The Osmotic Affect of Varying NaCI Solutions on a Potato Core


Osmosis is the passive transport of water through a semi-permeable membrane. To understand how osmosis works, an investigation was preformed using potato cores. Out of four potato cores that were used, one was dehydrated, one was put in a 0% NaCI solution, one in a 10% NaCI solution and the final one was placed in a 20% NaCI solution. The percentage represents the amount of solute (in this case NaCI) in the solution. The experiment was preformed over anhour period.

Subject: What is the osmotic affect of varying sodium chloride (NaCI) solutions on the physical characteristics of a potato core?

Hypothesis: If a potato core is placed in the 0% NaCI solution, then the potato core will increase in size because the potato core should be hypertonic to the solution. Also water that was in the test tube should be decreased, for it will have soaked into the core. If the potato core is put into the 10% NaCI solution, the core should have the reverse affect. The core should decrease in size because the core would be hypotonic to the solution. This theory goes for the 20% NaCI solution as well. The core placed in the 20% solution should decrease in size as well, for the core should also be hypotonic to the solution. With both the 10% and 20% NaCI solutions, the water in the test tube should have increased because diffusion would allow the water that was once in the core to move into the solution.


Day One Procedure: First, get a white potato, then you must get a 5mm wide cork borer and a small glass rod that is thin enough to remove the core from the borer. Take the borer and with the potato oriented upright vertically, push the borer into the top part of the core and out through the bottom. Then, remove the cork borer with the potato core still in it. Next, take the small thin glass rod and place it in the hole at the top of the core borer's handle. This should allow you enough force to push the core out of the borer. Repeat this process until you have four cores.

Next, using the millimeter side of a rule and a razor blade, cut each core so it's length is 40 mm. At that point, you get a piece of aluminum foil and frame it into a "boat" type shape. Place your aluminum boat on the scale. Measure the weight of your boat to the nearest tenth of a gram using a side arm balance scale. Place the core in the aluminum foil boat. Weigh the boat with the potato in it to the nearest tenth of a gram. Then subtract the weight of the aluminum boat by itself from the weight of the aluminum boat with the potato in it. This will give you the weight of the potato core. Record data in your day one chart. Repeat this step for the other three potato cores. Then, measure the diameter of each potato core in millimeters. After you have found the length, diameter and the weight, you may then calculate the volume of each core using the formula V=(3.14)(r2)(h).

Using a probe, place the potato core designated for the 0% solution into the test tube. Repeat this process for the cores of the 10% and 20% solutions. Then, using a 10 ml graduated cylinder, cover the core with 7 ml of 0% solution. Next, using a different 10 ml graduated cylinder, measure 7 ml of 10% solution and pour it into the test tube labeled 10%. Repeat this step again for the 20% solution. Place the three test tubes on the test tube rack and your aluminum boat with your potato core into the drying oven so it can dehydrate over a period of 72 hours.

Day Two Procedure: Collect your dehydrated potato core, and the three test tubes as well as a scale, three graduated cylinders and probe. Using a side arm balance scale, weigh your dehydrated core. After that, take the 0% solution and empty all the solution out into the first graduated cylinder. This will determine whether the core took in any water or released any water. Using a probe, carefully remove the potato core from the test tube. Place the core on the aluminum foil boat and onto the scale. Weigh the boat and core to the nearest gram. Then subtract the weight of the boat from the total weight of the boat and the core. This gives you the new weight of the core. Then, measure the core in millimeters. Next, measure the diameter of the core. Finally calculate the volume of the core. Repeat these steps for the 10% and 20% cores. Record results on your day two table.

The results from day one and day two will allow you to find the difference between the two days. Calculate the differences and place the information on a chart.


When determining if osmosis could be demonstrated through the affect of varying sodium chloride (NaCI) solutions on the physical characteristics of a potato core, three different variations in solutions were used, and the process was preformed over a course of 72 hours. Results show that there was a significant increase in size in the 0% solution core, having gained .1 gram in weight, 2 mm in length and .5 mm in diameter. The increasing size of the potato core in the 0% solution was most obvious, since the core was evidently hypertonic to the solution. This is because the 0% solution was known to be most 100% H20. On the other hand, the cores that were put into the 10% and 20% solutions do not show a substantial difference in any of their sizes. The data shows that overall the 0% solution increased in size, and the 10% and 20% solutions, decreased. Contrary to my hypothesis, both the 10% and 20% cores lost the same amount in size. By this I mean instead of the 20% core being even smaller than the 10% core, they are the same. Overall, my data does support my hypothesis though. For example, my hypothesis stated that the core in the 0% solution would increase in size. My hypothesis also stated that the cores in the 10% and 20% solution would decrease in size, which they did. Although the 20% did not decrease in size even more than the 10%, one can see how both the cores in the 20% solution and the 10% solution decreased in size compared to the core in 0% solution.


At the end of the lab, the class compared their results with one another. After comparing class members data, the class as a whole was not able to determine which set of data was right, if any. There could be many reasons why the results came back inconclusive when compared to others. One must consider the fact that on day two, there was no knowledge of whether or not it was the same scale used the day before. Thus, making the investigation possibly inaccurate. After all, when it comes to measurement, every tenth of a gram counts. Also, the instruments used in the measurements were very crude, and slight differences could have occurred because of the human margin of error.

Throughout the experiment, data was collected. This data was able to show the affects of osmosis on potato cores. The results collected in this experiment were able to conclude that more water was gained in the core with 0% solution because of the process of diffusion. Since the salt (solute) in the solution changed the concentration of the water molecules, the amount of water in the 10% and 20% cored decreased approximately the same amount. Because of this, logic suggests that no matter how much NaCI was put into the solution, the results would come out the same or almost the same as a result of diffusion. In order to have a more scientific study on the affects of osmosis on potato cores, it would be best to have the investigations preformed over a longer period of time, as well as have peer evaluation in order to compare your results.