Lab Report: Diffusion and Osmosis

Hypothesis and Prediction:
This lab was done to examine and comprehend how diffusion and osmosis works in diverse molarity of sucrose. Also how the solutions permeates through different mediums.
Methods:
Part A: Diffusion and Osmosis:
A 30 cm piece of 2.5 cm dialysis that has been soaked in water was obtained. The beginning of tubing was tied off, forming a bag with an open end that was rubbed between the fingers till separated. 15mL of the 15% glucose and 1% starch solution was placed into the bag and the ending of the dialysis bag was tied off, leaving some space for the development of the content within the dialysis bag. The color of the solution was recorded and was tested for the presence of glucose. Distilled water was poured into a 250 mL beaker (two-thirds of a cup) with about 4mL of Lugol’s solution (IKI). The color of the sucrose in the beaker was recorded and was tested for glucose. The dialysis bag was then submerged into the beaker of solution and left to stand for about 30 minutes (or until there was a color change in the dialysis bag or beaker).Once the bag was done soaking in the beaker, the final color of the solution in the bag and the beaker was recorded. The liquid in the bag and the beaker was then tested for the existence of glucose.

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Part B: Osmosis
Six strips of 30 cm presoaked dialysis tubing were obtained. For each strip, an end was tied and roughly 25 mL of different solutions (distilled water, 0.2 M sucrose, 0.4 M sucrose, 0.6 M sucrose, 0.8 M sucrose and 1.0 M sucrose) was poured into their individual bags. Most of the remaining air was then removed from each bag by bringing the bag between two fingers and tied off at the opened end. The outside of each bag was then bathed to wash away any sucrose that spilled when filling the bag. The exterior of each bag was then blotted and the initial mass of each bag was weighed and recorded. Distilled water was then filled into six 250 mL beakers. Each bag was then emerged into one of the six filled beaker and the beakers were labeled by which bag of solution was emerged in it. The bags stood in the beaker for half an hour. When the time was up, each bag was removed, blotted and the mass of each bag was recorded. The mass difference was calculated and then using the equation:
Percent change in mass = Final Mass – Initial Mass/Initial Mass x 100.
The individual and the class average of the percent change in mass were then graphed.
Part C: Water Potential
100 mL of the given solution was poured into six different labeled 250 mL beaker. The potato was then sliced into discs that were just about 3 cm thick. A cork borer (about 5 mm in diameter) was then used to cut four potato cylinders for each beaker, a total of 24 potato cores. Until the mass of cores were weighed by fours and recorded, the potato cores were kept in a covered beaker. Four cores were then put into each beaker of sucrose solution. Plastic wrap was then given to cover the beakers, preventing evaporation when left to stand overnight. The next day, the cores were then removed from the beakers and were blotted gently on a paper towel. Their total mass was then determined and recorded. The mass difference was calculated and then using the equation:
Percent change in mass = Final Mass – Initial Mass/Initial Mass x 100.
The individual and the class average of the percent change in mass were then graphed.
Part D: Calculation of Water Potential from Experimental Data
Determine the solute, pressure and water potential of the sucrose solution given and answer the questions about the possibility if zucchini cores were used with the sucrose solutions.
Results:
Part A: Diffusion and Osmosis
Table 1.1-Presence of Glucose in Water through a Dialysis Bag

 

Initial Contents

Solution Color

Presence of Glucose

 
 

Initial

Final

Initial

Final

BAG

15% GLUCOSE and 1% STARCH

Clear

Clear

Yes

Yes

BEAKER

H20 + IKI

Yellow
(an olive oil color)

Clear

No

Yes

Part B: Osmosis
Table 1.2: Individual Data of Change in Mass of Six Different Dialysis Bags

Contents in Dialysis Bag

Initial Mass

Final Mass

Mass Difference

Percent Change in Mass

a) distilled water

18.15 g

14.76 g

3.39 g

-18.68%

b) 0.2 M

19.40 g

17.33 g

2.07 g

-10.67%

c) 0.4 M

18.87 g

19.37 g

-0.5 g

2.65%

d) 0.6 M

19.83 g

19.68 g

-0.15 g

-0.5%

e) 0.8 M

21.91 g

20.05 g

-0.869 g

-8.2%

f) 1.0 M

18.78 g

18.07 g

-0.71 g

-3.7%

Table 1.3: Class Data of Percent Change in Mass of Dialysis Bags

 

