AQA GCSE Combined Science: Trilogy Biology Paper 1 (Foundation), 2024: Question 7

9 marks · Standard Demand difficulty · Short Answer

Investigate osmosis using partially permeable tubing by identifying the independent variable, explaining a control step, calculating percentage change in mass, plotting results, and using a graph to determine the salt concentration of solution Z.

Practise this question

Question

The question page describes an osmosis investigation using six equal lengths of partially permeable tubing filled with the same volume of solution Z, sealed, weighed, placed in different salt concentrations, left for 2 hours, and then reweighed. A diagram shows one sealed tube labelled as containing solution Z immersed in a beaker of salt solution, with labels for the beaker and partially permeable membrane. The sub-questions ask for the independent variable from multiple-choice options, explain why the outside of each tube was dried before measuring mass, calculate a missing percentage change in mass from a results table, complete a graph by plotting percentage change in mass against salt concentration and drawing a line of best fit, and then use the graph to determine the concentration of solution Z. The table gives salt concentrations from 0.0 to 1.0 mol/dm^3 with starting mass, mass after 2 hours, change in mass, and percentage change, and the graph axes are percentage change in mass against concentration of salt solution in mol/dm^3 with one point already plotted.
Question text

07 A student investigated the concentration of salt in solution Z.

The student used a method involving osmosis.

The student used tubing made of partially permeable membrane.

This is the method used.

1. Cut six pieces of tubing to the same length.

2. Tie one end of each piece of tubing.

3. Put the same volume of solution Z into each piece of tubing.

4. Tie the other end of each piece of tubing to form a sealed tube.

5. Record the mass of each tube.

6. Place each tube into a different concentration of salt solution.

7. After 2 hours, remove each tube from the salt solutions.

8. Record the mass of each tube.

Figure 8 shows one of the sealed tubes in a salt solution.

Figure 8

07.1 What was the independent variable for the investigation?

[1 mark]

Tick ( ) one box.

Change in mass of tube

Concentration of salt solution

Time in salt solution

Volume of solution Z

07.2 The student dried the outside of each tube with a paper towel before

recording the mass.

Why was it important to dry the tubes?

[1 mark]

Table 3 shows the results.

Table 3

Concentration of Mass of tube in grams

Percentage (%)

salt solution

3 change in mass

in mol/dm At start After 2 hours Change

0.0 15.54 16.50 0.96 X

0.2 15.16 15.78 0.62 4.1

0.4 15.00 15.35 0.35 2.3

0.6 15.29 15.37 0.08 0.5

0.8 14.95 14.75 –0.20 –1.3

1.0 14.77 14.40 –0.37 –2.5

07.3 Calculate value X in Table 3.

Give your answer to 1 decimal place.

*23* [3 marks]

Value X (1 decimal place) = %

07.4 Complete Figure 9.

You should:

• plot the percentage change in mass from Table 3 for salt concentrations of

only 0.2 mol/dm3 to 1.0 mol/dm3

• draw a line of best fit.

One of the results has been plotted for you.25

[3 marks]

Figure 9

07.5 Determine the concentration of salt in solution Z.

Use Figure 9.

[1 mark]

Concentration = mol/dm3

Mark scheme

Show the mark scheme The mark scheme is a table for Question 7 showing answers, extra information, marks, and AO/specification references. It gives 07.1 as concentration of salt solution, 07.2 as water or solution left on the tube increasing the mass and making results invalid, 07.3 as using 0.96 divided by 15.54 times 100 to obtain 6.17 percent and then 6.2 percent to one decimal place, 07.4 as all four points plotted correctly with a line of best fit, and 07.5 as a value from the student's line of best fit with an accepted range of 0.64 to 0.66 mol/dm^3 if no line is drawn. The total shown for Question 7 is 9 marks.

Question 7

AO /

Question Answers Extra information Mark

Spec. Ref.

07.1 concentration of salt solution 1 AO2

4.1.3.2

RPA2

AO /

Spec. Ref.

07.2 (water / solution on the tube / allow the results would not be 1 AO2

tubing) would affect / increase valid 4.1.3.2

the mass RPA2

AO /

Spec. Ref.

07.3 0.96 1 AO2

× 100 4.1.3.2

15.54

RPA2

6.17 (760618…) (%) 1

6.2 (%) allow answer written in Table 1 1

allow correct conversion to one

decimal place from students’

incorrect percentage change

calculation using figures from

0.0 Concentration of salt

solution in mol/dm3

AO /

Spec. Ref.

07.4 all 4 points plotted correctly allow a tolerance of ± ½ a small 2 AO2

square

allow 1 mark for 3 points plotted

correctly

ignore attempt to plot a point for

0 mol/dm3

line of best fit ignore extrapolation 1 AO3

4.1.3.2

RPA2

AO /

Spec. Ref.

07.5 value from student’s line of best allow a tolerance of ± ½ a small 1 AO3

fit square 4.1.3.2

RPA2 19

if no line of best fit drawn allow 1

mark for an answer in the range

0.64 to 0.66

Total Question 7 9

How to answer it

Osmosis and salt concentration in tubing

AQA GCSE Combined Science: Trilogy
What this question tests
You need to identify the independent variable, explain why the tube must be dried before weighing, calculate percentage change in mass, plot points on a graph, draw a line of best fit, and use the graph to estimate the concentration of solution Z. It mainly tests practical skills, graph work, and osmosis knowledge.

Part (a) 07.1 — Independent variable

✅ Correct answer

Concentration of salt solution

This is the variable the student changes between samples.

