Size of the Molecules: When a molecule is heavier, then it will obviously take a greater deal of time to diffuse across the semi permeable membrane as the molecules speed is decreased. This is also why the solute molecules from a hypertonic solution do not move into a hypotonic solution. In cells, the cell membrane is semi permeable. However, in experiments, the membrane may longer act as a sieve and the solute molecules can diffuse across the membrane. However, since the solute molecules attract water molecules, which form clusters around them, this causes the solute molecules to become heavier. Thus, they diffuse more slowly than the free water molecules in the hypotonic solution.
Permeability of the Membrane: In osmosis, the water molecules will move across the semi permeable membrane. This membrane in cells is generally the same size. However, plant cells have the presence of the cell wall, which contributes to the thickness of the cell membrane. Therefore, since the membrane is thicker, the rate of diffusion is greater as the water molecules must move over a greater distance. As a result, the rate of osmosis is decreased.
Apparatus
Before carrying out the procedure, I will first gather the required apparatus in order for my procedure to be performed properly. The necessary apparatus is listed below.
- I will select two No.2 cork borers and two No.3 potato cylinder plungers. These are required to form the potato tubors that are to be immersed in the different solutions.
- I will use a single fresh potato, where the cork borers will be inserted.
- I will use six petri dishes, each of the same size to store the different concentrations of solution. It is here where the potato tubors will be placed.
- I will use six beakers to store the different concentrations of solution.
- I will keep ready six measuring tubes, which I will use to measure the exact volume of each solution.
- I will use labels and place them over the beakers and petri dishes so that I will be able to distinguish between them and determine which solution is in which beaker and petri dish.
- I will use a razor blade to cut the skin of the potato tubors.
- I will use graph paper in order to measure the exact length of the potato tubors.
- I will use a weight scale to record the mass of the potato tubors.
- I will use distilled water as the hypotonic solution, which will be stored in its corresponding beaker.
- I will use five different molarities of sucrose solution, each stored in its corresponding beaker. The molarities that I will use are 0.125M, 0.25M, 0.5M, 0.75M and 1M.
Precautions
- I will make sure that I will use only one potato in order to form the potato tubors.
- I will make sure that I will use the same No. cork borer and plunger so that the size in potato tubors does not vary.
- When inserting the cork borer into the potato, I will place the potato on a hard board.
- In order to eliminate error, I will confirm the length of each potato tubor as four centimeters.
- I will make sure that each potato tubor is a proper and regular cylinder. Otherwise, the results of the experiment may be altered.
- I will make sure to remove the skin of the potato tubors so that it does not affect the result of the experiment.
- I will label each petri dish and beaker noting the concentration of solution that is to be stored in it.
- I will wash each beaker and petri dish properly to remove any traces of previous solutions.
- When noting the mass of the potato tubors, I will take into account the mass of the filter paper.
- I will use thongs to place the potato tubors onto the filter paper before taking the mass. I will also use thongs and refrain from touching the tubors when placing them in the petri dishes.
- I will make sure that I will pour an equal amount of solution for each beaker and then petri dish by taking the lower meniscus of the reading in the measuring tube.
- I will pour the correct concentration of solution in to its corresponding beaker and petri dish.
- I will take the exact time when I poured the solution in to its corresponding petri dish and close the petri dish.
- After twenty-four hours, when taking my readings of mass for the potato tubors, I will make sure to remove any excess water by rolling them over the filter paper using thongs.
Procedure
A cork borer and plunger used to make potato A weight scale.
cylinders.
I will now carry out my procedure using the apparatus that I have gathered. The steps of the procedure are listed below.
- I will first rinse and clean the petri dishes and beakers to remove any droplets of previous solutions.
- I will then label the petri dishes, noting the molarity of the solution that will be stored in it. Therefore, the concentrations should read ‘Water’, 0.125M, 0.25M, 0.5M, 0.75M, and 1M.
- I will label the six beakers, noting the concentration of solution to be stored inside.
- I will then measure the volume of the solution inside the beaker by using a measuring jar. I will use a volume of 30mL for each solution.
- I will then wash the fresh potato and place it on a piece of paper. I will then use the No.2 Cork borer and No 3. cylinder plunger to form potato tubors.
- I will plunge twenty-four tubors, four for each petri dish.
- I will then place the potato tubors onto a graph paper.
- I will use a washed razor blade to cut the potato tubors to exactly four centimeters. I will remove the potato skin in the process.