Group 1

Group 2

Group 3

Group 4

Group 5

Group 6

Group 7

Group 8

Total

Class Average

Distilled Water

-18.68%

 

-2.2%

 

-7.0%

 

-7.2%

 

-35.1

-8.8%

0.2 M

-10.67%

 

-22.3%

 

-5.2%

 

1.8%

 

-36.4%

-9.1%

0.4 M

2.65%

 

6.2%

 

2.5%

 

3.9%

 

15.3%

3.8%

0.6 M

 

-0.76%

 

-3.8%

 

-4.0%

 

-6.55%

-15.2%

-3.8%

0.8 M

 

-4.1%

 

-26.3%

 

-1.6%

 

-3.78%

-35.95%

-8.95%

1.0 M

 

-3.78%

 

-3.27%

 

-8.7%

 

-29.4%

-45.2%

-11.3%

Group 2, 4, 6 and 8 do not have any data for distilled water, 0.2M Sucrose, and 0.4M Sucrose and group 1, 3, 4 and 7 do not have any data for 0.6M Sucrose, 0.8M Sucrose and 1.0M Sucrose because of the lack of time. So, group 1 and 2 were paired up, 3 and 4, 5 and 6, and 7 and 8 to exchange data.
Part C: Water Potential
Table 1.4: Individual Data of Change in Mass of Potato Cores in Six Different Sucrose Solution

 
Contents in Beaker

Initial Mass

Final Mass

Mass Difference

Percent Change in Mass

Class Average % Change in Mass

a) Distilled Water

2.39g

2.95g

0.56g

23.4%

23.3%

b) 0.2M Sucrose

2.41g

2.69g

0.28g

11.6%

8.4%

c) 0.4M Sucrose

2.47g

2.38g

-0.09g

-3.6%

-3.7%

d) 0.6M Sucrose

2.33g

1.98g

-0.35g

-15.0%

-13.5%

e) 0.8M Sucrose

2.46g

2.05g

-0.41g

-16.7%

-19.9%

f) 1.0M Sucrose

2.49g

1.95g

-0.54g

-21.7%

-20.8%

Table 1.5: Class Data of Percent Change in Mass of Potato Cores in Six Different Sucrose Solution

 

Group 1

Group 2

Group 3

Group 4

Group 5

Group 6

Group 7

Group 8

Total

Class
Average

Distilled
Water

23.4%

 

18.9%

 

23.2%

 

27.5%

 

93%

23.3%

0.2M

11.6%

 

6.8%

 

5.0%

 

10.1%

 

33.5%

8.4%

0.4M

-3.6%

 

-3.7%

 

-7.0%

 

-0.4%

 

-14.7%

-3.7%

0.6M

 

-15.02%

 

-13.5%

 

-11.16%

 

-14.3%

-54%

-13.5%

0.8M

 

-16.67%

 

-22.5%

 

-20.33%

 

-20.2%

-79.7%

-19.9%

1.0M

 

-21.69%

 

-24.3%

 

-24.39%

 

