Chemistry 9701/53 — October/November 2024

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

Q1Medium-EasyPlanningAnalysis, Conclusions and Evaluation

A student uses the following method to determine the percentage by mass of the painkiller aspirin, $\text{C}_9\text{H}_8\text{O}_4(\text{s})$ , in some tablets.

step 1 Grind five tablets into a powder.

step 2 Use a weighing boat to accurately weigh by difference approximately $0.4\text{ g}$ of powdered tablets into a pear-shaped flask containing anti-bumping granules.

step 3 Add $25\text{ cm}^3$ of aqueous $1\text{ mol dm}^{-3}$ sodium hydroxide, $\text{NaOH}(\text{aq})$ , to the pear-shaped flask, forming mixture A.

step 4 Reflux mixture A for 20 minutes.

step 5 Allow mixture A to cool and then filter into a small beaker. Label the filtrate solution B.

step 6 Add $30\text{ cm}^3$ of alkaline aqueous iodine to solution B and leave to stand for 1 hour. A precipitate, C, $(\text{C}_6\text{H}_2\text{I}_2\text{O})_2(\text{s})$ , will form.

step 7 Filter the resulting mixture under reduced pressure. Wash the residue, C, with a small volume of cold distilled water.

step 8 Allow solid C to dry.

step 9 Weigh solid C and record its mass.

Alkaline aqueous iodine is irritating to the skin and eyes.

(a)

Identify an appropriate precaution, other than eye protection and a lab coat, that the student should take when using alkaline aqueous iodine.

1M Similar

DifficultyEasy

Worked solution

Answer

Wear chemically resistant gloves.

Final answer

Wear chemically resistant gloves.

Detailed explanation

Background Concept

Alkaline aqueous iodine is a mixture of iodine and sodium hydroxide (or potassium iodide/iodate in alkaline solution). Iodine is a corrosive substance that can stain skin and cause chemical burns. Alkaline solutions are corrosive and can cause severe skin and eye irritation or burns. Standard laboratory safety protocols require personal protective equipment (PPE) to prevent contact with hazardous chemicals.

Understanding the Question

The question asks for a precaution when using alkaline aqueous iodine, explicitly excluding eye protection (goggles) and a lab coat. We need to identify another piece of PPE or a safe working practice that mitigates the specific risks of this reagent.

Approach

Identify the hazards: corrosive/irritating to skin and eyes. Since eye protection and lab coat are excluded, the most direct additional protection for the hands is chemically resistant gloves.

Step-by-Step Reasoning

The reagent is alkaline aqueous iodine.
It is stated to be irritating to the skin and eyes.
Eye protection and a lab coat are already assumed/allowed.
The hands are most vulnerable to spills and splashes.
Therefore, wearing chemically resistant gloves (e.g., nitrile or rubber gloves) is the appropriate additional precaution to protect the skin from chemical burns and irritation.

Key Takeaways

Always match the precaution to the specific hazard. For corrosive/irritating liquids, gloves are essential in addition to goggles and a lab coat.

Common Mistakes

Writing "gloves" without specifying "chemically resistant" or "rubber". Standard cotton gloves offer no protection against alkaline solutions or iodine.
Suggesting "fume hood" or "good ventilation". While good practice, the primary hazard described is skin/eye irritation from direct contact, making gloves the most direct and expected answer.

Things to Be Careful About

Read the exclusion carefully: "other than eye protection and a lab coat". Do not repeat these.
Be specific: "gloves" is often accepted, but "chemically resistant gloves" is more precise and demonstrates understanding of the chemical hazard.

Techniques used

identify safety precautions for hazardous chemicals

(b)

Describe how the student should carry out step 2. Include a results table, with appropriate headings, for the student to fill in.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Method:

Measure the mass of the weighing boat plus the powdered tablets before transfer.
Transfer the powder to the flask.
Measure the mass of the weighing boat plus any residual solid after transfer.
The difference gives the mass of solid transferred.

Results Table:

Mass of boat + solid (before transfer) / g	Mass of boat + residue (after transfer) / g	Mass of solid transferred / g

Final answer

See working.

Detailed explanation

Background Concept

Weighing by difference is a standard analytical technique used to accurately determine the mass of a substance transferred from a container. Instead of trying to weigh the substance directly on the receiving vessel (which is prone to spillage and loss), you weigh the source container before and after transfer. The difference is the mass transferred, which is more accurate because it accounts for any material left behind.

Understanding the Question

Step 2 requires weighing approximately 0.4 g of powdered tablets into a pear-shaped flask. The question asks for a description of how to carry this out (the method) and a results table with appropriate headings and units to record the data.

Approach

Describe the weighing by difference method: weigh boat + solid, transfer, weigh boat + residue.
Design a table with clear, unambiguous headings that include the correct unit (g) for each column.

Step-by-Step Reasoning

M1 (Method): The student must measure the mass of the weighing boat with the solid before transfer. Then, after transferring the powder to the flask, measure the mass of the boat with the remaining (residual) solid. The difference is the mass transferred.
M2 (Table): The table needs three columns. The headings must be unambiguous (e.g., "Mass of boat + solid (before transfer)") and must include the unit "/ g". A third column for the calculated mass transferred is implied and earns the mark.

Key Takeaways

Weighing by difference is more accurate than direct weighing for powders. Tables must have clear headings with units. Units can be in the heading or the column, but must be present.

Common Mistakes

Writing "weigh the tablets" or "weigh the powder". This is not accurate enough; you cannot weigh a powder directly into a flask without a container, and you need to account for residue.
Forgetting the unit in the table heading. A table without units is not scientifically valid.
Writing "Mass of tablets" as a heading. This is ambiguous; it doesn't specify if it's before or after transfer.

Things to Be Careful About

The mark scheme requires the headings to be unambiguous. Use phrases like "(before transfer)" and "(after transfer)" or "residue / residual solid".
Ensure the table has enough rows to record at least one set of readings, though the mark scheme primarily rewards the structure and headings.

Techniques used

describe weighing by differenceconstruct results table with correct units

(c)

Complete Fig. 1.1 to show how step 4 is carried out in the laboratory.

Label your diagram fully.

Fig. 1.1

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Attach a vertical Liebig condenser to the top of the pear-shaped flask.
Label the water inlet at the bottom of the condenser and the water outlet at the top.

