Chemistry 9701/54 — October/November 2025

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

Q1MediumPlanningAnalysis, Conclusions and Evaluation

A student uses a technique called the Winkler method to determine the mass of oxygen dissolved in a sample of water from a lake.

Two solutions, X and Y, are prepared.

Solution X is $2.30\text{ mol dm}^{-3}$ aqueous manganese(II) sulfate, $\text{MnSO}_4\text{(aq)}$ .

Solution Y is alkaline aqueous potassium iodide, $\text{KI(aq)}$ .

(a)

Calculate the mass of solid hydrated manganese(II) sulfate, $\text{MnSO}_4\cdot\text{H}_2\text{O(s)}$ , needed to make $100.0\text{ cm}^3$ of solution X.

Give your answer to two decimal places.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

$n(\text{MnSO}_4\cdot\text{H}_2\text{O}) = c \times V = 2.30 \times 0.100 = 0.230\text{ mol}$

$M_r(\text{MnSO}_4\cdot\text{H}_2\text{O}) = 54.9 + 32.1 + 4(16.0) + 18.0 = 169.0$

$\text{mass} = 0.230 \times 169.0 = 38.87\text{ g}$

Answer

38.87 g

Final answer

38.87 g

Detailed explanation

Background Concept

Concentration in $\text{mol dm}^{-3}$ relates the amount of solute to the volume of solution: $n = c \times V$ . To find the mass of solid needed, use $\text{mass} = n \times M_r$ . A hydrated salt like $\text{MnSO}_4\cdot\text{H}_2\text{O}$ includes its water of crystallisation in the formula mass, so the molar mass is higher than that of the anhydrous salt.

Understanding the Question

We must calculate the mass of solid hydrated manganese(II) sulfate needed to prepare $100.0\text{ cm}^3$ of a $2.30\text{ mol dm}^{-3}$ solution. The volume is given in $\text{cm}^3$ and must be converted to $\text{dm}^3$ before using $n = cV$ .

Approach

Convert $100.0\text{ cm}^3$ to $0.100\text{ dm}^3$ .
Calculate the amount in moles: $n = c \times V$ .
Calculate the molar mass of $\text{MnSO}_4\cdot\text{H}_2\text{O}$ , including the one water of crystallisation.
Convert moles to mass: $\text{mass} = n \times M_r$ .

Step-by-Step Reasoning

$V = 100.0\text{ cm}^3 = 0.100\text{ dm}^3$
$n = 2.30 \times 0.100 = 0.230\text{ mol}$
$M_r(\text{MnSO}_4\cdot\text{H}_2\text{O}) = 54.9 + 32.1 + 4(16.0) + 18.0 = 169.0$
$\text{mass} = 0.230 \times 169.0 = 38.87\text{ g}$
The answer is given to two decimal places as required.

Key Takeaways

Always convert volume to $\text{dm}^3$ before using $n = cV$ . Include water of crystallisation when calculating the molar mass of a hydrated salt.

Common Mistakes

Forgetting the water of crystallisation, giving $M_r = 151.0$ instead of $169.0$ .
Using the volume in $\text{cm}^3$ directly without converting to $\text{dm}^3$ .
Rounding the molar mass incorrectly before multiplying.

Things to Be Careful About

The final answer must be given to two decimal places.
Use consistent $M_r$ values for each element (Mn ≈ 54.9, S ≈ 32.1, O = 16.0, H = 1.0).

Techniques used

convert volume from cm³ to dm³calculate moles from concentration and volumecalculate molar mass including water of crystallisationconvert moles to mass

(b)

The student is given a small beaker containing the mass of $\text{MnSO}_4\cdot\text{H}_2\text{O(s)}$ calculated in (a). Describe how the student should prepare exactly $100.0\text{ cm}^3$ of solution X.

Include the names and capacities of each piece of key apparatus used.

Write your answer using a series of numbered steps.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

Weigh out the 38.87 g of $\text{MnSO}_4\cdot\text{H}_2\text{O(s)}$ into a beaker.
Dissolve the solid in a small volume of distilled water in the beaker.
Transfer the solution to a $100\text{ cm}^3$ volumetric flask using a funnel.
Rinse the beaker and funnel with distilled water, adding the rinsings to the flask.
Add distilled water to the flask up to the $100\text{ cm}^3$ graduation mark, adding dropwise near the mark.
Stopper the flask and invert it several times to mix the solution thoroughly.

Final answer

Dissolve in distilled water; transfer to a 100 cm³ volumetric flask, rinsing the beaker; make up to the mark with distilled water; stopper and invert to mix.

Detailed explanation

Background Concept

Preparing a solution of exact concentration requires a volumetric flask, which has a single graduation mark calibrated to deliver a specific volume at a given temperature. Quantitative transfer — rinsing the beaker and funnel — ensures all the solute reaches the flask, so the final concentration matches the calculation.

Understanding the Question

The question asks for the procedure to make exactly $100.0\text{ cm}^3$ of solution X from the calculated mass of solid. Key apparatus: a $100\text{ cm}^3$ volumetric flask, a beaker, a funnel, and distilled water. The answer must be written as numbered steps with apparatus names and capacities.

Approach

Follow the standard sequence for preparing a solution of known concentration: dissolve → transfer → rinse → make up to the mark → mix.

Step-by-Step Reasoning

M1 (dissolve): The solid is dissolved in a small volume of distilled water in a beaker, so it fully dissolves before transfer.
M2 (transfer and rinse): The solution is poured into a $100\text{ cm}^3$ volumetric flask through a funnel. The beaker and funnel are rinsed with distilled water and the rinsings added, ensuring no solute is left behind (quantitative transfer).
M3 (make up and mix): Distilled water is added until the bottom of the meniscus sits on the graduation mark; the flask is stoppered and inverted several times to give a uniform solution.

