AQA GCSE Combined Science: Trilogy Biology Paper 1 (Foundation), 2020: Question 5
9 marks · Standard Demand difficulty · Short Answer
Investigate how changing the concentration of sugar solution affects the mass of potato pieces due to osmosis.
Practise this questionQuestion
Question text
05 A student investigated the effect of different concentrations of sugar solution on
pieces of potato.
This is the method used.
1. Cut five pieces of potato.
2. Record the starting mass of each piece of potato.
3. Place each piece of potato in a different concentration of sugar solution.
4. After 24 hours remove the pieces of potato from the solutions.
5. Record the final mass of each piece of potato.
6. Calculate the change in mass for each piece of potato.
05.1 What is the independent variable?
[1 mark]
Tick ( ) one box.
Change in mass of the pieces of potato
Concentration of the sugar solution
Length of time the pieces of potato are in the solution
Starting mass of the pieces of potato22
Table 3 shows the results.
Table 3
Concentration of Mass of potato at Mass of potato Change in mass
sugar solution start in grams after 24 hours in in grams
in mol/dm3 grams
0.0 7.94 10.14 2.20
0.1 7.95 9.10 1.15
0.2 7.96 8.21 0.25
0.3 7.93 7.53 –0.40
0.4 7.93 7.18 –0.75
0.5 7.95 7.00 –0.95
05.2 Explain why the potato in 0.0 mol/dm3 sugar solution increased in mass.
[2 marks]
05.3 Complete Figure 5.
Some of the results have been plotted for you.
You should:
• plot the data from Table 3
• draw a line of best fit through all the points.
[2 marks]
Figure 5
05.4 The mass of a piece of potato does not change when:
concentration of solution inside cells = concentration of solution outside cells
Determine the concentration of sugar solution inside the potato cells.
Use Figure 5.
[1 mark]
24 3
Concentration = mol/dm
Table 3 is repeated below.
Table 3
Concentration of Mass of potato at Mass of potato Change in mass
*23* sugar solution start in grams after 24 hours in in grams
in mol/dm3 grams
0.0 7.94 10.14 2.20
0.1 7.95 9.10 1.15
0.2 7.96 8.21 0.25
0.3 7.93 7.53 –0.40
0.4 7.93 7.18 –0.75
0.5 7.95 7.00 –0.95
05.5 Calculate the percentage change in mass for the potato in 0.2 mol/dm3 sugar solution.
Use Table 3.
Use the equation:
change in mass
percentage change in mass = × 100
mass of potato at start
Give your answer to 3 significant figures.
[3 marks]
Percentage change in mass (3 significant figures) = %
Mark scheme
Show the mark scheme
AO /
Question Answers Extra information Mark
Spec. Ref.
05.1 concentration of the sugar 1 AO1
solution 4.1.3.2
RPA 2
05.2 gained water 1 AO2
(water moves) by osmosis 1 AO1
or 4.1.3.2
allow converse statements RPA 2
(because) concentration of (because) concentration (of
water outside the potato is sugar solution) inside the potato
greater than inside the cells / is greater than outside the
potato potato / cells
05.3 all points correctly plotted allow ± ½ a square 1 AO2
4.1.3.2
RPA 2
line of best fit drawn as a curve ignore extrapolation of curve 1
through all the points
05.4 correct reading from their graph allow ± ½ a square 1 AO3
4.1.3.2
allow answer in range 0.23 to RPA 2
0.24 (mol/dm3) if no line drawn
0.25 × 100 (8.21-7.96) × 100
05.5 allow 1 AO2
7.96
7.96 4.1.3.2
RPA 2
= 3.14(070352) 1
3.14 (%) allow correct rounding to 3 sig 1
figs of an incorrectly calculated
percentage change
Total 9
How to answer it
Potato Osmosis Investigation
What this question tests
This question tests your understanding of required practical skills in osmosis: identifying the independent variable, explaining mass change using osmosis, plotting data on a graph, reading the concentration where there is no mass change, and calculating percentage change in mass correctly using data, units and significant figures.
💡 Key knowledge
- Osmosis is the movement of water through a partially permeable membrane.
- Water moves from a dilute solution to a more concentrated solution.
- If potato gains mass, water has entered the cells.
- If potato loses mass, water has left the cells.
- The point where change in mass = 0 shows equal concentration inside and outside the cells.
🧠 Exam technique
- Use the wording in the question carefully: “independent variable” means what is changed.
- For 2-mark explanations, include both what happened and why.
- On graphs, plot accurately and draw a smooth curve if the data changes in a pattern.
- In calculations, show substitution, working and final rounded answer.
Part (a) — 05.1 Independent variable
Identify what was changed in the investigation
✅ Correct answer
Concentration of the sugar solution
💡 Key knowledge
The independent variable is the variable the student deliberately changes. In this method, each potato piece was placed in a different concentration of sugar solution.
❌ Common errors
- Choosing change in mass — that is the dependent variable because it is measured.
- Choosing time — the time was kept the same at 24 hours, so it is a control variable.
