Potato Osmosis Practical: Mass Change and Water Potential

Potato (Solanum tuberosum) tubers. Photo: Ruslan V. Albitsky (pauk), Wikimedia Commons, CC BY-SA 3.0. Unmodified. This is a tissue-source photograph, not a photograph of the experiment described below.

GCSE practical foundations • A-level / AP Biology data skills

A potato osmosis experiment becomes much more useful when students explain the controls, calculate percentage change and evaluate a graph. This guide takes you from a fair test to a cautious water-potential estimate, using a clearly labelled teaching dataset.

In this lesson

• The investigation and its biological explanation
• A controlled classroom method
• Preparing dilutions without a calculation error
• Percentage mass change and an illustrative dataset
• Zero crossing and the water-potential estimate
• Evaluate the experiment rather than just describe it
• 20 revision MCQs
• 20-question interactive quiz

The investigation and its biological explanation

Question: how does external sucrose concentration affect the percentage mass change of potato tissue over a fixed period?

Living potato cells contain selectively permeable membranes. Water can enter or leave according to the difference in total water potential. A dilute external solution usually causes net water entry and mass gain; a sufficiently concentrated solution can cause net water loss and mass decrease. The response also depends on the tissue and conditions, so it should be measured rather than assumed.

The independent variable is sucrose concentration. The dependent variable is percentage mass change. Control temperature, immersion time, solution volume, tissue source, core dimensions and blotting method. Percentage change helps compare samples with slightly different starting masses; it does not eliminate differences in surface-area-to-volume ratio.

A controlled classroom method

1. Prepare a labelled dilution series, for example 0.00, 0.10, 0.20, 0.30, 0.40 and 0.50 mol L⁻¹ sucrose. Use at least three independent potato pieces per concentration. Use enough equal solution volume to cover the pieces fully.
2. Cut equal-diameter cylinders with a cork borer under teacher supervision. Remove skin and trim to equal lengths. Use the same potato or a carefully defined batch; distribute pieces across treatments rather than assigning one potato to each concentration.
3. Gently blot each piece using the same procedure and record its individual initial mass. Use a balance of suitable resolution and retain the sample identity.
4. Immerse pieces for the same predetermined time at a controlled temperature. A classroom pilot might use 45–60 minutes, but establish a workable period beforehand; do not claim this automatically guarantees equilibrium.
5. Remove each piece, blot consistently without squeezing and immediately record its final mass. Avoid evaporation while waiting to weigh.
6. Calculate percentage change for each piece, then calculate the mean at each concentration. Retain the individual values to evaluate spread and possible anomalies.

Practical care: a teacher should supervise sharp borers and knives. Wear appropriate eye protection when preparing solutions, follow local laboratory rules, and do not eat experimental material. Sucrose is a disaccharide, but this is a laboratory experiment, not a food activity.

Preparing dilutions without a calculation error

For a stock dilution, use C₁V₁ = C₂V₂. To prepare 20.0 mL of 0.40 mol L⁻¹ sucrose from a 1.00 mol L⁻¹ stock, V₁ = (0.40 × 20.0)/1.00 = 8.0 mL stock. Make up to a final volume of 20.0 mL with water; do not add 20 mL water to the 8 mL stock.

Prepare larger quantities if needed for replicates. Record concentration units explicitly: mol L⁻¹ is equivalent to mol dm⁻³. A mass percentage is a different concentration expression and cannot be inserted directly into a molarity-based water-potential formula.

Percentage mass change and an illustrative dataset

Illustrative teaching data; plotted values are not reported experimental measurements.

Percentage mass change = [(final mass − initial mass) / initial mass] × 100. A piece changing from 2.00 g to 2.16 g gains 0.16 g: (0.16/2.00) × 100 = +8.0%. A change from 2.00 g to 1.86 g is −7.0%. Keep the sign; reporting all changes as positive hides the direction.

The following values are invented teaching data chosen to practise calculations. They are not measurements from a PreachBio laboratory or a prediction for every potato.

Sucrose / mol L−1	Piece 1 / %	Piece 2 / %	Piece 3 / %	Mean / %
0.00	+11	+12	+13	+12
0.10	+7	+8	+9	+8
0.20	+2	+3	+4	+3
0.30	−3	−2	−1	−2
0.40	−8	−7	−6	−7
0.50	−12	−11	−10	−11

Plot concentration on the x-axis and mean percentage change on the y-axis. Show a horizontal zero line. If error bars are used, name their meaning, such as range or standard deviation; do not label them “error” without specifying the calculation. A reasonable local interpolation is more defensible than forcing a high-order curve through a small dataset.

