Methods of keeping the variables controlled:
I used a borer at a size of 4 in order to cut the pear into similar sizes. In order to make sure that they’re the same, I used a mm ruler (to get a more accurate measurements) and measure the sizes of each pear and all of them have an approximation of 2cm in length.
In order to keep the volume of distilled water (I use distilled water as it doesn’t have as much impurities as the tap water) and sugar solutions the same, I used the same measurement cylinder of 20cm³ in order to get a more correct measurement. I also use the pipette so that the volume will be more accurate. I then record the reading at eye level on the meniscus as if I don’t do this, the volume of water won’t be a control variable.
Independent variables:
- Different concentration of sugar levels (0.25M, 0.5M, 0.75M, 1M)
- Water (0M).
Dependent variables:
- Mass of the pear pieces after the osmotic activity. (This allows us to see whether if osmosis has taken place and to what extent).
Method:
First of all, I cut the pear into 15 identical pieces using the same borer size. I then measure each of the pear pieces using a mm ruler in order to make sure that they all have identical surface areas and lengths and this is a control variable and therefore it must be kept the same. I then place each of the pear pieces on the balance scale and record down their mass in g as my investigation involves with the masses of the pear pieces.
After measuring all the pear sizes, I fill in five identical test tubes with 10cm³ of distilled water (0M), and sugar solutions at 0.25M, 0.5M, 0.75M, and 1M. I then repeat it two more times so at the end, I get 15 test tubes as by repeating the test, I can draw out any anomalies and work out the average in order to make my data more reliable.
I then label the test tubes by A B C D E in the three tests so that I can get a neat recording of my investigation and avoids confusion.
After that, I place each of the pear pieces into the test tubes and leave it for 15 minutes in order to give it time for the osmotic activity to occur. While waiting, I neatly create my results table so that I am ready to fill in the results. I also clear away all the equipments that are not needed so that it will reduce the time of cleaning up.
After fifteen minutes, I remove the pear pieces from the test tubes and measure each of them on the same balance and then record each of the masses so that I can differentiate between the masses of the apple pieces before and after the investigation.
After the investigation, I also measure the volume of the solutions from each of the test tubes just to see whether if there is any difference between before and after the investigation.
Here is my results table that I have collected from my investigation:
Masses of pear pieces (g) BEFORE
Mean: 0.80 (2d.p)
Masses of pear pieces (g) AFTER
Mean: 0.84 (2d.p)
Test 1
BEFORE
Test 1
AFTER
Test 2
BEFORE
Test 2
AFTER
Test 3
BEFORE
Test 3
AFTER
From this set of results, I can see that some of the pear pieces’ sizes are smaller when I measure using a mm ruler. This could depend on the masses of the pear pieces as each mass vary although they are closely together so I don’t think it makes much difference with the outcome of the results. I can also see that some solutions in the test tubes have decreased. These can either be done with two factors:
- Either some solution from the test tube has gone into the pear piece as the osmotic activity is taking place.
- Either when transferring the solution into a measuring cylinder in order to record its measurements after the investigation, there is a chance of droplets of solution still left behind in the test tube and therefore not measured which is a problem.
Solution:
In order to keep the investigation unbiased, I would instead of pouring the solution into the cylinder, I could have use a pipette to draw out any excess remaining solution in the test tube and pour it in the measuring cylinder to ensure that no water is lost.
Judging from the data on the masses of the pear pieces before and after the investigation, I draw up a mean data from each of the data sets and conclude that the mass of the pear pieces has increase after the investigation.
Graphs results:
Test 1
Test 2
Test 3
Conclusion:
Judging from these three graphs, there is a difference in masses before and after the investigation in test 1, 2 and 3. I can see that the masses of the pear pieces before the investigation has a lower value than the masses of the pear pieces after the investigation. This shows that there is an osmotic activity between the solution and the pear pieces. The masses of the pear pieces after the investigation is due to the fact that the water molecules from the glucose solution is diffused to a region of high concentrated area which is the pear.
However, these graphs only shows the difference in mass of the pear pieces before and after the investigation which doesn’t really support my hypothesis of ‘the lower the concentration, the higher the mass of the pear’ so I decided to use a different method for it.
So for each concentration, I looked for a difference between the mass of the pear piece before and after the investigation.
For example, beaker A has a 0M. The mass of the pear before investigation is 0.76g and the mass after the investigation is 0.84g so I just subtracted it to find a difference between them which is 0.08g. I then did the same thing with test 2 and test 3 but with the same concentration and once I found out the difference of mass in the three tests, I then worked out the mean average which comes up as 0.053g(3.dp).
So here is the table of results of the mass of the pear pieces g:
From this table, it shows that the lower the concentration, the higher the mass of the pear in g. However, only one concentration that doesn’t matched up with my hypothesis which is the 0.25M. I think it may be due to the fact that there is a variation of the masses of the pear pieces as some values are higher than the others which could effect the prediction. I also notice that the highest concentration has the smallest number which is 0.007. I think this is because there might be some error while collecting the data as one of the data has a minus number in it so I considered that as an anomaly (beaker E, test 2 has a mean of -0.01) so if I discard that anomaly, the mean average for 1M would be 0.01g. I decided that this data is more accurate than 0.007 so I will change the number 0.007 to 0.01.
There are many problems and error in this investigation which could be improved.
Problem:
Different mass of the pear pieces
Solution:
This is very hard to keep in controlled as it’s hardly possible to be able to keep the mass exactly the same but in order to improve the data, I would try to scrape off some pear bits with a scalpel providing that I still maintain the same length.
Problem:
Accuracy when taking up readings
Solution:
In order to improve the accuracy of the results, I could use a mass balance which has smaller scales, perhaps up to 3.sf. I could also use a smaller cylinder, like a 10ml cylinder instead of 20 to get more accurate measurements. I must also be careful when dealing with the measurements of water as there may be some water left in the pipette or test tubes when the water is being poured into the cylinder for measurements. There may also be air bubbles in the measuring cylinder which can increase the volume of water. So in order to make sure that there is no air bubbles, I would give the cylinder a good shake (taking care that it is not spilt) to remove any excess of the air bubbles.
Problems:
Time taken to measure the length of the investigation may also not be exactly fifteen minutes duration as it takes time to take out water from each pear pieces so some pear pieces stays longer in the test tube so this could also effect the result.
Solution:
I think this is an unavoidable error as it’s hardly likely that I can manage to take each of the pear pieces out from the test tube at exactly 15 minutes so this can effect my result as some stays longer and therefore has more osmotic activity taken place than the others, unless if the concentration of water in the pear is the same with the concentration in the solution and this is the point when no more osmotic activity can take place in the cell as the solutions are all evenly distributed.
