AQA RP3 context · practical reasoning

Osmosis: compare tissue and solution

Investigate the effect of external sucrose concentration on the mass change of comparable plant-tissue pieces.

Each piece starts at 2.00 g. All pieces spend the same stated interval in solution. The example uses separate pieces from comparable tissue; it does not represent independent plants. Final masses are measured with a consistent blotting method.

Questions only · Matched example · 3 readings per condition.

Return to the workbench

Plan the comparison

What should be varied?

Vary external sucrose concentration between comparable tissue treatments.
Use progressively thinner tissue pieces as sucrose concentration increases.
Use tissue from a different plant source for each sucrose concentration.

Which calculation gives percentage mass change?

(Final mass − initial mass) ÷ final mass × 100.
Final mass ÷ initial mass × 100.
(Final mass − initial mass) ÷ initial mass × 100.

Which control approach is useful?

Weigh each piece straight from solution without removing surface liquid.
Use comparable tissue dimensions, exposure time and consistent blotting.
Leave the larger pieces immersed for longer to compensate for their size.

Calculate from evidence

Pieces are measured after consistent blotting.

Final mass / g (initial mass 2.00 g) · 3 readings per condition · Matched example
Sucrose concentration / mol dm⁻³	Reading 1	Reading 2	Reading 3
0	2.23	2.24	2.25
0.2	2.13	2.14	2.15
0.4	2.03	2.04	2.05
0.6	1.93	1.94	1.95
0.8	1.83	1.84	1.85

Percentage change = (mean final mass − initial mass) ÷ initial mass × 100. Estimate the zero crossing by local linear interpolation between the adjacent opposite signs. Give numerical answers to about three significant figures.

Calculate percentage mass change at 0.40 mol dm⁻³, using the mean final mass.

Estimate the zero-mass-change concentration by local linear interpolation, in mol dm⁻³.

Evaluate the comparison

Explain how the different blotting assumptions change the apparent zero-mass-change estimate. Suggest an improvement and explain why even a well-measured zero crossing does not establish identical solute composition inside and outside the tissue.

Use the alternate example below alongside the data above.

Final mass / g (initial mass 2.00 g) · 3 readings per condition · Surface solution remains on pieces
Sucrose concentration / mol dm⁻³	Reading 1	Reading 2	Reading 3
0	2.23	2.24	2.25
0.2	2.16	2.17	2.18
0.4	2.09	2.1	2.11
0.6	2.02	2.03	2.04
0.8	1.95	1.96	1.97

Original authored examples, not real experimental predictions, official marks or a laboratory protocol.