- I will now place each potato tubor onto filter paper and measure its mass on the weight scale. I will record these readings.
- I will now place four potato tubors into each petri dish.
- I will then pour 30mL of each solution into its corresponding petri dish, noting the time when the solution is poured.
- I will close each petri dish and keep them aside for twenty-four hours.
- After a day, I will check the potato tubors and note its appearance; if there are any differences.
- I will then note the mass of each potato tubor and will take the average.
- I will also take the length of each potato tubor and note the difference in length.
Safety
My experiment must also be conducted in such a manner to eliminate injuries. This especially occurs when handling the razor blade and forming the potato cylinders. So,
- I will place the potato on a platform or board and not on my hand when inserting the cork borer inside; if I do not, then the cork borer may cut through the hand.
- When cutting the potato tubors to equal length of 4 centimeters, I will make sure to hold it from the blunt end and be careful not to cut my fingers.
In my prior test, I have taken a single sucrose solution and distilled water to investigate how it alters the mass and length of the potato tubors. Therefore, my readings below are from my prior test and are only based on the effect of sucrose solution of concentration 0.125M and distilled water. My observations can be tabulated as follows:
In the following table, I have taken the average length and mass values in order to be more practical. I have found the average by adding the values together and dividing by the number of cylinders (4).
Now, using this information, I can now predict a graph of Molarity against Percentage Difference. In order to determine the percentage difference in mass and length, I will use the following formula.
Percentage Difference for length:
(Final Length – Initial Length) x 100
Initial Length
Percentage Difference for mass:
(Final Mass – Initial Mass) x 100
Initial Mass
After noting the percentage difference for mass and length, I can now predict what the graph of molarity against percentage difference should appear as.
From the graph, we see that as the molarity increases, the percentage difference decreases and with very high concentrations becomes a negative value. Therefore, from my predicted graph, I am stating that with concentrations higher than that of the potato cell sap, there will be a negative percentage difference in the length and mass of the potatoes. This supports the process of osmosis, which occurs in the cells; when the external solution is greater, water moves out of the cell. Therefore, there must be a decrease in size. Similarly, when the concentration of sucrose solution is greater, the potato tubors decrease in length and mass.
The percentage difference at first is positive, as the external solution must be less concentrated than the cell sap. Therefore, water moves into the potato cells in order for equalization to occur. This is why the potatoes will increase in length and mass and thus, a positive percentage difference.
When the two lines representing mass and length meet the x-axis, the percentage difference is equal to zero. Therefore, there is no more increase nor is there is a decrease in the potatoes. This is because the external solution’s concentration is equal to that of the cell sap. Therefore, there is no movement of water and so no change in mass or length. This is known as the isotonic solution and should lie between 0.25 and 0.5 M, as most potatoes have a cell sap lying in this range. Since the mass and length percentage differences occur at different concentrations, I will take the average to find the concentration of the cell sap. Therefore, the average isotonic solution is taken as the value of the cell sap’s concentration.
Obtaining Evidence
I have now conducted my experiments acknowledging the necessary safety precautions to take and have utilized the required equipment appropriately; I have placed the potato on a raised platform before forming the potato cylinders preventing the chance of any injury to my hands. Furthermore, I was extremely careful when using the razor blade to cut the potato cylinders to an exact length of four centimeters each. And finally, I have performed my experiment with the utmost care following the procedure step by step.
After performing the experiment, I have noted my observations and compared them to my predictions. I have placed four potato tubors in each petri dish with the respective liquid inside this dish. Therefore, for each concentration of solution, I have taken four readings as I have noted the mass and length readings for each of these tubors. The point of this is to obtain as much data as possible and then compile these readings together and obtain an average of these readings so that my final values are more accurate and reliable. Therefore, the results in my experiment are an average of the readings from the four potato tubors; so my readings are therefore more precise. Furthermore, the concentrations of sucrose solution that I have taken are 0.125M, 0.25M, 0.5M, 0.75M and 1M along with distilled pure water. These concentrations should appropriately reveal the relation that exists between the concentrations of sucrose solution and the amount of water that enters or exits the tubors; that is, it should properly reveal the phenomenon of osmosis. To obtain more accurate results, I repeated the experiment twice so that I get a fair final value.