Final answer

See diagram.

Detailed explanation

Background Concept

Reflux is a technique used to heat a reaction mixture for an extended period without losing solvent or volatile reactants/products. It involves boiling the mixture and condensing the vapours back into the flask. A Liebig (water) condenser is used, which consists of an inner tube (where vapours travel) and an outer jacket (through which cooling water flows).

Understanding the Question

Step 4 is "Reflux mixture A for 20 minutes". Fig 1.1 shows a pear-shaped flask with mixture A and anti-bumping granules, with heat applied. The top is open. The student must complete the diagram to show the reflux setup.

Approach

Identify the missing apparatus: a vertical condenser.
Draw it attached to the top of the flask.
Apply the principle of counter-current flow for cooling: cold water enters at the bottom (to fill the jacket completely) and exits at the top.

Step-by-Step Reasoning

M1: A condenser must be drawn vertically above the flask. It should be a Liebig condenser (inner tube, outer jacket). It must be attached to the flask (often with a clip, though just attaching it is usually sufficient for the diagram mark).
M2: The cooling water flow must be counter-current to the hot vapours. Cold water enters at the bottom inlet and exits at the top outlet. This ensures the condenser jacket is always full of cold water, providing maximum cooling efficiency. Labels "water in" at the bottom and "water out" at the top are required.

Key Takeaways

Reflux requires a vertical condenser. Water always flows in at the bottom and out at the top to ensure the jacket is full and cooling is efficient.

Common Mistakes

Drawing the condenser horizontally or at an angle. Reflux condensers must be vertical.
Reversing the water flow (in at top, out at bottom). This leads to poor cooling as the jacket may not fill completely, and the water warms up as it flows down, reducing efficiency.
Forgetting to label the water inlet and outlet. The diagram must be fully labelled.

Things to Be Careful About

The diagram must be clearly labelled. Use arrows to show water flow direction if text labels are ambiguous.
Do not draw a thermometer or other apparatus unless required. Only add what is necessary for reflux.

Techniques used

draw reflux apparatuslabel water flow direction in condenser

(d)

(i)

The student uses a measuring cylinder to measure the volume of alkaline aqueous iodine in step 6. Suggest why this is a suitable piece of apparatus to use.

1M Similar

(ii)

Suggest why the student leaves the mixture to stand for 1 hour in step 6.

1M Similar

(iii)

Explain why the residue is washed in step 7.

1M Similar

DifficultyEasy

Worked solution

Answer

To remove soluble impurities (such as excess iodine, sodium hydroxide, sodium iodide, or sodium hydrogencarbonate) from the surface of the precipitate.

Final answer

To remove soluble substances from the residue.

Detailed explanation

Background Concept

When a precipitate is filtered from a reaction mixture, it is coated with the mother liquor (the solution it was formed in). This mother liquor contains unreacted reagents, by-products, and solvent. To obtain a pure, dry solid, the precipitate must be washed to remove these soluble impurities.

Understanding the Question

Step 7 involves filtering the mixture and washing the residue (solid C) with cold distilled water. The question asks why this washing step is necessary.

Approach

Identify what is on the surface of the solid C after filtration. It will be coated with the reaction mixture.

Step-by-Step Reasoning

The reaction mixture contains excess alkaline iodine, NaOH, NaI, NaHCO₃, and water.
These are all soluble in water.
Washing the solid C with distilled water dissolves and removes these soluble impurities from the surface of the precipitate.
This ensures that when solid C is dried and weighed, the mass is due only to solid C, not to trapped soluble salts or reagents.

Key Takeaways

Washing a precipitate removes soluble impurities trapped in the filter cake, ensuring an accurate mass measurement of the pure product.

Common Mistakes

Saying "to clean the solid". Too vague. Must specify "soluble impurities" or name specific soluble substances.
Saying "to remove the solution". Technically true, but "soluble substances" or "impurities" is the required scientific terminology.

Things to Be Careful About

The mark scheme accepts specific named soluble substances (e.g., excess iodine, NaOH, NaI, NaHCO₃). Using the general term "soluble substances" or "soluble impurities" is also acceptable and often safer.

Techniques used

justify washing precipitate to remove impurities

(iv)

Explain why hot distilled water is not used in step 7.

1M Similar

(e)

The equation for the reaction between aspirin, $\text{C}_9\text{H}_8\text{O}_4(\text{s})$ , and $\text{NaOH}(\text{aq})$ , which takes place in step 4, is shown.

\text{C}_9\text{H}_8\text{O}_4(\text{s}) + 2\text{NaOH}(\text{aq}) \rightarrow \text{C}_7\text{H}_5\text{O}_3^-\text{Na}^+(\text{aq}) + \text{C}_2\text{H}_3\text{O}_2^-\text{Na}^+(\text{aq}) + \text{H}_2\text{O}(\text{l})

The equation for the reaction in which solid C, $(\text{C}_6\text{H}_2\text{I}_2\text{O})_2(\text{s})$ , is formed in step 6 is shown.

2\text{C}_7\text{H}_5\text{O}_3^-\text{Na}^+(\text{aq}) + 6\text{I}_2(\text{aq}) + 8\text{OH}^-(\text{aq}) \rightarrow (\text{C}_6\text{H}_2\text{I}_2\text{O})_2(\text{s}) + 8\text{I}^-(\text{aq}) + 2\text{NaHCO}_3(\text{aq}) + 6\text{H}_2\text{O}(\text{l})

The student’s results are shown in Table 1.1.

Table 1.1


mass of powdered tablets added to the pear-shaped flask in step 2	$0.409\text{ g}$
mass of dry $(\text{C}_6\text{H}_2\text{I}_2\text{O})_2(\text{s})$ recorded in step 9	$0.764\text{ g}$

(i)

Calculate the amount, in mol, of $(\text{C}_6\text{H}_2\text{I}_2\text{O})_2(\text{s})$ collected in step 9.

