For my final experiment I will:
- Take more readings at more frequent intervals. Every 30minutes for 4 hours.
- Measure my potatoes in a more accurate way
- Use a larger gradient of solution concentration ranging from 0M(Distilled water), 0.2M, 0.4M, 0.6M, 0.8M, 1M, 1.2M, 1.4M, 1.6M, 1.8M 2M. The Distilled water will act as a control to compare the results with.
- Use more potato chips for each concentration. E.g. 3chips per concentration.
- Measure my chips by mass instead of length, as it is more accurate.
- Use a cork borer to cut the chips to ensure that the chips are the same size and have the same dimensions.
- Remove excess water/solution by blotting the potato chip on a paper towel before weighing it.
- I will cover the solution with cling film to prevent a loss of solution by evaporation, as it will change the concentration.
- To be able to fit more potato chips in the solution I will use a beaker as if I used a test tube they would get stuck or not fit.
- Be careful to not cut myself using the cork borer.
Apparatus
- Sodium Chloride Solutions(0M, 0.2M, 0.4M, 0.6M, 0.8M, 1M, 1.2M, 1.4M, 1.6M, 1.8M, 2M)
- 11 Beakers
- Cork Borer
- Graduated Glass Pipette & Pi-Pump
- Labels & Markers
- Scales
- Volumetric Flask
- Potatoes
Method
- Cutting 33 potato chips using a single cork borer. This means that there would be 3 chips per concentration, which have been accurately measured.
- To ensure that the potato chips do not lose any valuable water , we cover them in damp tissue.
- Record the mass of each chip and ensure that they’re within a similar range. Alter the size of any which have a great variation.
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Fill each beaker with 150cm³of solution. Ensure that there is 150cm³in each beaker. After filling each beaker put a label onto each beaker stating how many Mols are in the beaker.
- Place 3 chips in every beaker
- Leave the potato chips in the beakers for 30minutes at a time. Then return and take the chips out of the beaker. When doing so blot them on to a paper towel to remove the excess liquid.
- Record the mass of each chip.
- Repeat the last two instructions every 30minutes until 4hours have passed.
Creating the stock solution.
Calculate the RAM of NaCl : (23+35.5) = 58.5 g
1-Accurately weigh out 58.5g of NaCl
2-Using the glass rod stir in 58.5g of NaCl into 150cm of Distilled Water until the crystals are dissolved.
3-Put the solution in a volumetric funnel
4-Wash the beaker at least 3 times using distilled water to remove any traces of sodium Chloride solution.
5-Add the rest of the solution using the glass pipette.
Volume of stock solution required = Concentration you want
Total volume you want
Diagram of a plant cell
OSMOSIS
Osmosis is the movement of water molecules from a region of low solute concentration (high water concentration) to a region of higher solute concentration (lower water concentration) across a partially permeable membrane until equilibrium is reached on each side of the membrane. If the solutions on either side of the partially permeable membrane are equally concentrated there will be no movement across the membrane. This is what is called an equilibrium state and the solutions are referred to as isotonic.
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Partially permeable membranes are thin sheets of material with give the ability to certain things to pass through them yet prevent other things from passing through. In this case the Cell Membrane will act as the partially permeable membrane.
- Cell membranes permit molecules such as Carbon Dioxide, Oxygen, water, Ammonia, Glucose and amino acids to pass through. Yet don’t permit Starch, or protein to pass through, as the molecules are too big.
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A high concentration of water is where there is a lot of water such as in a very dilute solution or in pure water therefore there is a high concentration of water.
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A low concentration of water is a solution; which is concentrated of substance such as sucrose or salt (sodium chloride) with little/no water.
The water (weak solution) diffuses through the small holes in the membrane; which allows smaller molecules to pass through. This is why it is only the weaker solution will passes through as the water molecules are smaller than that of the larger molecules.
TURGIDITY
Turgidity is the effect of plant cells being placed in water. This is due to the cell sap containing a strong solution. This causes the water to pass through into the cell by the process of osmosis. As the water enters the cell it swells up and the water begins to push against the cellulose cell wall. Eventually the cell will be saturated with water and will be unable to withstand any more of the water. This is turgidity.
PLASMOLYSIS
Cells are plasmolysed when the solution outside the cell is more concentrated than the solution in the vacuole. As the water molecules pass out the sap vacuole begins to shrink. The cells are now limp and they become flaccid. As more water leaves the cells the cytoplasm begins to peel away from the cell wall.
