A student separates petroleum. Fig. 1.1 shows how the student assembles the apparatus. (a) State the name of the piece of apparatus labelled A in Fig. 1.1. \_\_\_\_\_\_ (b) The student makes a mistake when assembling the apparatus. Describe this mistake. \_\_\_\_\_\_ (c) When it is correctly assembled, the apparatus shown in Fig. 1.1 is used to separate petroleum into useful substances. (i) State the name of this separation process. \_\_\_\_\_\_ (ii) State the property of…

(a) condenser · (b) The conical flask is sealed with a bung · (c)(i) fractional distillation · (c)(ii) boiling point · (d) piece of apparatus: electric heater; explanation: petroleum is flammable

Chemistry 5070/41 — October/November 2025

Cambridge O-Level · Alternative to Practical · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Experimental Contexts · Use of Techniques, Apparatus and Materials · Observations and Measurements · Planning Experiments and Investigations · Analysis, Conclusions and Evaluation · Qualitative Analysis

Q16MMedium-EasyUse of Techniques, Apparatus and MaterialsExperimental Contexts

A student separates petroleum.

Fig. 1.1 shows how the student assembles the apparatus.

Fig. 1.1

(a)

State the name of the piece of apparatus labelled A in Fig. 1.1.

______

1M Similar

(b)

The student makes a mistake when assembling the apparatus.

Describe this mistake.

______

1M Similar

(c)

When it is correctly assembled, the apparatus shown in Fig. 1.1 is used to separate petroleum into useful substances.

(i)

State the name of this separation process.

______

1M Similar

(ii)

State the property of these substances that allows them to be separated.

______

1M Similar

(d)

State the name of the piece of apparatus the student uses to safely heat the petroleum in Fig. 1.1.

Explain why this piece of apparatus is suitable.

piece of apparatus ______
explanation ______

2M Similar

Q214MMediumObservations and MeasurementsExperimental ContextsAnalysis, Conclusions and EvaluationUse of Techniques, Apparatus and MaterialsPlanning Experiments and Investigations

A student investigates the temperature changes when an acid neutralises aqueous sodium hydroxide.

The ionic equation for the reaction is shown.

\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}

The reaction is exothermic.

The temperature change is used to determine the concentration of hydrogen ions in an acid.

X is $1.60\text{ mol/dm}^3$ sodium hydroxide solution
Y is an acid of concentration $2.00\text{ mol/dm}^3$

The student:

step 1 fills the burette with Y
step 2 stands a plastic cup inside a beaker
step 3 uses a volumetric pipette to put $25.0\text{ cm}^3$ of X into the plastic cup
step 4 measures the initial temperature of X and records it in Table 2.1
step 5 adds $5.0\text{ cm}^3$ of Y from the burette to the plastic cup
step 6 stirs the mixture for 30 seconds and records in Table 2.1 the temperature of the mixture
step 7 repeats steps 5 and 6 until a total of $40.0\text{ cm}^3$ of Y is added.

Fig. 2.1 on page 6 shows two of the thermometer readings in $^\circ\text{C}$ during the experiment.

(a)

Complete Table 2.1.

You should:

record the student's results for $5.0\text{ cm}^3$ and $30.0\text{ cm}^3$ using Fig. 2.1
subtract the initial temperature of X from each temperature recorded to determine the temperature change.

All temperatures and temperature changes should be recorded to $+/-\ 0.5\,^\circ\text{C}$ .

