Potato Osmosis Lab
Yesenia, Lisa, Alex
Design an experiment to:
-identify the concentrations of the sucrose solutions
-use the solutions to determine the water potential of the plant tissues.
-Use the following questions to guide your investigation
Predict how molar concentration of sucrose will influence the mass of potato cell. By adding enough solute, the water potential outside the cell should equal the water potential inside; therefore causing no movement of water.
If we place six equal sized potato slices in six different solutions containing different molarities of sucrose and letting sit over the weekend, we can then identify the sucrose solutions by observing whether or not the potato absorbed the solutions or instead released water; therefore affecting its final mass and allowing us to calculate the molarity, and its water potential.
- Peel the potato skin off
- Cut the potato into six relative sized slices
- Record the initial mass of each slice before submerging into the sucrose solutions
- Set up six sucrose solutions that all contain different molarities. (0M, .2M, .4M, .6M, .8M, 1M.) All sucrose solutions will be prepared by the teacher. Solutions should also be dyed with different colors to differentiate them.
- Submerge the six relative sized slices into the six different sucrose solutions.
- Let the slices sit over the weekend.
- Calculate the new mass of each potato slice after it has been sitting over the weekend.
- Determine the water potential of each slice. Did it lose or gain water?
- Determine each of the solutions molarity mass, depending on the result of the experiment.
- Clean up lab.
Post Lab Questions
1. Which solution had a water potential equal to that of the plant cells? How do you know?
With our understanding about varying concentrations of sucrose, we compared the initial mass with the final mass and found the difference, in grams and discovered a difference of 0 grams in the lavender purple solution. This solution produced the highest water potential, which was a result of its low sucrose concentration. The evidence shows that the osmosis rate was at equilibrium, a result of both water potentials being equal.
2. Was the water potential in the different plants the same?
No, because each potato sample was placed in a different sucrose solution that had different concentrations changing the way the potato would react to one another.
3. What would your results be if the potato were placed in a dry area for several days before your experiment?
The potato would be a hypotonic environment and would have a lower water potential since there was hardly any eater, which will result in the flow of water within the solution to flow into the potato, increasing its volume and mass substantially.
4. When potatoes are in the ground, do they swell with water when it rains? If not, how do you explain that, and if so, what would be the advantage or disadvantage?
Potatoes do not absorb water directly; by means of adhesion and cohesion, water is enabled to flow from the roots to the actual potato. Due to the absorption of the water into the roots, the potato does not soak up all the water, as a person might expect, rather, the roots transfer the necessary nutrients that the plant needs as fuel and the result is an increase in energy. Since the water isn’t directly put into the potato, the potato will not swell substantially. It might swell a little bit, but for the most part, the potato is unable to suction enough of the water and has no purpose for the water, without the assistance from the roots.
Water flows from an area of high water potential to an area of low water potential and we observed this in the Potato Osmosis Lab.In the experiment, we observed how a solution with a greater molarity affected its ability to allow water to pass through.
By subtracting the initial mass from the final mass of the potato samples (g), we were able to observe how the solution affected the osmosis rate and how varying amounts of sucrose can affect the flow of the solution through the potato. Although our solutions were anonymous, our newfound knowledge gave us enough insight to make predictions about which solution had which amount of sucrose.
Calculations for Water Potential
-(1)(.2M)(.0831)(22C)= -.36564 bars
-(1)(.4M)(.0831)(22C)= -.7318 bars
-(1)(.6M)(.0831)(22C)= -1.09692 bars
-(1)(.8M)(.0831)(22C) = -1.46256 bars
-(1)(1M)(.0831)(22C)= -1.8282 bars
-(1)(0M)(.0831)(22C)= 0 bars