-12.9%

-83.3%

-20.8%

Group 2, 4, 6 and 8 do not have any data for distilled water, 0.2M Sucrose, and 0.4M Sucrose and group 1, 3, 4 and 7 do not have any data for 0.6M Sucrose, 0.8M Sucrose and 1.0M Sucrose because of the lack of time. So, group 1 and 2 were paired up, 3 and 4, 5 and 6, and 7 and 8 to exchange data.
Part D: Calculation of Water Potential from Experimental Data
Analysis:
Part A: Diffusion and Osmosis
From table 1.1 in this part of the lab, it is seen that IKI is flowing into the bag and glucose is flowing out of the bag. That is because of diffusion and osmosis. Knowing of this process is due to the color transformation of the bag, therefore showing that IKI has penetrated the bag. By testing the beaker for the existence of glucose, it was found that the glucose permeated through the dialysis bag, mixing with the IKI and H2O in the beaker. This is possible because as stated osmosis is a branched off form of diffusion, in which it is the diffusion of water through a selectively permeable membrane and glucose is one of the substance that is able to go through. IKI along with glucose is tiny enough to enter and exit the dialysis bag.
Part B: Osmosis
Both the individual and class data of percent change in mass is shown in graph 1.1. To receive the percent change in mass, the initial mass was subtracted from the final mass. The difference is then divided by the initial mass and 100 is then multiplied to the quotient. The product is then the percent change in mass. Osmosis is present due to the change in mass of the dialysis bag. The mass is different for each bag because of the sucrose in the bags different molarity. That establishes the amount of water that progresses in and out of the bag, which then changes the mass.
Part C: Water Potential
From testing the potato cores in different sucrose solution, graph 1.2 illustrates that on the best fit line, the molar concentration of sucrose, the sucrose molarity that shows the mass of the potato cores does not change, is 0.4M. So the lower the concentration of the molar concentration of sucrose, the percentage of the potato cores’ mass increases and anything with a higher concentration of the molar concentration of sucrose the percentage in the potato cores’ mass decreases. This is all because molecules of any sucrose with a higher concentration of 0.4M are too great to enter or exit into the potato cores.
Part D: Calculation of Water Potential from Experimental Data
It is given that the solute potential of the sucrose solution is calculated by using ψs= iCRT. Ψs is the solute potential, the variable “i” represents the ionization constant, variable “C” signifies the molar concentration, variable “R” standing for the pressure constant (R= 0.0831 liter bars/mole oK), T is the temperature oK (273+ oC of solution). Since it is the solute potential of sucrose that must be found, “i” is 1.0, due to the fact that sucrose does not ionize in water. From the information of “i”, “C” is determined to be 1.0 mole/liter. So the problem that has 1.0M sugar solution at 22 oC under atmospheric conditions would be answered like this:
Since the formula is ψs= iCRT, then when filling in for the variables the equation is now:
Ψs= -(1)(1.0mole/liter)(0.0831 liter bar/ mole oK) (273+22) à ψs =-24.51 bars
The water potential can then be figured out by the formula: ψ = ψp + ψs. By being able to solve for the solute potential, the product would then be used to solve for the water potential. And from the knowledge of knowing that the water pressure, ψp, is equal to zero the formula filled out would turn out as:
ψ=0 + (-24.51 bars) à ψ = -24.51 bars
From the graph of the percent change in mass of zucchini cores in different sucrose solution at 27 oC after 24 hours it can be concluded that the molar concentration of solute within the zucchini cell is 0.35 moles. From knowing the molar concentration of solute within the zucchini cell, the solute and water potential can be answered.
Solute Potential= -1(0.35moles/liter)(0.0831 liter bar/mole oK)(273 +27) à = -8.73 bars
Water Potential= 0 + (-8.73) à = -8.73 bars
This is an important piece of information because by knowing the water potential, it is possible to predict the direction of the flow of water. Water flows from an area of a higher water potential to and area of lower water potential; so if the information of the water potential of the solution in the beaker which the zucchini’s were soaked was given, the information of where the water flows would be known.
Conclusion:
This lab was to understand how diffusion and osmosis worked. The data that was received was consistent at some times. For part A and D of the lab, the results and calculations were consistent, but part B and C showed little consistency. That is because part B and C when comparing the percent change in mass with others, the numbers varied. The difference of the mass was changed, for it maybe misleading, into percentage, there eliminating any size factor and to compare the results. Though when comparing the percents with one another, some of the difference was too great to receive any accurate data. Some possibilities that may have altered the outcome of the results include the ends of the dialysis bags not being tied correctly, the inaccuracy of pouring the solutions, not a thorough cleaning of the outside of the dialysis bag and incorrect calculation and measurements. This lab can be modified to get a more consistent data by wearing gloves when working with the solutions, so when one is done pouring and tying the dialysis bag, gloves can be removed to reduce any chance of the solution being on the outside of the bag. A more accurate and precise measurement of the solution and the tightness and method of tying the ends of the bags can be arranged to be the same. With those alterations to the lab, the chance of a more consistent data is higher.