Pre-Lab: Diffusion and Osmosis in Model Systems
In parts 1 and 2 of this lab, you will have the opportunity to investigate the processes of diffusion and
osmosis in model membrane systems. You will also investigate the effect of solute concentration on
water potential as it relates to living plant tissues.
Objectives: At the completion of this AP laboratory, you should be able to:
1. Describe the mechanisms of diffusion and osmosis.
2. Describe how solute size and molar concentration affect the process of diffusion through a
selectively-permeable membrane.
3. Describe the relationship between solutions that are hypotonic, hypertonic, or isotonic.
4. Design an experiment to demonstrate water potential.
5. Relate osmotic potential to solute concentration and water potential.
6. Describe how pressure potential could affect the water potential of a solution.
7. Describe the effects of water gain or loss in animal and plant cells.
8. Calculate the water potential of living plant cells from experimental data.
Background Information
Part I: Osmosis.
In this part of the lab, you will use dialysis tubing filled with different molarities of sucrose to
investigate the relationship between solute concentration and the movement of water through a
selectively permeable membrane (the process of osmosis).
When a dialysis bag containing a sucrose solution is placed in fresh water, the bag accumulates
water as a result of osmosis. Because there is a higher concentration of water outside the bag than inside
the bag, water diffuses into the bag. The water outside the bag is said to be hypotonic to the solution in
the bag; the solution inside the bag is hypertonic relative to the water outside the bag. Water always
moves through a selectively permeable membrane (sucrose cannot move) from hypotonic to hypertonic
solutions.
In theory, if you were to add solute to the water outside the bag, you could decrease the water
concentration out there. It should be possible to add just the right amount of solute to the water so that
the concentration of dissolved substances outside the bag is the same as the concentration of dissolved
substances inside the bag. Such solutions are said to be isotonic, and no net movement of water will
occur. If you were to continue to add solute to the water outside the bag, you would decrease the
concentration of water outside the bag until the solution outside became hypertonic to the solution
inside. Then, the net movement of water would be in the opposite direction: from inside to outside, and
the bag would shrink.
Part 2: Determining the Water Potential of Potato Cells.
In this part of the lab, you will use pieces of potato tissue placed in different molar
concentrations of sucrose in order to determine the water potential of potato cells. First, however, let’s
explore what we mean by the idea of water potential!
In animal cells, movement of water into and out of the cell is influenced by the relative
concentration of solute on either side of the cell membrane. If water moves out of the cell, the cell will
shrink, (crenulate), and if the water moves into the cell it will swell, and may even burst (cytolize). In
plant cells, the presence of a rigid wall prevents cells from bursting as water enters the cells, but pressure
eventually builds up inside the cell and affects the process of osmosis.
In predicting which direction water will move through living plant tissues, a quantity known as
water potential is used. Water potential, abbreviated by the Greek letter  SI refers to the potential
energy of water. It has many components, but the two on which we will focus are osmotic potential
which expends on solute concentration, and pressure potential, which results from the exertion of
pressure (positive or negative) on a solution. We express this as:

water potential
P
+
= pressure potential
+
=
S
osmotic potential
Water will always move from an area of higher water potential (higher potential energy) to an
area of lower water potential (lower potential energy). Stated another way, water potential measures the
tendency of water to leave one place in favor of another place.
The water potential of pure water in a beaker open to the atmosphere is “0” ( = 0), because
both the osmotic and pressure potentials are taken to be zero (S = 0; P = 0). An increase in positive
pressure raises the pressure potential, and therefore raises the water potential. The addition of solute to
water lowers the osmotic potential (makes negative), and therefore lowers the water potential.
Therefore, any solution at atmospheric pressure (P = 0) will always have a negative water potential.
For example, a 0.1 M sucrose solution at atmospheric pressure (P= 0) has an osmotic potential (S) of
–2.3 bars due to the solute, and thus a total water potential of –2.3 bars (0 + -2.3 = -2.3). A bar is a
metric unit of pressure, measured with a barometer, which is about the same as one atmosphere.
