Diffusion and Osmosis Lab Lab Partners: Luke Woody, Thomas Burlingame Smith, Lydia Ruiz, and Jillian Jarvis Date: 11/6/15 1B:

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1 Diffusion and Osmosis Lab Lab Partners: Luke Woody, Thomas Burlingame Smith, Lydia Ruiz, and Jillian Jarvis Date: 11/6/15 1B: Question: The sucrose molarities of the six unknown solutions are the following: 1.0 M, 0.8 M, 0.6 M, 0.4 M, 0.2 M, and 0.0 M. Variables: Independent variable: Sucrose molarities of the six unknown solutions Dependent variable: Percent change in mass of the bags Constants: Tubing bags Same amount (ml) of distilled water surrounding the bag Materials: see lab handout Procedure: See lab handout Data Tables: Group Data of Dialysis Tubing Bag Results Color of Solution Initial Final Difference % Change in Solute Molarity Red g g 3.35 g 16.34% 1.0 M Orange g g 0.15 g 0.75% 0.0 M Yellow g g 2.50 g 0.46% 0.2 M Green g g 2.00 g 9.24% 0.6 M Blue g g 0.75 g 3.94% 0.4 M Purple g g 2.55 g 11.46% 0.8 M Mean Data of Dialysis Tubing Bag Results

2 Color of Solution Mean % Change in Molarity Red 7.4% 0.4 M Orange 0.3% 0.0 M Yellow 11.4% 0.8 M Green 9.0% 0.6 M Blue 3.3% 0.2 M Purple 12.9% 1.0 M Graph:

3 Analysis Questions: 1.)The greater the molarity of sucrose within the bag, the lower the water concentration in the bag. The water outside the bag will have a higher water potential because of a low solute concentration. Because of water's tendency to go from areas of high water concentration to areas of low water concentration, the water will diffuse into the bag, causing the mass to change. 2.)The dialysis bags with a higher sucrose concentration than 0.4M of solution from the beaker, then the bag would be hypotonic. Water would leave the beaker and diffuse into the bag due to the tendency to go to areas of lower concentration. If the dialysis bags have a lower sucrose concentration than 0.4M then the bag would be hypertonic. Water would be lost from the bag. If the beaker and bag had the same amount of solution, then they'd be isotonic. Water would travel equally between both the beaker and the bag. 3.)The mass of solution placed in each bag was done within a range rather than having the same mass measured out with maximum accuracy to each bag. percent takes in account the differences in mass between the content in the bags. 4.) 20 18=2 2/20 = 1/10=.1.1 x 100= 10% 5.)Hypertonic Conclusion: A trend is shown of a lower water potential for solution with a greater concentration of solutes relative to solutions with a smaller concentration of solutes.therefore, the water with the greatest tendency to flow would be found in the solutions with the least solute, and the changes in water potential relative to the solute in a solution, of course, exists on a continuum. In this way, it would be expected that the solution with the greatest concentration of solute would also exhibit the greatest percent increase of water mass that traveled inside the bag from the distilled water outside the bag.when the membrane is placed in a hypotonic solutions, the

4 water tends to move into the membrane while the reverse is true for a membrane placed in a hypertonic solution. The relative concentrations of water in the membrane and outside solution will tend toward being the same until an isotonic solution is formed and equilibrium is reached between the water leaving and going into the bag. From our results, we predicted that the solutions of the colors red, purple, green, blue, yellow, and orange were (for the concentration of sucrose) 1.0M, 0.8M,0.6M,0.4M, 0.2M, and was actually distilled water, respectively. This contrasted with the results obtained using the class average which instead showed that the solutions of the colors red, yellow, blue, and purple were actually 0.4 M, 0.8M,0.2M, and 1.0M, respectively. Error in our results could include uneven drying of the bags prior to weighing, which would have caused the bags that were dried least effectively to be measured as having gained more mass than the bag really had. The scale used was only calibrated at the beginning of the lab, and pressing too hard while weighing the bags may have lead to slight progressive skewing in measurements. Also, a miscommunication among my lab partners and I led to three bags being placed in their respective solution multiple seconds after the others. This would have allowed the three bags placed earlier to reach equilibrium with the environment, and these bags may have been skewed to weigh slightly more relative to the later placed bags. 1C: Question: What is the molar concentration of the solute within the potato cores? The molar concentration of the solute within the potato core is.15 M. Materials: see lab handout Procedure: see lab handout Variables: Independent Variable:The molarity of the solution in which the potato slices are placed. Dependent Variable: Percent change in mass of the potato slices.

5 Constants: the potato from which them potato slices are cut from and the volume of solution used. Data Tables: Group Data for Potato Cores Color of solution Initial of Potato Slice (grams) Final of Potato Slice (grams) Difference (grams) Percent Change in Molarity of Solution in Which Potatoes Were Placed Purple % 0.4 M Red % 0.6 M Blue % 0.8 M Orange % 1.0 M Green % 0.2 M Yellow % 0.0 M Class Data for Potato Cores Color of Solution Class Average Percent Change Molarity of Solution in Which Potatoes Were Placed Red 23.1% 0.4 M Orange 16.7% 0.0 M Yellow 41.3% 0.8 M Green 29.8% 0.6 M Blue 1.8% 0.2 M

Purple 36.6% 1.0 M Graph: 1D: Exercise 1D: Ψ s 1*.15*294*.0831= 3.67 bars since the derived molarity was at zero percent change, it is expected that the pressure potential would be zero.

6 Purple 36.6% 1.0 M Graph: 1D: Exercise 1D: Ψ s 1*.15*294*.0831= 3.67 bars since the derived molarity was at zero percent change, it is expected that the pressure potential would be zero. Therefore, 3.68 bars is also the value of water potential. Analysis: 1. Increase because water would be leaving the potato making the water potential of the air decrease as it slowly gains more water.

7 2. Hypertonic, assuming the pressure within the plant is zero, the factor affecting the water potential is the amount of dissolved ions to which the water molecules can become electrostatically bonded to. Because these bonds decrease the water potential, it is expected that the lower water potential within the cell is caused by a greater number of solute ions relative to the outside environment Dialysis bag 5.out of, the bag has a greater water potential, and, therefore, a greater concentration of water molecules which can easily flow out of the bag relative to the water molecules already outside of the bag. The rate of water leaving the bag will be greater than the water that can enter the bag. 6./7..25M concentration within the zucchini cells. 8.a. 6.2 bars 1*.25*.0831*300 b = It would make the solite potential smaller as the molar concentration of the solution would increase. 10. a.distilled water b.distilled water c.the water would flow from the area of greater water potential in the area of less water potential, and so, from the distilled water into the red

8 blood cell, and the red blood cell would expand due to not having a cell wall to add pressure against the addition of water. 11.The salt creates an environment outside the grass which is significantly hypertonic which then causes water to diffuse out of the more hypotonic cells of the grass.

2. investigate the effect of solute concentration on water potential as it relates to living plant tissues.

In this lab you will: 1. investigate the processes of diffusion and osmosis in a model membrane system, and 2. investigate the effect of solute concentration on water potential as it relates to living