Fig. 2.1

Table 2.1

volume of Y / $\text{cm}^3$	temperature / $^\circ\text{C}$	temperature change / $^\circ\text{C}$
0.0	20.5	0.0
5.0
10.0	26.0
15.0	29.5
20.0	31.5
25.0	30.5
30.0
35.0	27.5
40.0	26.5

4M Similar

DifficultyMedium-Easy

Worked solution

Answer

volume of Y / $\text{cm}^3$	temperature / $^\circ\text{C}$	temperature change / $^\circ\text{C}$
0.0	20.5	0.0
5.0	24.0	3.5
10.0	26.0	5.5
15.0	29.5	9.0
20.0	31.5	11.0
25.0	30.5	10.0
30.0	29.0	8.5
35.0	27.5	7.0
40.0	26.5	6.0

Final answer

Completed Table 2.1 with 24.0 °C and 29.0 °C recorded, and temperature changes 0.0, 3.5, 5.5, 9.0, 11.0, 10.0, 8.5, 7.0, 6.0 °C

Detailed explanation

Walkthrough

Read thermometer at $5.0\text{ cm}^3$ : From Fig. 2.1, the liquid level sits exactly at the fourth mark above $20$ , which is $24.0\,^\circ\text{C}$ .
Read thermometer at $30.0\text{ cm}^3$ : The liquid level is one mark below $30$ , which is $29.0\,^\circ\text{C}$ .
Calculate temperature changes: Subtract the initial temperature ( $20.5\,^\circ\text{C}$ $20.5^{\circ} C$ at $0.0\text{ cm}^3$ $0.0 cm^{3}$ ) from each temperature reading:
- For $0.0\text{ cm}^3$ : $20.5 - 20.5 = 0.0\,^\circ\text{C}$
- For $5.0\text{ cm}^3$ : $24.0 - 20.5 = 3.5\,^\circ\text{C}$
- For $10.0\text{ cm}^3$ : $26.0 - 20.5 = 5.5\,^\circ\text{C}$
- For $15.0\text{ cm}^3$ : $29.5 - 20.5 = 9.0\,^\circ\text{C}$
- For $20.0\text{ cm}^3$ : $31.5 - 20.5 = 11.0\,^\circ\text{C}$
- For $25.0\text{ cm}^3$ : $30.5 - 20.5 = 10.0\,^\circ\text{C}$
- For $30.0\text{ cm}^3$ : $29.0 - 20.5 = 8.5\,^\circ\text{C}$
- For $35.0\text{ cm}^3$ : $27.5 - 20.5 = 7.0\,^\circ\text{C}$
- For $40.0\text{ cm}^3$ : $26.5 - 20.5 = 6.0\,^\circ\text{C}$
Ensure all entries are recorded to $.0$ or $.5$ precision as instructed.

Key Takeaways

Thermometer readings and calculated differences in Paper 4 must be recorded with a consistent decimal place (e.g., $.0$ or $.5$ ) when specified.
Temperature change is always calculated relative to the initial starting temperature at $0.0\text{ cm}^3$ .

Common Mistakes

Writing whole numbers without the decimal place (e.g., writing $24$ instead of $24.0$ or $9$ instead of $9.0$ ).
Subtracting consecutive readings rather than always subtracting the initial temperature ( $20.5\,^\circ\text{C}$ ).

Things to Be Careful About

Double check the initial temperature value ( $20.5\,^\circ\text{C}$ ) used as the baseline for all subtractions.

Techniques used

read thermometer scale to the nearest 0.5 °Ccalculate temperature change by subtractionrecord data consistently with required precision

(b)

Plot a graph of the temperature change ( $y$ -axis) against the volume of Y ( $x$ -axis) on Fig. 2.2.

Draw a line of best fit through the points where the temperature change is increasing.

Draw a line of best fit through the points where the temperature change is decreasing.

Extend both lines so that they cross.

Fig. 2.2

5M Similar

DifficultyMedium

Worked solution

Answer

$y$ -axis labelled with temperature change / °C (or $^\circ\text{C}$ ).
Suitable linear scale on the $y$ -axis (e.g. $2\,^\circ\text{C}$ per major division/2 cm, running from $0$ to at least $12\,^\circ\text{C}$ ) occupying more than half the vertical grid.
All points plotted accurately to within $\pm\ \frac{1}{2}$ small square:
$(0.0, 0.0)$ , $(5.0, 3.5)$ , $(10.0, 5.5)$ , $(15.0, 9.0)$ , $(20.0, 11.0)$ , $(25.0, 10.0)$ , $(30.0, 8.5)$ , $(35.0, 7.0)$ , $(40.0, 6.0)$ .
Straight line of best fit drawn through increasing temperature change points and extended.
Straight line of best fit drawn through decreasing temperature change points and extended to intersect the first line.