Key Takeaways

The volumetric flask is the only apparatus that gives an exact volume. Rinsing is essential for accuracy. Mixing by inversion ensures homogeneity.

Common Mistakes

Using a measuring cylinder instead of a volumetric flask (not accurate enough).
Not rinsing the beaker, losing some solute and making the solution too dilute.
Adding water past the mark, which cannot be corrected.
Omitting the funnel or the inversion step.

Things to Be Careful About

The meniscus must sit exactly on the graduation mark.
The flask capacity must be stated as $100\text{ cm}^3$ .
The numbered-step format is required by the question.

Techniques used

dissolve a solid in distilled watertransfer solution quantitatively to a volumetric flaskmake up to the graduation markmix by inverting the flask

(c)

Solution Y is prepared as follows.


step 1	Place $100\text{ cm}^3$ of distilled water in a $250\text{ cm}^3$ beaker.
step 2	Add about $8\text{ g}$ of solid sodium hydroxide, $\text{NaOH(s)}$ , and stir to dissolve.
step 3	Cool the solution to room temperature using an ice-bath.
step 4	Repeat steps 2 and 3 until a total of $32\text{ g}$ of $\text{NaOH(s)}$ has been dissolved.
step 5	Dissolve about $14\text{ g}$ of potassium iodide, $\text{KI(s)}$ , into the solution formed in step 4.

(i)

Solution Y is corrosive.

Other than wearing safety goggles, state one safety precaution that the student should take when preparing solution Y.

1M Similar

(ii)

Suggest why the solution is cooled in step 3.

1M Similar

(d)

The student uses the following procedure to determine the mass of oxygen dissolved in a sample of water from the lake.


step 1	Collect a $250\text{ cm}^3$ sample of lake water in a bottle.
step 2	Add $1\text{ cm}^3$ of solution X and $1\text{ cm}^3$ of solution Y to the bottle.
step 3	Immediately stopper the bottle, ensuring as little air as possible is trapped.
step 4	Shake the bottle to mix its contents. A brown precipitate, manganese(III) hydroxide, $\text{Mn(OH)}_3\text{(s)}$ , is formed.
step 5	Add $1.5\text{ cm}^3$ of concentrated sulfuric acid to the contents of the bottle. The precipitate dissolves, and iodine is formed.
step 6	Dilute this solution to exactly $500.0\text{ cm}^3$ using distilled water to form solution Z.
step 7	Transfer $25.0\text{ cm}^3$ of solution Z into a conical flask, and titrate with $1.00 \times 10^{-3}\text{ mol dm}^{-3}$ aqueous sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3\text{(aq)}$ . Add $1\text{ cm}^3$ of starch solution near to the end-point.
step 8	Repeat step 7 as many times as necessary.

(i)

Suggest why it is important to avoid trapping air inside the bottle in step 3.

1M Similar

(ii)

Identify the piece of apparatus that the student should use to transfer the $25.0\text{ cm}^3$ of solution Z in step 7.

1M Similar

(iii)

Suggest why starch solution is added in step 7.

1M Similar

(e)

The student records the results shown in Table 1.1.

Table 1.1

	rough titration	titration 1	titration 2	titration 3
final burette reading / $\text{cm}^3$	13.60	12.75	26.20	14.50
initial burette reading / $\text{cm}^3$	0.00	0.05	13.15	1.35
titre / $\text{cm}^3$	13.60

(i)

Complete Table 1.1 and calculate the mean titre.

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Titration 1: $12.75 - 0.05 = 12.70\text{ cm}^3$

Titration 2: $26.20 - 13.15 = 13.05\text{ cm}^3$

Titration 3: $14.50 - 1.35 = 13.15\text{ cm}^3$

Concordant titres: 13.05 and 13.15 (within 0.10 cm³ of each other)

Mean titre = $\frac{13.05 + 13.15}{2} = 13.10\text{ cm}^3$

Answer

Titres: 12.70, 13.05, 13.15 cm³; mean titre = 13.10 cm³

Final answer

Mean titre = 13.10 cm³

Detailed explanation

Background Concept

A titre is the volume of titrant used, found by subtracting the initial burette reading from the final reading. Concordant titres — those within 0.10 cm³ of each other — are averaged to give a reliable mean; outliers are rejected.

Understanding the Question

Table 1.1 gives final and initial burette readings for a rough titration and three titrations. We must complete the titre column and calculate the mean titre.

Approach

Calculate each titre as final − initial.
Identify which titres are concordant (within 0.10 cm³).
Average only the concordant titres.

Step-by-Step Reasoning

Titre 1 = $12.75 - 0.05 = 12.70\text{ cm}^3$
Titre 2 = $26.20 - 13.15 = 13.05\text{ cm}^3$
Titre 3 = $14.50 - 1.35 = 13.15\text{ cm}^3$
Titration 1 (12.70) differs from the others by more than 0.10 cm³, so it is rejected as an outlier.
Titrations 2 and 3 (13.05 and 13.15) are concordant.
Mean = $\frac{13.05 + 13.15}{2} = 13.10\text{ cm}^3$

Key Takeaways

Reject outliers and average concordant titres. The rough titre is never included in the mean.

Common Mistakes

Averaging all three titres including the outlier (12.70).
Including the rough titre (13.60) in the mean.
Arithmetic errors in subtracting burette readings.

Things to Be Careful About

Concordance criterion is within 0.10 cm³.
The mean must be quoted to the same precision as the titres (two decimal places).

Techniques used

calculate titres from burette readingsidentify concordant titrescalculate a mean titre

(ii)

Explain why the student does not need to carry out any further titrations.

1M Similar

(iii)

Calculate the percentage error in the measurement of the titre for titration 3.

Show your working.

