III. Data:
A. Table 1:
Diffusion
B. Table 2:
Osmosis Investigation
Osmosis
C. Table 3:
Potato Cell Water Potential
Water
Potential
Sucrose Molarity (M)
D.
Plant Cell Plasmolysis
Cells in Distilled Water Cells in 15% NaCl
IV. Questions:
1.
Osmosis Diffusion
2. The diffusion which occurred in the setup of part A was the movement of glucose from inside the dialysis bag to the beaker. This was an example of molecule movement from area of high concentration to area of low concentration. I think osmosis did occur in this part of the experiment because the water potential in the dialysis bag was low due to the high solute concentration. Thus, water molecules in the beaker would move from area of higher water potential to area of lower water potential.
3. By definition, a selectively permeable membrane doesn’t allow certain molecules or particles to enter or leave the cell. Hence, the dialysis tubing did serve as a selectively permeable membrane because it restricted certain molecules or particles to diffuse through its microscopic holes. Base on the size of the molecules, water and glucose were small enough to fit through the microscopic holes, whereas starch and potassium iodide weren’t able to diffuse through.
4. In part B, the mass of the dialysis bags increased after 30 minutes because there was a net flow of water molecules going into the bags. At the conclusion of the experiment, there was more water in the dialysis bags because there was a high solute concentration of sucrose. Since solute potential is always a negative value and there is no physical pressure on the solution, the water potential will also be negative. Thus, water will have the tendency to move inside the dialysis bags.
5. The distilled water in the beakers was hypotonic in relation to the sucrose solutions found in the dialysis bags because it had a lower solute concentration. This also explains why the dialysis bags gained weight after 30 minutes. The hypertonic solution of the dialysis bags led to the flow of water from the beaker.
6. If the dialysis bags were placed in beakers containing a 0.6 M sucrose solution as opposed to distilled water, the results would change drastically. Depending on the sucrose concentration in the dialysis bags, the direction in which diffusion occurs will depict the mass change. For the dialysis bags containing less than 0.6 M of
sucrose, the sucrose molecules in the beakers will diffuse inside the dialysis bags where the concentration is low. The opposite effect applies to the dialysis bags containing more than 0.6 M of sucrose. Since there are more sucrose concentration inside the dialysis bags, sucrose molecules would diffuse out to the beakers where there’s less concentration. If the dialysis bag has the same concentration of sucrose as the solution, there would be no mass change because it is at equilibrium.
Diffusion of Sucrose Molecules
7. The point where the best fit line crosses the x-axis is at equilibrium. At that point, the initial mass and the final mass are the same, and there is no net flow of water molecules or sucrose molecules. Thus, the sucrose concentration of the potato
can be identified as 0.26 M.
8. If a cell is placed in a hypertonic solution, it has a less solute in solution than the surrounding fluid, and will therefore experience a net loss of water to its surrounding. This cell has a low water potential since there is a great deal of osmotic pressure causing water to leave the cell. Conversely, a cell sitting in a hypotonic solution has a high water potential, and since it will experience a net gain of water, there will be little osmotic pressure causing water to enter the cell.
9. The osmotic potential of the sucrose solution from chemistry lab is -6.37 bars.
ψ = -iCRT
ψ = -1(0.26M)(0.0831 liter bars/mole °Kelvin)(295 K)
ψ = -6.37 bars
10. The water potential of the potato cells is -6.37 MPa.
ψ = ψ + ψP
ψ = -6.37 + 0
ψ = -6.37 MPa
11. The water potential of the potato cells would not change if the cylinders were allowed to dry out because they are not in any kind of solution where there’s a difference in concentration. Although most of the water would be extracted out of the cells, it is mostly due to evaporation or absorption of water molecules. In addition, water potential is only affected by the solute potential and the pressure potential, both of which cannot be calculated in this case.
12. When cells are placed in a hypotonic solution where solute concentration is low, they will experience swelling. Due to osmosis, water will diffuse inside the cells where there is more solute concentration. Consequently, the swelling may cause
cytolysis in which the cell bursts open. Plant cells, however, are an exception because they have a tough cell wall which holds back the expansion of the cell.
On the contrary, cells in a hypertonic solution will shrink because of water leaving
to an area with higher solute concentration. The cells might experience lysis, which is the death of the cell due to the breakdown of cellular membrane. In an isotonic solution, the cells and their environment have an equal solute concentration. Thus, there’d be no net flow of water coming in or out of the cells because the condition is in equilibrium.
13. Humans can’t drink salt water for hydration because the water from our cells will diffuse outside where there is a high concentration of solute and low water potential. Consequently, we would become more dehydrated as the cells loose more water. In addition, the cells will become hypotonic to their surroundings and shrink, or plasmolyze.
V. Design an Experiment: Diffusion and Osmosis in Egg Shells
1. The objective of this experiment is to detect diffusion and osmosis in egg shells in three different solutions. An egg shell is made of mostly calcium carbonate and contains microscopic pores. The pores play an important role similar to a selectively permeable membrane, in which it only allows certain molecules or particles to pass through the shell. For this experiment, you will test raw hen eggs in distilled water, salt (NaCl) solution, and sucrose solution over a period of three days. You will need a total of three eggs and three empty containers. First, you’ll weigh and measure each egg and record the data in a table. Then, you’ll pour 250 mL of each solution into three containers, which you’ll need to label. Once you put an egg into each solution, you’ll leave them in the lab undisturbed for three days. On the third day, you will take the eggs out of the solutions and carefully dry them off with a paper towel. Then, you’ll weigh and measure the eggs for the final time to compare the results.
2. Hypotheses:
Osmosis will occur because the solute concentration inside the egg is higher.
Table salt (NaCl) will diffuse inside the egg because there is less concentration.
Sucrose molecules will not diffuse inside the egg because its solute concentration is higher than the solute concentration inside the egg.
3. Data:
Mass of the Eggs
Egg Measurements
4. Conclusion: When comparing the initial data to the final data, the changes in mass and measurement are evident that osmosis and diffusion of NaCl did occur inside the eggs. Sucrose, on the other hand, did not diffuse through the microscopic pores of the shell. Instead, water inside the egg diffused out to the sucrose solution where water potential was low due to its high solute concentration. Therefore, my three hypotheses were correct in which water molecules and NaCl could diffuse through the egg shell, but sucrose could not.