Results Table.
Key For Solutions
Strong Solution = 2 teaspoons
Weak Solution = ½ Teaspoon
Sketch Graph.
Conclusion
From the results of this experiment I can see that when you add sugar to the solution you put the potato chip in, it affects the amount that the chip extends or contracts by. In water and a weak sugar solution, the chip expands. In a strong sugar solution it contracts. In a medium strength solution the chip may stay the same size. If the potato chip was put into an isotonic solution then it would not expand or shrink. The isotonic point in my experiment was 0.35% so I would have found that the potato chips put in that strength solution would not have changed size.
Improvements.
If I had at least six solutions then I would stand a better chance of finding out at which concentration the chip doesn’t either expand or contract.
It would be better to measure the amount of sugar, rather than calling it a weak or strong solution. This can be done by using a molar solution and various dilutions of it.
To keep the test fair, every variable except the ones I am going to change and measure should stay exactly the same. These variables include:
- Temperature
- Light level
- The length of the time they are in the solutions for
- The amount of solution they are placed in
Predictions.
I predict that in my experiment:
- As the sugar solution gets more concentrated, the potato will tend to lose water and the mass will decrease.
- When the solution is weak, the potato will tend to absorb water and the mass will increase.
- There will be a point where the concentration inside the potato and outside are equal, so no change in mass will occur.
- The percentage change in mass will be proportional to the change in concentration of the sugar solution.
- There will be a limit to how much water the cells can absorb because of the pressure of the cell walls.
Aim Of My Practical.
The aim of my practical is to measure the change in mass of potato chips when they are soaked in a range of sugar solutions of different concentrations and to find the point when the amount of water moving into the cell is the same as that moving out.
Apparatus.
6 Boiling Tubes, Labels, Test Tube Rack, Cling Film, Cork Borer, Top Pan Balance, Filter Paper, Ruler, Measuring Cylinder and Six Solutions of known concentration (0 moles, 0.2 moles, 0.4 moles, 0.6 moles, 0.8 moles and 1mole).
Method.
One person sets up the apparatus whilst the other person sorts out the chunks of potato. Cut two pieces of potato for each boiling tube. Find the mass of each tube instead of the length, this way it is more accurate. Use a 1/100g top pan balance for good graph results. Label the tubes and fill them accordingly to their labels with the correct solutions. Use a 30ml-measuring cylinder to get the precise measurement. Use the same potato for each chip so you know you have a fair test. If another potato has to be used, make sure chips from two different potatoes are not put in the same boiling tube. When all the solutions and boiling tube are ready, cut up the potatoes. Make sure only two potato wedges are cut at a time so they don’t dry out. Once they have been cut, gently roll them on a paper towel to wipe off any excess liquid. Take each chip individually to the top pan balance. Once they have been weighed, immediately put them into the boiling tube. When the experiment is finished, after 24 hours, take out the potato chips in exactly the same order they were put in. As long as all parts of the experiment are carried out on the same bench at the same time then the temperature level should be the same.
Control Of Variables.
There are two types of variable. There is the independent variable and a dependent variable. An independent variable is something you can change and a dependent variable is something you measure. The independent variable in the experiment is the concentration of the sugar solution. The dependant variable in the experiment is the change in mass of the potato cell. All other variables, which might affect the results, must be kept the same to keep the test fair.
Two examples of controlled variables are:
- Surface area of the potato chip. If the surface area is bigger on some than the others then the rate of osmosis would differ. To solve this problem use a cork borer (the same one each time) and cut to 5cm to give precise chips and control the variable.
Safety Procedures.
- Do not leave the potato in the solution for longer than 24 hours as there is a chance of bacteria growing.
- Be careful when cutting with the scalpel so you do not cut yourself
- If you spill any water immediately clean it up so no-one slips on it
- Do not run in the laboratory
- Be careful of the sharp edges of the cork borer
Results Table For Average Mass Of Potato Chips.
Summary Of Results.
Conclusion
From my table and graph it can be seen that as the concentration of sugar solution increases the mass of the potato does decrease. The value of the isotonic point is 0.35 moles. At this point there is no change in the mass of potato because the amount of solvent (water) moving into the cell is the same as that moving out. This is because the concentration of the 0.35 molar sugar solution is equivalent to that of cell sap. The points of the graph lie almost on a straight line. I verified this by computer plotting a graph with and without the 0M reading which I considered possibly to be a significant error. This means that in this range the change in mass is proportional to the concentration in the sugar solution, as I predicted. In solutions weaker than the isotonic point the mass increases because water is moving into the cell by osmosis. In solutions more concentrated than the isotonic point water moves again from the weak solution to the more concentrated and so it moves out of the cell by osmosis.
