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Osmosis, Diffusion, and Water Potential Lab
Transcript of Osmosis, Diffusion, and Water Potential Lab
By Herga Zhang
Relationship of the two experiments.
Both lab deals with water potential.
Osmosis: water molecules moving from area of high water potential (low solute) to area of low water potential (high solute).
The dialysis tube from the first experiment represents cells, such as the potato cells tested in the second experiment.
In the first experiment, the type of solution in the cell was changed, while the molarity remained the same.
In the second experiment, the molarity was changed while the type of solution used remained the same.
The net movement of water molecule across a semi-permeable barrier (Osmosis, 2000)
Molecules move from area with high water potential to area of low water potential until equilibrium is reached.
Isotonic Solution- Solutions in which the concentration of solute is the same.
Hypertonic Solution- Solution with the higher concentration of solute.
Hypotonic Solution- Solution with the lower concentration of solute.
A type of diffusion.
The net movement of molecules from an area of high concentration to an area of low concentration. ("Diffusion and Osmosis Tutorial", n.d)
Does not require energy unless the movement of molecules is from low concentration to high concentration
ATP is used as energy
Protein carriers, called pumps, are used
Free energy per mole of water
Helps predicts the direction of the movement of water.
Composed of two components:
the solute potential
i = Ionization Constant
C = Molar Concentration
R = Pressure Constant
T = Temperature in Kelvins
the pressure potential
the pressure potential of pure water in an open beaker is always 0
addition of solute decreases pressure potential
Diffusion and Osmosis Lab
Materials and Procedure
Two dialysis tubes were made with 5 mL of each of the following solutions.
1 M Sodium Chloride
1 M Sucrose
1 M Glucose
Each dialysis tubes were placed into separate beakers filled with 30 mL of tap water
Initial weigh was recorded
Each pair of dialysis tube was weighed every 5 minutes for 30 minutes
Percent mass change was calculated
(final - initial)/initial
1. What's leaving and what's entering?
Entering: Water from beaker into tubes.
Increase in mass for every dialysis tube.
Water potential inside dialysis tube is lower than water potential outside.
There are solutes in solution of the dialysis tubes.
Water moves from area of high water potential to area of low water potential.
Leaving: NaCl and Glucose molecules
Solute potential is higher in dialysis tubes.
Molecules diffuse from area of high concentration to low concentration.
No movement from Sucrose
Molecules is too big to move across dialysis bag pores.
2. Which pair tested did not have a change in weigh?
None of the dialysis tube pairs' weigh remained the same for the whole experiment.
The weigh of the pair of NaCl dialysis tubes remained the same from the 25 minutes mark to the end of he experiment.
The dialysis tubes reached equilibrium with the solution in the beaker.
The concentration of solute and solvent is the same for the solution inside the tubes and outside of the tubes.
3. Predict what would happen to the mass of the bags if all of the bags were placed in 0.4 M sucrose solution.
Distilled Water bag:
Decrease in mass
Water potential is lower outside.
Water molecules move out.
1 M NaCl, 1M Sucrose, 1 M Glucose bag:
Increase in mass
Water potential is higher outside.
Water molecules move in.
4. The fluid given in hospital is a saline solution isotonic to the human body tissue. Why?
Hypertonic or hypotnic solution would result in a movement of salt and water into or out of the cell.
Too much or too less salt in the human body is unhealthy.
Too much water will cause the tissue cells to burst. Too less water would cause the cell to shrivel and die.
5. What would happen to the red blood cell of a patient given distilled water, rather than saline water?
Red blood cell would swell and eventually burst.
Water potential is higher outside the cell
Water moves into cell.
Cell membrane can no longer withstand the pressure inside the cell.
Water Potential Lab
Materials and Procedure
Two pieces of potato cylinders were made
Initial weigh was measured for each pair of cylinders
Potato cylinders were placed in each 250 mL beaker filled with 100 mL of various concentration of Sucrose solution.
0.0 M Sucrose
0.2 M Sucrose
0.4 M Sucrose
0.8 M Sucrose
1.0 M Sucrose
Beakers were covered with Parafilm
After 24 hours, mass and angle of bend were measures.
1. If the potato core is allowed to sit in open air and dehydrate, would the water potential increase or decrease?
Water potential decreases.
After dehydration, there are less water molecules in the potato core than before.
Less water molecule means lower water potential.
After dehydration, there is a higher solute concentration, meaning a more negative solute potential.
Since the pressure potential is equal to 0, the solute potential is the water potential.
2. If a plant cell has lower water potential than its surrounding environment and the pressure is 0, is the cell hypertonic or hypotonic. Will is gain or loss water?
Cell is hypertonic in terms of solutes.
A lower water potential = more negative solute potential = more solute.
Water molecules move from areas of high water potential to low water potential.
3. Water potential of solution inside the bag is -6.25 bar, while the water potential outside is -3.25 bar. In which direction will water flow?
Water move from outside the bag to inside the bag.
Water potential outside the bag is higher.
Water move from high to low.
4. Zucchini cores place in sucrose solutions.
Percent Change of Mass in the Zucchini Cores Placed in Different Concentration of Sucrose.
The graph shows a down trend. As the concentration of sucrose increased, the percent change in mass decrease.
A. According to the graph, the sucrose molarity concentration equivalent to the molarity of the zucchini cell is 0.36 M.
Water Potential of Zucchini Cell
Water potential = pressure potential + solute potential.
Pressure potential = 0
Solute potential = -iCRT
-(1)(0.36)(0.0831 liter bar/mole K)(300 K) = -8.9748 bars
Water potential of zucchini cell =
5. Design an experiment to identify the concentration of an unknown sucrose solution.
Place two potato cores in the unknown sucrose solution.
Record the initial and final mass and the percent change in mass.
Compare the percent change in mass to the data obtained in this experiment.
If the concentration of the sucrose is the same as one of the sucrose solutions in this experiment, the percent change in mass should be similar, if not the same.
After obtaining the sucrose molarity, use the water potential equation to find the water potential of the plant cell considering the water potential of the plant cell will equal the water potential of the solution at equilibrium.