I will submerge 7 potato chips in 10 ml of different concentrations of glucose solution.
Test Tube A 10ml of 0M glucose (distilled water)
Test Tube B 10ml of 0M glucose (tap water)
Test Tube C 10ml of 0.2M glucose solutions
Test Tube D 10ml of 0.4M glucose solutions
Test Tube E 10ml of 0.6M glucose solutions
Test Tube F 10ml of 0.8M glucose solutions
Test Tube G 10ml of 1.0M glucose solutions
Then I will leave these test tubes for 40 minutes for an adequate and significant amount of water to be able to diffuse into the potato cells. Then I will extract all the potato chips from the solutions and dry off the liquid that has not yet diffused into the cells to make it a fair test, as the water will add to the mass of the potato chip. I will then weigh the mass and measure the length of the chips and record them in a table with the original masses and lengths. These values can then be compared and brought to a conclusion. I will record the values in a table.
Prediction:
I predict that the potato chip submerged in the 1.0M concentration glucose will shrink and lose mass the most because the concentration of solute is much higher in the solution than inside the potato so a lot of water inside the potato will move, by osmosis, out into the glucose solution. I also predict that the potato chips submerged in the 0M solutions, which are distilled water and tap water, will gain mass and expand. This will happen because distilled water contains no solute at all so there will be a significant difference between the concentration of solutes in the potato cells and in the solution. The tap water might not be totally free of solute but I believe that the concentration of solute will still be significantly lower than inside the potato. We know that the amount of water that diffuse through a membrane depends on how different the solute concentration is on both sides and that the water diffuses from a low concentration to a high concentration of solute. Assuming our solute concentration inside the potato cells was 0.2M then we would expect the potato cell to attract more water particles because the solute concentration is lower on the outside. If the solute concentration were equal, like in test tube C, where both concentrations are 0.2M, we would expect that there would be none or little osmosis occurring. If the concentration were higher on the outside of the cell as in test tube D, E, F and G, we would expect the water to diffuse out of the cell into the solution. Because the rate or amount of osmosis that occurs also depends on how different the concentrations are, we expect that the higher the difference the more water that is going to be lost to the solution.
Results:
Analysis:
The results show me that the effect of osmosis on potato chips does depend on the difference in the concentration of solute between the potato cell and the surrounding liquid. It also shows that water diffuses from a low solute concentration to a higher one and that the amount of water diffusing or the rate of diffusion depends on the difference in concentration. A greater difference (e.g. 1M and 0.5M concentrations) between the concentrations of solute will cause more water to diffuse through and a smaller difference (e.g. 0.5M and 0.6M concentrations) will mean that less water diffuses from the lower to the higher concentration. The evidence confirms my predictions. The graphs also show me that the concentration of solute inside the potato is around 0.2M. There was only one anomaly in the results and that was the length of the potato chip in tap water was longer than in the distilled water which is not very realistic because we expected there to be a little solute in the tap water which means that the concentration of solute in tap water would be higher than that of the distilled so less water should diffuse into the potato chip but what happened was the chip increased by 1 mm longer than the chip in the distilled water. This might have happened because I measured the chip inaccurately or that my amount of solution was a little more because I forgot to measure it accurately as time was pressing. There was an obvious sequential order from the lowest concentration to the highest and the results would change in the same order. The pattern is that the next concentration is always shorter or lighter than the last one. There is this descending order because of the concentration difference. We know that the amount of water that diffuse through a membrane depends on how different the solute concentration is on both sides and that the water diffuses from a low concentration to a high concentration of solute. Assuming our solute concentration inside the potato cells was 0.2M then we would expect the potato cell to attract more water particles because the solute concentration is lower on the outside. If the solute concentration were equal, like in test tube C, where both concentrations are 0.2M, we would expect that there would be none or little osmosis occurring. If the concentration were higher on the outside of the cell as in test tube D, E, F and G, we would expect the water to diffuse out of the cell into the solution. Because the rate or amount of osmosis that occurs also depends on how different the concentrations are, we expect that the higher the difference the more water that is going to be lost to the solution. So the G chip would be smallest in size and mass then F, E, D, C, B, and A. My results fit in exactly with my predictions and my hypothesis.
Evaluation:
The experiment generally proceeded very smoothly and without any major problems. The results were quite reliable because I kept all the variables constant. The only major thing that I could have improved was that I didn't measure the volume of one of the solutions in the tube. Because time was pressing I did not measure the volume of the tap water accurately therefore leading to the anomaly in the results. I could improve the experiment by instead of putting the chips into the solutions at the same time, which I can't because I don’t have 7 hands, to put them in the solution at intervals and take them out in the same order at the same intervals this way they will all be exposed to the solution for the same amount of time. The entire test tube rack could also be put in a water bath to nearly get rid of all temperature changes in the atmosphere that could affect the experiment but it wouldn’t matter too much because there isn’t that significant a change in temperature and even if a change was recorded, all of the test tubes would be exposed to the same conditions so there won’t be any unfair results. A relationship between the surface area and the mass of the chip with a proportion to its volume is also an idea worth investigating. Also, during our experiment we were not provided with glucose, instead we were given a salt solution ranging from 0.0M to 1.0M. It served the purpose because there isn’t any very significant difference that will threaten the credibility of our results. Apart from the little disturbances the experiment went perfectly fine.