Evaluation.
The practical did work and it was a sound experiment. It was easy to perform with no real problems. To improve the experiment we could use something like a chip cutter so all the chips are the same size and width so it’s an even fairer test. More top pan balances would have been useful so we could have done it quicker. There were two cork borers that took quite a while to cut the potato so it took a long time to cut the potato. If the potato had got stuck in the cork borer, you needed to push it out with a pencil. Then the end would need to be cut off because it’s squashed and will expand in the water. I did predict that the percentage change in mass would be proportional to the concentration of sugar solution and so I might have expected a straight-line graph. The 0 Molar reading was slightly off the expected straight line. This was probably due to a small error in my reading. More readings, especially were the mass had increased, would have made my positioning of the line of best fit much more accurate and consequently my results would have been more reliable. When wiping excess fluid off the chips it’s very hard to take exactly the same amount of water off each one. It may possibly be better to take the chips out with tweezers rather than fingers and shake them dry. If I extended the range of concentrations beyond those that I used, then I predict that there would be a point were the relationship that I have observed would no longer be proportional. This is because there is a limit to how much water a plant cell can absorb or lose without plasmolysing or reaching a point of maximum turgor pressure because of the resistance of the cell wall.
The isotonic point automatically calculated on the computer-generated graph (with the suspected significant error excluded) coincides precisely with my original judgement in the hand-plotted graph. This serves to confirm my interpretation of the results as
Reliable.
As an example of further work that could be done to provide additional relevant evidence would be to study the effects of osmosis on animal cells, perhaps blood cells.
One of the major differences between animal cells and plant cells is that animal cells do not have a cell wall and therefore there is the possibility that the cell may ‘burst’.
Sodium chloride solution can be made at a concentration that is isotonic to blood. This is called physiologic saline solution and is 0.85% NaCl.
Osmotic solutions (in relationship to the fluids of the body) can either be hypertonic, hypotonic, or isotonic. Hypertonic solutions have a greater concentration of solutes than water. A hypertonic solution that will be used in this investigation is 10% NaCl solution. Hypotonic solutions have a greater concentration of solvent compared to the concentration of solute. For example, distilled water, which has more water molecules and fewer solute molecules, will be used in this lab exercise as a hypotonic solution. Isotonic solutions have equal solute to solvent ratio. Therefore, an example of an isotonic solution would be a physiologic saline solution
.
When blood cells are placed in these different osmotic solutions, it causes the blood cells to either shrink, burst, or stay the same. Blood cells placed in an isotonic solution will stay the same since there is an equal amount of solvent/solute both inside and outside the cell. However, when blood cells are placed in a hypotonic solution, the cells will burst because there is a lesser amount of water inside the cell and a greater amount of water outside the cell. So water rushes inside the cell (to establish equilibrium), but there is no mechanism that tells the cell equilibrium has been established. Therefore water continues to rush into the cell until the cell eventually bursts. This bursting process is called hemolysis (hema=blood; lysis=to burst). Blood cells placed in a hypertonic solution will have a reverse reaction to the hypotonic solution. Water will rush out of the cell thus causing the cell to shrink. This shrinking process is scientifically termed crenation.
The information above is taken from the Georgia Perimeter College website.
I would organise my investigation as follows:
I would take samples of blood on microscope slides and add different concentrations of sailine solution with a pipette in measured amounts.
Then I would observe and record any changes by using a microscope and a notebook.
To carry out this experiment I would need:
- 4 test tubes
- Test tube rack
- Dropper bottle of sheep blood
- Dropper bottle of 0.9% NaCl solution
- Dropper bottle of 10% NaCl solution
- Wash bottle of distilled H2O
- Marker pen
- 6" vinyl ruler
- Rubber Stoppers for test tubes
- Four clean microscope slides
- Four cover slips
I predict that as the concentration gets stronger, the paler the colour of liquid. As the concentrations get weaker, the liquid will go a very deep red colour. This is because as the concentration gets weaker there is more H20 than salt and the blood cell absorbs more water through osmosis. Eventually there will be no space left in the cell to absorb any more water and it will pop. The weaker solutions of saline will be a red colour because of the haemoglobin leaking out. This is known as haemolysis.
The picture above is of red blood cells bursting when the skin us burnt.
From this picture you can see that it is very difficult to tell the difference between the burst and un-burst cells. This experiment would need a very strong microscope to see the blood cells and would need an expert to tell which are burst and which are not.
