IB Biology HLTopic 1 — Life’s Building Blocks & VarietyPaper 1 & 2Practical skill~13 min read
Osmosis (Skills)
The potato practical looks simple: cut some cylinders, weigh them, put them in sugar solutions, weigh them again. What examiners actually reward is the thinking around it — why you controlled what you controlled, why you used percentage change instead of raw mass, and what the point where your line crosses zero really tells you.
📚 What you need to know
Preliminary research and trials come before the real experiment: they identify variables, show how to control them, and check you have enough materials.
Quantitative data supports firmer conclusions; qualitative observations can back them up.
Plant tissue in a hypotonic solution gains mass; in a hypertonic solution it loses mass; in an isotonic solution the mass is unchanged.
Bathing tissue in a range of solutions lets you estimate the osmotic concentration of the tissue itself.
Always calculate percentage change in mass, not raw change, so cylinders of different starting mass can be compared.
The point where the line of best fit crosses the x-axis is the solute concentration inside the tissue.
Standard deviation measures spread around the mean; standard error shows how close the sample mean is likely to be to the true mean.
Overlapping error bars suggest no significant difference; non-overlapping bars suggest a significant one.
Planning before you touch a potato
Planning is what makes a conclusion valid. Before designing the experiment, preliminary research should settle four things: what results you will collect, how you will measure them precisely and accurately, how many repeats you need, and which variables you will test and control.
After that come preliminary studies — small trial runs. These are worth doing because they reveal extra variables you had not thought of, show you the best way to control them, and tell you what quantities of solution you need so you do not run out halfway through. An experiment run without either step is likely to be invalid, simply because the variables affecting the results were never identified.
If a question asks you to “suggest an improvement”, reach for these first: more repeats, a narrower range of concentrations around the crossing point, and controlling temperature in a water bath. They are almost always creditworthy.
The potato practical
The idea is straightforward. A hypotonic solution has a lower osmotic concentration than the tissue, so the tissue gains mass. A hypertonic solution has a higher one, so the tissue loses mass. Somewhere between the two sits an isotonic solution where the mass does not change — and that concentration matches the inside of the cells.
Apparatus
Two potatoes of the same variety, a cork borer (about 5 mm), a white tile and a scalpel
A ruler or vernier calipers, and a balance reading to two decimal places
Sucrose solutions at 0, 0.20, 0.40, 0.60, 0.80 and 1.00 mol dm-3, 10 cm3 of each
Boiling tubes in a rack, a 10 cm3 measuring cylinder and paper towels
🧩 Method, in order
Cut the cylinders. Use the cork borer on a white tile to produce cylinders of the same diameter — at least five for each solution, so you have real repeats.
Trim them to the same length with a scalpel and ruler, then blot them dry to remove surface moisture.
Weigh each one and record the initial mass. Record the initial length too if you are measuring length change as well.
Measure 10 cm3 of each solution into a labelled boiling tube, including one tube of distilled water as your 0 mol dm-3 control.
Add one cylinder per tube and leave for a fixed time, for example 30 minutes, ideally in a water bath at a set temperature.
Remove, blot dry and reweigh. Blotting matters — surface liquid would inflate the final mass and ruin the result.
Control variables that actually earn marks: same potato variety, same cylinder diameter and length, same volume of solution, same time in solution, same temperature, and the same blotting technique before every weighing.
Analysing the results
Raw change in mass is useless on its own, because a heavier cylinder will gain more grams than a light one even if the osmosis is identical. Converting to a percentage removes that problem and lets every cylinder be compared fairly.
The one calculation you must know
percentage change in mass = (final mass − initial mass) ÷ initial mass × 100
Sucrose / mol dm-3
Initial mass / g
Final mass / g
Change / g
% change
0.00
4.85
5.45
+0.60
+12.4
0.20
4.92
5.27
+0.35
+7.1
0.40
5.01
5.10
+0.09
+1.8
0.60
4.95
4.77
-0.18
-3.6
0.80
5.04
4.64
-0.40
-7.9
1.00
4.90
4.35
-0.55
-11.2
Plot percentage change in mass against sucrose concentration, draw a line of best fit, and read off where it crosses the x-axis. At that concentration there is no net movement of water, which means the solution is isotonic with the potato cells — so that value is the solute concentration inside the tissue.
Read the crossing point off the line of best fit, not off the nearest data point. None of your tubes will land exactly on zero.
