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AQA GCSE Biology · 8461
AQA 8461 · 4.1.3.2 Check the specification (PDF) (opens in a new tab)
Define osmosis as the diffusion of water from a dilute solution to a concentrated solution through a partially permeable membrane.
Use simple compound measures of rate of water uptake.
Calculate percentage gain and loss of mass of plant tissue.
Plot, draw, and interpret graphs related to osmosis experiments.
Investigate the effect of salt or sugar solution concentrations on the mass of plant tissue.
A cell membrane separates the liquid inside a cell from the liquid around it. Water molecules can pass through this membrane, but some dissolved substances cannot. A dilute solution has less dissolved substance and more water than a concentrated solution. Osmosis is the diffusion of water from a dilute solution to a concentrated solution through a partially permeable membrane. Water moves both ways, but its net movement is towards the more concentrated solution.
Water moves both ways across the membrane, but net movement is from the dilute solution to the concentrated solution.
If a potato cell is placed in a dilute sugar solution, water enters it by osmosis. In a more concentrated sugar solution, water leaves it. This movement changes the mass of potato tissue. If the solutions on either side of the membrane have the same tendency to draw in water, movement continues both ways but there is no net movement and the tissue’s mass does not change overall.
AQA’s required practical investigates how a range of salt or sugar solution concentrations affects the mass of plant tissue. One approach is to cut potato pieces to the same size and measure each piece’s initial mass using a balance. Put the pieces into clearly labelled tubes containing different concentrations of the same solute. Use enough solution to cover each piece, leave them for the same length of time, then remove them, blot them in the same way and measure their final masses. Blotting removes surface liquid that would otherwise add to the measured mass. A school procedure might use a cork borer to make equal-length cylinders; thin potato pieces can also be used.
The independent variable is solution concentration. The dependent variable is the change in the tissue’s mass. Keep the type and size of tissue, volume of solution, temperature, immersion time and blotting method as similar as possible. These controls help ensure that a difference in mass is due to concentration rather than another factor. Repeating measurements at each concentration and calculating a mean helps reveal unusual results.
Calculate change in mass as . A positive result means the tissue gained water overall; a negative result means it lost water overall. Pieces may start with slightly different masses, so percentage change makes their results easier to compare:
For example, a piece with an initial mass of 2.00 g and a final mass of 2.24 g has gained 0.24 g, or of its initial mass. A piece that falls from 2.00 g to 1.80 g has a percentage change of .
A rate describes change per unit time. If the 0.24 g gain occurred over 15 minutes, the average rate of mass gain was . This is a simple measure of net water uptake over that interval; a single initial and final mass cannot show whether the rate stayed constant throughout.
Plot solution concentration on the horizontal axis and percentage change in mass on the vertical axis. Plot the results accurately and draw a suitable best-fit line or curve through the pattern. The graph shows the usual interpretation.
Positive values indicate mass gain, negative values mass loss, and the zero crossing indicates no net osmosis.
Data for Effect of sucrose concentration on potato mass
| Series | Sucrose concentration (mol dm⁻³) | Percentage change in mass (%) |
|---|---|---|
| Illustrative potato results | 0 | 12 |
| Illustrative potato results | 0.2 | 6 |
| Illustrative potato results | 0.4 | 0 |
| Illustrative potato results | 0.6 | -6 |
| Illustrative potato results | 0.8 | -12 |
Points above zero show a mass gain: the surrounding solution was dilute enough for net water movement into the tissue. Points below zero show a mass loss: the surrounding solution drew water out. Where the best-fit line crosses zero, the tissue would have no overall change in mass. This estimates the concentration of the external solution at which there is no net osmosis. It need not tell you the exact concentration of every dissolved substance inside the cells.
Water diffuses from a dilute solution to a concentrated solution through a partially permeable membrane. Net water movement into plant tissue increases its mass; net movement out decreases it.
Vary salt or sugar solution concentration. Measure plant tissue mass before and after immersion, keeping tissue size, solution volume, temperature, time and blotting consistent.
; .
Plot concentration on the horizontal axis and percentage change in mass on the vertical axis. Above zero means gain; below zero means loss. The zero crossing estimates the concentration at which there is no net osmosis.
For percentage change, divide by the initial mass, not the final mass. Keep the sign: a negative answer means mass was lost.
Label graph axes with both the quantity and its unit. Percentage change in mass has the unit %; concentration may be given in mol dm⁻³.
At zero percentage change, there is no net movement of water. Water molecules still move in both directions.
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Osmosis
The diffusion of water from a dilute solution to a concentrated solution through a partially permeable membrane.
Partially permeable membrane
A membrane that allows some substances, including water, to pass through but prevents others from passing through.
Concentration gradient
The difference between two concentrations; in osmosis, it helps determine the direction of net water movement.
Percentage change in mass
The mass gained or lost, divided by the initial mass and multiplied by 100.
Put your knowledge into practice — try past paper questions for Biology
Osmosis
The diffusion of water from a dilute solution to a concentrated solution through a partially permeable membrane.
Partially permeable membrane
A membrane that allows some substances, including water, to pass through but prevents others from passing through.
Concentration gradient
The difference between two concentrations; in osmosis, it helps determine the direction of net water movement.
Percentage change in mass
The mass gained or lost, divided by the initial mass and multiplied by 100.