Original application practice · 14 indicative marks
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Comparable plant-tissue cylinders were immersed for the same interval in sucrose solutions at the same temperature. The table gives mean masses from three independent cylinders at each concentration. These original classroom data describe this tissue and method; no water-potential calibration is supplied.
| Sucrose concentration / mol dm⁻³ | Initial mean mass / g | Final mean mass / g |
|---|---|---|
| 0 | 2.00 | 2.18 |
| 0.2 | 2.00 | 2.06 |
| 0.4 | 2.00 | 1.94 |
| 0.6 | 2.00 | 1.80 |
| 0.8 | 2.00 | 1.68 |
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Original self-review guidance, not an official mark scheme. Other scientifically valid responses may express the same ideas.
(1.94 − 2.00) ÷ 2.00 × 100 = −3.0%.
Alternatives: Equivalent working that uses initial mass as the denominator is valid. −3% and −3.00% express the same result here.
A positive 3% describes the size of the loss only; percentage change needs the negative sign.
Percentage change = (1.94 − 2.00)/2.00 × 100 = −3.0%. The initial mass is the reference. The negative value describes net mass loss over the immersion interval.
Mean mass change falls from +0.18 g at 0 to −0.32 g at 0.8 mol dm⁻³. Cylinders gain mass at 0 and 0.2, but lose mass from 0.4 upwards.
Alternatives: Correct percentage changes, final masses compared with the 2.00 g starting mass, or other accurate comparisons can support the description.
An osmosis explanation alone does not describe the results.
Mean final mass decreases from 2.18 g at 0 to 1.68 g at 0.8 mol dm⁻³, against an initial 2.00 g throughout. The tissue gains mass at 0 and 0.2, then loses mass from 0.4 upwards. The sign changes between 0.2 and 0.4 mol dm⁻³.
The external solution has a lower water potential than the tissue cells initially. Water moves out by osmosis across partially permeable membranes, reducing tissue mass.
Alternatives: The explanation may use water-potential gradients or describe water moving from higher to lower water potential. It must identify the outward direction.
The mass readings are consistent with this explanation; they do not measure water potential directly.
The concentrated external solution is consistent with a lower water potential than the tissue cells initially. Water moves out across partially permeable membranes down the water-potential gradient. Net loss of water can therefore reduce tissue mass, although the mass measurement alone does not identify every possible contribution.
The changes are +0.06 g at 0.2 and −0.06 g at 0.4 mol dm⁻³, so the interpolated zero is halfway between them: 0.30 mol dm⁻³.
Alternatives: A correctly drawn graph and interpolated crossing near 0.30 mol dm⁻³ is acceptable. The estimate assumes a roughly linear relationship between these points.
The estimate is not a directly measured concentration inside the cells.
The changes at 0.2 and 0.4 are equal and opposite, so linear interpolation places zero at 0.30 mol dm⁻³. This is an estimate between the measured points, assuming an approximately linear relationship there. Additional nearby concentrations and repeat evidence would refine it.
The zero-change estimate concerns this tissue under the assay conditions. Other solutes and pressure affect water potential, so it is not a sucrose assay. A tissue mean cannot establish the composition of every cell, and the interpolated value has unreported uncertainty.
Alternatives: Valid limits include unequal contributions of different cells, changes during immersion, or uncertainty in the mass measurements. A useful improvement can support a limit but must explain why the exact claim is unsupported.
Distinguish rejecting an exact claim from showing that osmosis did not occur. Water can still move in both directions at zero net change.
The estimate is consistent with no net mass change in this tissue and method, rather than proof of equal sucrose concentrations. Cells contain other solutes and can differ, and water potential also includes pressure effects. Test more concentrations near 0.30 with independent repeats and report their spread to assess the estimate.
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