Zero crossing and the water-potential estimate

The mean crosses zero between 0.20 M (+3%) and 0.30 M (−2%). Linear interpolation over that interval gives:

C_zero = 0.20 + [3/(3 + 2)] × (0.30 − 0.20) = 0.26 mol L⁻¹.

This is the external concentration associated with approximately zero net tissue mass change under the chosen conditions. It is not automatically the exact concentration of the cell sap. Tissue cells have pressure potential; water and solute exchange, damaged cells and incomplete equilibration also complicate interpretation.

A-level / AP extension

For an approximately ideal sucrose solution open to the atmosphere, Ψ_p ≈ 0 and Ψ ≈ Ψ_s = −iCRT. With i = 1, C = 0.26 mol L⁻¹, R = 0.008314 L MPa mol⁻¹ K⁻¹ and T = 298 K:

Ψ ≈ −0.26 × 0.008314 × 298 = −0.644 MPa. Under near-equilibrium conditions this estimates the tissue's effective water potential. It does not separately measure its solute and pressure components. Treat the value as an estimate and state the temperature and assumptions.

Evaluate the experiment rather than just describe it

Reliability: repeat with independent pieces and report spread. Validity: change concentration while controlling other influential variables. Measurement: use consistent blotting and a balance appropriate for the small mass changes. Extra retained surface solution produces a falsely high final mass.

Different potato tissues have biological variation. Very short immersion may not allow a meaningful comparison; very long immersion can introduce tissue damage or microbial changes. A mass increase is evidence of net change, not proof that only water moved. Investigate unusual results before excluding them and state any justified exclusion.

Improve the zero-crossing estimate by testing more concentrations around the sign change, such as 0.24, 0.26 and 0.28 M, with replicates. More distant concentrations alone do not make the crossing more precise. A final conclusion should report the observed trend, the estimated zero point and the practical limitations.

Frequently asked questions

Why use percentage mass change rather than mass change alone?

Dividing by the initial mass provides a relative change for each sample. It improves comparisons between slightly different starting masses, while dimensions and tissue source still need control.

Why blot before and after weighing?

Surface solution adds mass unrelated to uptake by the tissue. A standard gentle blotting procedure reduces this bias; excessive pressure may remove tissue water or damage cells.

Does zero mass change mean no water molecules move?

No. Water can exchange in both directions. Approximately zero change means there is little net change over the measured interval.

Why is the zero-crossing concentration not necessarily cell sap concentration?

Equal total water potentials can coexist with different solute concentrations because the cell has pressure potential. Whole-tissue measurements also have experimental and biological limitations.

Should the graph be forced through the origin?

No. Potato tissue may gain mass in pure water, so its percentage change at zero sucrose need not be zero. Use the actual data.

How should a result outside the repeated values be handled?

Check sample identity, blotting, balance readings and preparation errors. Repeat where possible. Do not discard a point merely because it disagrees with the expected trend.

20 revision MCQs with explained answers

Choose one answer before opening the explanation. The interactive quiz below uses a different set of questions.

1. What is the independent variable in this investigation?
   1. Sucrose concentration
   2. Final potato mass
   3. Percentage mass change
   4. Balance reading
   Answer and explanation

   A. Sucrose concentration — Concentration is deliberately changed between treatments.

2. What is the dependent variable used for comparison?
   1. Sucrose stock concentration
   2. Percentage mass change
   3. Cork-borer diameter
   4. Room label
   Answer and explanation

   B. Percentage mass change — The tissue response is expressed as a relative mass change.

3. Which equation calculates percentage mass change?
   1. 100 × final / initial
   2. Final minus initial without division
   3. 100 × (final − initial) / initial
   4. 100 × initial / final
   Answer and explanation

   C. 100 × (final − initial) / initial — The initial mass is the baseline for a percentage change.

4. A sample changes from 2.00 g to 2.16 g. What is the percentage change?
   1. −8%
   2. +16%
   3. +0.08%
   4. +8%
   Answer and explanation

   D. +8% — The gain is 0.16 g; 0.16/2.00 × 100 = 8%.

5. A sample changes from 2.00 g to 1.86 g. What is the percentage change?
   1. −7%
   2. +7%
   3. −14%
   4. −0.07%
   Answer and explanation

   A. −7% — The change is −0.14 g divided by the initial 2.00 g.

6. Why control core diameter and length?
   1. They make temperature irrelevant
   2. They affect surface-area-to-volume ratio and exchange
   3. They set sucrose molecular formula
   4. They eliminate all biological variation
   Answer and explanation

   B. They affect surface-area-to-volume ratio and exchange — Tissue geometry can alter the rate and extent of measured exchange.

