So that you might better understand the
procedure for calculating water potential, here
is a practice problem.
Once you know the solute concentration, you
can calculate solute potential using the
Solute potential ()
of particles the molecule will make in
water; for NaCl this would be 2; for
sucrose or glucose, this number is 1
concentration (from your experimental
constant = 0.0831 liter bar/mole K
in degrees Kelvin = 273 + °C of
solution. Assume room temperature is =
to 22 oC.
The molar concentration of a sugar solution
in an open beaker has been determined to be
0.3M. Calculate the solute potential at 27
degrees. Round your answer to the nearest
The pressure potential of a solution open to the
air is zero. Since you know the solute potential
of the solution, you can now calculate the water
Notes to help answer questions.
1. Draw a straight line (or use the
regression button on your logger pro) and
determine where the line crosses the x axis.
The point at which the line crosses the x axis
represents the molar concentration of sucrose
with a water potential that is equal to the
potato tissue water potential. At this
concentration there is no net gain or loss of
water from the potato cells.
2. By graphing the data of the percent change
in mass of the potato cores, it can be
determined that the water potential of potato
cells is equal to the water potential of an
unknown sugar solution molarity. Determine that
3. Using the explanation above the water
potential of the potato cells can be determined.
4. If the potato is allowed to dehydrate by
sitting out in the open air the water potential
would decrease (be more negative) because the
concentration of solutes within the cells would
increase as potato cells dehydrate. Therefore,
the osmotic pressure and water potential both