Free Osmosis questions for IGCSE Biology (0610), part of Movement into & out of cells. Each question has the answer and the marking points an examiner would look for.
Generate a full Osmosis worksheet →All Movement into & out of cells questionsPotato cylinders are placed in sucrose solutions for one hour. The percentage changes in mass are shown.
0.0 mol/dm³: +7.5% 0.2 mol/dm³: +1.5% 0.4 mol/dm³: −4.5% 0.6 mol/dm³: −10.5% 0.8 mol/dm³: −16.5%
What is the best estimate of the concentration of the cell sap, in mol/dm³?
Answer: C
A potato cylinder gains mass in a sugar solution.
What does this show about the solution?
Answer: A
A student cuts six potato cylinders, weighs them, and places each in a different concentration of sucrose solution for 30 minutes. She then dries and reweighs each cylinder. Table 3.1 shows the results.
| concentration / mol/dm³ | initial mass / g | final mass / g | percentage change in mass |
|---|---|---|---|
| 0.0 | 2.42 | 2.74 | |
| 0.2 | 2.19 | 2.33 | |
| 0.4 | 2.47 | 2.36 | −4.5 |
| 0.6 | 2.05 | 1.83 | −10.7 |
| 0.8 | 2.49 | 2.14 | −14.1 |
| 1.0 | 2.03 | 1.72 | −15.3 |
Calculate the percentage change in mass for 0.0 and 0.2 mol/dm³. Complete Table 3.1.
Plot a graph of percentage change in mass (y-axis) against concentration (x-axis). Draw a smooth curve.
Use your graph to estimate the concentration of sucrose that has the same water potential as the potato cells.
Explain why the cylinder in 0.0 mol/dm³ gained mass.
State two variables that should be kept constant.
Explain why percentage change in mass is used rather than change in mass.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 13.2% and 6.4% (final − initial) ÷ initial × 100 M1 +13.2 and +6.4 A1 | 2 |
| (b) | See graph Axes labelled with quantity and unit, percentage change in mass on the y-axis and concentration / mol/dm³ on the x-axis; suitable scales so the points occupy at least half the grid B1 All points plotted correctly to within half a small square (B1 for at least four correct) B2 Single, thin, well-judged best-fit line (or smooth curve) B1 | 4 |
| (c) | 0.32 (accept ±0.03) Value where the curve crosses 0% B1 | 1 |
| (d) | Water entered by osmosis, from higher water potential outside to lower water potential inside, through the partially permeable membrane osmosis B1 water moves from high water potential (dilute) to low water potential B1 through partially permeable (cell) membrane B1 | 3 |
| (e) | Size/surface area of cylinders; time; temperature; volume of solution Any two: size/surface area, time, temperature, volume of solution, potato type B2 | 2 |
| (f) | The cylinders had different starting masses, so percentage allows a fair comparison Different initial masses B1 | 1 |
A student cuts cylinders of potato, weighs them, places them in sucrose solutions of different concentration for 2 hours, then blots them dry and reweighs them. Table 4.1 shows the mean percentage change in mass.
| concentration of sucrose / mol per dm³ | 0.0 | 0.2 | 0.4 | 0.6 | 0.8 |
|---|---|---|---|---|---|
| mean percentage change in mass | +12.3 | +5.2 | -1.8 | -8.8 | -15.8 |
Define osmosis.
A cylinder had a mass of 2.50 g before it was placed in 0.6 mol per dm³ sucrose. Calculate its mass after 2 hours.
Use Table 4.1 to estimate the concentration of sucrose solution that would cause no change in mass. Explain how you decided.
Explain the change in mass in 0.8 mol per dm³ sucrose solution.
Explain why the student calculated percentage change in mass rather than change in mass.
Explain why the cylinders were blotted dry before reweighing.
Suggest why plant cells do not burst when placed in pure water, but red blood cells do.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | net movement of water through partially permeable membrane, from dilute to concentrated net movement of water molecules B1 from a region of higher water potential (dilute solution) to lower water potential (concentrated solution) through a partially permeable membrane B1 | 2 |
| (b) | 2.28 g 2.50 × (1 − 8.8/100) C1 2.28 g A1 | 2 |
| (c) | 0.35 mol per dm³ 0.35 mol per dm³ (allow ± 0.05) B1 the change in mass is zero between the positive and negative values / where the line would cross zero B1 | 2 |
| (d) | water leaves by osmosis; solution lower water potential; cells flaccid / plasmolysed solution has a lower water potential than the potato cells B1 water leaves the cells by osmosis B1 through the partially permeable membrane, so mass decreases (cells become flaccid / plasmolysed) B1 | 3 |
| (e) | cylinders had different starting masses the cylinders had different starting masses, so percentage change allows a fair comparison B1 | 1 |
| (f) | remove surface solution to remove solution on the surface, which would add to the mass B1 | 1 |
| (g) | cell wall resists expansion; turgid; RBC no wall plant cells have a cell wall that resists further expansion / cell becomes turgid B1 red blood cells have no cell wall B1 | 2 |
A student places potato cylinders, each of initial mass 2.50 g, in sucrose solutions of different concentrations for 1 hour, then reweighs them.
| sucrose concentration / mol/dm³ | 0.0 | 0.2 | 0.4 | 0.6 | 0.8 |
|---|---|---|---|---|---|
| final mass / g | 2.82 | 2.60 | 2.48 | 2.27 | 2.15 |
Define osmosis.
Calculate the percentage change in mass at 0.8 mol/dm³.
Explain why the potato in 0.0 mol/dm³ gained mass.
Explain why percentage change in mass is used rather than change in mass.
Estimate the concentration of sucrose that has the same water potential as the potato cells. Explain your answer.
Describe the appearance of a plant cell placed in a concentrated sugar solution, and explain why.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | net movement of water through partially permeable membrane from dilute to concentrated solution net movement of water molecules from a region of higher water potential (dilute solution) to lower water potential (concentrated solution) B1 through a partially permeable membrane B1 | 2 |
| (b) | −14.0% (2.15 − 2.50)/2.50 × 100 M1 −14.0% A1 | 2 |
| (c) | water potential higher outside; water enters by osmosis; cells become turgid water potential of the water is higher than in the potato cells B1 water enters the cells by osmosis through the partially permeable membrane B1 cells become turgid B1 | 3 |
| (d) | cylinders may differ in starting mass; allows comparison to allow comparison when the starting masses are different B1 | 1 |
| (e) | about 0.36 mol/dm³ about 0.36 mol/dm³ B1 where there is no change in mass (no net movement of water) B1 | 2 |
| (f) | plasmolysed: membrane pulls away from wall; water leaves vacuole by osmosis cell becomes plasmolysed / flaccid with the cytoplasm pulling away from the cell wall B1 water leaves by osmosis B1 | 2 |
Dialysis (Visking) tubing is a partially permeable membrane.
A tube filled with 20 % sucrose solution is placed in water. Predict what happens to the level of liquid in a capillary tube attached to it.
Explain your answer.
Explain why sucrose molecules do not leave the tubing.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | rises the level rises B1 | 1 |
| (b) | water enters by osmosis water moves into the tubing by osmosis B1 from higher water potential (pure water) to lower water potential (sucrose solution) B1 | 2 |
| (c) | too large to pass through membrane sucrose molecules are too large to pass through the pores B1 | 1 |
Riyaz Shaikh has mentored IGCSE, IB and A Level students for 15+ years, one-on-one, online and in Mumbai. The hardest questions on this site are the ones he uses to find a student's weak points.
Meet Riyaz →Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in