1M Similar

(f)

The following equations show the reactions that take place during the procedure in (d).

steps 2, 3 and 4

4\text{Mn}^{2+}\text{(aq)} + 8\text{OH}^-\text{(aq)} + \text{O}_2\text{(aq)} + 2\text{H}_2\text{O(l)} \rightarrow 4\text{Mn(OH)}_3\text{(s)}

step 5

2\text{Mn(OH)}_3\text{(s)} + 2\text{I}^-\text{(aq)} + 6\text{H}^+\text{(aq)} \rightarrow \text{I}_2\text{(aq)} + 6\text{H}_2\text{O(l)} + 2\text{Mn}^{2+}\text{(aq)}

step 7

\text{I}_2\text{(aq)} + 2\text{S}_2\text{O}_3^{2-}\text{(aq)} \rightarrow 2\text{I}^-\text{(aq)} + \text{S}_4\text{O}_6^{2-}\text{(aq)}

(i)

Calculate the amount, in mol, of iodine, $\text{I}_2\text{(aq)}$ , in $25.0\text{ cm}^3$ of solution Z.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

$n(\text{S}_2\text{O}_3^{2-}) = \frac{13.10 \times 1.00 \times 10^{-3}}{1000} = 1.31 \times 10^{-5}\text{ mol}$

From $\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}$ :

$n(\text{I}_2) = \frac{n(\text{S}_2\text{O}_3^{2-})}{2} = 6.55 \times 10^{-6}\text{ mol}$

Answer

$6.55 \times 10^{-6}\text{ mol}$

Final answer

6.55 × 10⁻⁶ mol

Detailed explanation

Background Concept

The titration reaction is $\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}$ , so 1 mol of iodine reacts with 2 mol of thiosulfate. Moles of thiosulfate are found from $n = c \times V$ using the mean titre.

Understanding the Question

Given the mean titre of 13.10 cm³ of $1.00 \times 10^{-3}\text{ mol dm}^{-3}$ thiosulfate, find the amount of iodine in 25.0 cm³ of solution Z.

Approach

Convert the titre to dm³.
Calculate moles of thiosulfate: $n = c \times V$ .
Halve to find moles of iodine using the 2:1 stoichiometric ratio.

Step-by-Step Reasoning

$V = 13.10\text{ cm}^3 = 0.01310\text{ dm}^3$
$n(\text{S}_2\text{O}_3^{2-}) = 1.00 \times 10^{-3} \times 0.01310 = 1.31 \times 10^{-5}\text{ mol}$
From the equation, $n(\text{I}_2) = \frac{1}{2} n(\text{S}_2\text{O}_3^{2-}) = 6.55 \times 10^{-6}\text{ mol}$

Key Takeaways

Use the mean titre (13.10 cm³), not individual titres. Apply the stoichiometric ratio from the balanced equation.

Common Mistakes

Forgetting to halve the thiosulfate moles for iodine.
Using an individual titre instead of the mean.
Failing to convert cm³ to dm³.

Things to Be Careful About

The ratio is 1 I₂ : 2 S₂O₃²⁻.
Keep consistent units (dm³ for volume in $n = cV$ ).

Techniques used

calculate moles of thiosulfate from titreuse stoichiometric ratio to find moles of iodine

(ii)

Use your answer to (f)(i) and the equations given to calculate the amount, in mol, of dissolved oxygen, $\text{O}_2\text{(aq)}$ , in $500.0\text{ cm}^3$ of solution Z.

1M Similar

DifficultyMedium

Worked solution

Working

From the equations: $1\text{ mol O}_2 \rightarrow 4\text{ mol Mn(OH)}_3 \rightarrow 2\text{ mol I}_2$ , so $n(\text{O}_2) = \frac{1}{2} n(\text{I}_2)$ .

$n(\text{O}_2)\text{ in }25.0\text{ cm}^3 = \frac{6.55 \times 10^{-6}}{2} = 3.275 \times 10^{-6}\text{ mol}$

$n(\text{O}_2)\text{ in }500.0\text{ cm}^3 = 3.275 \times 10^{-6} \times \frac{500}{25} = 6.55 \times 10^{-5}\text{ mol}$

Answer

$6.55 \times 10^{-5}\text{ mol}$

Final answer

6.55 × 10⁻⁵ mol

Detailed explanation

Background Concept

The three equations form a chain linking dissolved O₂ to the iodine titrated. From step 2–4: $4\text{Mn}^{2+} + 8\text{OH}^- + \text{O}_2 + 2\text{H}_2\text{O} \rightarrow 4\text{Mn(OH)}_3$ , so 1 mol O₂ gives 4 mol Mn(OH)₃. From step 5: $2\text{Mn(OH)}_3 + 2\text{I}^- + 6\text{H}^+ \rightarrow \text{I}_2 + 6\text{H}_2\text{O} + 2\text{Mn}^{2+}$ , so 2 mol Mn(OH)₃ give 1 mol I₂. Combining: 4 mol Mn(OH)₃ give 2 mol I₂, i.e. 1 mol O₂ gives 2 mol I₂. Thus $n(\text{O}_2) = \frac{1}{2} n(\text{I}_2)$ .

Understanding the Question

We know the moles of I₂ in 25.0 cm³ of solution Z from (f)(i). We must find the moles of dissolved O₂ in the full 500.0 cm³ of Z, using the stoichiometry of the equations.

Approach

Use the chain stoichiometry to relate O₂ to I₂.
Calculate moles of O₂ in 25.0 cm³.
Scale up from 25.0 cm³ to 500.0 cm³ (factor of 20).

Step-by-Step Reasoning

From the equations, $n(\text{O}_2) = \frac{1}{2} n(\text{I}_2)$ .
$n(\text{O}_2)\text{ in }25.0\text{ cm}^3 = \frac{6.55 \times 10^{-6}}{2} = 3.275 \times 10^{-6}\text{ mol}$
Scaling to 500.0 cm³: $3.275 \times 10^{-6} \times \frac{500}{25} = 6.55 \times 10^{-5}\text{ mol}$

Key Takeaways

Chain stoichiometry across multiple equations requires tracking the mole ratios carefully. The dilution factor is the ratio of total volume to aliquot volume.