Final answer

Graph plotted with suitable scale, points correctly plotted, and two intersecting straight lines of best fit drawn

Detailed explanation

Walkthrough

Axis Setup: The $x$ -axis is pre-labelled with volume of Y / cm³ from $0$ to $40$ . Label the vertical axis as temperature change / °C. Choose a linear scale such that the maximum value ( $11.0\,^\circ\text{C}$ ) occupies more than half the grid height (for instance, $2\,^\circ\text{C}$ every 2 cm / 10 small squares, reaching $12$ or $14\,^\circ\text{C}$ at the top).
Plotting Points: Plot all 9 pairs of coordinates from Table 2.1 accurately to within half a small square using small crosses or encircled dots.
Lines of Best Fit:
- Draw a straight line of best fit through the rising section of points (from $0.0\text{ cm}^3$ to $20.0\text{ cm}^3$ ).
- Draw a second straight line of best fit through the falling section of points (from $25.0\text{ cm}^3$ to $40.0\text{ cm}^3$ ).
Intersection: Extend both straight lines until they intersect clearly. The intersection point corresponds to the stoichiometric neutralisation endpoint.

Key Takeaways

In thermometric titrations, two distinct linear trends appear: an increase while the reaction occurs, and a decrease once all reactant has been consumed due to heat loss and dilution by cold titrant.
The equivalence point is found at the intersection of these two straight lines.

Common Mistakes

Drawing a single curved line over all points instead of two distinct straight lines.
Forgetting units on the $y$ -axis label.
Choosing a compressed scale where the plotted data spans less than half of the available grid.

Things to Be Careful About

Ensure straight lines are drawn with a ruler and extended cleanly to cross each other.

Techniques used

select a suitable linear scale for the y-axisplot points accurately on a griddraw two straight lines of best fitextend lines to determine point of intersection

(c)

Find the point on the graph where the two lines cross.

Determine the volume of Y at this point.

volume = ______ $\text{cm}^3$

1M Similar

(d)

X is $1.60\text{ mol/dm}^3$ sodium hydroxide solution.

The formula of sodium hydroxide is $\text{NaOH}$ .

Calculate the number of moles of hydroxide ions, $\text{OH}^-$ , in $25.0\text{ cm}^3$ of X.

number of moles = ______

1M Similar

(e)

The volume of Y in (c) is the volume needed to completely neutralise $25.0\text{ cm}^3$ of X.

Use your answers to (c) and (d) to calculate the concentration of hydrogen ions, $\text{H}^+$ , in Y.

\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}

concentration = ______ $\text{mol/dm}^3$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

From the equation $\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}$ , $1\text{ mol}$ of $\text{OH}^-$ reacts with $1\text{ mol}$ of $\text{H}^+$ .

\text{moles of }\text{H}^+ = \text{moles of }\text{OH}^- = 0.0400\text{ mol}

\begin{aligned} \text{concentration of }\text{H}^+ &= \frac{\text{moles}}{\text{volume of }\mathbf{Y}\text{ in }\text{dm}^3} \\ &= \frac{0.0400}{\frac{20.0}{1000}} \\ &= \frac{0.0400 \times 1000}{20.0} \\ &= 2.00\text{ mol/dm}^3 \end{aligned}