When analyzing the results, I noticed that there is an obvious pattern that falls into my predictions. I have noticed that when the potato tubors were placed in a concentrated sucrose solution, they decreased in mass and length. In fact, when the sucrose solution was more concentrated, the decrease in mass and length was greater. Therefore, there was a greater percentage difference in mass and length. I have also realized however that with the weakest sucrose solution, there was actually a gain in mass and length in the potato tubors rather than a loss. There was an even greater gain in mass and length when the tubors were placed in distilled water.
For this experiment, I have used an equal volume of solution for each petri dish as 30mL. Therefore, before pouring the required solution into its respective petri dish, I have measured its volume correctly in a measure flask. I have taken more accurate readings and avoided parallax by taking the readings at eye level. I have also taken precautions to measure the length of the potato tubors at an exact four centimeters so as to avoid any inaccuracies in my experiment. Then by placing the tubors in the petri dish, I have poured the respective solution into the dish noting the time until all the tubors are submerged in the solution. I have then closed the dish and set it aside. I have repeated this for the other concentrations of solution.
The data that I have obtained from my experiment is shown below. I have tabulated it to properly reveal the changes in length and mass after the tubors have been placed in their respective solutions for about 24 hours.
I have obtained these readings after 24 hours of allowing the tubors to be placed in their petri dishes. I have then collected then, dried them using filter paper and weighed each individual tubor on a weight scale. A picture of this scale is shown below.
The new mass of each tubor can then be compared with its mass reading before being placed in the solution in the petri dish. Also the tubors will be measured to note its new length. This new length can then be compared with its original length of four centimeters and thus with this information compiled together, I will be able to analyze what has really happened with the help of what I have learnt before on the topic of osmosis.
Again to be noted, I have taken the necessary safety precautions in my experiment to avoid any injury to myself so that the experiment is both safe and practical. Thus, when using the razor blade to cut the tubors to an even length of four centimeters, I have focused and careful not to cut myself. Furthermore, when inserting the cork borer into the potato, I have made sure to place the potato on a raised platform and made sure the potato was not cupped in my hand to prevent any injury to my hand.
Analyzing Evidence
With the data obtained from my experiment, it is now necessary to fully understand them and analyze the results. More than that, it is important to prove why the results are given as so. To do so, it may help to construct a bar graph to clearly present the data obtained from the experiment. The bar graph on the next page clearly reveals the change in length for the potato tubors in the different concentrations of solution; it presents the initial length of the tubors and its final length.
From this graph, it is obvious that with a less concentrated solution, the potato tubor actually gains length. For example, when the potato tubors are placed in distilled water, from four centimeters, they increase in length to 4.43 centimeters, an increase of 43 millimeters. This relies on the fact of osmosis; if a cell is placed in a less concentrated solution, then in order for equalization to occur, that is for the concentration to be equal on both sides, then water must move into the cell so that its cell sap becomes diluted and so that the concentration gradient is eliminated. In this case, since the potato cells inside the tubor obviously have cell sap, a mixture of salts and sugars with water, it forms a weak solution. Even though it is weak, it is still more concentrated than the distilled water. Thus, in order for equalization to occur, some of the distilled water must move across the cell membrane of the potato cell and enter it. In this case, the cells will become bloated or turgid; they become firm and full. Due to this, the potato tubor will increase in length as the potato cells have become firm and turgid; they occupy more space and so the tubor increases in length. In fact, it is seen that the potato tubor increases in length the most when placed in distilled water because the concentration gradient is greater; the difference in concentration between the external and internal solution is greater and so more water enters the potato cells filling its vacuole more rapidly. This can be supported by comparing the change in length of the potato tubors when placed in a 0.125M sucrose solution; when placed in the petri dish containing sucrose solution of this concentration, the potato tubors only increase by an average of 31 millimeters giving a total length of 4.31 centimeters. Since these tubors are shorter than those placed in distilled water, it obviously means that less water has entered the potato cells; nevertheless, water has entered the cells. Water enters the cells in order to cause equalization; in this case, the potato cell sap must be more concentrated than the 0.125M sucrose solution thus water moves into the cell to equalize the concentration of the internal and external solutions. However, since the tubor is not as long as when placed in distilled water, it must mean that the concentration gradient is less; there is a smaller difference in concentration since less water enters the potato tubor. Thus, there is less change in length. Furthermore, when the potato tubors are placed in a 0.25M sucrose solution, the potato tubors increase in length but by an even smaller margin. This must mean that there is an even smaller concentration gradient; the concentration difference between the external solution and the cell sap is smaller even though the cell sap is more concentrated than the 0.25M sucrose solution. Thus, less water must enter the cell in order to cause equalization of the external and internal solutions. And when the concentration is of 0.5M sucrose solution, we see that the potato tubors no longer increase in length but in fact become shorter. This means that the potato cell sap is less concentrated than its external solution; therefore in order for equalization to occur, water must move out of the cells and thus they become flaccid or plasmolyzed. In this case, the cell become shriveled up and so the potato tubor becomes smaller in size, thus the decrease in length. And since there is a sudden change between these concentrations of 0.25M and 0.5M, that is, at 0.25M the potato tubors gained length while at 0.5M the tubors decreased in length, it shows that obviously the isotonic solution exists between these two values; the isotonic solution is where the external solution is equal to the internal solution and is where there is no net flow of water molecules. Thus, the isotonic solution must exist between 0.25M and 0.5M sucrose solutions as at one point there will be no increase in length of the potato tubors. Also from the graph, it is seen that with higher concentrations of sucrose solution such as those of 0.75M and 1M, there is a greater change in length as there is now a greater concentration gradient forms; thus more water must move out of the potato tubors and so more cells become even more flaccid. Thus, with higher concentrations of sucrose solution, the potato tubors become shorter and shorter.