[ $M_r$ : $(\text{C}_6\text{H}_2\text{I}_2\text{O})_2$ , 687.6]

1M Similar

DifficultyEasy

Worked solution

Working

\text{amount of } (\text{C}_6\text{H}_2\text{I}_2\text{O})_2 = \frac{\text{mass}}{M_r} = \frac{0.764}{687.6}

\text{amount} = 1.111 \times 10^{-3} \text{ mol}

Answer

$1.111 \times 10^{-3} \text{ mol}$

Final answer

1.111e-3 mol

Detailed explanation

Background Concept

The amount of substance (in moles) is calculated from its mass and relative molecular mass ( $M_r$ ) using the equation: $n = \frac{m}{M_r}$ . This is a fundamental conversion in chemistry.

Understanding the Question

Given the mass of dry solid C ( $0.764 \text{ g}$ ) and its $M_r$ ( $687.6$ ), calculate the amount in mol.

Approach

Apply the formula $n = m / M_r$ .

Step-by-Step Reasoning

Mass $m = 0.764 \text{ g}$
$M_r = 687.6 \text{ g mol}^{-1}$
$n = \frac{0.764}{687.6} = 0.0011111... \text{ mol}$
Round to 3 significant figures: $1.11 \times 10^{-3} \text{ mol}$ (or keep more digits for intermediate calculations: $1.111 \times 10^{-3} \text{ mol}$ )

Key Takeaways

Always use the formula $n = m / M_r$ and ensure units are consistent (mass in g, $M_r$ in g/mol).

Common Mistakes

Forgetting to divide by $M_r$ and just writing the mass.
Using the wrong $M_r$ value.
Rounding too early, which can cause errors in subsequent calculations.

Things to Be Careful About

Keep at least 4 significant figures in intermediate calculations to avoid rounding errors.
The mark scheme accepts $1.11 \times 10^{-3}$ or $1.111 \times 10^{-3}$ .

Techniques used

calculate moles from mass and molar mass

(ii)

Use your answer to (i) to calculate the mass, in g, of $\text{C}_9\text{H}_8\text{O}_4(\text{s})$ in the powdered tablets added to the flask in step 2.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

From the equations:
1 mol $(\text{C}_6\text{H}_2\text{I}_2\text{O})_2$ is produced from 2 mol $\text{C}_7\text{H}_5\text{O}_3^-\text{Na}^+$ .
2 mol $\text{C}_7\text{H}_5\text{O}_4$ (aspirin) produces 2 mol $\text{C}_7\text{H}_5\text{O}_3^-\text{Na}^+$ .

Therefore, 2 mol $\text{C}_9\text{H}_8\text{O}_4$ produces 1 mol $(\text{C}_6\text{H}_2\text{I}_2\text{O})_2$ .

\text{amount of } \text{C}_9\text{H}_8\text{O}_4 = 2 \times 1.111 \times 10^{-3} = 2.222 \times 10^{-3} \text{ mol}

$M_r$ of $\text{C}_9\text{H}_8\text{O}_4 = (9 \times 12.0) + (8 \times 1.0) + (4 \times 16.0) = 180.0 \text{ g mol}^{-1}$

\text{mass of } \text{C}_9\text{H}_8\text{O}_4 = 2.222 \times 10^{-3} \times 180.0 = 0.400 \text{ g}

Answer

$0.400 \text{ g}$

Final answer

0.400 g

Detailed explanation

Background Concept

Stoichiometry allows us to relate the amounts of different substances in a chemical reaction using the balanced equations. By linking two consecutive reactions, we can find the overall molar ratio between the initial reactant and the final product.

Understanding the Question

We need to find the mass of aspirin ( $\text{C}_9\text{H}_8\text{O}_4$ ) that produced the $1.111 \times 10^{-3} \text{ mol}$ of solid C. We are given two equations:

$\text{C}_9\text{H}_8\text{O}_4 + 2\text{NaOH} \rightarrow \text{C}_7\text{H}_5\text{O}_3^-\text{Na}^+ + \text{C}_2\text{H}_3\text{O}_2^-\text{Na}^+ + \text{H}_2\text{O}$
$2\text{C}_7\text{H}_5\text{O}_3^-\text{Na}^+ + 6\text{I}_2 + 8\text{OH}^- \rightarrow (\text{C}_6\text{H}_2\text{I}_2\text{O})_2 + 8\text{I}^- + 2\text{NaHCO}_3 + 6\text{H}_2\text{O}$

Approach

Find the molar ratio between aspirin and the intermediate ( $\text{C}_7\text{H}_5\text{O}_3^-\text{Na}^+$ ).
Find the molar ratio between the intermediate and solid C.
Combine to find the overall ratio between aspirin and solid C.
Calculate moles of aspirin, then mass.

Step-by-Step Reasoning

From equation 1: 1 mol $\text{C}_9\text{H}_8\text{O}_4$ produces 1 mol $\text{C}_7\text{H}_5\text{O}_3^-\text{Na}^+$ .
From equation 2: 2 mol $\text{C}_7\text{H}_5\text{O}_3^-\text{Na}^+$ produces 1 mol $(\text{C}_6\text{H}_2\text{I}_2\text{O})_2$ .
Therefore, 2 mol $\text{C}_9\text{H}_8\text{O}_4$ are needed to produce 1 mol $(\text{C}_6\text{H}_2\text{I}_2\text{O})_2$ .
Moles of $\text{C}_9\text{H}_8\text{O}_4 = 2 \times \text{moles of } (\text{C}_6\text{H}_2\text{I}_2\text{O})_2 = 2 \times 1.111 \times 10^{-3} = 2.222 \times 10^{-3} \text{ mol}$ .
$M_r$ of $\text{C}_9\text{H}_8\text{O}_4 = 180.0 \text{ g mol}^{-1}$ .
Mass $= 2.222 \times 10^{-3} \times 180.0 = 0.400 \text{ g}$ .

Key Takeaways

When linking multiple reactions, trace the common intermediate to find the overall stoichiometric ratio.

Common Mistakes

Assuming a 1:1 ratio between aspirin and solid C. Must use the equations to find the correct ratio (2:1).
Forgetting to multiply by 2 for the moles of aspirin.
Using the wrong $M_r$ for aspirin (must calculate or be given: 180.0).

Things to Be Careful About

The mark scheme explicitly states: "amount of $\text{C}_9\text{H}_8\text{O}_4 = 2 \times 1\text{(e)(i)}$ ". This confirms the 2:1 ratio.
Keep track of significant figures. $0.400 \text{ g}$ is 3 sig figs.

