All questions and explanatory responses
1. Calculate the percentage mass change at 0.4 mol dm⁻³. Show the reference mass and interpret the 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.
- Use final minus initial: 1.94 − 2.00 = −0.06 g.
- Divide by the initial 2.00 g and multiply by 100.
- Obtain −3.0%; the tissue lost mass overall.
Common confusion: Dividing by the final mass changes the reference quantity and gives a different percentage.
2. Describe how mean mass change varies across the concentration range, using at least two numerical comparisons.
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.
- State the transition from mass gain at low concentrations to loss at higher concentrations.
- Compare 2.18 g at 0 with 1.68 g at 0.8 mol dm⁻³.
- Locate the change in sign between 0.2 and 0.4 mol dm⁻³.
Common confusion: A description reports the pattern and evidence. Adding only an osmosis mechanism does not describe these results.
3. Suggest and explain why the tissue loses mass in the 0.8 mol dm⁻³ solution.
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.
- Relate the result to a lower external water potential under the stated interpretation.
- Explain net movement of water across partially permeable cell membranes.
- Connect outward water movement to the observed net loss of mass.
Common confusion: Do not describe sucrose as being pumped out of the tissue or confuse osmosis with active transport.
4. Estimate the sucrose concentration giving zero mean mass change by linear interpolation between the adjacent points. State a limit of the estimate.
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.
- Use +0.06 g at 0.2 and −0.06 g at 0.4 mol dm⁻³.
- Locate zero halfway between them: 0.30 mol dm⁻³.
- State that local linearity and the measured means limit this estimate.
Common confusion: Zero net mass change does not establish identical solute composition inside and outside the cells.
5. A student claims that 0.30 mol dm⁻³ is the exact concentration of sucrose inside every cell. Evaluate this claim and suggest one useful improvement.
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.
- Explain that zero net change estimates a balance of water potentials under these conditions.
- Reject the claim of an exact, identical sucrose concentration in every cell.
- Suggest closer concentrations and independent repeats with spread or uncertainty reported.
Common confusion: A mean for cylinders cannot establish the internal composition of every cell.
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