Common Mistakes

Getting the O₂:I₂ ratio wrong (e.g. 1:1 instead of 1:2).
Forgetting to scale up from 25 cm³ to 500 cm³.
Using the wrong dilution factor.

Things to Be Careful About

The dilution factor is 20 (500/25).
The O₂:I₂ ratio is 1:2, so halve the iodine moles.

Techniques used

combine stoichiometry across multiple equationsscale moles for dilution

(iii)

Dissolved oxygen content, $\text{mg dm}^{-3}$ , is the mass of oxygen dissolved in water.

Use your answer to (f)(ii) to calculate the dissolved oxygen content in the lake water collected in step 1.

[If you were unable to obtain an answer to (f)(ii), then use amount of $\text{O}_2\text{(aq)}$ in $500.0\text{ cm}^3$ of solution Z = $7.12 \times 10^{-5}\text{ mol}$ . This is not the correct answer.]

1M Similar

DifficultyMedium

Worked solution

Working

$n(\text{O}_2)\text{ in }1\text{ dm}^3\text{ of lake water} = 6.55 \times 10^{-5} \times \frac{1000}{250} = 2.62 \times 10^{-4}\text{ mol}$

$\text{mass of O}_2 = 2.62 \times 10^{-4} \times 32.0 = 8.38 \times 10^{-3}\text{ g} = 8.38\text{ mg}$

Answer

$8.38\text{ mg dm}^{-3}$

Final answer

8.38 mg dm⁻³

Detailed explanation

Background Concept

Dissolved oxygen content in $\text{mg dm}^{-3}$ is the mass of oxygen (in mg) dissolved in 1 dm³ of the original lake water. The lake water sample was 250 cm³, and all its oxygen ended up in the 500.0 cm³ of solution Z, so the moles of O₂ in Z equal the moles originally in the lake sample.

Understanding the Question

Using the moles of O₂ in 500.0 cm³ of Z from (f)(ii), calculate the dissolved oxygen content in $\text{mg dm}^{-3}$ of the lake water.

Approach

Scale the moles from the 250 cm³ lake sample to 1 dm³ (×1000/250).
Convert moles to mass using $M_r(\text{O}_2) = 32.0$ .
Convert grams to milligrams (×1000).

Step-by-Step Reasoning

$n(\text{O}_2)\text{ in }500\text{ cm}^3\text{ of Z} = 6.55 \times 10^{-5}\text{ mol}$ . This is the O₂ in the original 250 cm³ lake sample.
$n(\text{O}_2)\text{ in }1\text{ dm}^3 = 6.55 \times 10^{-5} \times \frac{1000}{250} = 2.62 \times 10^{-4}\text{ mol}$
$\text{mass} = 2.62 \times 10^{-4} \times 32.0 = 8.38 \times 10^{-3}\text{ g} = 8.38\text{ mg}$
Content = $8.38\text{ mg dm}^{-3}$

Key Takeaways

The dilution to 500 cm³ does not change the total amount of O₂. Unit conversions: 1 dm³ = 1000 cm³, 1 g = 1000 mg.

Common Mistakes

Using 500 instead of 250 for the original sample volume.
Forgetting to convert grams to milligrams.
Using $M_r$ of O (16.0) instead of O₂ (32.0).

Things to Be Careful About

The original sample was 250 cm³; the dilution to 500 cm³ keeps the total O₂ constant.
The final unit is $\text{mg dm}^{-3}$ .

Techniques used

convert moles to massscale to per dm³convert g to mg

Q2MediumPlanningAnalysis, Conclusions and Evaluation

A student uses the following method to investigate the kinetics of the reaction between iodine and tin to produce tin(IV) iodide, $\text{SnI}_4$ .


step 1	Rinse a block of tin with distilled water and then rinse it with propanone.
step 2	Place $50\text{ cm}^3$ of a $0.400\text{ mol dm}^{-3}$ solution of iodine dissolved in methylbenzene in a $100\text{ cm}^3$ beaker.
step 3	Suspend the block of tin from a three decimal place balance as shown in Fig. 2.1. Start a timer.
step 4	Record the balance reading every 100 seconds.

Fig. 2.1

(a)

(i)

Suggest why the student rinses the block of tin with propanone after rinsing it with distilled water in step 1.

1M Similar

DifficultyEasy

Worked solution

Answer

Propanone is volatile and miscible with water, so it increases the rate of drying and removes the distilled water from the surface of the tin.

Final answer

Propanone is volatile and miscible with water, so it increases the rate of drying and removes the distilled water.

Detailed explanation

Background Concept

When washing an object with an aqueous solution, residual water remains on the surface. To dry it quickly, a solvent that is miscible with water and highly volatile (evaporates easily at room temperature) is used. Propanone (acetone) fits both criteria: it dissolves in water, mixing with the residual droplets, and its low boiling point (56 °C) means it evaporates rapidly, taking the water with it.

Understanding the Question

The question asks why propanone is used as a final rinse after distilled water. The goal is to remove water from the tin block before it is suspended in the reaction mixture.

Approach

Identify the physical properties of propanone that make it an effective drying agent for an aqueous rinse, and explain how these properties remove water.

Step-by-Step Reasoning

Distilled water leaves residual droplets on the tin.
Propanone is miscible with water, meaning it mixes with and displaces the water droplets.
Propanone is highly volatile (evaporates very quickly at room temperature).
As the propanone evaporates, it takes the water with it, leaving the tin dry and ensuring the initial mass measurement is not affected by water.

Key Takeaways

Using a volatile, water-miscible solvent like propanone is a standard technique to rapidly dry glassware or solid samples after an aqueous wash, preventing mass errors from residual water.

