Osmosis Virtual Lab: Potato Strips in Sugar Solutions

This osmosis virtual lab follows the AQA GCSE required practical on osmosis in plant tissue and the potato part of the AP Biology diffusion and osmosis investigation. Six potato strips soak in sugar solutions from 0.0 to 1.0 mol/dm³. Students record the starting and final masses, calculate the percentage change in mass, plot it against concentration, and read off the concentration where the mass does not change. They come away with a full results table, a graph with a clear zero crossing, and an estimate of the concentration of the potato cell sap. Every number below was read from the simulation.

Curriculum links

• AQA GCSE Biology: specification 4.1.3.2, osmosis, and its required practical on the effect of a range of concentrations of salt or sugar solutions on the mass of plant tissue.
• AP Biology: Unit 2 (Cell Structure and Function), topic 2.8, tonicity and osmoregulation (ENE-2), and the AP diffusion and osmosis investigation. The extension below adds the AP water potential calculation.
• Simulic is not affiliated with or endorsed by the College Board or AQA.

Before the lab (5 min)

Ask: "Potato strips soak in sugar solutions of 0.0, 0.2, 0.4, 0.6, 0.8 and 1.0 mol/dm³. Which strips will gain mass?" Students choose all six, only 0.0 and 0.2, only 0.8 and 1.0, or none, and give a reason. On a class link this is question 1; the simulation unlocks after they answer.

Method in the simulation

1. Choose Potato strip experiment in the mode menu under the simulation.
2. Check the sliders: Temperature 25 °C, Cell sap 0.30 mol/dm³, Strip width 5 mm. These are the defaults.
3. Press Reset. The strips soak at 5 minutes per second until "Soaked for 180 min", about 36 seconds. The table fills in as they soak.
4. Copy the starting mass and final mass ("Mass now") for each beaker into your table.
5. Calculate % change in mass = (final mass − starting mass) ÷ starting mass × 100. Check your answer against the Change column.
6. Tap each beaker and record the state of its cells: turgid, flaccid or plasmolysed.
7. Plot % change in mass against sugar concentration and read where the line crosses zero.
8. Press Reset for a second run. The masses change a little each run, because the simulation adds small weighing errors. Take the mean of the two runs.

Sugar (mol/dm³)	0.0	0.2	0.4	0.6	0.8	1.0
Starting mass (g)
Final mass (g)
Change in mass (%)
Cells

Expected results

Each strip starts at about 1.3–1.4 g. One test run gave:

Sugar (mol/dm³)	0.0	0.2	0.4	0.6	0.8	1.0
Start → final (g)	1.33 → 1.51	1.38 → 1.47	1.33 → 1.28	1.33 → 1.21	1.35 → 1.16	1.34 → 1.11
Change (%)	+13.5	+6.5	−3.8	−9.0	−14.1	−17.2
Cells	turgid	turgid	flaccid	plasmolysed	plasmolysed	plasmolysed

The mean values without weighing errors are +13.1, +5.8, −4.0, −9.9, −14.1 and −17.1 %. The curve is not a straight line: it flattens at both ends. The cell wall stops turgid cells from swelling much further, and plasmolysed cells have less and less water left to lose.

The line crosses zero at about 0.32 mol/dm³: 0.31 to 0.33 in our runs, and almost always between 0.30 and 0.34. At that concentration there is no net osmosis, so it estimates the concentration of the potato's cell sap. The Cell sap slider is set to 0.30, so students can check how close their estimate is.

Questions for students

1. Prediction: which strips will gain mass?
2. Which variables must be the same for all six strips?
3. Default settings, 180 min. Calculate the % change in mass of the strip in 1.0 mol/dm³.
4. At what sugar concentration does your graph cross zero?
5. Why do we compare percentage change in mass instead of change in mass?

Answers for teachers:

1. Only the strips in 0.0 and 0.2 mol/dm³.
2. Strip width, soaking time and temperature.
3. About −17 % (accepted from −20 % to −14.5 %).
4. About 0.32 mol/dm³ (accepted from 0.27 to 0.37).
5. The strips start with different masses, so percentages compare them fairly.

Common misconceptions

• "Sugar moves into the potato." The membrane holds sugar back. Only water moves, by osmosis.
• "Water moves to where there is more water." Water moves from a dilute solution (higher water potential) to a more concentrated one.
• "Plant cells burst in pure water." The cell wall stops them. The cells become turgid, and the gain levels off.

Extension

• Water potential (AP). Use the zero crossing as the cell sap concentration and calculate the solute potential ψs = −iCRT, with i = 1, R = 0.0831 L·bar/(mol·K) and T = 298 K. For 0.32 mol/dm³: ψs = −(1)(0.32)(0.0831)(298) ≈ −7.9 bar. At this point the water potential of the cells equals that of the solution.
• Rate of osmosis. Set Strip width to 10 mm and repeat. The time constant rises from τ ≈ 20 min to τ ≈ 80 min, and after 180 min the strip in 1.0 mol/dm³ has changed only about −15 %. Ask why thicker strips take longer to reach a steady mass.

FAQ

Why does every run give slightly different masses?

The simulation adds small weighing errors and gives each strip a slightly different starting mass, as in a real lab. That is why the question set accepts a range and why two runs and a mean are worth doing.

Can students find an unknown cell sap concentration?

Partly. You can set a different cell sap value in the starting values of your class link, for example 0.26 mol/dm³. But the Cell sap slider still shows the value in the simulation, so treat it as a check, not a secret.

Is there a membrane model too?

Yes. The same simulation has Diffusion across a membrane and Osmosis of water modes, which show the particles. They work well as a short introduction before the potato strips.

Related

Diffusion across membranes and osmosis in cells Photosynthesis rate – light, CO₂, temperature and the compensation point

For a full lesson structure around a measured relationship like this one, see how to create a virtual lab activity. To use the potato prediction as a class discussion, see predict–observe–explain with simulations.