Answer

$2.00\text{ mol/dm}^3$

Final answer

2.00 mol/dm3

Detailed explanation

Walkthrough

From the neutralisation equation $\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}$ , the reacting ratio is $1:1$ .
Therefore, moles of $\text{H}^+$ in the neutralisation volume of Y equals the moles of $\text{OH}^-$ from part (d) ( $0.0400\text{ mol}$ ).
Use the formula:

\text{concentration} = \frac{\text{moles}}{\text{volume (in }\text{dm}^3\text{)}}

Substitute the volume from part (c) ( $20.0\text{ cm}^3 = 0.0200\text{ dm}^3$ ):

\text{concentration} = \frac{0.0400}{0.0200} = 2.00\text{ mol/dm}^3

(Allow error carried forward from (c) and (d) via $\frac{\text{(d)} \times 1000}{\text{(c)}}$ ).

Key Takeaways

Neutralisation between strong acids and alkalis involves a $1:1$ ratio between $\text{H}^+$ and $\text{OH}^-$ ions.
Concentration is calculated by dividing moles by volume in $\text{dm}^3$ .

Common Mistakes

Inverting the division (dividing volume by moles).
Forgetting to convert volume from $\text{cm}^3$ to $\text{dm}^3$ .

Things to Be Careful About

Ensure the volume used is your value from (c) and the moles are from (d).

Techniques used

relate stoichiometry of neutralisationcalculate concentration from moles and volumeapply error-carried-forward from previous parts

(f)

Suggest why Y is added to X using a burette and not a measuring cylinder.

______

1M Similar

(g)

Suggest why the temperatures are recorded after 30 seconds and not immediately after mixing.

______

1M Similar

Q314MMediumExperimental ContextsQualitative AnalysisObservations and MeasurementsUse of Techniques, Apparatus and Materials

A student does a series of tests using aqueous solution Q and aqueous solution R.

(a)

Solution Q is acidic. Solution Q contains one cation and one anion.

The tests the student does on Q are shown in Table 3.1.

Some of the observations for these tests are also shown.

Table 3.1

test number	test	observations
1	add five drops of universal indicator	colour change
2	add a piece of magnesium ribbon	magnesium ribbon disappears
3	add dilute nitric acid followed by aqueous silver nitrate	white precipitate

(i)

Describe the expected colour change in test 1.

______

1M Similar

(ii)

A gas is produced in test 2.

Predict one other observation made by the student in test 2.

______

1M Similar

(iii)

Predict the identity of the gas produced in test 2.

______

1M Similar

(iv)

Describe a test to confirm the identity of this gas.

Include the result of a positive test.

test ______
result ______

2M Similar

(v)

Identify the cation and the anion in Q.

cation = ______ anion = ______

2M Similar

(b)

The tests the student does on R are shown in Table 3.2.

Some of the observations for these tests are also shown.

Table 3.2

test number	test	observations
4	flame test	blue-green flame
5	add aqueous ammonia drop by drop until a change is seen add excess aqueous ammonia	light blue precipitate blue precipitate dissolves

(i)

Describe how the student does a flame test.

______

2M Similar

(ii)

Describe one additional observation expected in test 5 based on the observations in tests 4 and 5.

______

1M Similar

(iii)

Describe one additional test to confirm the identity of the cation in R.

Include the result of a positive test.

test ______
result ______

3M Similar

(iv)

Identify the cation in R.

cation = ______

1M Similar

Q46MMediumPlanning Experiments and InvestigationsExperimental Contexts

Alcohols are used as fuels to heat water.

Plan an experiment to determine which alcohol, methanol or ethanol, releases more thermal energy per gram of alcohol burned.

Your plan should describe the use of an alcohol burner, as shown in Fig. 4.1, to heat water. You should use water, methanol, ethanol and common laboratory apparatus. No other chemicals should be used.

Fig. 4.1

Your plan should include:

the additional apparatus needed
the method to use and the measurements to take
how the measurements are used to determine which alcohol releases more thermal energy per gram burned.

You may draw a diagram to help answer the question.