Common Mistakes

Stating "propanone dries the tin" without explaining why (i.e. missing the points about volatility and miscibility with water).
Saying "propanone is a solvent" without specifying it dissolves water and evaporates quickly.

Things to Be Careful About

Ensure the answer explicitly mentions both the removal/miscibility with water and the rapid evaporation (volatility). The mark scheme specifically looks for "increase the rate of drying" or "removal of water".

Techniques used

explain solvent properties for dryingassess solvent miscibility and volatility

(ii)

Suggest why water is not used as the solvent for iodine.

1M Similar

(iii)

Suggest why a three decimal place balance is more suitable than a two decimal place balance for this experiment.

1M Similar

DifficultyEasy

Worked solution

Answer

The change in mass of the tin block over the course of the reaction is very small, so a balance with higher precision (three decimal places) is needed to measure the mass change accurately.

Final answer

The mass change is very small, so a more precise balance (three decimal places) is needed to measure it accurately.

Detailed explanation

Background Concept

In kinetics experiments where the progress of reaction is monitored by measuring the mass of a reactant decreasing (or a product increasing), the precision of the balance determines the resolution of the data. If the total mass change over the experiment is only ~0.36 g, a balance that reads to 0.01 g (two decimal places) would give a relative uncertainty of ~3%, and might not even resolve small early-time changes. A balance reading to 0.001 g (three decimal places) gives a relative uncertainty of ~0.3%, allowing for a much more accurate determination of the rate of reaction.

Understanding the Question

The question asks why a three-decimal-place balance is preferred over a two-decimal-place balance. We must consider the magnitude of the mass change expected in this experiment.

Approach

Estimate or observe the total mass change from the data (from 4.979 g to 4.620 g, a change of 0.359 g) and relate this to the precision of the balance.

Step-by-Step Reasoning

The total mass of tin reacted is 0.359 g over 500 seconds.
The mass change at each 100-second interval is roughly 0.07 g.
A two-decimal place balance (precision 0.01 g) would only give readings like 4.91 g, losing significant information about the rate.
The mass change is very small, so higher precision is required to obtain accurate data for the rate calculation.

Key Takeaways

Apparatus precision must be matched to the magnitude of the expected change. Small changes require high-precision instruments to minimize percentage uncertainty.

Common Mistakes

Saying "it is more accurate" without explaining why (i.e. the mass change is small).
Confusing precision with accuracy.

Things to Be Careful About

The mark scheme specifically looks for the phrase "mass change is (too / very) small". Link the small magnitude of change directly to the need for higher decimal places.

Techniques used

justify apparatus precision based on expected magnitude of change

(iv)

Suggest a control experiment that could be used to verify that the loss in mass of tin is caused by reaction with iodine and not any other factor.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Repeat the experiment using only methylbenzene (without any iodine) and measure the mass change of the tin block over the same time period.

Final answer

Repeat the experiment without any iodine (using only methylbenzene).

Detailed explanation

Background Concept

A control experiment is used to establish a baseline and rule out alternative factors that could cause the observed effect. In this experiment, the loss in mass of the tin block is attributed to the reaction with iodine. However, tin could also react with dissolved oxygen, or the tin could physically erode, or the methylbenzene could dissolve some tin.

Understanding the Question

The question asks for a control experiment to verify that the mass loss is specifically due to the reaction with iodine, and not some other factor (like reaction with air or solvent effects).

Approach

Remove the independent variable (iodine) from the experiment while keeping all other conditions (solvent, temperature, time, tin block) identical.

Step-by-Step Reasoning

The variable being tested is the presence of iodine.
To control for other factors, repeat the entire procedure but replace the iodine/methylbenzene solution with pure methylbenzene.
If the tin block loses mass in pure methylbenzene, that mass loss must be subtracted from the results of the iodine experiment to get the true mass lost to the reaction with iodine.
If there is no mass loss in methylbenzene alone, it confirms iodine is the cause.

Key Takeaways

Controls involve removing the key variable of interest to see if the effect persists. This isolates the cause-and-effect relationship.

Common Mistakes

Suggesting a control with water (water is not used, so this doesn't control the methylbenzene variable).
Suggesting a control with a different metal (this tests the metal, not the iodine).
Not specifying "without iodine" or "with only methylbenzene".

Things to Be Careful About

The control must use the same solvent (methylbenzene) and same conditions, just without the reactant being investigated (iodine). The mark scheme accepts "repeat without any iodine".

Techniques used

design a control experiment to isolate variables

(b)

The student’s results are shown in Table 2.1.

Complete Table 2.1.

Table 2.1

time / s	balance reading / g	total mass of tin reacted / g
0	4.979	0.000
100	4.910
200	4.859
300	4.761
400	4.688
500	4.620

1M Similar

DifficultyEasy

Worked solution

Working

Total mass of tin reacted = initial mass - balance reading at time $t$
Initial mass = 4.979 g

At 100 s: $4.979 - 4.910 = 0.069$ g
At 200 s: $4.979 - 4.859 = 0.120$ g
At 300 s: $4.979 - 4.761 = 0.218$ g
At 400 s: $4.979 - 4.688 = 0.291$ g
At 500 s: $4.979 - 4.620 = 0.359$ g

Answer

time / s	total mass of tin reacted / g
100	0.069
200	0.120
300	0.218
400	0.291
500	0.359

Final answer

0.069, 0.120, 0.218, 0.291, 0.359

Detailed explanation

Background Concept

In a kinetics experiment monitoring mass loss, the raw data is often the mass of the remaining reactant. To analyze the rate of reaction, it is more useful to plot the total mass reacted (or volume of gas evolved, etc.) against time. This is calculated by subtracting the current reading from the initial reading.

Understanding the Question

The student has recorded the balance reading (mass of tin remaining) at 100-second intervals. The table requires the 'total mass of tin reacted' to be calculated for each time point.

Approach

Subtract each balance reading from the initial balance reading (at time = 0) to find the cumulative mass lost.