From my experiment, I have also obtained data on the change of mass in the potato tubors when placed in the different concentrations of solution. In order to present this information clearly, I have shown this on a bar graph, which indicates the initial and final mass for the tubors placed in each of these different solutions. This is shown on the next page.
When examining this graph, it is necessary to take into account the scientific backing of osmosis and these results may also be compared with those of the length readings to support what has happened. Now, it is seen that from the bar graph, the least concentrated solution attributes to the greatest increase in mass. That is, when the potato tubors are placed in distilled water, the mass of the potato tubors changes from 1.22 grams to 1.43 grams. This is because since the concentration is greater inside the cell, water must enter the cells in order to cause equalization, thus the cells must become turgid. As they become turgid, it is almost as if the potato tubors have sucked in water and begin to swell up. Since they swell up with water, they will increase in mass. This is the same reason why the potato tubors also increase in length when placed in distilled water. Similarly, when the potato tubors are placed in sucrose solution of concentration 0.125M, they gain mass. In this case, the mass changes from 1.19 grams to 1.40 grams. This is an increase of 0.21 grams, the same increase in mass as those tubors placed in distilled water. This signifies an error in my experiment as generally, when the tubors are placed in 0.125M sucrose solution, the mass of the tubors should increase but not as much as it increases when placed in distilled water; this is because when placed in 0.125M sucrose solution, there is a smaller concentration gradient as this concentration is closer to the concentration of the potato cell sap and so less water must enter the potato cells in order for equalization to occur. Since however there is the same increase in mass, there must have been an error, for example, perhaps not all the factors affecting osmosis were kept constant. For example, perhaps the temperature was not kept constant as so perhaps while the petri dishes were kept aside for 24 hours, they were alterations in the temperature and so the process of osmosis took place at a faster rate and so perhaps there was a greater change in mass than expected. From the graph, we also see that when the concentration is 0.25M, the mass increases but now by a smaller margin, from 1.18 grams to 1.38 grams. This is because now since the concentration of the sucrose solution is nearing that of the concentration of the potato cell sap, less water is required to enter the potato cell’s vacuole to cause equalization of the concentrations of the internal and external solutions and so there will be a smaller increase in mass due to the fact that the tubor has not taken in as much water as before.
When comparing the length readings of the potato tubors, I have noticed that when taking into account the concentrations of 0.25M and 0.5M sucrose solutions, there was a sudden change in the change of length of the potato tubors; when placed in a solution of concentration 0.25M, the potato tubors increased in length while when placed in a solution of concentration 0.5M, the tubors decreased in length. This means that between these concentrations must lie the isotonic solution, a solution that has a concentration equal to the potato cell sap; in this case, there will be no net flow of water molecules and thus the potato tubor will neither increase or decrease in length. This can be supported now by examining the change in mass with different solutions. When placed in a solution of concentration 0.25M, the potato tubors increase in length. However, when placed in a solution of concentration 0.5M, the potato tubors decrease in mass, from 1.17 grams to 1.08 grams. This must mean that the concentration of 0.5M is greater than the concentration of the potato cell sap so water must move out of the cell in order for equalization to occur. However, since when placed in 0.25M, the potato tubors increase in mass, the isotonic solution must therefore lie between 0.25M and 0.5M sucrose solutions. Therefore, this is further proof to this statement mentioned above.
