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 question

Question

The question describes an investigation into how different concentrations of sugar solution affect pieces of potato. It lists a method: cut five potato pieces, record starting mass, place each in a different sugar concentration, leave for 24 hours, record final mass, and calculate the change in mass. Part 05.1 asks for the independent variable with four tick-box options: change in mass, concentration of sugar solution, length of time in solution, and starting mass. A results table shows sugar concentrations from 0.0 to 0.5 mol/dm3 with starting mass, final mass after 24 hours, and change in mass. Part 05.2 asks why potato in 0.0 mol/dm3 increased in mass. Part 05.3 asks the student to plot the data and draw a line of best fit on a blank graph with change in mass on the y-axis and concentration of sugar solution on the x-axis. Part 05.4 asks to use the graph to determine the concentration of sugar solution inside the potato cells when the mass does not change. Part 05.5 asks to calculate percentage change in mass for the potato in 0.2 mol/dm3 solution using the equation percentage change in mass = change in mass ÷ mass at start × 100.
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 The mark scheme gives the accepted answer for 05.1 as concentration of the sugar solution. For 05.2 it accepts gained water, or that water moves by osmosis, or that the water concentration outside the potato is greater than inside the cells; converse wording is allowed. For 05.3 it awards a mark for all points correctly plotted and a mark for drawing a curve of best fit through the points, ignoring extrapolation. For 05.4 it awards a mark for a correct reading from the graph, allowing about 0.23 to 0.24 mol/dm3 if no line is drawn. For 05.5 it shows the calculation 0.25 × 100 ÷ 7.96, giving 3.14%, with allowance for equivalent correct working using 8.21 minus 7.96 over 7.96 and correct rounding to 3 significant figures.

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

Standard Demand

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.

Question overview

💡 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

1 mark: correct choice from the list.

💡 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.

2 marks: one for gaining water / water moving in, and one for osmosis or correct concentration idea.

💡 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
0.02.20
0.11.15
0.20.25
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 all points plotted correctly (allow ± ½ a square).
1 mark for a curved line of best fit through the points.

🧠 Exam technique

  1. Read each coordinate as (concentration, change in mass) .
  2. Plot each point carefully to within half a square.
  3. Do not join the dots with straight lines.
  4. 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.

1 mark: correct reading from the graph (allow ± ½ a square).

💡 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

  1. Use the value for change in mass at 0.2 mol/dm³: 0.25 g
  2. Use the starting mass at 0.2 mol/dm³: 7.96 g
  3. Substitute into the equation:
    percentage change in mass = (change in mass ÷ mass at start) × 100
  4. Calculation:
    (0.25 ÷ 7.96) × 100 = 3.14070352...
  5. Round to 3 significant figures:
    3.14%
1 mark for correct substitution.
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.