7. Why blot the pieces consistently?
   1. To remove all cell water
   2. To increase sucrose concentration
   3. To reduce surface-solution mass differences
   4. To kill every cell
   Answer and explanation

   C. To reduce surface-solution mass differences — External droplets can distort the final mass.

8. Why use independent replicates?
   1. To guarantee the expected answer
   2. To replace all controls
   3. To change the independent variable
   4. To assess variability and improve confidence in the mean
   Answer and explanation

   D. To assess variability and improve confidence in the mean — Repeated independent samples reveal the spread of responses.

9. Which variable should remain controlled?
   1. Immersion time
   2. Treatment concentration
   3. Percentage change
   4. Observed direction of osmosis
   Answer and explanation

   A. Immersion time — Different durations would confound the concentration comparison.

10. What generally explains mass gain in dilute solution?
    1. Loss of all membranes
    2. Net water entry into tissue
    3. ATP directly pumping water
    4. Cell walls becoming sugar
    Answer and explanation

    B. Net water entry into tissue — Water may enter down its total water-potential gradient.

11. Which axis should show sucrose concentration?
    1. Both axes without units
    2. Neither axis
    3. x-axis
    4. y-axis only
    Answer and explanation

    C. x-axis — The independent variable belongs on the horizontal axis.

12. What does the graph's zero crossing estimate?
    1. Exact number of potato cells
    2. Maximum sucrose sweetness
    3. Dry mass of the potato
    4. External concentration giving approximately zero net mass change
    Answer and explanation

    D. External concentration giving approximately zero net mass change — It locates the treatment where gains and losses balance approximately.

13. What concentration is estimated between 0.20 M at +3% and 0.30 M at −2%?
    1. 0.26 M
    2. 0.20 M
    3. 0.50 M
    4. 0.03 M
    Answer and explanation

    A. 0.26 M — Three fifths of the interval lies between the positive point and zero.

14. What is the particle factor for dilute sucrose in Ψs = −iCRT?
    1. 298
    2. 1
    3. 2 always
    4. 6
    Answer and explanation

    B. 1 — Sucrose does not dissociate into multiple ions in this approximation.

15. What temperature must be used in the potential calculation?
    1. Fahrenheit
    2. The percentage change
    3. Kelvin
    4. Celsius without conversion
    Answer and explanation

    C. Kelvin — The gas-law expression requires absolute temperature.

16. What is the approximate water potential of 0.26 M sucrose at 298 K?
    1. +0.644 MPa
    2. −6.44 MPa
    3. Zero
    4. −0.644 MPa
    Answer and explanation

    D. −0.644 MPa — Use −1 × 0.26 × 0.008314 × 298.

17. Why is pressure potential of an open solution approximately zero here?
    1. Atmospheric pressure is the reference
    2. The solution has no molecules
    3. Sucrose destroys pressure
    4. All solutions have negative pressure
    Answer and explanation

    A. Atmospheric pressure is the reference — There is no added pressure relative to the atmospheric reference.

18. What does zero net mass change imply about molecular movement?
    1. The tissue must be dead
    2. Water exchange can still occur
    3. Every water molecule has stopped
    4. Only solute molecules move
    Answer and explanation

    B. Water exchange can still occur — A zero net change does not mean molecular immobility.

19. Which improvement best refines the zero crossing?
    1. Removing all negative results
    2. Forcing the graph through zero at 0 M
    3. More replicated concentrations close to the sign change
    4. Only one more point at 2.0 M
    Answer and explanation

    C. More replicated concentrations close to the sign change — Local data around the crossing improve the estimate.

20. How should an obvious unusual result first be approached?
    1. Delete it automatically
    2. Change it to the expected value
    3. Ignore all other replicates
    4. Investigate possible causes and repeat where possible
    Answer and explanation

    D. Investigate possible causes and repeat where possible — Transparent evaluation is more defensible than selective removal.