Furthermore, when referring to the graph, it is seen that when the potato tubors are placed in solutions of greater concentration such as 0.75M and 1M, the increase in mass is greater. This is because with an increase in the external solution’s concentration, there is a greater difference in the concentration between the internal and external solutions and thus a greater amount of water must leave the potato cells in order for equalization to occur. Therefore, more cells will lose water and become flaccid. Therefore, overall the potato tubors will decrease in its mass by a greater amount. This further supports the results in my length readings as they fall into the same pattern.
Therefore, an overall statement may be made about the tubors when placed in these concentrations. When placed in distilled water, the potato tubors will increase in both its mass and length by the greatest amount. When placed in a sucrose solution of concentration 0.25M, the potato tubors will increase in both mass and length but when placed in a sucrose solution of concentration 0.5M, the tubors decrease in its mass and length. This indicates that the isotonic solution lies between concentrations of 0.25M and 0.5M, as at a certain concentration between these, the potato tubors will neither increase nor decrease in its length or mass. And finally when the potato tubors are placed in a sucrose solution of higher concentrations such as 0.75M and 1M, they will both decrease in mass and length, decreasing the most when placed in a sucrose solution of concentration 1M as then the concentration gradient is greater and so more water must escape the tubors in order for equalization to occur.
These results may also be presented in terms of the percentage difference in mass and length. First, I will consider the length changes in the potato tubors when placed in solutions of different concentrations. The percentage difference will signify the change in the length and thus clearly shows how this difference changes with a change in concentration. When presented on a graph, it will also be able to present whether there is a positive or negative percentage difference in length. However, before presenting the results on a graph, it is necessary to use the percentage difference formula to calculate the percentage difference in length of the potato tubors when placed in these different concentrations.
To find the percentage difference in length, we must use the following formula:
Therefore, using the data compiled from my experiment, I will be able to find out the percentage difference in length of the tubors placed in solutions of different concentrations.
Thus, when the potato tubors are placed in distilled water,
4.43 – 4.00 x 100
4.00
= 10.75% is the percentage difference in length.
When the potato tubors are placed in sucrose solution of concentration 0.125M,
4.31 – 4.00 x 100
4.00
= 7.75% is the percentage difference in length.
When the potato tubors are placed in sucrose solution of concentration 0.25M,
4.24 – 4.00 x 100
4.00
= 6.00% is the percentage difference in length.
When the potato tubors are placed in sucrose solution of concentration 0.5M,
3.78 – 4.00 x 100
4.00
= -5.50% is the percentage difference in length, indicating a negative percentage difference; this means that the potato tubors have decreased in length.
When the potato tubors are placed in sucrose solution of concentration 0.75M,
3.46 – 4.00 x 100
4.00
= -13.50% is the percentage difference in length, indicating a negative percentage difference; this means that the potato tubors have decreased in length.
And finally when the potato tubors are placed in sucrose solution of concentration 1M,
3.26 – 4.00 x 100
4.00
= -18.50% is the percentage difference in length, indicating a negative percentage difference; this means that the potato tubors have decreased in length.
This information may now be tabulated and then presented on a graph clearly showing the change in percentage difference with a change in the concentration of solution.
Since not all the points fall into a straight line, I have decided to take a line of best fit for more accurate results and to present the graph in the form of a straight line so that it clearly shows the relation existing between the percentage difference in length to the concentration. The graph is shown on the next page and it shows at when the potato tubors are placed in solutions of distilled water to a sucrose solution of concentration before 0.4375M, there is a positive percentage difference in length. This means that in this range, the potato tubors will in fact increase in length. Nevertheless, the increase in length is the greatest when the tubors are submerged in distilled water, as only then is there the greatest concentration gradient. Furthermore, from the graph it is obvious that the graph crosses the x-axis at a concentration of 0.4375M. This means that at this concentration, there is no longer an increase in length and thus this concentration must be the isotonic solution; the isotonic solution is where the concentration of the external solution is equal to the concentration of the internal solution and thus there will no longer be any net flow of water. Thus, there will not be any increase in length. So, in this case, since at this concentration, the percentage difference is 0, it means that this concentration is equal to the concentration of the potato cell sap and thus acts as the isotonic solution. We also see from the graph that when the potato tubors are placed in solutions of concentration above 0.4375M to 1M, there is a negative difference in length. This means that the potato tubors will decrease in length. Nevertheless, the decrease in length is the greatest when the concentration of sucrose solution is 1M as then the concentration gradient is the most so there is a greater amount of water required to escape the potato cells. This is seen from the graph, as at 1M, there is the greatest percentage difference of –18.75%.
