Laboratory 1

Diffusion and Osmosis

Paul Jones

Exercise 1A: Diffusion

Table 1.1

Presence of Glucose in Water Through Dialysis Bag

Initial Contents Solution ColorPresence of

Glucose

Initial Final Initial Final

Bag15% Glucose & 1%

StarchClear Indigo blue

Yes(1000+++)

Yes(500++)

Beaker H2O + IKIGolden orange

Golden orangeNo

(100)Yes

(175)

Analysis of Results

1. Glucose is leaving the bag, while Lugol’s solution is entering the bag. The indigo blue color of the bag proves Lugol’s is entering the bag. We proved that glucose diffused by testing the beaker, and found there to be glucose.

2. The results show that IKI moved from the beaker to the bag, and this is how the color of the dialysis bag changed. This happened because it wants to reach equilibrium. Conversely, glucose moved out of the bag. Thus, everything except starch (for it is too large) diffused to reach equilibrium.

3. In order to find how much water diffused into the dialysis bag, one would need to calculate the percentages of each substance, and then recalculate the percentages afterwards. Then, one would need to compare the percentages with one another.

4. Water, glucose, IKI, membrane pores, and then starch. I know this because starch did not leave the dialysis bag because the outside of the bag did not turn blue. Thus, membrane pores are smaller than starch. Then, I compared the atomic mass of the molecules in question.

5. If the experiment started with the glucose and IKI solution inside the bag and only starch and water outside, glucose and IKI would want to move out of the bag to reach equilibrium, and thus changing the color of the contents of the beaker. The starch, however, would still not move through because it is too large.

Exercise 1B: Osmosis

Table 1.2: Dialysis Bag Results: Individual Date

Percent Change in Mass of Dialysis Bags –Group Data

Contents in Dialysis Bag

Initial Mass Final MassMass

Difference

Percent Change in

Mass

0.2 M Sucrose 25.3 grams 26.3 grams 1.2 grams 4.7%

0.4 M Sucrose 25.3 grams 26.4 grams 1.1 grams 4.3%

Table 1.3: Dialysis Bag Results: Class DataPercent Change in Mass of Dialysis Bags–Class Data

Dialysis Bag Group 1 Group 2 Group 3 Group 4 Class AverageDistilled Water 2.7% -1% 0%0.2 M Sucrose 6.9% 4.7% 5.8%0.4 M Sucrose 4.3% 6.27% 7.3%0.6 M Sucrose 14.17% 5.86% 10%0.8 M Sucrose 6.64% 11% 13%1.0 M Sucrose 17% 15.19% 16.4%

10a.) Molarity 10b.) Class average

Graph 1.1

Molarity Compared to Class Average

0%

2%

4%

6%

8%

10%

12%

14%

16%

18%

0 0.2 0.4 0.6 0.8 1 1.2

Molarity

Class Average

Analysis of Results1. As the molarity of sucrose increases in a solution, the percent change of

the mass increases as well, and thus, they as one increases, so does the other. This can be explained by the fact that sucrose weighs more than water.

2. In distilled water, the mass increases because sucrose has more mass. Similarly, a 0.2M bag would increase because of the net move into the bag. Conversely, in a 0.4M bag would remain the same because equilibrium will have been reached. The trend dictates that in a 0.6M through 1.0M the mass would decrease inside the bag because of diffusion.

3. We used percent change because each bag had a different start and end mass, and thus we could not accurately compare them. Additionally, the graph would be difficult to read.

4.

Initial Mass Final Mass Mass DifferencesPercent Change in

Mass

20.0 grams 18.0 grams -2.0 grams -10%

5. Hypotonic

Exercise 1C: Water Potential

Table 1.4: Potato Core: Individual Date

Potato Core Results–Group Data

Contents in Beaker

Initial Mass

Final Mass

Mass Difference

Percent Change in Mass

Class Average Percent

Change in Mass

0.2 M Sucrose 2.7 grams 2.7 grams 0.0 grams 0.0% 0%

0.4 M Sucrose 2.5 grams 2.0 grams 0.5 grams -20.0% -20.0%

Table 1.5: Potato Core Results: Class Data

Potato Core Results–Class Data

Contents in

Beaker

Group 1

Group 2

Group 3

Group 4

Group 5

Class Average

0.0 M Distilled Water

20.0% 22.0% 21.0%

0.2 M Sucros

e0.0% 0.0% 0.0%

0.4 M Sucros

e-20.0% -24.0% -22.0%

0.6 M Sucros

e-30.0% -27.0% -28.5%

0.8 M Sucros

e-28.0% -43.0% -38.5%

1.0 M Sucros

e-30.0% -20% -42.0%

Graph 1.2: Percent Change in Mass of Potato Cores at Different Molarites of Sucrose

Percent Change in Mass of Potato Cores at Different Molarites of Sucrose

-60.00%

-50.00%

-40.00%

-30.00%

-20.00%

-10.00%

0.00%

10.00%

20.00%

30.00%

0 0.2 0.4 0.6 0.8 1 1.2

Sucrose Molarity within Beaker

Percent Change in Mass of

Potato Cores

Molar concentration of sucrose = .43M

Exercise 1D: Calculation of Water Potential from Experimental Data2. (-1)(0.43 mole / liter)(0.0831 liter bar / mole °K)(295°K) = -10.541235 bars

Questions1. The water potential of the potato core would decrease because all of the

water was allowed to evaporate from the potato core. 2. The cell is hypertonic because the solute concentration is higher than the

water concentration. Thus, the cell will gain water.3. The pressure potential is equal to 0.4. Dialysis bag.5. The concentration of sucrose is greater outside the bag, and water will

always move to reach equilibrium if it can, thus, water will diffuse out of the bag.

Graph 1.3

Percent Change in Mass of Zucchini Cores at Different Molarites of Sucrose

-40%

-30%

-20%

-10%

0%

10%

20%

30%

0 0.2 0.4 0.6 0.8 1 1.2

Sucrose Molarity within Beaker

Percent Change in Mass of

Zucchini

b. 0.35M8a. -1(0.35 mole / liter)(0.0831 liter bar / mole °K)(295°K) = -8.58018b. 0 + -8.5801 = -8.5801 bars10a. Distilled Water 10b. Distilled Water10c. The RBC will expand for water will flow from areas of high concentration

to areas of low concentration.

Exercise E: Onion Cell Plasmolysis1.

2.

The onion cells appear to have shriveled up.3. The cells of the onion seem to have expanded back to their original size.

Newly added freshwater made the NaCL diffuse, reach concentration gradient, and come to more acceptable levels for the onion (ie; closer to normal).

Analysis of Results1. Plasmolysis is the process in plant cells where the plasma membrane

pulls away from the cell wall due to the loss of water through osmosis.2. The salt solution shrank the cytoplasm.3. The ice combines with the salt to create a salty H2O solution. This solution

penetrates the cells of grass and cause it to plasmolisize.