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Investigation Into Distinguishing Between Old and Fresh Potatoes

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Investigation Into Distinguishing Between Old and Fresh Potatoes By Means of Osmosis Scenario: When buying root vegetables at a supermarket, how can you be sure that they turn over the produce correctly and that all that you buy are equally fresh? Aim: To determine a way of distinguishing between old and new potatoes by immersing potato pieces in different concentrations of sucrose solutions. This will enable equivalent concentrations of the two ages of potato to be found and therefore the two types identified. Introduction: Osmosis is a type of diffusion involving water molecules only. Osmosis can occur when there are two solutions separated by a partially permeable membrane which only allows certain molecules through. The solution on one side has a higher water potential (lower solute concentration) than the solution on the other side. This solution is described as the most hypertonic. Because, as in diffusion, water molecules move from areas of low to high solute concentration, the majority of water molecules pass through the membrane down the water potential gradient from high to low. Water molecules will continue to move across the membrane until the water potential is the same on either side. ...read more.


Due to lack of time and resources I was unable to use 11 different sucrose solutions and had to cut down to 6. Lack of resources also meant that I was unable to repeat the experiment 3 times as it would have meant using 33 boiling tubes. Therefore, I decided that to reduce experimental error as much as possible, I would put three pieces of potato in each tube and achieve three results in that way. Also due to lack of time, I was unable to leave the potato pieces in the solutions for 1 hour and I had reduce the time to 30 minutes. Results: Table Showing the Amount of Weight Gained by a Piece of Fresh Potato in Various Concentrations of Sucrose Sucrose Concentration (mol dm3) 0 0.2 0.4 0.6 0.8 1.0 Weight Before (g) 1.67 1.64 1.62 1.56 1.58 1.45 1.64 1.69 1.69 1.60 1.70 1.66 1.75 1.59 1.56 1.72 1.71 1.70 Weight After (g) 1.69 1.69 1.71 1.55 1.60 1.45 1.62 1.62 1.58 1.54 1.52 1.46 1.58 1.42 1.39 1.49 1.48 1.47 Weight Difference (g) + 0.02 + 0.05 + 0.09 - 0.01 + 0.02 + 0.00 - 0.02 - 0.07 - 0.11 - 0.06 - 0.18 - 0.20 - 0.17 - 0.17 - 0.17 - 0.23 - ...read more.


As we were given the solutions ready made for us to use, I could ensure that this area of unreliability is accounted for by preparing the sucrose solutions myself. Because I was unable to do as I had planned and use three separate boiling tubes for the three pieces of potato in each concentration, this meant that if one result was wrong, they all were. Another possible error leading to poor reliability was the drying of the potato before reweighing them. It was impossible for us to consistently dry each piece of potato so as to allow for fair comparisons, so the calculated weight loss or gain may not be exact. This undermines the validity of the statement "The new potato has an equivalent concentration of 0.21 mol dm-3 while the old potato has a concentration of 0.23 mol dm-3" that I made in my conclusion. This is because if, for instance, I had left just 0.01g more water outside of the potato than at the beginning of the experiment, the whole line of best fit would have been higher than it should have been. It would have therefore met the x axis at a different place to give a higher reading of the equivalent concentration. My measurements were very accurate throughout the experiment as I used an accurate balance for the weighing and always rounded to two decimal places. ...read more.

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