When a solution, such as that inside a potato cell, is separated from pure water by the selectively
permeable cell membrane, water will move by osmosis from the surrounding area where water potential
is higher ( = 0) into the cell where water potential is lower due to its dissolved solutes ( is negative).
In a case where the solute cannot leave the cell, the movement of water into the cell causes the cell to
swell, and the cell membrane pushes against the cell wall. This creates positive turgor pressure in the
cell.
Eventually, enough positive turgor pressure builds up to oppose the more negative osmotic
pressure of the cell. This will continue until the water potential of the cell equals the water potential of
the pure water outside the cell ( cell =  outside the cell = 0). At this point, a dynamic equilibrium is
reached and NET movement of water will cease.
If you add solute to the water outside the potato cells, you decrease the water potential of the
solution surrounding the cells. It should be possible to add just the right amount of solute to the water
so that the water potential outside the cell is the same as the water potential inside the cell. Therefore,
net water movement will cease. At this point, the water potential of the solution is equal to the water
potential of the turgid potato cells. (It is this information that you will use to calculate the water
potential of cells in the potato cores, which have been soaking in different molarities of sucrose.)
Eventually, if enough solute is added to the solution in the beaker, the water potential of the cell
will be greater (less negative) than the water potential of the solution in the beaker because of the turgor
that still exists within the cells. Water will diffuse out of the potato cells in response to a pressure
gradient, from the area of higher water potential (inside) to the area of lower water potential (outside). If
enough water was lost, the cell membrane would shrink and fall away from the cell wall, and the cell
would become plasmolyzed.
By weighing some potato cubes and immersing them in different sucrose solutions, then reweighing, you should be able to pinpoint the sucrose solution which corresponds to the water potential
of the potato tissue!
Lab: Diffusion and Osmosis in Model Systems
Refer to the Pre-Lab for lab OBJECTIVES and BACKGROUND INFORMATION. These model
systems for testing principles of diffusion and osmosis are probably new to you, so you are not being
asked to make a hypothesis for this lab.
Part I. Osmosis.
(a) Materials:
Pre-soaked dialysis tubing
Sucrose solutions, 10 mL each (available: 0 M, 0.2 M, 0.4 M, 0.6 M, 0.8 M, 1.0 M)
Plastic Cups
Distilled water
(b) Procedures:
1. Obtain dialysis tubing (keep in fresh distilled water untilled used)
2. Open one of the dialysis tubing by rubbing it between your fingers. Close one end of the tubing
using the provided clap (make sure it is a water tight seal).
3. Repeat with the other strips of dialysis tubing (one end should still be opened).
4. Pour 10 ml of test solution into the open end of the tubing.
5. Remove most of the air from the bag by drawing the dialysis bag between two fingers. To close the
bags, tie a knot at the other end, as before (step 2), thereby sealing the solution within the bag. BE
SURE that you have left 1.5 to 2 times as much empty space as that taken up by the volume of the
solution in the bag. (This leaves enough unfilled space within the bag to accommodate the possible
accumulation of water.)
6. Fill cups with distilled water.
7. Carefully rinse off and blot dry the outside of each dialysis bag. Measure and record the initial mass
of each of your two bags in Table 1. (Are you keeping track of which bag contains which solution?)
8. Place each bag in one of the beakers of distilled water and label the beakers to indicate the molarity
of the solution in the dialysis bags.
9. Let stand for 30 minutes.
10. Remove the bags from the water and carefully blot and re-weigh them. Record the final masses in
Table 1 and calculate % change in mass. Pool class data.
11. Graph the % change in mass as a function of dialysis bag sucrose solution molarity.
Part 2. Determining the Water Potential of Potato Cells
(a) Materials:
Fresh potato (Russett, Red, or Sweet)
Razor blade or scalpel
Metric ruler
Scale
Dixie Cups
Sucrose solutions (available: 0 M, 0.2 M, 0.4 M, 0.6 M, 0.8 M, 1.0 M)
Cling wrap
(b) Procedures:
1. Obtain 25 ml of each sucrose solution. Pour each solution into a separate, labeled Dixie cup.
2. Make four potato cubes! Cut each potato into a 1cm3 square (make sure there is no skin). Keep the
potato cubes covered until it is your turn to use the scale.
