- #1

AN630078

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- 25

- Homework Statement
- Hello, I have been revising osmosis and its mechanisms in plant cells which I continually find to be a source of confusion concerning water pressure.

I came across a past paper with some questions testing one's knowledge of the topic and wondered if someone could offer any feedback upon my given solutions and reasoning, possibly dispensing some further guidance.

A student carried out an investigation to test the hypothesis that the cells of sweet potatoes have a lower water potential than the cells of white potatoes.

One gram of potato tissue was placed in a test tube and 10 cm^3 of a 0.1 mol dm^–3 sucrose solution was added.

After 30 minutes, one drop of blue colouring was added to the test tube and the contents of the tube mixed.

A pipette was used to remove some of the solution from the test tube and one drop was placed in the middle of a second test tube containing the original 0.1 mol dm–3 sucrose solution.

The movement of the blue coloured drop was observed.

The procedure was repeated for a further six different concentrations of sucrose solution. The results are shown in the table attached.

(a) State how evaporation would affect the water potential of the solution in the test tubes.

(b) Explain the movement of the drop placed in the 0.2 mol dm^–3 sucrose solution.

(c)Explain what the student could conclude about the water potential of white potatoes.

(d) The student concluded that the sweet potato and white potato have the same water potential.

Explain why this is not a valid conclusion.

(e) Suggest three ways in which the student's investigation could be improved

(f) Calculate the concentration of 0.2 mol dm^03 of sucrose in g cm^-3. 1 mole of sucrose =270g

- Relevant Equations
- Ψ=P+π

a) Evaporation will remove water from the test tubes as it turns into water vapour, meaning that the solution will have a greater solute concentration and thus an increased osmotic potential which results in a more negative osmotic potential. Consequently this lowers the solution's water potential, which will become more negative than it was initially.

b) Water has moved from the surrounding solution into both the sweet potato and the white potato by osmosis, which has lowered the volume of water in the solution and increased the sucrose concentration remaining in the solution. Consequently, the solution has a greater density than it did previously causing the drop placed in the second 0.2 mol dm^–3 sucrose solution to move downwards.

c) If two solutions are separated by a partially permeable membrane water will move by osmosis from a higher to lower water potential. Thus, one can conclude that the water potential is lowered and becomes more negative as the osmotic potential increases as water diffuses into the potatoes via osmosis through the permeable membrane between sucrose concentrations of 0.1-0.4 mol dm^-3 making the potato heavier and more saturated. This is evident in the subsequent drops of solution falling downwards since they have a greater density. However, at sucrose concentrations of 0.5-0.7 mol dm^-3 the water potential becomes more positive and is raised as the hydrostatic pressure of the cell wall, the turgor pressure, increases as the potatoes become more saturated. As water passes through the cell wall and cell membrane the total amount of water present inside the cell increases which exerts an outward pressure that is opposed by the rigidity of the cell wall. The internal water pressure will exceed the water pressure in solution, meaning water will leave the potato cells through the permeable membrane via osmosis.

d)This is not a valid result as although the results of the experiment are the same for the direction of motion of the drops of solution obtained, as such one could conclude the water potential is the same. However, the water potential is not necessarily equal in magnitude between the two types of potatoes. Similarly, neither the osmotic or turgor potentials would quantitively exhibit the same values, at least this investigation has not measured the water potential precisely but rather found whether it has become more or less negative as a result of increased sucrose concentrations.

Furthermore, external physical variables that will affect the rate of osmosis like temperature or surface area have not been controlled. A greater surface area could for example reduce the penetration of the solution to the centre of the potato piece while an increase in temperature could increase the rate of osmosis because at high temperature the particles in the solution have more kinetic energy, therefore colliding more frequently and at greater speeds, increasing the rate of successful collisions.