DifficultyMedium

Worked solution

Answer

Apparatus:

Measuring cylinder (or burette / volumetric pipette)
Thermometer
Balance
Copper can / beaker / calorimeter
Clamp and stand / tripod and gauze

Method and Measurements:

Measure a fixed volume of water (e.g. $100\text{ cm}^3$ ) using a measuring cylinder and pour it into a copper can / beaker.
Measure and record the initial temperature of the water using a thermometer.
Measure and record the initial mass of the alcohol burner containing methanol using a balance.
Place the burner beneath the container of water, light the wick, and heat the water (e.g. for a set time or until a temperature rise of around $20\text{ }^\circ\text{C}$ is reached).
Extinguish the flame, stir the water, and record the highest final temperature reached.
Reweigh the alcohol burner and record its final mass.
Repeat steps 1 to 6 using ethanol, keeping the volume of water, the distance between the flame and the container, and the container identical.

Determination:

Calculate the temperature rise: $\Delta T = \text{final temperature} - \text{initial temperature}$ .
Calculate the mass of alcohol burned: $\Delta m = \text{initial mass} - \text{final mass}$ .
Calculate the temperature change per gram of alcohol burned for each alcohol:

\text{temperature change per gram} = \frac{\Delta T}{\Delta m}

The alcohol with the greater temperature change per gram releases more thermal energy per gram.

Final answer

Plan describing: measuring a fixed volume of water; recording initial/final water temperatures and initial/final burner masses for methanol and ethanol; calculating temperature change per gram burned (temperature change / mass of fuel burned); the fuel giving the greater temperature change per gram releases more energy.

Detailed explanation

Walkthrough

To compare the energy released per gram of two liquid fuels (methanol and ethanol), a standard simple calorimetry investigation is planned:

Apparatus required:
- A measuring cylinder or pipette to measure a defined volume of water.
- A container (such as a beaker or copper can/calorimeter) to hold the water.
- A thermometer to measure water temperature.
- A balance (precision balance) to measure the mass of the spirit burner before and after combustion.
- A clamp stand or tripod and gauze to position the container over the burner.
Measurements to record:
- Mass of burner + alcohol before burning and after burning (to find the mass of alcohol burned, $\Delta m$ ).
- Temperature of water before heating and maximum temperature after heating (to find the temperature rise, $\Delta T$ ).
Control variables (Fair test):
- Same volume/mass of water used for each alcohol.
- Same type of container (material and size).
- Same starting temperature of water (or same height/distance of beaker above the wick).
Processing results:
- For each alcohol, calculate:

\text{Temperature change per gram} = \frac{\text{temperature change (}^\circ\text{C)}}{\text{mass of alcohol burned (g)}} = \frac{\Delta T}{\Delta m}

Compare the two values: the alcohol with the larger value of $\frac{\Delta T}{\Delta m}$ releases more thermal energy per gram.

Key Takeaways

In simple combustion calorimetry, the thermal energy released is proportional to the temperature rise of the water (assuming heat capacity and mass of water are kept constant: $Q = mc\Delta T$ ).
Comparing energy per gram requires dividing the temperature increase by the actual mass of fuel consumed, not just burning for a fixed duration.

Common Mistakes

Measuring the volume of alcohol burned instead of the mass burned (the question asks for energy per gram).
Forgetting to take the final mass of the burner or assuming all alcohol in the burner was consumed.
Not keeping the volume of water constant between trials.
Stating that the alcohol with the highest temperature rise wins without dividing by the mass of alcohol burned.

Things to Be Careful About

Always state both the initial and final readings for both temperature and mass.
Clearly state the mathematical relationship used to draw the final conclusion: $\frac{\Delta T}{\Delta m}$ .

Techniques used

identify required laboratory apparatusdescribe a step-by-step calorimetry procedurespecify initial and final measurements of mass and temperaturecontrol experimental variablescalculate and compare temperature rise per gram of fuel