3. Mass 4 of the cubes together; record this in Data Table 2 as the initial mass. Place all the cubes
together into individual sucrose solutions (label!).
4. Cover the cups of sucrose with cling wrap to prevent evaporation, and let stand overnight.
5. Remove the 4 potato cubes from one sucrose solution; blot them gently on a paper towel, and
reweigh them (the four as a unit). Use the same balance as you did the day before, if at all possible.
Repeat with the other set of 4 cubes.
6. Record the final weights of both sets of potato cubes in Table 2, and calculate % change in mass.
Pool class data.
7. Graph the % change in mass of the potato cylinders as a function of sucrose solution molarity.
Graphs To Include in Your Report
1. Graph 1: Part 1 data.
2. Graph 2: Part 2 data.
Data Analysis, Water Potential of the Potato Cells
1. Determine the iso-osmolar concentration of the potato tissue and the sucrose solution by examining
Graph 2. Note the point on the curve corresponding to zero change in potato mass (y-axis); drop a
perpendicular line to find the corresponding point on the x-axis. This point will represent the
concentration of sucrose iso-osmolar to the potato issue. At this concentration there is no net gain or
loss of water from the tissue.
(From graph) Iso-osmolar concentration = ____________M.
2. The osmotic potential (S) of iso-osmolar sucrose solution can be calculated from the molarity of
the solution by applying this formula:
S = iCRT
where:
i = ionization constant (for sucrose, we can use a value of 1 since sucrose does not ionize in water).
C=
the iso-osmolar concentration (usually expressed in molarity; read off the graph)
R=
pressure constant (handbook value R = 0.0821 bar/mole deg K for water)
T=
temperature, degree Kelvin (273 + degrees C) of solution.
Sample problem: What is S of a 1.0 molar sucrose solution at 30 degrees C under standard pressure
conditions?
What is the osmotic potential of the sucrose solution found (by your graph) to be iso-osmolar to the
potato tissue? SHOW WORK.
Answer: ___________ bars.
3. Since the pressure potential (P) of our solutions was zero, we can calculate the water potential of
the solution from the equation for water potential given in the pre-lab. The water potential of the
solution at equilibrium will be equal to the water potential of your potato cells. What, then is the water
potential of your potato cells? SHOW WORK.
Answer: ___________ bars.
Questions
1. You have calculated the water potential of your potato cells. Earlier you learned that pure water at
standard pressure and temperature has a water potential of 0. If you threw your potato into a jar of
pure H20, would the potato shrink, or swell? Explain.
2. We said the water potential () of the potato cells is equal to the water potential of the solution, at
equilibrium. Is the osmotic potential (S) of your potato cells the same as that of the solution? Why
or why not?
3. If a potato cylinder is allowed to dehydrate by sitting in the open air, would the potato cells’ water
potential become higher or lower? Explain.
4. (See figures below.) A dialysis bag is attached to the end of a glass tube to make a simple
osmometer. If the pressure potential (P) is 0. The bag is 1.0 M sucrose solution, with  of –35
bars.
(a) What is the osmotic potential (S) of the sucrose solution in the bag?
(b) If the osmometer is placed in a beaker of pure water and osmosis occurs, what will be the
values of , P, and S at equilibrium for both the solution in the bag and the water in the
beaker?
Data
Part 1 Data Table
Contents of
Dialysis Bags
Initial
Mass
Final
Mass
Percent Change
in Mass
Class Ave.
% Change
Initial
Mass
Final
Mass
Percent Change
in Weight
Class Ave.
% Change
0.0 M Sucrose
0.2 M Sucrose
0.4 M Sucrose
0.6 M Sucrose
0.8 M Sucrose
1.0 M Sucrose
Part 2 Data Table
Sucrose Solutions
Soaking Potato Cubes
0.0 M Sucrose
0.2 M Sucrose
0.4 M Sucrose
0.6 M Sucrose
0.8 M Sucrose
1.0 M Sucrose
HINT: % Change = Final Mass – Initial Mass
Initial Mass