💡 Key knowledge

  • The independent variable is the one you deliberately change.
  • Here, the student placed the tubes into different concentrations of salt solution.
  • The mass change is the dependent variable because it is measured.

🧠 Exam technique

  • Look for the variable that changes in the method.
  • In AQA practical questions, “what was changed?” usually means the independent variable.

❌ Common errors

  • Choosing change in mass because it is what they measure.
  • Choosing time in salt solution because it is mentioned in the method, even though it is kept the same.
  • Choosing volume of solution Z because it is controlled, not changed.

Part (b) 07.2 — Why dry the outside of the tubes?

✅ Correct answer

So water/solution on the outside of the tube would not affect the mass and make the result invalid.

💡 Key knowledge

  • Only the mass of the tubing and the solution inside should be measured.
  • If liquid remains on the outside, the tube appears heavier than it really is.
  • This would reduce the reliability of the mass change result.

🧠 Exam technique

For one-mark “why” questions, keep the reason short and direct: extra water would change the mass.

❌ Common errors

  • Saying only “to make it dry” without explaining the effect on mass.
  • Writing “to prevent evaporation” — that is not the key point here.
  • Vague answers like “to make it fair” may not gain the mark unless explained.

Part (c) 07.3 — Calculate percentage change in mass

For solution concentration 0.0 mol/dm³

📐 Calculations — step by step

  1. Find the change in mass: 16.50 − 15.54 = 0.96 g
  2. Use the percentage change formula:
    percentage change = change ÷ starting mass × 100
  3. Substitute the values:
    0.96 ÷ 15.54 × 100
  4. Calculate:
    = 6.17...%
  5. Round to 1 decimal place:
    6.2%

✅ Correct answer

X = 6.2%

Marks are awarded for using the correct method and giving the rounded answer.

💡 Key knowledge

  • Always use the starting mass in the denominator.
  • Percentage change is based on the original value, not the final value.
  • Keep units in grams until the final percentage answer, which has %.

❌ Common calculation traps

  • Using 16.50 instead of 15.54 as the starting mass.
  • Doing 15.54 ÷ 0.96 × 100 in the wrong order.
  • Forgetting to round to 1 decimal place.
  • Writing only the change in mass, not the percentage.

Part (d) 07.4 — Plot the graph and draw a best-fit line

✅ Correct answers

  • Plot the four points for 0.2 to 1.0 mol/dm³.
  • Draw a smooth line of best fit through the points.

The point for 0 mol/dm³ was already plotted, but the mark scheme says you may ignore attempts to plot it again.

💡 Key knowledge

  • Use the x-axis for concentration of salt solution in mol/dm³.
  • Use the y-axis for percentage change in mass.
  • The graph should show the overall trend, not dot-to-dot joins.

🧠 Exam technique

  • Read the axis labels carefully before plotting.
  • Marks are given for accuracy: the mark scheme allows a small tolerance of ± ½ small square.
  • If you miss one point but plot three correctly, you can still gain partial credit.
  • For a line of best fit, do not force the line through every point if it creates a poor trend.

❌ Common errors

  • Plotting the wrong coordinates, often by mixing up mass change and percentage change.
  • Joining points with straight segments instead of a best-fit curve/line.
  • Ignoring the axes or using the wrong scale.
  • Forgetting that the graph is used later to estimate the concentration of solution Z.
Examiner insight: Full marks required all 4 points plotted correctly for the data set and a sensible line of best fit. The key feature is a smooth trend showing how percentage change in mass varies with salt concentration.

Part (e) 07.5 — Determine the concentration of salt solution Z

✅ Correct answer

About 0.65 mol/dm³

Acceptable range if no line is drawn: 0.64 to 0.66 mol/dm³.

💡 Key knowledge

  • Solution Z is found by using the point where the mass change would be 0%.
  • That is where the line of best fit crosses the x-axis.
  • This means the solution has the same concentration as the contents of the tubing, so there is no net movement of water.

🧠 Exam technique

  • Use the graph, not the table, to estimate the value.
  • Draw a line across from 0% change in mass to the line of best fit, then drop down to the x-axis.
  • Give a sensible value to 2 decimal places if needed from the graph, but 0.65 mol/dm³ is the key estimate.

❌ Common errors

  • Choosing the concentration where the mass change is closest to zero from the table only, without using the graph.
  • Reading the wrong axis.
  • Giving a value far from the intersection point because the line of best fit was not drawn carefully.

Practical science link: what is happening?

💡 Key knowledge

  • Osmosis is the net movement of water through a partially permeable membrane.
  • Water moves from a dilute solution to a more concentrated solution.
  • If the outside salt solution is very dilute, water enters the tubing and mass increases.
  • If the outside salt solution is more concentrated, water leaves the tubing and mass decreases.

🧠 How to turn this into marks

  • Use the word osmosis when explaining mass change.
  • Include the idea of a partially permeable membrane.
  • Link the direction of water movement to the concentration difference.

Quick mark summary

✅ Answers

  • 07.1: concentration of salt solution
  • 07.2: water/solution on the tube would affect the mass
  • 07.3: 6.2%
  • 07.5: about 0.65 mol/dm³

💡 What gained top marks

  • Correctly identifying variables.
  • Clear percentage change calculation with rounding.
  • Accurate plotting and a sensible best-fit line.
  • Reading concentration from the graph where the line crosses 0% change.

Topics

Biology · B1: Cell Biology

Question and mark scheme from the AQA GCSE Combined Science: Trilogy examination, Biology Paper 1 (Foundation), 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.