Step-by-Step Reasoning

Initial mass at $t = 0$ s is 4.979 g. Reacted mass = 0.000 g.
At $t = 100$ s: reacted = $4.979 - 4.910 = 0.069$ g.
At $t = 200$ s: reacted = $4.979 - 4.859 = 0.120$ g.
At $t = 300$ s: reacted = $4.979 - 4.761 = 0.218$ g.
At $t = 400$ s: reacted = $4.979 - 4.688 = 0.291$ g.
At $t = 500$ s: reacted = $4.979 - 4.620 = 0.359$ g.

All values should be given to three decimal places to match the precision of the balance.

Key Takeaways

When processing kinetics data, converting 'remaining' to 'reacted' (or vice versa) is a common first step to make the graph's gradient directly represent the rate of reaction.

Common Mistakes

Subtracting the reacted mass from the current reading (double counting).
Not maintaining three decimal places in the final answer.
Calculation errors in simple subtraction.

Things to Be Careful About

Ensure the values are to 3 decimal places. The mark scheme awards 1 mark for the correct set of values.

Techniques used

calculate derived quantities from raw data

(c)

(i)

Use the results from Table 2.1 to plot a graph on the grid in Fig. 2.2 to show the relationship between total mass of tin reacted and time.

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

Fig. 2.2

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Plot the following points using crosses ( $\times$ ) on the grid:

(0, 0.000)
(100, 0.069)
(200, 0.120)
(300, 0.218)
(400, 0.291)
(500, 0.359)

Draw a straight line of best fit that passes as close as possible to all the points, with roughly equal numbers of points above and below the line.

Final answer

See diagram: points plotted at (0,0), (100,0.069), (200,0.120), (300,0.218), (400,0.291), (500,0.359) with a straight line of best fit.

Detailed explanation

Background Concept

Graphical analysis is a fundamental tool in kinetics. Plotting the amount of product formed (or reactant consumed) against time allows the gradient of the line to represent the rate of reaction. A straight line indicates a constant rate (zero-order reaction with respect to the monitored quantity, though here it's likely first-order overall but the mass loss is linear over this short period or the surface area is constant).

Understanding the Question

The student must plot the calculated data from part (b) on the provided grid (Fig 2.2) and draw a line of best fit. The y-axis is 'total mass of tin reacted / g' and the x-axis is 'time / s'.

Approach

Determine the coordinates for each data point.
Plot them accurately on the grid.
Draw a straight line of best fit.

Step-by-Step Reasoning

Point 1: $x=0$ , $y=0.000$ . Plot at the origin.
Point 2: $x=100$ , $y=0.069$ . Plot slightly below the 0.070 line.
Point 3: $x=200$ , $y=0.120$ . Plot exactly on the 0.120 line.
Point 4: $x=300$ , $y=0.218$ . Plot slightly below the 0.220 line.
Point 5: $x=400$ , $y=0.291$ . Plot slightly above the 0.290 line.
Point 6: $x=500$ , $y=0.359$ . Plot just below the 0.360 line.
Draw a straight line of best fit. The line should pass through or near most points. Point 3 (200, 0.120) appears slightly below the general trend, making it a potential anomaly, but the line should still be drawn to minimize overall deviation.

Key Takeaways

When plotting graphs, use a sharp pencil, plot crosses (not dots) so the center is visible, and ensure the line of best fit is a straight ruler-drawn line that balances the points on either side.

Common Mistakes

Plotting the balance reading (mass remaining) instead of mass reacted.
Using dots instead of crosses.
Drawing a 'join-the-dots' line instead of a line of best fit.
Incorrect axis scaling or misreading the grid.

Things to Be Careful About

The mark scheme gives M1 for correct plotting (center of crosses on the line) and M2 for a straight line of best fit. Ensure the line is straight (use a ruler) and doesn't just connect the points.

Techniques used

plot data points on a graphdraw a line of best fit

(ii)

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

Suggest one reason why this anomaly may have occurred during this experimental procedure.

Assume all measurements of mass are accurate.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Anomalous point: The point at time = 200 s (mass reacted = 0.120 g) is most anomalous (it lies noticeably below the line of best fit).

Reason: The balance reading was taken before 200 s had elapsed (the timer was not started at the same time as immersion, or the reading was taken too early).

Final answer

Anomalous point at 200 s. Reason: The balance reading was taken before 200 s had elapsed (or the timer was started too late / reading taken too early).

Detailed explanation

Background Concept

An anomalous result (or outlier) is a data point that does not fit the general trend of the data. In kinetics, this can arise from human error in timing, equipment malfunction, or environmental fluctuations. Identifying and explaining anomalies is a key evaluation skill.

Understanding the Question

The student must circle the most anomalous point on the graph and suggest one reason for it. The question states to assume all mass measurements are accurate, so the error must be in the timing or procedure.

Approach

Visually inspect the plotted points and the line of best fit to find the point furthest from the line.
Propose a timing-related error that would cause the mass reacted to appear lower than expected at that time.

Step-by-Step Reasoning

Looking at the data: the rate is roughly constant. From 0-100s, mass lost = 0.069g. From 300-400s, mass lost = 0.073g. From 400-500s, mass lost = 0.068g.
At 200s, the mass lost is 0.120g. If the rate was constant (~0.0007 g/s), at 200s we would expect ~0.140g. The value 0.120g is significantly lower.
This means less mass had reacted than expected at 200s. This could happen if the timer was started after the tin was immersed, or if the reading at 200s was taken before 200s had actually elapsed. Essentially, the time recorded is too large for the amount of mass reacted, or the mass reacted is too small for the time recorded.
Another possibility: the timer was not started immediately when the tin was lowered into the solution.

Key Takeaways

Anomalies in time-based experiments often relate to timing errors (starting/stopping the clock too early or too late). Always check if the error is consistent with a timing mismatch.