Techniques used

use stoichiometric ratios to find moles of reactantcalculate mass from moles and molar mass

(iii)

Use your answer to (ii) to calculate the percentage by mass of aspirin, $\text{C}_9\text{H}_8\text{O}_4(\text{s})$ , in the tablets.

If you were unable to obtain an answer to (ii) you may use $0.374\text{ g}$ for the mass of $\text{C}_9\text{H}_8\text{O}_4(\text{s})$ . This is not the correct value.

1M Similar

DifficultyEasy

Worked solution

Working

\% \text{ by mass of aspirin} = \frac{\text{mass of aspirin}}{\text{mass of powdered tablets}} \times 100

\% \text{ by mass} = \frac{0.400}{0.409} \times 100 = 97.8\%

Answer

$97.8\%$

Final answer

97.8%

Detailed explanation

Background Concept

Percentage by mass is a way to express the concentration of a component in a mixture. It is calculated as the mass of the component divided by the total mass of the mixture, multiplied by 100.

Understanding the Question

Calculate the percentage by mass of aspirin in the tablets. We have the mass of aspirin calculated in (ii) ( $0.400 \text{ g}$ ) and the mass of powdered tablets added in step 2 ( $0.409 \text{ g}$ ).

Approach

Apply the percentage by mass formula.

Step-by-Step Reasoning

Mass of aspirin $= 0.400 \text{ g}$
Mass of powdered tablets $= 0.409 \text{ g}$
$\% \text{ by mass} = \frac{0.400}{0.409} \times 100 = 97.799...\%$
Round to 3 significant figures: $97.8\%$

Key Takeaways

Percentage by mass = (mass of component / total mass) x 100. Ensure both masses are in the same units.

Common Mistakes

Forgetting to multiply by 100.
Using the wrong mass for the tablets (e.g., mass of 5 tablets instead of the mass actually weighed).
Rounding errors from previous steps.

Things to Be Careful About

The mark scheme accepts $97.8\%$ . Using $97.79\%$ or $97.80\%$ may also be accepted depending on rounding, but 3 sig figs is standard.
The question provides an alternative value ( $0.374 \text{ g}$ ) if the student couldn't calculate (ii). This is for error carried forward (ecf).

Techniques used

calculate percentage by mass

(f)

Another student follows the same method but does not allow solid C to dry completely in step 8.

State and explain the effect that this has on the calculated percentage by mass of aspirin, $\text{C}_9\text{H}_8\text{O}_4(\text{s})$ , in the tablets.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

If solid C is not completely dry, its measured mass will be greater than the true mass (due to water).
This leads to a calculated amount of solid C that is too high.
Consequently, the calculated mass and percentage by mass of aspirin will be greater than the true value.

Final answer

The mass of solid C would be greater, so the calculated percentage by mass of aspirin would be greater.

Detailed explanation

Background Concept

Error analysis in quantitative experiments requires tracing the effect of a procedural error through the entire calculation chain to determine whether the final result is too high or too low.

Understanding the Question

The student does not allow solid C to dry completely in step 8. We need to state and explain the effect on the calculated percentage by mass of aspirin.

Approach

Effect on mass of solid C.
Effect on moles of solid C.
Effect on moles of aspirin.
Effect on mass of aspirin.
Effect on percentage by mass.

Step-by-Step Reasoning

Effect on mass: If solid C is not dry, it contains water. The measured mass ( $0.764 \text{ g}$ in the example) will be higher than the true mass of solid C alone.
Effect on moles: Since $n = m / M_r$ , a higher mass leads to a higher calculated amount of solid C.
Effect on aspirin: The stoichiometry links moles of solid C to moles of aspirin (2:1 ratio). Higher moles of solid C -> higher calculated moles of aspirin -> higher calculated mass of aspirin.
Effect on percentage: Percentage = (calculated mass of aspirin / mass of tablets) x 100. Since the numerator is too high and the denominator is unchanged, the percentage will be greater than the true value.

Key Takeaways

Trace the error through the calculation: higher measured mass of product -> higher calculated reactant mass -> higher percentage.

Common Mistakes

Saying "the percentage would be lower". This is incorrect; the error propagates to make it higher.
Not explaining the chain of reasoning. Must state both the effect on the mass of C and the effect on the percentage.
Saying "the mass of aspirin would be greater" without mentioning the percentage. The question asks for the effect on the percentage.

Things to Be Careful About

The mark scheme requires TWO points: (1) mass/amount of solid C would be greater, and (2) calculated percentage would be greater. Both are needed for the mark.
Use clear language: "greater than the true value" or "too high".

Techniques used

analyze effect of incomplete drying on calculated result

Q2MediumPlanningAnalysis, Conclusions and Evaluation

Crystal violet, $\text{C}_{25}\text{H}_{30}\text{N}_3\text{Cl}(\text{s})$ , is a purple dye.

Some light is absorbed when it passes through $\text{C}_{25}\text{H}_{30}\text{N}_3\text{Cl}(\text{aq})$ .

Absorbance is the proportion of light absorbed at a particular wavelength. This is measured using a colorimeter.

A graph of absorbance against wavelength for $\text{C}_{25}\text{H}_{30}\text{N}_3\text{Cl}(\text{aq})$ is shown in Fig. 2.1.

Fig. 2.1

A student investigates how to determine the concentration of aqueous crystal violet, $\text{C}_{25}\text{H}_{30}\text{N}_3\text{Cl}(\text{aq})$ , using colorimetry.

(a)

Suggest the best wavelength of light to use in the colorimeter when measuring the concentration of $\text{C}_{25}\text{H}_{30}\text{N}_3\text{Cl}(\text{aq})$ .

1M Similar

DifficultyEasy

Worked solution

Answer

588 nm

Final answer

588 nm

Detailed explanation

Background Concept

When using a colorimeter to measure the concentration of a colored solution, the Beer-Lambert law states that absorbance is directly proportional to concentration. To obtain the most accurate and sensitive measurements, the wavelength of light used should correspond to the wavelength of maximum absorbance ( $\lambda_{\text{max}}$ ) of the substance. At this wavelength, even small changes in concentration produce the largest possible changes in absorbance, minimizing percentage errors.