Interactive quiz: 20 new application questions

One correct option per question. Select your answers, check your score, then use the explanations to revise. Your answers stay in this browser; this quiz does not send them to a server.

1. A piece starts at 1.50 g and finishes at 1.65 g. What is its percentage change?

A. +10%B. +15%C. −10%D. +0.10%

2. A piece starts at 2.50 g and finishes at 2.25 g. What is its percentage change?

A. −0.10%B. −10%C. +10%D. −25%

3. Two pieces gain 0.10 g, but their starting masses are 1.0 g and 2.0 g. How do their percentage gains compare?

A. Both 5%B. 5% and 10%C. 10% and 5%D. Both 10%

4. How much 1.0 M stock is needed for 50 mL of 0.2 M solution?

A. 20 mLB. 50 mLC. 250 mLD. 10 mL

5. After measuring 10 mL stock for a 50 mL final dilution, what should be done?

A. Make up to 50 mL final volume with waterB. Add 50 mL of waterC. Remove all waterD. Add 100 mL stock

6. A student leaves droplets on the final sample only. What bias is likely?

A. Sucrose concentration is doubledB. Final mass is too highC. Final mass is always too lowD. Initial mass becomes zero

7. Each concentration uses tissue from a different potato. What is the main concern?

A. All potatoes have identical sapB. The balance cannot measure waterC. Potato differences are confounded with concentrationD. Replicates become unnecessary

8. Three percentage changes are +4%, +6% and +5%. What is the mean?

A. +15%B. +6%C. +4%D. +5%

9. At 0.10 M the mean is +4%; at 0.20 M it is −4%. What is the linear zero estimate?

A. 0.15 MB. 0.10 MC. 0.30 MD. 0.05 M

10. At 0.30 M the mean is +1%; at 0.40 M it is −3%. What is the linear zero estimate?

A. 0.10 MB. 0.325 MC. 0.375 MD. 0.35 M

11. Why should the zero estimate be reported cautiously?

A. All cells contain pure waterB. Mass change is unrelated to water exchangeC. The tissue may not be at equilibrium and pressure mattersD. Sucrose always has positive water potential

12. At fixed concentration, a higher absolute temperature does what in the ideal Ψs formula?

A. Makes Ψs exactly zeroB. Removes concentration from the formulaC. Necessarily makes Ψs positiveD. Increases the negative magnitude

13. Using R = 0.008314 and T = 300 K, estimate Ψs for 0.40 M sucrose.

A. About −0.998 MPaB. About +0.998 MPaC. About −0.10 MPaD. Exactly −4 MPa

14. Which reported unit matches a molarity-based concentration axis?

A. SecondsB. mol L−1C. g without a volumeD. MPa only

15. A graph at 0 M shows +12%. Should it be forced through (0,0)?

A. Yes, to remove mass gainB. Only if the result is positiveC. No, use the measured responseD. Yes, every graph starts at zero

16. Which conclusion is justified by a negative percentage mass change?

A. Every cell lost exactly the same massB. All tissue cells diedC. No molecules entered the tissueD. The tissue had a net mass loss over the interval

17. Why does percentage change not replace dimensional control?

A. Exchange depends on surface-area-to-volume ratioB. Percentages always use final massC. Geometry changes sucrose formulaD. All percentage changes are identical

18. Which information makes error bars interpretable?

A. The expected textbook answerB. The named spread measure and number of replicatesC. A colour name aloneD. An unlabeled line

19. If mass is still changing quickly at the end, what limitation affects the potential estimate?

A. All sugar has become starchB. The zero line is mathematically impossibleC. Near-equilibrium may not have been reachedD. The balance must be broken

20. What should accompany a final practical conclusion?

A. Only the preferred answerB. A claim of exact cell sap concentrationC. A claim that water stopped movingD. Trend, zero-point estimate, controls and limitations

Continue learning

• Water potential: definitions, signs and root uptake
• Carbohydrates: sucrose, starch and glycosidic bonds

Sources and further reading

These independently written notes and teaching examples can be checked against the following curriculum and scientific references.

• AQA A-level Biology: cells, osmosis and practical 3
• OpenStax: passive transport and osmosis
• Practical Biology: osmosis investigations
• College Board: AP Biology course overview