Data:
Analysis:
Initially, there was starch and glucose present in the bag and water and IKI (iodine) present in the beaker. However, after the bag was immersed in the solution inside the beaker, glucose was present in the beaker solution. The starch remained in the bag because starch is a macromolecule and thus too big to diffuse through the semi-permeable membrane of the dialysis bag. The glucose, however, diffused into the outside solution because it can pass through the membrane. The dialysis bag turned black because iodine (IKI) diffused into the bag from the outside solution. An error in this lab may be the fact that solute concentrations of the solutions used were not accurate as given in the procedures.
Conclusion:
If a semi-permeable bag filled with glucose and starch solution is placed in a beaker of distilled water, then the glucose will move out of the bag, but starch will not because the molecules of starch are not small enough to pass through the membrane. Water will move into the bag to produce an isotonic solution.
Experiment 2: Osmosis
Hypothesis:
If semi-permeable bags of sucrose solution are placed in distilled water, then the mass of each bag will decrease because sucrose will diffuse out of the bag and water will diffuse into the bag until equilibrium is met, but if a semi-permeable bag of distilled water is placed in distilled water, then the mass will remain the same because the two solutions are isotonic to each other and there will be no net movement of water.
Materials:
- Six 30-cm strips of presoaked dialysis tubing
- Solutions of 0.2-M, 0.4-M, 0.6-M, 0.8-M, 1.0-M sucrose
- Distilled water
- Six 250-mL beakers
- Scale
- Stopwatch
Procedures:
- Take each strip of dialysis tubing and form six bags.
- Pour 25 ml of distilled water and each of the solutions into separate bags, keeping the air out.
- Fill each beaker with two-thirds of distilled water
- Record the mass of each bag.
- Immerse each bag in a beaker of distilled water and label the beakers to indicate the molarity of the solution in the dialysis bag.
- Wait 30 minutes.
- Record the mass of each bag.
- Determine the percent change in mass for each bag:
% change = final mass – initial mass x 100
initial mass
Data:
Analysis:
The masses of the bags with the distilled water, 0.2-M sucrose, and 0.4-M sucrose solutions decreased and each gave a negative percent change in mass. These are errors and were probably caused by inexact measurements of solutes and solutions. The mass of the distilled water should have stayed the same because there would have been no net movement of water because the two liquids are both distilled water. The mass of all the sucrose-containing bags should have increased because even though sucrose comes out of the bags, the amount of water that diffuses into the bag has more mass than the sucrose. The masses of the bags with 0.6, 0.8, and 1.0-M sucrose solutions increased. From looking at the data, the percent change in mass of the 0.8-M sucrose solution should have been between the percentages of the 0.6 and 1.0-M sucrose solutions, but it isn’t. This may be due tot he fact that after the 0.8-M sucrose solution bag was immersed in the distilled water, a good amount of the sucrose solution spilled from the bag. If this experiment was to be held again, I would make sure to fasten bags even more tightly.
Conclusion:
If semi-permeable bags of sucrose solution are placed in distilled water, then the mass of each bag will increase because sucrose will diffuse out of the bag but water will diffuse into the bag. However, if a semi-permeable bag of distilled water is placed in distilled water, then the mass will remain the same because the two solutions are isotonic to each other and there will be no net movement of water.
Experiment 3: Water Potential of Potato Cells
Hypothesis:
If potatoes are placed in solutions of sucrose, then they will decrease in mass because water will diffuse out of them to produce equilibrium, but if potatoes are placed in distilled water, then they will swell because solutes will diffuse out of the potatoes and water will diffuse into them to create an isotonic environment.
Materials:
- Solutions of 0.2-M, 0.4-M, 0.6-M, 0.8-M, 1.0-M sucrose
- Distilled water
- Six 250mL beakers
- 24 potato cylinders of 3cm in length (without skin)
- Scale
- Plastic wrap
- Thermometer
- Paper towels
Procedure:
- Pour 100 mL of each solution and distilled water into separate beakers and label them.
- Record the mass of four potato cylinders together.
- Place four cylinders into each beaker and cover them with plastic wrap.
- Let stand overnight.
- Next day, record the temperatures of the liquid in the beakers.
- Remove the cores from the beakers, blot them on paper towel, and record the masses.
- Calculate the percent change of the potatoes in each beaker:
% change = final mass – initial mass x 100
initial mass
Data:
Analysis:
The mass of the potato cylinders increased in the distilled water. This is due to the fact that potatoes have solutes (such as starch) and so water diffused into the potatoes. The mass of the 0.2-M sucrose also increased because the 0.2-M sucrose solution was hypotonic to the solutions inside the potatoes and thus water diffused into the potatoes. The masses of the rest of the potatoes all decreased. This means that these solutions were hypertonic to the solution of the potatoes and therefore water diffused out of the potatoes, decreasing their mass. An error to take into account is the excess water on the potato cylinders after they were taken out of the solutions. Or, the other way, when blotting the potato cells after taking them out of the solutions, too much solution may have been blotted.
According to the graph, the molar concentration of sucrose at which the mass of the potato cores does not change is 0.4. To find this out, I drew a line of best fit in the graph. 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. However, from my data, the potatoes decreased in mass in the 0.4-M solution. So, either my graph is wrong, or my data of the 0.4-M potato final mass is wrong.
Using the formula, I found the osmotic potential of the 0.4-M sucrose solution to be – 9.84 bars. Because osmotic potential is equal to the water potential when the pressure potential of the solution is zero, the water potential is also – 9.84 bars.
Conclusion:
If potatoes are placed in distilled water, then their mass will increase because the water is hypotonic to the potato solution. If potatoes are placed in sucrose solutions of 0.2 and 0.4-M, then their mass will increase because the solutions are also hypotonic to the potato solution. Howver, if potatoes are placed in sucrose solutions of 0.6, 0.8, and 1.0-M, then they will decrease in mass because water will diffuse out of the potatoes because they are immersed in a hypertonic environment.