Understanding the Question

The question asks for the best wavelength to use when measuring the concentration of aqueous crystal violet. We are given a graph (Fig. 2.1) of absorbance against wavelength. We need to read the wavelength at which the absorbance is at its peak.

Approach

Locate the highest point on the absorbance vs. wavelength curve in Fig. 2.1 and read the corresponding value on the x-axis (wavelength in nm).

Step-by-Step Reasoning

Examine Fig. 2.1, which plots absorbance on the y-axis against wavelength in nm on the x-axis.
Identify the peak of the curve. The curve rises from 400 nm, has a shoulder around 540 nm, and reaches a maximum peak just before 600 nm.
Reading the x-axis at the peak, the maximum absorbance occurs at approximately 588 nm (accept 589.7 nm or 590 nm depending on reading precision).
This wavelength ( $\lambda_{\text{max}}$ ) is the optimal choice for the colorimeter.

Key Takeaways

Always choose the wavelength of maximum absorbance ( $\lambda_{\text{max}}$ ) for colorimetric analysis to ensure maximum sensitivity and minimize measurement error.

Common Mistakes

Choosing a wavelength on the shoulder of the curve (e.g., 540 nm) instead of the true peak.
Reading the y-axis value (absorbance) instead of the x-axis value (wavelength).

Things to Be Careful About

Ensure the wavelength is read from the x-axis in nm.
Small variations in reading the graph (e.g., 588 to 590 nm) are acceptable as long as it is clearly at the peak.

Techniques used

identify wavelength of maximum absorbance

(b)

Solution D is $500.0\text{ cm}^3$ of $2.50 \times 10^{-2}\text{ mol dm}^{-3}$ $\text{C}_{25}\text{H}_{30}\text{N}_3\text{Cl}(\text{aq})$ .

(i)

Calculate the mass of $\text{C}_{25}\text{H}_{30}\text{N}_3\text{Cl}(\text{s})$ needed to prepare solution D.

Give your answer to three significant figures.

[ $M_r$ : $\text{C}_{25}\text{H}_{30}\text{N}_3\text{Cl}(\text{s})$ , 407.5]

1M Similar

DifficultyMedium-Easy

Worked solution

Working

n = c \times V = 2.50 \times 10^{-2} \text{ mol dm}^{-3} \times 0.5000 \text{ dm}^3 = 1.25 \times 10^{-2} \text{ mol}

\text{mass} = n \times M_r = 1.25 \times 10^{-2} \text{ mol} \times 407.5 \text{ g mol}^{-1} = 5.09375 \text{ g}

Answer

5.09 g

Final answer

5.09 g

Detailed explanation

Background Concept

To prepare a solution of a known concentration, we first calculate the number of moles of solute required using the equation $n = c \times V$ , where $c$ is concentration in mol dm $^{-3}$ and $V$ is volume in dm $^3$ . Once the moles are known, the mass required is found using $m = n \times M_r$ , where $M_r$ is the relative formula mass.

Understanding the Question

We need to calculate the mass of solid crystal violet ( $M_r = 407.5$ ) needed to make 500.0 cm $^3$ (which is 0.5000 dm $^3$ ) of a $2.50 \times 10^{-2}$ mol dm $^{-3}$ solution. The final answer must be given to three significant figures.

Approach

Convert the volume from cm $^3$ to dm $^3$ .
Calculate the moles of crystal violet needed.
Multiply the moles by the molar mass to find the mass in grams.
Round to three significant figures.

Step-by-Step Reasoning

Volume $V = 500.0 \text{ cm}^3 = 500.0 / 1000 = 0.5000 \text{ dm}^3$ .
Moles $n = c \times V = (2.50 \times 10^{-2}) \times 0.5000 = 1.25 \times 10^{-2} \text{ mol}$ .
Mass $m = n \times M_r = (1.25 \times 10^{-2}) \times 407.5 = 5.09375 \text{ g}$ .
Rounding to three significant figures gives 5.09 g.

Key Takeaways

Always ensure volume is in dm $^3$ when using $n = cV$ . Pay attention to significant figure requirements in the final answer.

Common Mistakes

Forgetting to convert cm $^3$ to dm $^3$ (using 500 instead of 0.5), which gives a mass 1000 times too large.
Rounding too early in the calculation.
Not adhering to the requested number of significant figures (5.094 g or 5.1 g would be incorrect).

Things to Be Careful About

The volume is 500.0 cm $^3$ , which has four significant figures, but the concentration has three. The final answer is correctly given to three significant figures as 5.09 g.

Techniques used

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

(ii)

The student is given a small beaker containing the mass of $\text{C}_{25}\text{H}_{30}\text{N}_3\text{Cl}(\text{s})$ calculated in (i).

Describe how the student should prepare $500.0\text{ cm}^3$ of solution D.

Include the name and capacity of the key apparatus which should be used and describe how the student should ensure the volume is exactly $500.0\text{ cm}^3$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

Add a small volume of distilled water to the beaker containing the solid and stir to dissolve it completely.
Transfer the solution to a 500 cm $^3$ volumetric flask, rinsing the beaker and stirring rod with distilled water and adding the washings to the flask.
Add distilled water to the volumetric flask until the bottom of the meniscus is exactly on the calibration mark.
Stopper the flask and invert it several times to mix the solution thoroughly.

Final answer

See working

Detailed explanation

Background Concept

Preparing a standard solution of a precise concentration requires a volumetric flask. The process involves dissolving the solute in a small amount of solvent, transferring it quantitatively to the flask, and making up to the exact calibration mark. 'Quantitative transfer' means ensuring no solute is left behind in the beaker.

Understanding the Question

The student has calculated the mass of crystal violet (5.09 g) and has a small beaker containing it. They must prepare exactly 500.0 cm $^3$ of solution. We need to describe the procedure, naming the key apparatus (500 cm $^3$ volumetric flask) and explaining how the exact volume is achieved.

Approach

Outline the steps in chronological order: dissolving, quantitative transfer, making up to the mark, and mixing. Emphasize the use of distilled water and the specific apparatus.

