These preliminary results can’t be trusted because all the variables will have been different. For example, when we tested the chip in the salt solution, we didn’t measure how much salt was put into the water. Also, we didn’t measure the length of the chip; we just made an approximate decision. Furthermore, we did not use a stop clock to time the whole process, we just used a clock and finished after 20 minutes had passed. This therefore does not make the preliminary results very reliable and can only be used for seeing if salt concentration does affect the rate of osmosis. Therefore a new experiment needs to be carried out, which I will control all the variables apart from the salt concentration which will vary.
The Variables I will change and try to control
The variable that I will change in this experiment is the salt concentration (distilled water and salt water) as this is the main factor. The variables I will try and control are:
- The species of potato,
- Volume/concentration of solution (salt water/distilled water),
- The weight of the chip (using balance),
- Surface area (using a Cork borer, scalpel and a ruler)
- Temperature (using a thermometer).
- The timing of the whole experiment (chips need to stay in solution for exactly 20 minutes).
Method
- I will firstly cut three chips all of the same size using a cork borer, a scalpel and a ruler. Being careful with the scalpel.
-
I will then pour out the required amount of distilled water and salt water, using a 10cm3 measuring cylinder, into a test tube (for example, 8cm2 of salt water and 12cm2 of distilled water).
- This will be repeated two more times to make the experiment a fair test.
- I will then weigh each of the three chips on a balance and record them into a table.
- Then I will place each of the three chips into the three test tubes from earlier.
- The chips will then be left for 20 minutes. This will be timed on a stop watch.
- When 20 minutes has passed, I will take the chips out of the test tubes and dry them off. Remembering which chip came from which test tube.
- I will then re-weigh each of the chips on a balance to try and see if any change has occurred.
- I will then record these results into the table from earlier to see if any of the weights have changed.
- From these results, I will draw up a graph to try and find any patterns between the sets of results.
- Then I will write an analysis showing what the graph is telling us and to see if the results match up with my predictions.
Results
Here is a table of results to show the masses of potato chips before and after they have been placed in different salt concentrations. I have also stated the average percentage change to make the test farer (if all three results are correct).
Analysis of results
From the graph, you can see that as salt concentration increases, the average percentage change decreases. For example, with a salt concentration of 0%, the average percentage change was +11.44, whereas with a concentration of 5%, the average percentage change was -16.48.
On the graph, I have marked three points. These are A, B & C. Below I will explain what happened in each of these different points.
Part A:
All the chips at this point are decreasing in mass. This matches with my predictions earlier on in the investigation. This is because the concentration of water outside the cell will eventually become less than the inside of the chip, because water can osmose into the chip quicker than water can osmose out of it, therefore resulting in water osmosing out of the chip resulting in mass lost from the chip.
Part B
All the chips at this point are decreasing in mass. This again matches my predictions from earlier on in the investigation. The mass of the chip decreases because the salt concentration is higher outside the chip than inside; therefore water will be able to osmose out through the membrane of the chip quicker than it can osmose into the chip.
Part C
This point is called the isotonic point. This is because the concentration outside the chip is exactly the same as the concentration on the inside of the chip therefore the mass of the chip will stay the same.
