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AS · Practice questions · Osmosis and water potential

Higher to lower, never back.

Six original Cambridge-style questions on osmosis, including predicting water movement between cells and analysing a potato cylinder investigation.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Think in water potential.

01
Recall
[2 marks]

Define osmosis in terms of water potential.

  • The net movement of water molecules from a region of higher water potential to a region of lower water potential. ✓
  • Through a partially permeable membrane. ✓
02
Explain
[3 marks]

Three plant cells, X, Y and Z, are next to each other in a row. Their water potentials are X −300 kPa, Y −550 kPa and Z −700 kPa. Describe and explain the movement of water between them.

  • Water moves from X to Y, because −300 kPa is higher than −550 kPa. ✓
  • Water moves from Y to Z, and also from X to Z, because Z has the lowest (most negative) water potential. ✓
  • The movement is by osmosis, through the partially permeable cell surface membranes. ✓
03
Explain
[4 marks]

Red blood cells and onion epidermis cells are each placed in a concentrated sucrose solution. Describe and explain what happens to each type of cell.

  • The solution has a lower water potential than the cell contents, so water leaves the cells by osmosis. ✓
  • Red blood cells shrink and become crenated (wrinkled), because there is no cell wall. ✓
  • In the onion cells the vacuole and cytoplasm shrink and the protoplast pulls away from the cell wall: plasmolysis. ✓
  • The cell wall is fully permeable, so sucrose solution fills the space between the wall and the membrane. ✓
04
Calculation
[2 marks]

A potato cylinder had a mass of 2.40 g before soaking and 2.13 g after soaking in a sucrose solution. Calculate the percentage change in mass.

  • Change = 2.13 − 2.40 = −0.27 g. ✓
  • Percentage change = −0.27 ÷ 2.40 × 100 = −11.25%, so −11% (a loss). ✓
05
Data
[4 marks]

Potato cylinders were soaked in sucrose solutions. Percentage change in mass: 0.0 mol dm−3 +12%; 0.2 mol dm−3 +5%; 0.4 mol dm−3 −3%; 0.6 mol dm−3 −9%. A table gives the water potential of 0.30 mol dm−3 sucrose as −850 kPa and of 0.35 mol dm−3 as −990 kPa. Estimate the water potential of the potato tissue.

  • Between 0.2 and 0.4 mol dm−3 the change falls from +5% to −3%, a fall of 8% over 0.2 mol dm−3. ✓
  • Zero is reached 5 ÷ 8 of the way: 0.2 + 0.125 = 0.325 mol dm−3 (or read from the line of best fit). ✓
  • No change in mass means the solution and tissue have the same water potential. ✓
  • 0.325 mol dm−3 lies halfway between the table values, so the tissue water potential is about −920 kPa. ✓
06
Analysis
[3 marks]

Suggest three ways to improve the reliability or accuracy of the potato cylinder investigation.

  • Cut all cylinders from the same potato, with the same cork borer and to the same length, and remove the skin. ✓
  • Blot every cylinder in the same way before each weighing, so surface liquid does not add mass; keep the time and temperature constant. ✓
  • Use three or more cylinders per concentration and calculate a mean; use more concentrations close to where the line crosses zero. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.