Section B – C: The concentration of water molecules is lower outside the cells, therefore the water molecules diffuse from a higher to a lower concentration from the large central vacuole and through the partially permeable cell membrane and cell wall out of the cell. There is a concentration gradient between the cells and the solution. Water diffuses out of the cells, which causes a loss in mass of the potato piece.
Section C – D: The concentration of water molecules is lower outside the cells; therefore the water molecules diffuse from the large central vacuole, through the partially permeable cell membrane and cell wall out and cells. As the concentration increases, the percentage mass loss still decreases, but by a smaller amount: the graph therefore begins to curve.
Section D – E: The concentration of water molecules is still lower than the outside, however, the amount of mass loss no longer increases as the cells have lost all their water and become plasmolysed. There are no water molecules remaining for osmosis to take place, therefore osmosis stops. This is shown as the graph becomes level. The solution will travel through the fully permeable cell wall and surround the cell membrane.
Preliminary Plan
To determine:
- Range of concentrations
- The time for which the potato pieces shall be immersed.
A preliminary experiment shall be carried out.
Preliminary Plan
Apparatus
Diagram
Method
- Extract potato cores from potato using size 9 cork borer. This ensures that the curved surface area is constant.
- Divide the potato core into 2cm pieces using a ruler and cut with a scalpel, to ensure that the surface areas are the same. Remove any potato skin, as this will effect osmosis.
- Measure the mass of each piece, making sure that the balance is tared before each measurement.
-
Measure the water required for each solution to the bottom of the meniscus using a 30cm3 measuring cylinder. This is the measuring cylinder closest to the volume of water that will be measured; therefore results will be more accurate.
-
Measure the 1.0M salt solution required for each solution to the bottom of the meniscus in a 10cm 3 measuring syringe. The total volume of the salt solution and water will be 30 cm3 so the pressure on each potato piece will be constant. The concentrations to be used are:
- 0.0M solution
- 0.25M solution
- 0.5M solution
- 0.75M solution
- 1.0M solution
- Place potato pieces into large boiling tubes.
- Leave samples for 24 hours.
- Remove potato pieces from boiling tube and measure mass again. Remove excess water by rolling it onto a paper towel. The water clinging to the potato would increase the mass and alter results. However, the potato should not be squeezed, as this will remove water from the cells of the potato, and cause a lower mass to be recorded.
- Calculate the percentage mass change. The percentage change of each piece of potato will be measure as each piece has a different mass. If the percentage mass change is found, the resulting figures can be compared and plotted on to a graph. I will calculate the percentage change in mass using this formula:
Mass after – Mass before x 100 = Percentage Mass Change
Mass before
- Repeat the experiment three times to produce a more accurate average value to plot onto the graph.
Preliminary Results
Improvements
From my preliminary experiment, I have decided to make the following changes.
- To prevent excess water loss from the potato pieces, I will prepare the solutions before extracting the potato core. If this is not done, the potato will begin to dry out, therefore experiment results will not be accurate.
- The potato pieces should be prepared, i.e. measuring and weighing in consistent order so that each piece spends the same amount of time in similar conditions.
-
As a 30cm3 measuring cylinder was not available and the next size of 50cm3 was too large for accurate measuring of the volumes I require, if the volume exceeds volume cylinder of 25cm3, I will measure the difference using a measuring syringe. I will also use a measuring cylinder as well as a measuring syringe to measure the salt solution. This is because the measuring syringe only has a maximum value of 10cm3, so multiple measurings for the larger volumes will decrease accuracy. This will increase the accuracy of measuring the salt solution and water.
The concentrations chosen for this experiment were appropriate. The time period that was chosen is also suitable and produced measurable results.
Considering the above changes, the method is now as follows:
Final Plan
Apparatus
- Balance 2dp
- Boiling tubes x5
- Potato
- Distilled water
- Cork borer
- Scalpel
- Ruler
-
10cm3 Measuring syringe
Diagram
Method
-
Measure the water required for each solution to the bottom of the meniscus using a 30cm3 measuring cylinder. This is the measuring cylinder closest to the volume of water that will be measured; therefore results will be more accurate.
-
Measure the 1.0M salt solution required for each solution to the bottom of the meniscus in a 10cm 3 measuring syringe. The total volume of the salt solution and water will be 30 cm3 so the pressure on each potato piece will be constant. The concentrations to be used are:
- 0.0M solution
- 0.25M solution
- 0.5M solution
- 0.75M solution
- 1.0M solution
- Extract potato cores from potato using size 9 cork borer. This ensures that the curved surface area is constant.
- Divide the potato core into 2cm pieces using a ruler and cut with a scalpel, to ensure that the surface areas are the same. Remove any potato skin, as this will effect osmosis.