A graph may also be constructed to present the percentage difference in mass. However, before doing so, it is necessary to make the necessary calculations to determine the percentage difference in mass at the different concentrations of solution. This can be accomplished using the formula shown below.
Thus, using this formula I may now determine the percentage difference in mass at the different concentrations of solution.
When the potato tubors are placed in distilled water,
1.43 – 1.22 x 100
1.22
= 17.21% is the percentage difference in mass.
When the potato tubors are placed in a 0.125M sucrose solution,
1.40 – 1.19 x 100
1.19
= 17.65% is the percentage difference in mass. (This is an obvious anomalous reading in my experiment as it would be expected that the potato tubors placed in this concentration would have a smaller percentage difference as compared to those tubors placed in distilled water as in the case of distilled water, there is a greater concentration gradient and so more water would be expected to enter the potato tubors than compared to that entering the tubors placed in 0.125M sucrose solution. Thus, there must have been an error in my experiment such as an alteration in the temperature or perhaps a sudden change in the permeability of the cell membrane of the potato cells.)
When the potato tubors are placed in a 0.25M sucrose solution,
1.38 –1.18 x 100
1.18
= 16.95% is the percentage difference in mass.
When the potato tubors are placed in a 0.5M sucrose solution,
1.08 – 1.17 x 100
1.17
= -7.69% is the percentage difference in mass. The negative sign indicates a negative percentage difference, which means at this concentration the potato tubors decrease in their mass.
When the potato tubors are placed in a 0.75M sucrose solution,
0.93 – 1.21 x 100
1.21
= -23.14% is the percentage difference in mass. The negative sign indicates a negative percentage difference, which means at this concentration the potato tubors decrease in their mass.
When the potato tubors are placed in a 1M sucrose solution,
0.87 – 1.21 x 100
1.21
= -28.10% is the percentage difference in mass. The negative sign indicates a negative percentage difference, which means at this concentration the potato tubors decrease in their mass.
Before presenting this information on a graph, I may now tabulate the data I have obtained after using the original data from my experiment.
With this information obtained, I may now clearly represent this on a graph to present carefully the trend that exists between these different concentrations and what its relation is to the percentage difference in mass. This graph is shown on the next page.
Since not all the points fall into a straight line, I have decided to take a line of best fit for more accurate results and to present the graph in the form of a straight line so that it clearly shows the relation existing between the percentage difference in mass to the concentration. From this graph, it is quite obvious that when the potato tubors are placed in distilled water to a sucrose solution of concentration just less than 0.4825M, there is a positive percentage difference in mass, indicating that the potato tubors will increase in mass. This is due to the fact that the potato cell sap is more concentrated than the surrounding solution and thus in order for equalization to occur water moves into the cells. Thus, there is an increase in mass. Nevertheless, only those potato tubors placed in distilled water have the greatest percentage difference, as there is the greatest concentration gradient; there is a greater difference in concentration comparing the external solution to the potato cell sap. Thus, more water is required to move into the cells for equalization and thus, there is a greater percentage difference. Now, the graph crosses the x-axis at 0.4825M. This means that at this concentration, there is neither an increase nor decrease in mass. Thus, this must be the isotonic solution for the potato, where its potato cell sap is equal in concentration to the surrounding solution. Therefore, since at this point there is no percentage difference in mass of the potato tubors, 0.4825M must be the concentration of the potato cell sap. Also, when the potato tubors are placed in a sucrose solution of concentration just above 0.4825M to a sucrose solution of concentration 1M, there is a negative percentage difference, which means that the potato tubors are decreasing in mass. This is because the external solution is more concentration than the internal solution, so water moves out of the potato cells in order for equalization to occur. Thus, there is a loss in mass and so a negative percentage difference. Nevertheless, when the concentration of the sucrose solution is 1M, there is the greatest negative percentage difference as the potato tubors decrease in mass the most; this is because since there is a greater difference in concentration between the internal and external solutions, more water is required to escape the potato cells in the tubor and thus there is a greater loss in mass.