Common Mistakes

Circling a point that is actually on the line of best fit.
Suggesting 'human error' without being specific (the mark scheme requires a specific reason like 'taken too early').
Suggesting the balance was inaccurate (the question says to assume mass measurements are accurate).

Things to Be Careful About

The reason must link to the specific anomaly: the mass reacted is lower than expected for that time, so the time must have been recorded too late or the reading taken too early.

Techniques used

identify anomalies in dataevaluate experimental errors

(d)

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

State the coordinates of both points you used in your calculation. These must be selected from your line of best fit.

Give your gradient to three significant figures.

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

gradient = ...............................................................

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Select two points on the line of best fit (not the data points themselves).
Example coordinates from the line of best fit:

Point 1: (0, 0.000)
Point 2: (500, 0.360)

Gradient = $\frac{\Delta y}{\Delta x} = \frac{0.360 - 0.000}{500 - 0} = \frac{0.360}{500} = 0.000720 \text{ g s}^{-1}$

(Other acceptable points on the line, e.g., (100, 0.072) and (400, 0.288) giving gradient = $\frac{0.288 - 0.072}{400 - 100} = \frac{0.216}{300} = 0.000720$ )

Answer

coordinates 1: (0, 0.000)
coordinates 2: (500, 0.360)
gradient = 0.000720 g s $^{-1}$

Final answer

coordinates 1: (0, 0.000), coordinates 2: (500, 0.360), gradient = 0.000720 g s^-1

Detailed explanation

Background Concept

The gradient of a 'mass reacted against time' graph represents the rate of reaction (change in mass per unit time). To calculate the gradient accurately, one must use two points on the line of best fit, not the actual data points (which contain experimental error). The points should be far apart to minimize percentage uncertainty in the gradient calculation.

Understanding the Question

The student must determine the gradient of the line of best fit from part (c). They must state the coordinates of two points used (must be from the line, not the data) and calculate the gradient to 3 significant figures.

Approach

Read off two points that lie exactly on the drawn line of best fit.
Calculate $\frac{y_2 - y_1}{x_2 - x_1}$ .
Round to 3 significant figures.

Step-by-Step Reasoning

The line passes through the origin (0, 0) approximately.
At $x = 500$ , the line is at approximately $y = 0.360$ (the data point is 0.359, the line is slightly above).
Using (0, 0.000) and (500, 0.360):
Gradient = $\frac{0.360 - 0.000}{500 - 0} = 0.000720$ g/s.
Using (100, 0.072) and (400, 0.288):
Gradient = $\frac{0.288 - 0.072}{400 - 100} = \frac{0.216}{300} = 0.000720$ g/s.
The value 0.000718 (using data points 0 and 500) is also acceptable if the line is drawn very close to it, but 0.000720 is a cleaner 3-sig-fig answer from the line.
Units are g s $^{-1}$ (or g/s).

Key Takeaways

Always use points on the line of best fit for gradient calculations, not the raw data points. Choose points that are far apart and easy to read from the grid.

Common Mistakes

Using raw data points (e.g., (0,0) and (500, 0.359)) to calculate the gradient. While this gives 0.000718, the mark scheme requires points from the line of best fit.
Not stating the coordinates.
Incorrect significant figures (must be 3).
Wrong units.

Things to Be Careful About

The coordinates must be clearly stated in (x, y) format. The gradient calculation must use these exact coordinates. 3 significant figures: 0.000720 (not 0.00072 or 0.0007200).

Techniques used

calculate gradient from a line of best fit

(e)

Another student makes various concentrations of solutions of iodine dissolved in methylbenzene by dilution of the $0.400\text{ mol dm}^{-3}$ $\text{I}_2$ solution.

The student repeats the experiment at a different temperature using these solutions.

Table 2.2

volume of $0.400\text{ mol dm}^{-3}$ $\text{I}_2$ solution used / $\text{cm}^3$	volume of methylbenzene used / $\text{cm}^3$	$[\text{I}_2]$ / $\text{mol dm}^{-3}$	relative rate of reaction
100.0	0.0	0.400	4.76
		0.300	3.57
		0.200	2.35
		0.100	1.15

(i)

Complete Table 2.2 by adding the volumes of solutions that are mixed to make $100.0\text{ cm}^3$ of a solution of iodine dissolved in methylbenzene for each required concentration.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Total volume = 100.0 cm $^3$ . Initial concentration $C_1 = 0.400$ mol dm $^{-3}$ .
Using $C_1V_1 = C_2V_2$ :

For $[\text{I}_2] = 0.300$ : $V_1 = \frac{0.300 \times 100.0}{0.400} = 75.0$ cm $^3$ . Volume of methylbenzene = $100.0 - 75.0 = 25.0$ cm $^3$ .
For $[\text{I}_2] = 0.200$ : $V_1 = \frac{0.200 \times 100.0}{0.400} = 50.0$ cm $^3$ . Volume of methylbenzene = $100.0 - 50.0 = 50.0$ cm $^3$ .
For $[\text{I}_2] = 0.100$ : $V_1 = \frac{0.100 \times 100.0}{0.400} = 25.0$ cm $^3$ . Volume of methylbenzene = $100.0 - 25.0 = 75.0$ cm $^3$ .

Answer

volume of 0.400 mol dm $^{-3}$ I $_2$ solution / cm $^3$	volume of methylbenzene / cm $^3$
75.0	25.0
50.0	50.0
25.0	75.0

Final answer

75.0, 25.0; 50.0, 50.0; 25.0, 75.0

Detailed explanation

Background Concept

To investigate the effect of concentration on rate, a series of solutions with different concentrations must be prepared by dilution. The total volume must be kept constant (100.0 cm $^3$ ) to ensure that the volume of the reaction mixture is the same in all experiments (a control variable). The dilution formula is $C_1V_1 = C_2V_2$ .

