Potato Investigation Presented with raw potato how would you carry out an experiment to determine the concentration of the potato cell in molar strength? The experiment was set up to be as fair as possible. Then the potato chips were cut into 15 cylinders using a cork cutter and then cut to a length of exactly 4cm each with a scalpel. Then three were weighed at one time on a top pan balance to find an average mass, this was done because it did not involve confusion over which potato cylinder was which after the experiment. Once these averages were found and recorded the potato chips were placed into their test tube solutions of 0.0M(distilled water), 0.25M, 0.5M, 0.75M and 1.0M solutions of sucrose. Then the contents, each containing three potato cylinders were left for 50minutes. When the time was over I removed the potato with a pair tweezers and placed them on a paper towel to try and dry of any excess water. Then I weighed the potato again together in their threes on a top pan balance and found the average mass. Prediction I believe from my past knowledge of a similar experiment involving raw eggs that the potato cylinders placed in the distilled water will be hyperosmotic, therefore the water will diffuse by osmosis into the potato from an area of high concentration down the concentration gradient. This will result in a gain of mass for the potato. However in the 1.0M solution I think that the opposite will occur as there will be a higher concentration of water within the potato than the solution, containing lots of solute, this will make the potato hpo-osmotic and it will lose its water content as the water diffuses out.Finally, a difference between the egg experiment and the potatoes that the cells will not reach equilibrium (become iso-osmotic). This is because plants have a cellulose cell wall which cannot expand beyond a certain point as it is turgid, therefore the contents will never be able to be equal to the outside or it will not be able to give out all its water as it has a rigid structure. It is the pressure created by the cell wall which stops the cell reaching equilibrium. I think that the contents of the cell will be about 0.3M concentration as it will be in between 0.25M and 0.5MThe experiment has yielded a graph, (see attached) this graph gives me an excellent set of data. The graph shows that in the distilled water the potato gains in mass by approximently 7.4% then it gains 4.4% in the 0.25M solution. In the 0.5 M solution there is a change, the gradient of the graph decreases steeply and the potato loses mass to 4.4% then in the 0.75M solution the gradient decreases as it loses yet more mass to 10.4%. The first observation I made was after the potato cylinders were placed in their solutions I could see a difference. The ones in the 0.0M and 0.25M solutions were floating and the potatoes in the 0.5M, 0.75M and 1.0M solutions were at the bottom of the test tube, this lead me to drawing my next conclusion. The graph shows that the potato in the 0.0M solution and in the 0.25M is hyperosmotic, as I said in my prediction, this means that there is a higher water potential in the distilled water and 0.25M solution than in the potato, which has a high concentration of solutes. Therefore this is why the water diffuses by osmosis down the concentration gradient from high concentration to low concentration, resulting in the potato gaining mass. The opposite occurs in the solutions where the molarity is higher 0.5M, 0.75M and 1.0M, the potato in these is hypo-osmotic because there is a higher concentration of water inside the potato than in the solutions, therefore the water diffused out of the potatoes by osmosis, down the concentration gradient and into the solutions outside, this resulted in the loss of mass. Another important fact is that the graph makes a shape that will result in the potato not being able to take in any more water or lose any more. Like I said in my prediction the cell wall causes the pressure that prevents this from happening, therefore the cell will never become iso-osmotic like an egg. On my graph I have marked the point where the graph line crosses the place on the axis where the potato neither loses mass nor gains mass. This happens at 0.36M, therefore the contents of the potato cells in molar strength is 0.36M. This result is very close to my prediction of 0.3M and it shows that the contents of the cell are between 0.25M and 0.5M which is why in these concentrations they gained mass and lost mass. I think that the experiment went very well, there were no odd results and they produced a good graph. However there were a few areas where there could be improvement. Firstly, when I dried off the excess water on the potato cylinders after the experiment and before I weighed them, I used a paper towel. This might have either taken some water out of the potato or it might of left some excess water on the potato. This part of the experiment is difficult to come up with an accurate and fair method, as other ways would also lead to some slight mistakes. Also the potato itself was not definitely from the same potato and was not exactly the same size, although I did try to cut them to 4cm each, this could have effected the amount of water gained or lost. Another way of improving the results would have been to leave the experiment running longer, this would have enabled me to find the saturation point (when the potato can no longer take in any more water) and dehydration point (when the potato cannot lose any more water) and therefore get a more accurate result. Finally, I could extend the experiment to a more exact level by looking at the potato cylinders under a microscope, then I would be able to see the cells in greater detail and draw some more observational results.