Investigating the concentration of the solutions i

nside the vacuole of a potato cell.

Osmosis Investigation
We are trying to find out what the concentration of the solutions is inside the vacuole of a potato cell.

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We will investigate how the mass of the potato chips change in different sucrose solution concentrations.


We will use osmosis to do this.


I will be measuring and controlling many different variables, these include; –
The dependent Variable – Weight of potato chip
Independent Variable – Concentration of sucrose
Control Variables consist of; –
1.Temperature
2.Surface area of potato
3.Volume of sucrose solution
4.Same time for each potato chip submerged in solution
5.Potato chips all from same potato
The effects on the dependent variables all differ; here is how the main variables in my experiment work out.


Variable Effect on Dependent Variable
Surface area of potato chipThe greater the surface area the more water it will be able to absorb
Same chip from the same potatoThe genetic make-up will be the same and therefore the partly permemble membrane will be similar
Time left in the solutionThe longer spent in solution the more that is absorbed
Coating on the potato chipsIf not rinsed off and dried after a certain period of time the chips will form a coat of sugar thus decreasing surface area
Quantitative prediction
I predict that the effect of changing the sucrose concentration will be that as the concentration of the sucrose solution increases, first of all the mass of the chip will increase, and then the change in mass will gradually decrease until mass is lost and this mass loss will gradually increase in amount.


Because when the sucrose concentration is low, the concentration of water outside the cells of the potato chips will be greater than that inside, and therefore water will osmosis into the cells of the chip which will gain mass. As the concentration of sucrose increases the concentration of water outside the call will eventually become less than inside the cells of the chip and mass will be lost.


Dependent
Variable


Inside the cells Outside the cells
Dilute solution Concentrated solution

Higher water concentration Low water concentration
Key:
= Sucrose particle
= water molecule
= osmosis
= partially permeable membrane
In the higher sucrose concentration solution, the net movement of water (osmosis) is to the outside of the cell, and the chip will lose mass, the cells will become plasmolysed.


The chip in a low concentration of sucrose solution, is the opposite of the diagram above, in that the water osmoses into the cells of the chip, mass it gained, through osmosis of water into the plant cells, the cells will become turgid.


Because I am measuring osmosis, I will try to keep constant the other that affect osmosis, those other than the concentration of sucrose and water, I will be using potato chips from the same variety of potato (assuming that the genetic make-up and therefore partially permeable membranes will be more similar in make-up). I will use the same volume of solution at each concentration of sucrose solution, that being 20ml3. I will leave the potato chips in solution for the same time at each solution concentration. I will use the same method at each concentration, and for each potato chip individually, in that I will dry the excess water off the chips in a similar manner, and use the same accuracy of the weighing scales, prepare the chips for use in the same way and other aspects. I will keep the surface area of the chips as constant as possible, this being that I will use chips of the same cross sectional area and length. I found in my preliminary work that the chips in the higher concentration of sucrose solution tended to float thus excluding a certain part of their cross-sectional area from the solution, I do not know how to stop this. I will carry out all of my experimental work at room temperature and thus temperature will not affect my experiment. The time will not effect my results because I plan to leave my chips in solution for 30 minutes at each concentration, I discovered that significantly enough results were obtained after this length of time in my preliminary work.


What I plan to do.


I plan to cut out 12 potato chips into cubes of uniform diameter, using a scalpel, and they won’t be at a specific size, although they will all be of a unison length and width with the same surface area.

N.B. They will be the same size as I used in my preliminary work.


I will place 2 of each of these chips into each concentration of sucrose solution, and leave them in for a time limit of 30 minutes once the time is up I will rinse of the sugar coating that will form over the potato chips thus minimizing the surface area. Then weigh and record their weight.


I will have weighed them before potting them into the solution, and I will weigh them when they come out. I will then calculate percentage change in mass.


Change in mass (g)
X 100
Original mass (g)

Apparatus I plan to use
My apparatus I plan to use consists of:
1x Scalpel,
1x Pair of tweezers,
12x Test tubes,
1xDigital scales accurate to 1/100
of a gram,
1xCutting tile.


I plan to make 12 results over a range of 1.0M. I will then make various calculations and plot a graph.


Reliability
Reliability is a major factor in this experiment so I will do every thing possible to keep this an accurate and fair test.


Many ways to do this include using: –
The same set of scales to weigh each potato chip.


Repeating the experiment, at least twice so I have 12 results all together.


I will keep the sizes of the potato chips consistent and I will remember to compensate mass % gain.


Peel all skin off. For a wider surface area to make it fairer as not all potatoes will have the same amount of skin.


Try to keep the potato chips fully submerged, from my preliminary results I have found out that this is a major problem in the experiment and there is not a way to keep them fully submerged while maintaining the same surface area on each potato chip.


Try getting chips from same potato for same genetic DNA makeup, the length of storage of potato if different must not be within a day older or newer due to dryness that would affect the experiment.


Calculate weight to one hundredth of a gram.


Preliminary work
I planned my method following what I found out from my preliminary experiment, the main thing I found out what was that the chips I had used were to big and this created many problems such as –
Bigger chips needed more potatoes thus not having same DNA makeup.

Sometimes half of the chip wasn’t submerged in the solution or the chip would be to big and would then touch sides of test tube therefore minimizing surface area.


Results
Here is a table of results to show the masses of potato chips before and after they have been placed in varying concentrations of sucrose solutions. I will also show the percentage changes in mass for each chip.




Analysis
From the graph we can see that as the sucrose solution increases in molar the mass change in grams changed accordingly, we see that the mass change decreases in a pattern along with the increase in sucrose concentration in molar. As the sucrose concentration increases in molar strength the mass change in grams decreases with a steady ratio.
We can see that the point when osmosis is occurring and has occurred (with potato 1) is at 0.415 molar when no mass is lost of added, this is where the sugar levels are the same inside the vacuole of the potato as outside of the potato in the sucrose solution.


Potato 2 follows the same pattern just with different statistics, since the potato chips from both potatoes didn’t have exact same characteristics and thus not making it as an accurate as possible test. I recorded the state of osmosis at 0.19 molar sucrose concentration.
To Conclude
The pattern on the graph is directly proportionality to the change in mass as the change in molar of sucrose solution changes as explained in my analysis.


Using the mean number from the % changed in mass from both potatoes you can see that as sucrose concentration increases, firstly, at a concentration of 0.0M, which is de-ionised water, the percentage mass change is a large gain of 4.57%. This falls to an increase of only 1.29% at a concentration of 0.2M. The mass change = 0.0 at a concentration of 0.32M. The percentage change in mass continues to fall with speed, and at a concentration of 0.4M, the mass decreases by

x

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