Moreover, variables related to the potatoes themselves being their age, source and treatment prior to the investigation are not controlled, all of which will result in a variation in water potential. The permeability of the potato can be determined by the age of the potato, as it ages it becomes less permeable meaning the rate of osmosis is reduced.

e) The student's investigation could be improved by performing repeats, for example using several pieces of potato for each sucrose solution which permits the identification of any anomalous results from errors in procedure or irregularities in the potatoes used. The student should also calculate an average result for each concentration since this is likely to be more accurate and closer to the true value than one individual reading. Moreover, the student should implement certain control variables upon temperature, surface area, potato age, origin etc.

f) I am not sure how to convert here. I know that normally to convert from mol dm^-3 to g cm^-3 one would multiply by the relative formula mass, but that is not known nor what I really think the question is asking here. Would it be that 0.2 mol dm^-3 = 270g/5= 54 g dm^-3.

Then to convert dm^-3 to cm^-3 multiply this result by 1000;

54 g dm^-3 = 54000 g cm^-3.

Would this be correct?

I would greatly appreciate any feedback or suggestions to improvements I could make

b) Water has moved from the surrounding solution into both the sweet potato and the white potato by osmosis, which has lowered the volume of water in the solution and increased the sucrose concentration remaining in the solution. Consequently, the solution has a greater density than it did previously causing the drop placed in the second 0.2 mol dm^–3 sucrose solution to move downwards.

c) If two solutions are separated by a partially permeable membrane water will move by osmosis from a higher to lower water potential. Thus, one can conclude that the water potential is lowered and becomes more negative as the osmotic potential increases as water diffuses into the potatoes via osmosis through the permeable membrane between sucrose concentrations of 0.1-0.4 mol dm^-3 making the potato heavier and more saturated. This is evident in the subsequent drops of solution falling downwards since they have a greater density. However, at sucrose concentrations of 0.5-0.7 mol dm^-3 the water potential becomes more positive and is raised as the hydrostatic pressure of the cell wall, the turgor pressure, increases as the potatoes become more saturated. As water passes through the cell wall and cell membrane the total amount of water present inside the cell increases which exerts an outward pressure that is opposed by the rigidity of the cell wall. The internal water pressure will exceed the water pressure in solution, meaning water will leave the potato cells through the permeable membrane via osmosis.

d)This is not a valid result as although the results of the experiment are the same for the direction of motion of the drops of solution obtained, as such one could conclude the water potential is the same. However, the water potential is not necessarily equal in magnitude between the two types of potatoes. Similarly, neither the osmotic or turgor potentials would quantitively exhibit the same values, at least this investigation has not measured the water potential precisely but rather found whether it has become more or less negative as a result of increased sucrose concentrations.

Furthermore, external physical variables that will affect the rate of osmosis like temperature or surface area have not been controlled. A greater surface area could for example reduce the penetration of the solution to the centre of the potato piece while an increase in temperature could increase the rate of osmosis because at high temperature the particles in the solution have more kinetic energy, therefore colliding more frequently and at greater speeds, increasing the rate of successful collisions.

Moreover, variables related to the potatoes themselves being their age, source and treatment prior to the investigation are not controlled, all of which will result in a variation in water potential. The permeability of the potato can be determined by the age of the potato, as it ages it becomes less permeable meaning the rate of osmosis is reduced.

e) The student's investigation could be improved by performing repeats, for example using several pieces of potato for each sucrose solution which permits the identification of any anomalous results from errors in procedure or irregularities in the potatoes used. The student should also calculate an average result for each concentration since this is likely to be more accurate and closer to the true value than one individual reading. Moreover, the student should implement certain control variables upon temperature, surface area, potato age, origin etc.

f) I am not sure how to convert here. I know that normally to convert from mol dm^-3 to g cm^-3 one would multiply by the relative formula mass, but that is not known nor what I really think the question is asking here. Would it be that 0.2 mol dm^-3 = 270g/5= 54 g dm^-3.

Then to convert dm^-3 to cm^-3 multiply this result by 1000;

54 g dm^-3 = 54000 g cm^-3.

Would this be correct?

I would greatly appreciate any feedback or suggestions to improvements I could make