These two graphs may be superimposed on one. This is shown on the graph on the next page. In this case, both graphs cross the x-axis at different points, indicating that the potatoes have different cell sap concentrations. Nevertheless, the points at which the graphs cross the x-axis is relatively the same and so it is necessary to take an average of these points to find the average isotonic solution. This is done as follows. The point at which the graph indicating the percentage difference in mass crosses the x-axis is 0.4825M and the point at which the graph indicating the percentage difference in mass crosses the x-axis is 0.4375M. By taking these two values into account, I may now found a reliable and more accurate value for the concentration of the potato cell sap.
0.4825 + 0.4375
2
= 0.46M
Thus, the average isotonic solution is 0.46M and so the average concentration of the potato cell sap was 0.46M.
Therefore, I have now satisfied the purpose of my experiment and have found the concentration of the potato cell sap, 0.46M, relying on the process of osmosis.
Evaluation
I have now completed my experiment in its entirety and have determined the average concentration of the potato cell sap used in my experiment to be 0.46M. I believe this value to be rather reliable as my method itself is both practical and suitable to determine the concentration of the solution. It is a reliable method as I have taken into account the original masses and lengths of the potato tubors and compares them with their new masses and lengths appropriately. It also allows the potato tubors to be submerged fully in their respective solutions for up to 24 hours so that enough time is given for the process of osmosis to occur. Nevertheless, there are several improvements that could be suggested in order to make the method even more reliable. However, before suggesting any improvements to the methods, it is necessary to pinpoint certain errors in the experiment.
One blatant error in my experiment is when taking the mass of the potato tubor placed in the petri dish containing a sucrose solution of concentration 0.125M. In this case, the potato tubors had a greater percentage difference than that of distilled water, which should generally not be the case. Thus, this error could have been for several reasons, one being that the other factors affecting osmosis were not kept constant. Such factors may be the temperature; with a greater temperature the water molecules would move across the cell membrane of the cell more quickly and thus the rate at which osmosis occurs is increased. With a decrease in temperature, then the rate of osmosis also decreases and so perhaps the temperature was not kept at a constant during the entire experiment. Furthermore, perhaps the tubors used were irregular, that is, perhaps they were not perfect cylinders. In this case, the amount of water that enters may differ and thus the total mass of the cylinders will differ. Thus, there are bound to be some errors. Other errors may be that perhaps the same potato was not used and thus this would contribute to an error as the cell sap concentration differs between different potatoes. Another common error is taking the wrong amount of solution in the measuring jar; that is falling into the common parallax error and taking more than or less than 30mL of solution. These are the most noticeable errors.
I may suggest how my experiment could have been improved. The first way to improve my experiment would be to take into account these noticeable errors and overcome them to make my experiment more reliable. Now, in order to keep the temperature constant, I will keep the petri dishes containing the tubors inside a stored area, where the temperature is kept constant always; this will prevent the chance of any change in the rate of osmosis and thus will prevent any inaccuracies in the mass and length readings. Furthermore, in order to prevent the tubors from being irregular, I will use only those potato tubors that are almost a perfect cylinder. In this case, the readings in mass and length are more accurate. Another common error to overcome is that of parallax; when using the measuring jar to confirm 30mL of solution to be used, I will make sure to take this reading at eye level so that I am actually using an exact volume of 30mL exactly. Also the parallax error can be avoided when cutting the potato tubors to an exact four centimeters; its length should be noted at eye level to confirm the length of the tubor is four centimeters. And finally, in order to prevent the concentration of potato cell sap from altering, I will use the same potato.
Though my experiment is suitable and appropriate, it can be improved further more my using a different range of concentrations of sucrose solution to better reveal the change in mass and length of the potato tubors when placed in these solutions but more importantly to reveal the exact concentration of the potato cell sap. Thus, in this case, it may perhaps be better to use a range of concentrations between 0.25M and 0.5M. Thus, this will help to narrow down the exact concentration of the cell sap. The experiment may also been carried out using an egg rather than potato tubors or perhaps another substance could have been used to more appropriately show what is happening when the substance is placed in different sucrose solutions.
Even though, in conclusion, I have found out that when a substance is placed in a more concentrated solution, it loses water in order for equalization to occur and thus in the case of the potato tubors, they will decrease in mass and length. Furthermore, when placed in a less concentration solution, the tubors will gain water in order for equalization to occur and so they will increase in mass and length. And also from my experiment, I have determined the potato cell sap concentration to be 0.46M.