Required practical 2 · Biology · Biology Paper 1, both tiers
Osmosis in plant tissue
Investigate the effect of a range of sugar or salt solution concentrations on the mass of plant tissue.
The exam trap. Compare percentage change, not grams. Where the line crosses zero, the solution matches the cells' concentration.
Variables
- Independent (you change)
- Concentration of the sucrose solution (mol/dm³)
- Dependent (you measure)
- Percentage change in mass of the potato cylinder
- Control (you keep the same)
- Length and diameter of the cylinders (same cork borer)
- Volume of solution in each tube
- Time left in solution
- Temperature
- Type and source of potato
Apparatus
- Potato and cork borer
- Scalpel, ruler and white tile
- Boiling tubes and rack
- Sucrose solutions, for example 0.0, 0.2, 0.4, 0.6 and 0.8 mol/dm³
- Measuring cylinder
- Top-pan balance
- Paper towels
Method, and why each step matters
Cut potato cylinders with a cork borer and trim them to the same length with a scalpel.
Why Equal size means equal surface area, so differences come from concentration only.
Blot each cylinder and record its starting mass on a balance.
Why Surface water would add mass that is not part of the tissue.
Put each cylinder in a boiling tube with the same volume of a different concentration of sucrose solution.
Why The concentration is the independent variable; the volume is controlled.
Leave all the tubes for the same time, for example 24 hours, at room temperature.
Why Osmosis continues over time, so time must be the same.
Remove each cylinder, blot it dry in the same way and reweigh it.
Why Blotting the same way each time removes surface solution consistently.
Calculate the percentage change in mass for each and plot it against concentration.
Why Percentage change allows a fair comparison because starting masses differ.
Risks
| Hazard | How to reduce the risk |
|---|---|
| The scalpel and cork borer are sharp. | Cut downwards onto a tile, away from fingers. |
A worked set of results
| Concentration in mol/dm³ | Start mass in g | End mass in g | Change in mass in % |
|---|---|---|---|
| 0.0 | 2.50 | 2.95 | +18 |
| 0.2 | 2.48 | 2.63 | +6 |
| 0.4 | 2.52 | 2.42 | −4 |
| 0.6 | 2.50 | 2.20 | −12 |
| 0.8 | 2.45 | 2.06 | −16 |
- Percentage change = (end mass − start mass) ÷ start mass × 100. Keep the sign.
- Where the line of best fit crosses zero change, the solution has the same concentration as the potato cells (about 0.32 mol/dm³ here).
- Below that concentration water enters the cells by osmosis; above it water leaves.
Mistakes that cost marks
Comparing raw changes in grams when the cylinders started at different masses.
Instead Always compare percentage change.
Naming both variables when asked for the independent variable.
Instead The independent variable is only the one you change: the concentration.
Saying sugar moves into the cells.
Instead It is water that moves, from the more dilute to the more concentrated solution.
Practice questions on this practical
- A student investigated osmosis in potato tissue.In Diffusion, osmosis and active transport
- A class repeated the potato osmosis investigation with more concentrations.In Diffusion, osmosis and active transport
- A student wants to find out how the concentration of salt solution affects the mass of carrot tissue.In Diffusion, osmosis and active transport