / 10.8 = 4.629... \rightarrow 4.63$ (2 d.p.)
- He: $50 / 15.3 = 3.267... \rightarrow 3.27$ (2 d.p.)
- Ne: $50 / 34.5 = 1.449... \rightarrow 1.45$ (2 d.p.)
- Ar: $50 / 39.7 = 1.259... \rightarrow 1.26$ (2 d.p.)
- Kr: $50 / 70.4 = 0.7102... \rightarrow 0.71$ (2 d.p.)

Key Takeaways

Always check the required precision (s.f. or d.p.) before finalizing table values. The volume effusing is the difference between the start and end volumes: $60 - 10 = 50$ cm³.

Common Mistakes

Forgetting that the volume is 50 cm³, not 70 cm³ (70 cm³ is the total in the syringe, but only 50 cm³ effuses during the timed interval). Rounding intermediate values too early.

Things to Be Careful About

$\sqrt{\frac{1}{M}}$ must be to 3 s.f., not 3 d.p. Rate must be to 2 d.p. Ensure units are correct (cm³ s⁻¹).

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.

1M Similar

(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 (×) to plot each data point. Draw a suitable line of best fit.

Fig. 2.2

2M Similar

(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

(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.

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(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.

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(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.

$M =$ .........................................................

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(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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(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.

effect on the rate of effusion .....................................................................................................

explanation ...............................................................................................................................

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