Step-by-Step Reasoning

M1 (Dissolving): The solid cannot be dissolved directly in 500 cm $^3$ of water because the final volume would exceed 500 cm $^3$ . Add a small volume (e.g., 50-100 cm $^3$ ) of distilled water to the beaker and stir until the solid is fully dissolved.
M2 (Transfer and washings): Pour the dissolved solution into a 500 cm $^3$ volumetric flask. To ensure all the solute is transferred (quantitative transfer), rinse the beaker and stirring rod with small amounts of distilled water and add these washings to the flask.
M3 (Make up to mark and mix): Carefully add distilled water to the flask until the bottom of the meniscus sits exactly on the calibration mark etched on the neck of the flask. This ensures the total volume is exactly 500.0 cm $^3$ . Finally, stopper the flask and invert it multiple times to ensure the solution is homogeneous.

Key Takeaways

Volumetric preparation requires a volumetric flask, quantitative transfer (including washings), and making up to the calibration mark with the correct solvent (distilled water).

Common Mistakes

Saying 'add 500 cm $^3$ of water to the solid' (this is incorrect; the final volume must be 500 cm $^3$ , not the volume of water added).
Forgetting to mention washings or the volumetric flask.
Not specifying 'distilled' water (tap water contains ions that could affect the experiment).

Things to Be Careful About

The question specifically asks to 'include the name and capacity of the key apparatus' (500 cm $^3$ volumetric flask) and 'describe how the student should ensure the volume is exactly 500.0 cm $^3$ ' (make up to the calibration mark / bottom of meniscus on the mark).

Techniques used

describe volumetric preparation of a solution

(c)

A small sample of solution D was diluted to form solution E, $2.50 \times 10^{-4}\text{ mol dm}^{-3}$ $\text{C}_{25}\text{H}_{30}\text{N}_3\text{Cl}(\text{aq})$ .

The student prepares solutions 2 to 6 as shown in Table 2.1.

The total volume needed for each of solutions 2 to 6 is $20.00\text{ cm}^3$ .

Each solution is placed into a colorimeter and the absorbance is measured.

(i)

Complete Table 2.1 to show the volumes of solution E and distilled water needed to prepare each of the solutions from 2 to 6. Give all volumes to two decimal places.

Table 2.1

solution	volume of $2.50 \times 10^{-4}\text{ mol dm}^{-3}$ $\text{C}_{25}\text{H}_{30}\text{N}_3\text{Cl}(\text{aq})$ (solution E) / $\text{cm}^3$	volume of distilled water / $\text{cm}^3$	$[\text{C}_{25}\text{H}_{30}\text{N}_3\text{Cl}(\text{aq})]$ / $\text{mol dm}^{-3}$	absorbance
1	0.00	20.00	0.00	0.000
2			$0.50 \times 10^{-4}$	0.191
3			$1.00 \times 10^{-4}$	0.270
4			$1.50 \times 10^{-4}$	0.545
5			$2.00 \times 10^{-4}$	0.711
6			$2.50 \times 10^{-4}$	0.860

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

solution	volume of $2.50 \times 10^{-4}$ mol dm $^{-3}$ C $_{25}$ H $_{30}$ N $_{3}$ Cl(aq) (solution E) / cm $^3$	volume of distilled water / cm $^3$
2	4.00	16.00
3	8.00	12.00
4	12.00	8.00
5	16.00	4.00
6	20.00	0.00

Final answer

See working

Detailed explanation

Background Concept

When preparing a series of dilutions from a stock solution to create a calibration curve, the total volume of each diluted solution is kept constant. The volume of stock solution needed for a desired concentration can be found using the dilution equation $C_1V_1 = C_2V_2$ , where $C_1$ and $V_1$ are the concentration and volume of the stock solution, and $C_2$ and $V_2$ are the concentration and total volume of the diluted solution. The volume of water added is simply the total volume minus the volume of stock solution.

Understanding the Question

Solution E has a concentration of $2.50 \times 10^{-4}$ mol dm $^{-3}$ . We need to prepare solutions 2 to 6 with total volumes of 20.00 cm $^3$ and target concentrations ranging from $0.50 \times 10^{-4}$ to $2.50 \times 10^{-4}$ mol dm $^{-3}$ . We must calculate the volumes of solution E and distilled water for each, to two decimal places.

Approach

For each solution, use $V_1 = (C_2 \times V_2) / C_1$ to find the volume of solution E, then subtract this from 20.00 cm $^3$ to find the volume of water.

Step-by-Step Reasoning

Solution 2: $C_2 = 0.50 \times 10^{-4}$ . $V_1 = (0.50 \times 10^{-4} \times 20.00) / (2.50 \times 10^{-4}) = 4.00$ cm $^3$ . Water = $20.00 - 4.00 = 16.00$ cm $^3$ .
Solution 3: $C_2 = 1.00 \times 10^{-4}$ . $V_1 = (1.00 \times 10^{-4} \times 20.00) / (2.50 \times 10^{-4}) = 8.00$ cm $^3$ . Water = $20.00 - 8.00 = 12.00$ cm $^3$ .
Solution 4: $C_2 = 1.50 \times 10^{-4}$ . $V_1 = (1.50 \times 10^{-4} \times 20.00) / (2.50 \times 10^{-4}) = 12.00$ cm $^3$ . Water = $20.00 - 12.00 = 8.00$ cm $^3$ .
Solution 5: $C_2 = 2.00 \times 10^{-4}$ . $V_1 = (2.00 \times 10^{-4} \times 20.00) / (2.50 \times 10^{-4}) = 16.00$ cm $^3$ . Water = $20.00 - 16.00 = 4.00$ cm $^3$ .
Solution 6: $C_2 = 2.50 \times 10^{-4}$ . This is the same as solution E, so $V_1 = 20.00$ cm $^3$ and water = $0.00$ cm $^3$ .

Key Takeaways

In dilution series, the sum of the stock solution volume and solvent volume must equal the total required volume. Always check that the volumes add up correctly.

Common Mistakes

Calculating the volume of water instead of the stock solution, or vice versa.
Forgetting to give volumes to two decimal places as requested (e.g., writing 4 instead of 4.00).
Not ensuring the volumes sum to exactly 20.00 cm $^3$ .

Things to Be Careful About

The table already provides the target concentrations; use those directly. The volumes must be given to two decimal places (e.g., 4.00, not 4).

Techniques used

calculate dilution volumes using C1V1 = C2V2

(ii)

Identify the dependent variable.

