Method
Put the 12 test tubes into 6 pairs and label one pair 1M, one pair 0.8M, one pair 0.6M, one pair 0.4M, one pair 0.2M, and finally one pair 0M. Each pair then needs to have 10cm³ of the relevant solution put into it. For example, the 1M pair of test tubes need 1M sucrose solution put into them, the 0.8M pair need 0.8M solution put into them, and so on. However, because there is only 1M sucrose solution available, the other test tubes need to have diluted solution put into them. To dilute the solution, you need to add water, but you need to add the appropriate amount of water to make each solution appropriately concentrated. For example, to make 0.8M solution, you add 80% sucrose solution, and 20% water. In this experiment, that would be 8cm³ sucrose solution, 2cm³ water. These measurements are specific because to make sure the test is a fair one, each test tube has to have the same amount of solution in it.
After this, cut up the banana into six pieces so that they are all as near to the same size as possible and small enough to fit into the test tubes. Put one piece of banana into one of the test tubes from each pair and then add one drop of Methylene blue stain to each tube that has a slice of banana in it and then gently shake each of these tubes gently to mix in the stain. Leave the experiment for a minimum of 20 minutes to allow the banana slices to soak.
After 20 minutes or so, take the square tipped needle and attach it to a small syringe. Take some of the stained solution from one pair and add one drop to the other test tube of that pair, which has the same concentrated solution. To insert the drop u place the tip of the needle into the middle of the solution and then push out one drop of the stained solution.
Now observe whether the drop sinks, stays in the middle or floats to the top.
Results
Table
Pictures
Conclusion
From looking at my results, I can see that for the concentrations of 1M, 0.8M and 0.6M, the drop of stained solution floated, showing that the water potential of the banana sample was originally higher than that of the solution outside. However, the water molecules then moved down the concentration gradient by osmosis so that the outside solution’s water potential increased, which made the density smaller. Therefore when placed in the original solution, the drop would float.
With the concentrations 0M and 0.2M, the stained drop sank because the original solution’s water potential was higher than the water potential of the banana. Therefore, the water molecules moved down a concentration gradient by osmosis into the banana cell through the semi permeable membrane. This left the outside solution with a more negative water potential, meaning that it had an increased density. When placed in the original solution therefore, the drop will sink.
The 0.4M solution however shows an example where the water potential inside the banana was equal to that of the solution outside the banana. I can conclude this because the stained 0.4M solution hovered in the other 0.4M solution. This shows that their densities were equal, showing that the water potential of each solution was the same. Therefore, osmosis did not take place between the banana and the first solution, because there was no downward concentration gradient so the water molecules did not favour going to one side of the semi-permeable membrane or the other.
The banana can therefore be said to have had a sucrose concentration of 0.4M.
Sources of Error
- Surface area of banana may have differed
- Banana may have been bruised or otherwise damaged
- Volumes of solution may have differed
- Number of drops of methylene blue stain may have differed
- May have dropped more than just one drop of stained solution into the other tube.
- May not have injected stained solution directly in the middle of the other solution, therefore affecting whether it hovered/sank/floated.
- Different time for each banana slice in the solutions, allowing for more osmosis to take place.
- Dilution process may have resulted in mistake of resulting concentration