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Investigating Osmosis: The Molarity Of A Potato Cell.

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Investigating Osmosis: The Molarity Of A Potato Cell Introduction Knowing that osmosis (diffusion of water) will occur across a semi-permeable membrane whenever there is a difference between the water concentrations on the two sides of the membrane, and knowing that when this happens to cells, they will either become turgid if water flows into them, or plasmolysed if water flows out of them, therefore changing their volume. I want to test the hypothesis that: If the concentration of a solution into which a cylinder of potato is placed is greater than a certain level the cylinder will decrease in mass, and if the concentration is less than that level it will increase in 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 hypertonic, 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, this will make the potato hypotonic and it will lose its water content as the water diffuses out. ...read more.


Difference in mass Time left inside the testube (minutes) Before After g % 0 2.29 2.27 -0.02 -0.87 15 0.2 2.31 2.23 -0.08 -3.46 15 0.4 2.23 2.13 -0.10 -4.48 15 0.6 2.25 2.14 -0.11 -4.89 15 0.8 2.20 2.07 -0.13 -5.91 15 1 1.71 1.57 -0.14 -8.19 15 Second attempt: Molarity of solution (Molar) Mass of potato (g) Difference in mass Time left inside the testube (minutes) Before After g % 0 2.48 2.58 +0.10 +4.03 15 0.2 2.52 2.56 +0.04 +1.59 15 0.4 2.49 2.46 -0.03 -1.20 15 0.6 2.51 2.43 -0.08 -3.19 15 0.8 2.50 2.40 -0.10 -4.00 15 1 2.49 2.34 -0.15 -6.02 15 Conclusion The first observation I made was after the potato cylinders were placed in their solutions I noticed that the potato cylinders in the 0.0M and 0.2M solutions were floating and the ones in the 0.4M, 0.6M, 0.8M and 1.0M solutions were at the bottom of the test tube, this led me to drawing my first conclusion. The graph (see attached) shows that the potato in the 0.0M solution and in the 0.2M is hypertonic, as I said in my prediction, this means that there is a higher water potential in the distilled water and 0.2M solution than in the potato, which has a high concentration of solutes. ...read more.


This might have either taken some water out of the potato or it might have left some excess water on the potato. It is difficult to come up with an accurate and fair method to do this part of the experiment 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 affected 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. Also the relatively low number of solutions tested (6) suggests that there is a range of possible values for the molarity of the potato cell, therefore that we cannot accurately predict values for the change in mass at different concentrations. Finally, the results obtained in this experiment support my hypothesis, and I was also able to discover the approximate molarity of a potato cell. Daniel Rios-Perez ...read more.

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Here's what a teacher thought of this essay

3 star(s)

*** This candidate has collected some valid data, analyzed it fairly well and come to a valid conclusion. The understanding of osmosis seems poor in some sections of the report.
To improve:
Planning: The background theory on osmosis was confused in several places and was not explained clearly in all paragraphs.
The planned experiment could be used to collect valid data but there was insufficient consideration of the key variables and the written method lacked key details. No risk assessment was carried out and this is awarded marks under newer schemes.
Carrying out:
Some data was collected but there very different sets of data between the two experiments. The candidate needed to take more care to ensure that table headings contain full descriptions and units for each column.
Analysis and Evaluation:
The graph was not included. The data seems to have been interpreted correctly and a sensible value obtained for the molarity of the solution inside the potato cells. The conclusion does refer back to the prediction and contains an adequate amount of background theory. The evaluation could be improved by including more detailed suggestions for improving the method or extending the range of the investigation.

Marked by teacher Stevie Fleming 29/05/2013

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