Reading the graph in words
A positive percentage change means the potato gained water, so the solution had a lower osmotic concentration than the potato. Those cylinders feel hard, because the cells have become turgid and turgor pressure is pushing on the walls.
A negative percentage change means the opposite: the solution was more concentrated than the potato, so water left the cells. The cylinder in the strongest sucrose solution loses the most mass, because that is where the concentration gradient between the cells and the solution is steepest. Those cylinders feel floppy, the cells are flaccid, and under a microscope some may be plasmolysed.
A cylinder that neither gains nor loses mass was in an isotonic solution: no concentration gradient, so no net movement of water.
Standard deviation and standard error
Repeats let you calculate a mean mass for each concentration, but a mean on its own hides how consistent your data were. Standard deviation fixes that by measuring the spread of the values around the mean — a small standard deviation means tightly clustered results, which is exactly what you want when comparing data sets.
Standard error answers a slightly different question: how close is your sample mean likely to be to the true population mean? It is the standard deviation divided by the square root of the sample size, so a larger sample gives a smaller standard error and a more trustworthy mean.
On a graph, standard error is shown as error bars above and below each plotted mean. If the bars for two means overlap, the difference between those means is probably not significant. If they do not overlap, the difference probably is.
The shaded band on the left is the range the two error bars share. Any shared range at all means you cannot claim a significant difference.
You are not asked to memorise the formulae for standard deviation or standard error. You are asked to use them — to say what a large spread means, or what overlapping error bars allow you to conclude.
Worked examples
WORKED EXAMPLE
A potato cylinder had an initial mass of 6.20 g. After 30 minutes in sucrose solution its mass was 5.83 g. Calculate the percentage change in mass and state what it shows. [3]
Step 1 — find the changechange = 5.83 – 6.20 = -0.37 gStep 2 — divide by the initial mass and multiply by 100(-0.37 ÷ 6.20) × 100 = -5.9677…-5.97% (3 s.f.)Step 3 — say what it means
The cylinder lost water by osmosis, so the solution was hypertonic to the potato tissue.
Keep the minus sign. A negative answer is the whole point of the question.
WORKED EXAMPLE
Explain why percentage change in mass is used rather than the change in mass in grams. [2]
Point 1 — the problem with raw mass
The cylinders do not all start at exactly the same mass, so a larger cylinder would gain or lose more grams even if the osmosis were identical.
Point 2 — the fix
Expressing the change as a percentage of the initial mass makes the results comparable between cylinders and between concentrations.
2 marksThe same argument explains why percentage change in length is used for length data.
WORKED EXAMPLE
A student’s line of best fit crosses the x-axis at 0.49 mol dm-3. State what this value represents and explain how they could improve the estimate. [3]
Step 1 — what the value is
At 0.49 mol dm-3 there is no net movement of water, so the solution is isotonic with the potato cells. This is therefore an estimate of the solute concentration inside the tissue.
Step 2 — improvement 1
Test more concentrations clustered between 0.40 and 0.60 mol dm-3 so the crossing point is defined by more nearby data.
Step 3 — improvement 2
Increase the number of repeats at each concentration to reduce the standard error of each mean.
3 marks“Do it more accurately” scores nothing. Say exactly what you would change.
💡 Exam tip
Show your working line by line. Method marks survive an arithmetic slip; a bare wrong answer does not.
Give percentage changes with a sign and to a sensible number of significant figures.
Label graph axes with the quantity and the unit, and remember the y-axis needs negative values below zero.
When asked for a conclusion, name the crossing point and what it represents biologically.
Blotting dry is a genuine mark: it removes surface liquid that would otherwise be weighed as gained water.
Link “reliable” to repeats and “valid” to controlled variables. Examiners treat those words precisely.
⚠ Common mix-up
Dividing by the final mass. Percentage change always uses the initial mass as the denominator.
Reading the crossing point off a data point. Use the line of best fit.
Confusing standard deviation with standard error. One describes the spread of data, the other the reliability of the mean.
Assuming overlapping error bars prove the means are equal. They only show the difference is not significant.
Calling extra repeats a control variable. Repeats improve reliability; control variables keep the test fair.
Forgetting the distilled water tube. It is the 0 mol dm-3 control and anchors the top of the graph.
Up next: Water Potential — the same osmosis you have just measured, rewritten in the language and units the IB uses for HL.
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