1M Similar

(d)

(i)

Plot a graph of absorbance against $[\text{C}_{25}\text{H}_{30}\text{N}_3\text{Cl}(\text{aq})]$ on the grid in Fig. 2.2.

Use a cross ( $\times$ ) to plot each data point.

Draw a straight line of best fit.

Fig. 2.2

2M Similar

DifficultyMedium

Worked solution

Answer

Final answer

See diagram

Detailed explanation

Background Concept

A calibration graph (or working curve) relates the measured absorbance to the known concentration of a series of standard solutions. According to the Beer-Lambert law, absorbance is directly proportional to concentration, so the graph should be a straight line passing through the origin (0,0). A line of best fit should have an equal number of points above and below it, and should not necessarily pass through every point.

Understanding the Question

We are given a blank grid (Fig. 2.2) and a table of concentrations (x-axis) and absorbances (y-axis) for solutions 1 to 6. We need to plot the data points using crosses (×) and draw a straight line of best fit.

Approach

Plot the six data points from the table on the given grid.
Draw a straight line that best represents the trend, ensuring it passes through or near the origin (0,0) and the point (2.50, 0.860).

Step-by-Step Reasoning

Plotting points:
- (0.00, 0.000)
- (0.50, 0.191)
- (1.00, 0.270) — Note: This point is slightly below the expected line (anomalous).
- (1.50, 0.545)
- (2.00, 0.711)
- (2.50, 0.860)
Line of best fit: Draw a straight line that goes through (0,0) and roughly through the other points, balancing the points above and below the line. The line should have a positive gradient.

Key Takeaways

Calibration graphs for colorimetry should be linear and pass through the origin. The line of best fit minimizes the overall distance to all data points.

Common Mistakes

Plotting points incorrectly (e.g., swapping x and y axes).
Drawing a curve or connecting the dots with straight line segments instead of a single straight line of best fit.
Forcing the line through the anomalous point (solution 3) instead of allowing it to be an outlier.

Things to Be Careful About

Use crosses (×) for plotting, not dots or circles.
The line must be straight, not curved.
Ensure the axes are used correctly: concentration on the x-axis, absorbance on the y-axis.

Techniques used

plot a calibration graph and draw a line of best fit

(ii)

Circle the point on the graph you consider to be most anomalous.

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

Assume no error was made in the measurement of absorbance.

2M Similar

DifficultyMedium

Worked solution

Answer

Anomalous point: Solution 3 (concentration $1.00 \times 10^{-4}$ mol dm $^{-3}$ , absorbance 0.270).

Reason: The volume of distilled water added to the mixture was too large (or the volume of solution E / crystal violet solution added was too small).

Final answer

Solution 3; volume of water too large

Detailed explanation

Background Concept

In a calibration graph, most points should fall on or very close to the line of best fit. A point that deviates significantly from this line is considered anomalous. Anomalies can arise from experimental errors, such as incorrect volumetric measurements, contamination, or incomplete dissolution.

Understanding the Question

We need to identify the most anomalous point on the graph we just plotted and suggest a reason for it, assuming the absorbance measurement itself was correct. This means the error occurred during the preparation of the solutions.

Approach

Look at the plotted points and the line of best fit. Identify the point furthest from the line.
Consider how the solutions were prepared (dilutions of solution E with water). An error in measuring the volumes of solution E or water would change the actual concentration.

Step-by-Step Reasoning

Identifying the anomaly: Looking at the data, solution 3 has a concentration of $1.00 \times 10^{-4}$ mol dm $^{-3}$ . Based on the linearity (e.g., solution 2 has 0.50 x 10^-4 with absorbance 0.191, so 1.00 x 10^-4 should have ~0.382), the measured absorbance of 0.270 is significantly lower than expected. This point is the most anomalous.
Reasoning about the error: If the absorbance is lower than expected for a given concentration, the actual concentration of crystal violet in the cuvette was lower than intended. This could happen if:
- Too much distilled water was added to the mixture (making it more dilute than calculated).
- Too little of solution E (the stock crystal violet solution) was added to the mixture.
The mark scheme accepts either of these equivalent volumetric errors.

Key Takeaways

Anomalous results in dilution series often stem from volumetric errors. If the measured response is lower than expected, the solution was more dilute than intended.

Common Mistakes

Circling a point that is actually on the line (e.g., solution 4 or 5).
Suggesting 'human error' or 'inaccurate colorimeter' (the question states to assume no error in absorbance measurement).
Suggesting the solid didn't dissolve (this would affect solution D, not the dilutions from solution E).

Things to Be Careful About

The question asks for one reason. Give a clear, specific volumetric error.
Ensure the reason is consistent with the direction of the anomaly (absorbance too low -> concentration too low -> too much water or too little stock solution).

Techniques used

identify anomalous data and suggest procedural error

(iii)

State the relationship between $[\text{C}_{25}\text{H}_{30}\text{N}_3\text{Cl}(\text{aq})]$ and absorbance.

1M Similar

(iv)

Suggest how the student could improve the reliability of the data obtained in the experiment in (c).

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Repeat the preparation and measurement of solution 3 (the solution giving the anomalous result) to obtain a mean absorbance or confirm the result.

Final answer

Repeat solution 3

Detailed explanation

Background Concept

Reliability of experimental data is improved by repeating measurements and calculating a mean. When an anomalous result is identified, repeating that specific experiment helps to determine if the anomaly was a one-off error or a systematic issue.

Understanding the Question

We identified solution 3 as having an anomalous absorbance. The question asks how to improve the reliability of the data obtained in the experiment.

Approach

The most direct improvement when an anomaly is found is to repeat the procedure for that specific anomalous point.

Step-by-Step Reasoning

Solution 3 gave an absorbance (0.270) that did not fit the calibration line.
To improve reliability, the student should repeat the preparation of solution 3 (measuring the volumes of solution E and water again) and remeasure its absorbance.
This will provide a second (and potentially third) data point for that concentration, allowing the student to calculate a mean and identify if the original result was truly anomalous or if there was a persistent error.

Key Takeaways

Repeating anomalous results is a key method for improving data reliability and identifying random errors.

Common Mistakes

Suggesting 'use a more accurate colorimeter' (this doesn't address the volumetric error that caused the anomaly).
Suggesting 'repeat the whole experiment' (too vague; specifying the anomalous solution is better).
Suggesting 'use a larger range of concentrations' (this improves the calibration curve but doesn't fix the reliability of the anomalous point).