WATER POTENTIAL
The likelihood of a solution gaining / losing water molecules from another solution depends on its water potential. A solution that is dilute will have a high quantity of free water molecules therefore having a higher water potential than a concentrated solution.
Pure water has the highest possible water potential because the water molecules are able to flow through to any other aqueous solution regardless of how dilute it is. If the water potential is greater inside of the cell than outside of the cell, then there will be a net movement of water out of the cell. If the water potential is greater outside the cell than inside the cell then osmosis will be a spontaneous movement into the cell.
Distilled Water/0Mols
The potato chips increased in size by 0.35g. This is due to the lack of concentration of solutes in the solution of 0Mols. The water moved into the potato cells increasing the amount of sell sap. Therefore increasing the mass of the potato chip.
0.2Mols
The potato chips in the solution of 0.2Mols decreased in mass. There was a decrease of 0.1g, this could mean either two things. The first thing is the most scientifically explainable reason. The cell sap solution was more concentrated than that in the solution surrounding the potato chip. As concentration increases, the mass decreases. For that reason the particles in the potato chip moved out of the cells causing a loss in mass. The other thing is that maybe the readings I took were inaccurate especially since during the course of the day the mass of the potato kept increasing and decreasing.
0.4Mols
The potato chips in the solution of 0.4Mols have significantly decreased in mass by 0.54g. This concludes that the concentration within the potato cell is a great deal higher than the solution surrounding it in the beaker. There is also a clear decrease in mass, there are no chips or readings whereby the mass increases.
0.6Mols
The potato chips in the solution of 0.6Mols decrease by 0.54g in mass, although this is the same as the chips in the solution of 0.4Mols, here we find that within the first 30minutes the chips decrease in mass by 0.36g; which is a great amount.
0.8Mols
The potato chips in the solution of 0.8Mols decrease in mass by 0.57g. This once again is not a greatly different to the above results but as 0.6Mols the greatest loss of mass is within the first 30minutes where 0.35g is lost. The concentration of solute outside the potato cell is greater than within.
1.0Mols
The potato chips in the solution of 1.0Mols decrease in mass by 0.58g. The greatest period of loss in mass was once again in the first 30minutes where 0.51g was lost. The concentration of water is higher within the cell.
1.2Mols
The potato chips in the solution of 1.2Mols decrease in mass by 0.57g. This may be due to inaccurate readings as this is less than the previous readings at 1.2Mols. Yet once again the greatest mass loss was within the first 30minutes and 0.48g was lost.
1.4Mols
The potato chips in the solution of 1.4Mols decrease in mass by 0.55g.
1.6Mols
The potato chips in the solution of 1.6Mols decrease in mass by 0.56g.
1.8Mols
The potato chips in the solution of 1.8Mols decrease in mass by 0.58g.
2.0Mols
The potato chips in the solutions of 2.0Mols decrease in mass by 0.53g.
Analysing the Results
- Because the highest possible water potential is in water, my results show that the only solution in which the mass increased was in 0M (Distilled Water). This agrees with the theory as all other results show a decrease in mass. As the Molar increased it seemed as though there was negative water potential.
- The lower the Molar concentration of solution the more turgid the cells are. The cells began to swell up with the water from the surrounding solution. The potato chips felt harder in the 0Mols, 0.2Mols and 0.4Mols solutions.
- As the Molar concentration increased the cells began to plasmolyse. This is shown by the loss of mass in the cells. This is distinct in the cells that were placed in the solutions of 0.4Mols and above. This is because there was a greater concentration of solute outside the cell.
- The isotonic point where there was no net change in the mass of the potato chip was .As shown on my graph. This means that solutions on either side of the membrane have equal concentrations of solute in them.
Evaluation
I felt my experiment went well the main problem occurred when I had to cut 36 potato chips using the cork borer. This was time consuming and although I tried to keep the chips damp after awhile I saw a few of them going black/grey.
If the experiment was put over a longer period of time and there were more concentrations, maybe increasing it to 4Mol/dm³, the results would have been better.
Working as part of a greater team would also help and shorten the time that was wasted when I was recording the mass of some potato chips and the rest were left in their solutions for too long (maybe affecting the result).
I could next time maybe look at how other factors affect the movement of osmosis such as temperature. This could be achieved by using facilities such as a water bath.