- Measure the mass of each piece; making sure that the balance is tared before each measurement.
- Place potato pieces into large boiling tubes containing appropriate solution.
- Leave samples for 24 hours.
- Remove potato pieces from boiling tube and measure mass again. Remove excess water by rolling it onto a paper towel. The water clinging to the potato would increase the mass and alter results. However, the potato should not be squeezed, as this will remove water from the cells of the potato, and cause a lower mass to be recorded.
- Calculate the percentage mass change. The percentage change of each piece of potato will be measure as each piece has a different mass. If the percentage change is measured, the resulting figures can be compared and plotted on to a graph. I will calculate the percentage change in mass using this formula:
Mass after – Mass before x 100 = Percentage change in mass
Mass before
- Repeat the experiment three times to produce a more accurate average value to plot onto the graph.
Results – repeat 1
Results – repeat 2
Results – repeat 3
Analysis
This is what the graph of Average Percentage Mass Change against Concentration shows.
Section A - B: At A, when the x-axis = 0, the salt solution is 0M. The concentration of water molecules is greater outside the cells; therefore net osmosis takes place through the partially permeable cell membrane into the cells. Water moves through the fully permeable cell walls, diffuses through the partially permeable cell membrane and through the cytoplasm and into the large central vacuole. This causes the mass of the potato to increase by an average of 12.6%. From A to B, osmosis still takes place into the cells, but as the concentration increases, the average percentage mass change increases by less. For example, when the concentration is 0.0625M, the mass increases by less by 7.5%.
B is the isotonic point that occurs when the concentration is 0.15M; this is when the concentration of water molecules inside and outside the cells is equal. The concentration gradient is therefore zero, so net osmosis does not take place. This means that there is no change in the mass of the potato.
Section B – C: After point B, the concentration of water molecules becomes lower outside the cells, therefore net osmosis takes place out of the cells. Water moves from the large central vacuole, diffuses through the cytoplasm, through the partially permeable cell membrane and through the fully permeable cell wall. This causes the mass of the potato to decrease. At C the mass decreases by an average of 19% when the concentration is 0.39M.
Section C – D: Net osmosis still occurs out the cells, but as the concentration increases, the percentage mass loss decreases but by a smaller amount This is shown as the line begins to level off. The cells have lost enough water for the cells to begin to plasmolyse.
Section D – E: Although the concentration of water molecules is still higher than inside the cells, there is no more water inside the vacuole to diffuse out. The cells are completely plasmolysed, so the solution moves through the fully permeable cell wall and surrounds the cell membrane. This causes a slight increase in mass. The mass loss has reached its maximum value; therefore the mass loss becomes constant. This is shown as the graph becomes level at an average percentage mass loss of –25.7.
This graph does agree with the graph the in the prediction, as the shape of the line is constant with my prediction.
Evaluation
The results obtained for the investigation appear to be fairly accurate as the points are quite close to my line of best fit. Accuracy is supported as even before the average is found, it can be seen that the repeats are close together. However, as the concentration increases, the repeats become less accurate. As shown on the graph, the points for the higher concentrations fluctuate more.
The investigation however is not completely fair. Fluctuations in the points could be because the solutes in the potato will not be completely equal. For example, around the potato eyes where shoots will grow, the concentration of solutes may be higher. This would increase osmosis into the cells. This variation could be improved by using smaller lengths of potato so all the samples could be obtained from a single potato core. Further inaccuracies could be caused by errors in measuring the water and salt solution. This would cause more or less osmosis to take place than would occur. This could be improved by using a burette to measure the liquids. Also the way in which the potatoes were dried after being submerged differed. Some may have been dried more thoroughly than others this would have caused results to be higher or lower than they otherwise would be. Also the excess water was not removed as the potato was cut. This problem could be improved by rolling the potato under a tile of certain mass. This way, the same force would be applied to each piece.
There are no anomalous results obtained in the investigation.
This evidence is not sufficient to draw a firm conclusion as more than five points are needed to draw an accurate best fit curve. More data is needed to support the curve. To provide additional data to determine that the shape of the graph and increase reliability, the experiment could be repeated but using concentrations of smaller intervals, for example testing concentrations of 0.125M, 0.25M, 0.375M, 0.5M, 0.625M, 0.75M, 0.875M, 1.0M. This would provide more points on which to base the line of my graph. Also, to prove the graph is correct, the isotonic point could be tested. By using, according to the graph, the isotonic point of 0.15M there should be no change in the mass of a potato piece. However, the isotonic point would vary from potato to potato, so the piece used to test the isotonic point should be from the same potato tuber.