Things to Be Careful About

The improvement must be specific and relevant to the anomaly identified in the previous part. Mentioning 'solution 3' or 'the anomalous result' shows clear understanding.

Techniques used

suggest repetition to improve reliability

(e)

The student carries out a further experiment to examine the kinetics of the reaction between crystal violet, $\text{C}_{25}\text{H}_{30}\text{N}_3\text{Cl}(\text{aq})$ , and aqueous sodium hydroxide, $\text{NaOH}(\text{aq})$ .

\text{C}_{25}\text{H}_{30}\text{N}_3\text{Cl}(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{C}_{25}\text{H}_{30}\text{N}_3\text{OH}(\text{aq}) + \text{Cl}^-(\text{aq})

The disappearance of the purple colour as the reaction proceeds can be monitored by measuring how the absorbance of light by the mixture changes using a colorimeter.

The student mixes $5\text{ cm}^3$ of solution 6 with $5\text{ cm}^3$ of $\text{NaOH}(\text{aq})$ , a large excess, and immediately starts the stopwatch.

The resulting mixture is then placed in a colorimeter. The absorbance of this mixture is measured every 100 seconds after starting the stop-watch.

Fig. 2.3 shows a graph of the student’s results.

Fig. 2.3

(i)

Suggest why it is not possible for the student to measure the absorbance of the mixture at $t = 0\text{ s}$ .

1M Similar

(ii)

Use the graph in Fig. 2.3 to find the half-life, $t_{1/2}$ , starting at $100\text{ s}$ .

State the coordinates of both points on the line of best fit used in your calculation.

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

half-life .............................. s

2M Similar

DifficultyMedium

Worked solution

Answer

Coordinates 1: (100, 0.360)
Coordinates 2: (315, 0.180) (accept 310 to 320)
half-life: 215 s (accept 210 to 220)

Final answer

215 s

Detailed explanation

Background Concept

The half-life ( $t_{1/2}$ ) of a reaction is the time taken for the concentration of a reactant to fall to half of its initial value. Since absorbance is directly proportional to concentration (from part d), we can use absorbance to determine the half-life. We find the time at which the absorbance has halved from a starting point.

Understanding the Question

We are given a graph of absorbance against time for the reaction. We need to find the half-life starting at $t = 100$ s. We must state the coordinates of the two points used and calculate the half-life.

Approach

Find the absorbance at $t = 100$ s.
Calculate half of this absorbance value.
Find the time on the graph where the absorbance equals this half-value.
The half-life is the difference between this new time and the starting time (100 s).

Step-by-Step Reasoning

Step 1: At $t = 100$ s, read the absorbance from the line of best fit. The graph shows an absorbance of approximately 0.360. So, coordinates 1 = (100, 0.360).
Step 2: Half of 0.360 is 0.180.
Step 3: Find the time where absorbance = 0.180 on the line of best fit. Reading across from 0.180 on the y-axis to the curve and down to the x-axis gives approximately 315 s (accept 310 to 320 s). So, coordinates 2 = (315, 0.180).
Step 4: Calculate the half-life: $t_{1/2} = 315 \text{ s} - 100 \text{ s} = 215 \text{ s}$ (accept 210 to 220 s based on reading variations).

Key Takeaways

When using a graph to find half-life, always state the coordinates of the two points used for the calculation. Half-life is a time interval ( $\Delta t$ ), not an absolute time.

Common Mistakes

Reading the absorbance directly from the data points instead of the line of best fit.
Forgetting to subtract the starting time (100 s) to get the half-life duration (giving 315 s instead of 215 s).
Not stating the coordinates clearly.

Things to Be Careful About

Use the line of best fit, not the raw data points (if plotted, though here the curve is given).
Ensure coordinates are in the format (time, absorbance) or clearly labeled.
The half-life must be in seconds.

Techniques used

determine half-life from a concentration-time graph

(iii)

Another student repeats the experiment at a different temperature and measures two half-life values. The values obtained are $420\text{ s}$ and $425\text{ s}$ .

Use these values to deduce the order of the reaction with respect to $\text{C}_{25}\text{H}_{30}\text{N}_3\text{Cl}(\text{aq})$ .

Explain your answer.

1M Similar

DifficultyMedium

Worked solution

Answer

Order: First order.

Explanation: The half-life is constant (within experimental error) regardless of the concentration (or temperature, in this context, the key is the constant half-life at a given concentration for a first-order reaction).

Final answer

first order; half-lives are constant

Detailed explanation

Background Concept

The order of a reaction with respect to a reactant can be determined by examining how the half-life changes with concentration:

Zero order: Half-life decreases as concentration decreases ( $t_{1/2} \propto [A]$ ).
First order: Half-life is constant and independent of concentration ( $t_{1/2} = \ln 2 / k$ ).
Second order: Half-life increases as concentration decreases ( $t_{1/2} \propto 1/[A]$ ).

Understanding the Question

Another student repeated the experiment at a different temperature and found two half-life values: 420 s and 425 s. These values are very close (constant within experimental error). We need to deduce the order with respect to crystal violet and explain.

Approach

Recognize that constant half-life is the defining characteristic of a first-order reaction. State the order and provide the reasoning.

Step-by-Step Reasoning

The two half-life values obtained (420 s and 425 s) are approximately equal.
A constant half-life (independent of the initial concentration of the reactant) is a characteristic feature of a first-order reaction.
Therefore, the reaction is first order with respect to crystal violet.
The explanation is simply that the half-lives are constant (within experimental error).

Key Takeaways

Constant half-life = first order. This is a fundamental relationship in chemical kinetics that allows quick determination of reaction order from half-life data.

Common Mistakes

Saying 'the rate is constant' (that would be zero order).
Not explaining why it is first order (must mention constant half-life).
Confusing the effect of temperature on half-life with the definition of reaction order (the key is that for a first-order reaction, at any given concentration, the half-life is constant; the question implies these are half-lives at different starting concentrations or just generally constant behavior).

Things to Be Careful About

The question asks to 'deduce the order' AND 'explain your answer'. Both parts are required for the mark.
Use the exact terminology: 'first order' and 'half-lives are constant'.

Techniques used

deduce reaction order from constant half-life