- Choosing starting mass — that should also be controlled as much as possible, not changed on purpose.
Part (b) — 05.2 Why did the potato in 0.0 mol/dm³ increase in mass?
Explain the result using osmosis
✅ Correct answer
The potato gained water.
Water moved into the potato by osmosis.
Because the concentration of water outside the potato was greater than inside the potato cells.
💡 Key knowledge
- 0.0 mol/dm³ sugar solution is effectively pure water.
- Outside the potato, the solution is more dilute than inside the cells.
- So water moves into the potato cells through their partially permeable membranes.
- This causes the potato mass to increase.
🧠 Exam technique
For full marks, do not just say osmosis . Say clearly: water moved into the potato by osmosis .
Strong answers include the concentration difference: higher water concentration outside than inside .
❌ Common errors
- Saying sugar moved into the potato — the mark scheme wants movement of water.
- Saying simply it absorbed solution — too vague.
- Mixing up the direction of movement and saying water moved out.
Part (c) — 05.3 Complete the graph
Plot the remaining data and draw the line of best fit
💡 Data to plot from Table 3
| Concentration of sugar solution / mol dm⁻³ | Change in mass / g |
|---|---|
| 0.0 | 2.20 |
| 0.1 | 1.15 |
| 0.2 | 0.25 |
| 0.3 | −0.40 |
| 0.4 | −0.75 |
| 0.5 | −0.95 |
✅ Correct answer
- All points correctly plotted.
- A curve drawn as the line of best fit through all the points.
1 mark for a curved line of best fit through the points.
🧠 Exam technique
- Read each coordinate as (concentration, change in mass) .
- Plot each point carefully to within half a square.
- Do not join the dots with straight lines.
- Draw a smooth curve that follows the overall trend of the data.
Examiner insight: students often lost the second mark by drawing a ruler line or by linking points one-to-one instead of drawing a smooth best-fit curve.
❌ Common errors
- Plotting change in mass on the wrong axis.
- Missing the negative values below zero.
- Using a jagged line through each point instead of a smooth curve.
- Careless plotting more than half a square away.
Part (d) — 05.4 Concentration inside the potato cells
Read the value where the change in mass is zero
✅ Correct answer
Concentration ≈ 0.23 to 0.24 mol/dm³
A sensible quoted value is 0.24 mol/dm³.
💡 Key knowledge
The potato does not change mass when the concentration inside the cells equals the concentration outside. This is the point where the graph crosses change in mass = 0 .
🧠 Exam technique
You must use the line of best fit, not just guess from the table. Look for where the curve crosses the x-axis.
Top answers usually read the graph carefully and give a sensible value between the two surrounding data points: 0.2 mol/dm³ → +0.25 g and 0.3 mol/dm³ → −0.40 g .
❌ Common errors
- Writing 0.25 mol/dm³ because of the change in mass value — wrong quantity.
- Choosing exactly 0.2 or 0.3 mol/dm³ from the table rather than interpolating from the graph.
- Reading the y-axis instead of the x-axis.
Part (e) — 05.5 Percentage change in mass
Calculate the percentage change for 0.2 mol/dm³ sugar solution
📐 Calculations
- Use the value for change in mass at 0.2 mol/dm³: 0.25 g
- Use the starting mass at 0.2 mol/dm³: 7.96 g
- Substitute into the equation:
percentage change in mass = (change in mass ÷ mass at start) × 100 - Calculation:
(0.25 ÷ 7.96) × 100 = 3.14070352... - Round to 3 significant figures:
3.14%
1 mark for correct calculation.
1 mark for correct rounding to 3 significant figures.
✅ Correct answer
Percentage change in mass = 3.14%
🧠 Exam technique
- Use change in mass, not final mass.
- Use the starting mass in the denominator because that is what the formula says.
- Include the percentage sign.
- Round only at the end.
Examiner insight: many students lost marks by using 8.21 g instead of 0.25 g, or by dividing by the wrong mass.
❌ Common calculation traps
- Using 8.21 instead of 0.25 for the change in mass.
- Forgetting to subtract to find change in mass if needed.
- Dividing by final mass instead of starting mass.
- Giving too many decimal places instead of 3 significant figures.
Full-mark answer summary
✅ Answers by part
- 05.1: concentration of the sugar solution
- 05.2: potato gained water; water moved in by osmosis because water concentration outside was greater than inside the cells
- 05.3: all points plotted correctly; smooth curved line of best fit
- 05.4: about 0.23–0.24 mol/dm³
- 05.5: 3.14%
💡 Final revision takeaway
This is a classic osmosis practical question. To score highly, you need to know the variable types, explain water movement clearly, plot a graph accurately, read where the graph crosses zero, and complete percentage calculations with the correct formula and rounding.
Topics
Biology · Chemistry · B1: Cell Biology · C3: Quantitative Chemistry
Question and mark scheme from the AQA GCSE Combined Science: Trilogy examination, Biology Paper 1 (Foundation), 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.