# Water Potential and Osmosis Potatoes Investigation

• Biology
• AN630078
In summary, the conversation discusses the process of osmosis and its effects on the water potential of solutions and the weight of potatoes. It is observed that the water potential becomes more negative as the osmotic potential increases and that the turgor pressure of potato cells can also affect the water potential. However, there are external variables that can affect the accuracy of the results, such as temperature, surface area, and potato age and origin. To improve the investigation, repeats should be performed and control variables should be implemented. There is also a discussion about converting between mol dm^-3 and g cm^-3, with the correct conversion being 54 g dm^-3 = 54 g cm^-3.
AN630078
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

#### Attachments

• Osmosis experiment potatoes table.png
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Think physically. If you have 54 g in 1 dm3, do you think you have 54000 g in 1 cm3?

AN630078
mjc123 said:
Think physically. If you have 54 g in 1 dm3, do you think you have 54000 g in 1 cm3?
Thank you for your reply. No I suppose not, I am just uncertain how to calculate this. I multiplied by 1000 since I know that the to convert between dm^-3 and cm^-3 you would do so. Would I actually only multiply by 10, since this is how to convert from dm to cm?

No, there are 1000 cm3 in 1 dm3. But do you multiply by 1000 or divide by 1000? As I said, think physically. Will there be a greater amount of solute in 1 cm3 than in 1 dm3, or less?

AN630078 said:
I multiplied by 1000 since I know that the to convert between dm^-3 and cm^-3 you would do so.
No, you multiply by 1000 to convert dm3 to cm3. For the reciprocal quantities (dm-3 to cm-3) you divide by 1000.

Another approach would be to put the units in explicitly - the conversion factor is 1000 cm3/dm3. Now
54 g/dm3 * 1000 cm3/dm3 = 54000 g cm3/dm6. The units are obviously wrong. But
54 g/dm3 / 1000 cm3/dm3 = 0.054 g/cm3. The dm3 cancels out.

AN630078
mjc123 said:
No, there are 1000 cm3 in 1 dm3. But do you multiply by 1000 or divide by 1000? As I said, think physically. Will there be a greater amount of solute in 1 cm3 than in 1 dm3, or less?No, you multiply by 1000 to convert dm3 to cm3. For the reciprocal quantities (dm-3 to cm-3) you divide by 1000.

Another approach would be to put the units in explicitly - the conversion factor is 1000 cm3/dm3. Now
54 g/dm3 * 1000 cm3/dm3 = 54000 g cm3/dm6. The units are obviously wrong. But
54 g/dm3 / 1000 cm3/dm3 = 0.054 g/cm3. The dm3 cancels out.
Thank you for your reply. Sorry yes, I see my mistake I think.
If there are 1000cm^3 in 1 dm^3 then there will be a greater volume of solute in 1cm^3 than 1dm^3?
So could the concentration be found by;
1 g dm^3=270 mol dm^3 /5=54g dm^3
Then divide this result by 1000 to convert dm^3 to cm^3:
54 g dm^3/1000=0.054 g cm^3

Would this be correct? Also would you be able to comment upon my other solutions at all? Thank you for your help

mjc123 said:
No, there are 1000 cm3 in 1 dm3. But do you multiply by 1000 or divide by 1000? As I said, think physically. Will there be a greater amount of solute in 1 cm3 than in 1 dm3, or less?No, you multiply by 1000 to convert dm3 to cm3. For the reciprocal quantities (dm-3 to cm-3) you divide by 1000.

Another approach would be to put the units in explicitly - the conversion factor is 1000 cm3/dm3. Now
54 g/dm3 * 1000 cm3/dm3 = 54000 g cm3/dm6. The units are obviously wrong. But
54 g/dm3 / 1000 cm3/dm3 = 0.054 g/cm3. The dm3 cancels out.
Sorry I am supposed to be converting from 0.2 mol dm^-3 to g cm^-3 (the reciprocal quantities) so would I still divide by 1000?
54 g dm^-3 = 0.054 g cm^-3 ?

AN630078 said:
If there are 1000cm^3 in 1 dm^3 then there will be a greater volume of solute in 1cm^3 than 1dm^3?
You really think that? Go to the lab or the kitchen. Put about 1 L of water (doesn't have to be exactly measured, just approximate) in a jar. Put about 1 cc of water in another jar. Look at them. Now think: if they were solutions of the same concentration, which would contain more solute? It's obvious, isn't it? (Physically do it, not a thought experiment.)
AN630078 said:
Then divide this result by 1000 to convert dm^3 to cm^3:
No! To convert dm^-3 to cm^-3! You really must make the distinction! It may have been a typo, but do you really understand it?
AN630078 said:
Sorry I am supposed to be converting from 0.2 mol dm^-3 to g cm^-3 (the reciprocal quantities) so would I still divide by 1000?
What do you think? Have you grasped the point yet?

## 1. What is water potential?

Water potential is a measure of the tendency of water to move from one area to another. It is affected by factors such as solute concentration, pressure, and temperature.

## 2. How is water potential related to osmosis?

Osmosis is the movement of water from an area of high water potential to an area of low water potential. This means that water will naturally move from an area with a high concentration of water molecules to an area with a lower concentration of water molecules.

## 3. How is the water potential of a potato determined in this investigation?

In this investigation, the water potential of a potato is determined by placing potato slices in solutions with different concentrations of solutes. The potato slices will either gain or lose water depending on the concentration of the solution, and the water potential can be calculated based on the change in weight of the potato slices.

## 4. What is the purpose of using potatoes in this investigation?

Potatoes are used in this investigation because they are made up of cells with semi-permeable membranes, which allow for the movement of water through osmosis. This makes them a good model for studying the effects of water potential and osmosis.

## 5. How does temperature affect water potential and osmosis in this investigation?

Temperature can affect water potential and osmosis by altering the rate of molecular movement. Higher temperatures can increase the rate of osmosis, while lower temperatures can decrease it. This can impact the results of the investigation and should be controlled for in the experimental design.

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