Core practical: osmosis in potatoes

In osmosis, water has net movement through a partially permeable membrane from a dilute solution to a more concentrated solution. Potato cell membranes provide this boundary. Water entering the tissue raises its mass; water leaving lowers it.

The investigation asks: how does sucrose concentration affect the percentage change in mass of potato tissue? A known initial mass is needed because osmosis changes the amount of water in the potato, which changes its mass.

Apparatus and arrangement

Use potato cylinders, a cork borer, ruler and cutting tile, a suitable cutting tool, labelled test tubes or small beakers, a rack, a range of sucrose solutions, a measuring cylinder or pipette, an electronic balance, a timer and paper towel. A useful range is 0.0 to $1.0\,\mathrm{mol}\,\mathrm{dm}^{-3}$ sucrose at regular $0.2\,\mathrm{mol}\,\mathrm{dm}^{-3}$ intervals; a preliminary trial can show whether the range surrounds the concentration that gives no mass change.

For each concentration, prepare at least three separately labelled tubes with the same volume of solution, such as $10.0\,\mathrm{cm}^{3}$. One potato cylinder goes into each tube and must be fully covered by solution. Separate tubes keep the repeats identifiable.

Variables

• Independent variable: sucrose concentration, in $\mathrm{mol}\,\mathrm{dm}^{-3}$.
• Dependent outcome: percentage change in potato mass, calculated from initial and final mass in grams.
• Controlled variables: potato cylinder diameter and length, source of potato tissue, volume of solution, immersion time, temperature and the blotting method.

The same cork borer fixes the diameter, and a ruler is used to cut equal lengths. Use tissue from the same potato where possible. Measure equal solution volumes, give every sample the same recorded immersion time, keep all tubes together at the same room temperature, and blot each cylinder in the same gentle way before every weighing.

Reproducible method

1. Label three tubes for every sucrose concentration and add $10.0\,\mathrm{cm}^{3}$ of the correct solution to each.
2. Prepare equal-diameter potato cylinders with the same length. Make enough for three repeats at every concentration.
3. Gently blot one cylinder, measure its initial mass on a zeroed balance and record the mass with its sample ID.
4. Place it in its assigned solution so that it is fully submerged. Insert the cylinders in a recorded order at regular intervals, then remove them in the same order so that each has the same immersion time.
5. Leave each sample for the fixed time, for example 45 minutes, at the same temperature.
6. At the end of each sample's timed interval, remove the cylinder, blot away surface solution using the same procedure, and immediately measure and record its final mass.
7. Calculate percentage change for each cylinder, then calculate the mean percentage change at each sucrose concentration. Check the repeats before excluding any anomalous result; do not remove a result merely because it is inconvenient.
8. Plot mean percentage change in mass on the y-axis against sucrose concentration on the x-axis. Use an appropriate best-fit line or curve and read the concentration where it crosses 0% change.

Record raw measurements before processing them:

Safety and measurement quality

The cutting tool and cork borer can cut or puncture skin. Potato cylinders should be prepared by a teacher or cut under supervision on a stable cutting tile, with hands kept out of the cutting path and the tool directed away from the body. Clean spilled solution promptly to prevent slips, keep liquids away from the balance, and do not eat potato or sucrose solution used in a laboratory.

Blotting matters because liquid clinging to a cylinder would be weighed as though it were part of the potato. A zeroed balance with 0.01 g resolution, consistent blotting and immediate weighing reduce variation that is unrelated to osmosis.

From mass measurements to evidence

A $0.30\,\mathrm{g}$ mass change is proportionally larger for a $2.00\,\mathrm{g}$ cylinder than for a $5.00\,\mathrm{g}$ cylinder. Percentage change expresses the change relative to starting mass, giving a fairer comparison.

Use the initial mass in the denominator and the same mass unit throughout. A positive result is a gain; a negative result is a loss.

Worked example: a loss

A cylinder changes from $3.20\,\mathrm{g}$ to $2.88\,\mathrm{g}$.

1. Find the signed change: $2.88-3.20=-0.32\,\mathrm{g}$.
2. Divide by initial mass and multiply by 100%:

1. Check the sign: final mass is smaller, so the negative answer is sensible. Water had net movement out of the potato.

Worked example: a gain

For $2.50\,\mathrm{g}$ changing to $2.70\,\mathrm{g}$, the change is $+0.20\,\mathrm{g}$:

The positive sign shows net water movement into the potato. Neither result means that sucrose moved by osmosis; osmosis is water movement.

Interpreting and improving the evidence

As external sucrose concentration increases, the mean percentage change usually becomes less positive and then more negative. In dilute external solution, water concentration is higher outside the potato cells, so net osmosis into the cells produces a mass gain. In concentrated external solution, water concentration is lower outside, so net osmosis out of the cells produces a mass loss.

The x-value where the best-fit line or curve crosses 0% change estimates the sucrose concentration of an external solution with the same overall osmotic effect as the potato cell contents. This is often called the potato tissue's equivalent sucrose concentration. At this point, water molecules still cross cell membranes in both directions, but there is no net movement and therefore no overall mass change.

A conclusion is stronger when repeats at each concentration are close together and the pattern is supported across the range. Deal with weaknesses by linking each one to a specific improvement:

Estimating a zero crossing between results

Suppose mean mass change is $+4\%$ at $0.2\,\mathrm{mol}\,\mathrm{dm}^{-3}$, and $-4\%$ at $0.4\,\mathrm{mol}\,\mathrm{dm}^{-3}$. If the line between them is straight, zero is halfway between these changes, at $0.3\,\mathrm{mol}\,\mathrm{dm}^{-3}$. If the gains and losses differ in size, zero is not the midpoint: use the fraction of the vertical change required to reach zero. A best-fit curve may give a slightly different estimate when the relationship is not linear.

Reading the straight-line model

To plot the worked results, put sucrose concentration on the horizontal axis and percentage mass change on the vertical axis. Use regular scales, label both quantities and units, and mark $(0.2,+4)$ and $(0.4,-4)$. The straight line between them slopes down and crosses zero at $0.3\,\mathrm{mol}\,\mathrm{dm}^{-3}$. Reading a vertical value at a chosen concentration turns the graph back into numerical information.

A straight-line relationship has the form $y=mx+c$: $m$ is its gradient and $c$ its vertical intercept. Here $x$ is the numerical concentration in $\mathrm{mol}\,\mathrm{dm}^{-3}$, and $y$ is the signed percentage mass change. For this illustrative line:

The gradient is −40 percentage points per $\mathrm{mol}\,\mathrm{dm}^{-3}$: a concentration rise of $0.1\,\mathrm{mol}\,\mathrm{dm}^{-3}$ gives a fall of four percentage points within this model. Using $(0.2,+4)$, $c=y-mx=4-(-40\times0.2)=12$, so $y=-40x+12$. The vertical intercept would be 12% at $x=0$, but that extends beyond these two measured concentrations; it is a model prediction, not an observation. Real osmosis results may curve, so use a straight-line equation only where a linear fit is appropriate.