Understanding the Question

The student has a stock solution of 0.400 mol dm $^{-3}$ I $_2$ in methylbenzene. They need to make 100.0 cm $^3$ of solutions with concentrations 0.300, 0.200, and 0.100 mol dm $^{-3}$ by mixing the stock solution with pure methylbenzene.

Approach

Calculate the volume of stock solution ( $V_1$ ) needed for each concentration using $C_1V_1 = C_2V_2$ , then subtract from 100.0 cm $^3$ to find the volume of methylbenzene.

Step-by-Step Reasoning

For 0.300 mol dm $^{-3}$ :
$V_1 = \frac{0.300 \times 100.0}{0.400} = 75.0$ cm $^3$ of stock solution.
Methylbenzene = $100.0 - 75.0 = 25.0$ cm $^3$ .
For 0.200 mol dm $^{-3}$ :
$V_1 = \frac{0.200 \times 100.0}{0.400} = 50.0$ cm $^3$ of stock solution.
Methylbenzene = $100.0 - 50.0 = 50.0$ cm $^3$ .
For 0.100 mol dm $^{-3}$ :
$V_1 = \frac{0.100 \times 100.0}{0.400} = 25.0$ cm $^3$ of stock solution.
Methylbenzene = $100.0 - 25.0 = 75.0$ cm $^3$ .

All volumes should be given to 1 decimal place (or 3 sig figs) to match the precision of the 100.0 cm $^3$ total volume.

Key Takeaways

When preparing dilution series, keep the total volume constant to control for volume effects. Use $C_1V_1 = C_2V_2$ to find the solute volume, and subtract from total to find the solvent volume.

Common Mistakes

Forgetting to subtract from 100.0 to find the methylbenzene volume.
Using the wrong total volume.
Not maintaining 3 significant figures / 1 decimal place.

Things to Be Careful About

The table already has the first row (100.0 and 0.0). Fill in the remaining three rows. Ensure the sum of volumes in each row is exactly 100.0 cm $^3$ .

Techniques used

calculate dilution volumes to prepare standard solutions

(ii)

Identify the independent variable in this experiment.

1M Similar

(iii)

The student concludes that the rate equation for the reaction between iodine and tin is as follows.

\text{rate} = k [\text{I}_2]^2

State whether the results support the student’s conclusion.

Explain your answer using values from Table 2.2.

2M Similar

DifficultyMedium

Worked solution

Answer

No, the results do not support the conclusion.

Explanation:
If rate $= k[\text{I}_2]^2$ , then halving the concentration should reduce the rate to $\frac{1}{4}$ (0.25) of its original value.

When $[\text{I}_2]$ is halved from 0.400 to 0.200 mol dm $^{-3}$ , the rate changes from 4.76 to 2.35.
Ratio of rates: $\frac{2.35}{4.76} \approx 0.49$ (or $\frac{1}{2}$ ).
Since halving the concentration halves the rate (rate is proportional to $[\text{I}_2]$ ), the reaction is first order with respect to iodine, not second order.
The correct rate equation is rate $= k[\text{I}_2]$ .

Working for verification:

$\frac{0.400}{0.200} = 2$ ; $\frac{4.76}{2.35} = 2.03 \approx 2$ . (First order)
If second order: $2^2 = 4$ , rate ratio should be 4.

Answer

No. The rate is proportional to the concentration of iodine (first order), not the square of the concentration. For example, halving $[\text{I}_2]$ from 0.400 to 0.200 mol dm $^{-3}$ halves the rate (4.76 to 2.35), indicating first order.

Final answer

No. The rate is proportional to [I2] (first order). Halving [I2] from 0.400 to 0.200 mol dm-3 halves the rate (4.76 to 2.35).

Detailed explanation

Background Concept

The rate equation expresses how the rate of reaction depends on the concentration of reactants: rate $= k[A]^m[B]^n$ . The orders $m$ and $n$ must be determined experimentally. If doubling $[A]$ doubles the rate, the reaction is first order in A ( $m=1$ ). If doubling $[A]$ quadruples the rate, it is second order ( $m=2$ ).

Understanding the Question

A student proposes rate $= k[\text{I}_2]^2$ (second order). We must use the data in Table 2.2 to test this conclusion.

Approach

State whether the results support the conclusion (Yes/No).
Calculate the ratio of concentrations and the ratio of rates for a pair of data points.
Compare the ratios to determine the actual order.

Step-by-Step Reasoning

Look at $[\text{I}_2] = 0.400$ (rate = 4.76) and $[\text{I}_2] = 0.200$ (rate = 2.35).
Concentration ratio: $\frac{0.400}{0.200} = 2$ (concentration halved).
Rate ratio: $\frac{4.76}{2.35} = 2.026 \approx 2$ (rate halved).
Since halving the concentration halves the rate (or doubling doubles the rate), the rate is directly proportional to $[\text{I}_2]$ .
This means the reaction is first order with respect to iodine.
The proposed equation rate $= k[\text{I}_2]^2$ implies second order, which is incorrect.
Therefore, the results do not support the student's conclusion.

Key Takeaways

To determine order, compare how the rate changes when concentration changes. Use ratios: if rate ratio = (concentration ratio) $^n$ , then $n$ is the order.

Common Mistakes

Saying 'Yes' without checking the math.
Calculating the ratio incorrectly (e.g., $\frac{2.35}{4.76} = 0.5$ , then saying $0.5 = 2^2$ which is wrong).
Not using actual values from the table to explain.
Stating 'it's first order' without showing the calculation from the table.

Things to Be Careful About

The mark scheme requires: M1 for 'no' (or equivalent), AND M2 for explanation using actual values from columns 3 and 4. You must show the calculation or comparison using the numbers in the table (e.g., 'halving concentration from 0.400 to 0.200 halves the rate from 4.76 to 2.35').

Techniques used

determine order of reaction from concentration-rate dataevaluate a proposed rate equation