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The Effect of Different Concentrations of Water on Potato Tubes

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Introduction

The Effect of Different Concentrations of Water on Potato Tubes Introduction I am going to investigate the effect of water concentration in a boiling tube on a potato tube. Plan To be accurate with my results I will have five tubes from the same potato in each of the water baths. This is because I can then have enough results to distinguish any anomalous points, and thus disregard them. This is because I would be able to take an average from the five results each time, making my results more accurate. Therefore, I will have five potatoes in five boiling tubes, thus I will have 25 potato tubes. Osmosis is a phenomenon that operates in the same way as diffusion of particles, but with water molecules. Therefore if I change the concentration of water in the water baths, and keep the concentration of water in the potato as constant as I can, then there should be a movement of water molecules. Therefore I predict that if the water concentration in the potato molecules is less then the concentration of water in the water bath, water will come in through the selectively permeable membrane into the potato, thus increasing the potato's area. This could also work the other way, if the water concentration is smaller then in the potato tube, water will comes through the semi permeable membrane into the solution outside. ...read more.

Middle

3 Apparatus * Boiling rack * 5 Boiling tubes * Cork borer * Distilled water * Potato * Scalpel * 2M Sodium Chloride * Tile * Volumetric pipette * Water resistant permanent marker * Weighing scales Method 1. Take an Estima potato of large size. With a cork borer, take out twenty-five pieces of equal cross sectional area. When using the cork barer use it on a tile so that the table does not get damages. Measure the weight of these tubes using a weighing machine making sure the potato tubes are dry. Set up the experiment as shown below. 2. In a boiling tube, add 40ml of water. Label this beaker, 1. Make the other four solutions as in Fig.3 below using a volumetric pipette. Place five potato tubes into each of the beakers. 3. Leave the beakers for twelve hours. 4. Use the same weighing machine to note the change weight of each potato tube, making sure they are dry. Record these measurements in a table like Fig. 2 5. Plot a graph of these results, and comment on your observations. Volume Water (cm3) Volume Sodium Chloride (cm3) Molar Solution of NaCl in the Boiling Tube (dm-3) Potato Tube Number Weight Difference (g) 40 (x5) 0 (x5) 0 (x5) 1-5 30 (x5) 10 (x5) 0.5 (x5) 6-10 20 (x5) 20 (x5) 1 (x5) 1-15 10 (x5) 30 (x5) 1.5(x5) ...read more.

Conclusion

This is because I feel those two points give me the accuracy to give me a firm conclusion. I also have enough results to give me a reliable conclusion and get rid of any anomalous points. To make the results clearer I will now show the total average mass change of each solutions and the average percentage change in mass for each solution. 0M NaCl Average Weight Change 0.22 (H2O) Average Percentage Change 24% 0.5M NaCl Average Weight Change -0.24 Average Percentage Change -26% 1M NaCl Average Weight Change -0.35 Average Percentage Change -33% 1.5M NaCl Average Weight Change -0.46 Average Percentage Change -43% 2M NaCl Average Weight Change -0.56 Average Percentage Change -47% Using this information, I have drawn graphs. Firstly, I have drawn a graph for each solution. This shows the mass change in each boiling tube, over 24 hours. I have then drawn a more conclusive graph showing the percentage change of each solution. I think that this is more conclusive because if a potato were heavier then normal, it would suggest it had a higher percentage of water in it. If I only drew graphs for the mass change, this potato would lose more water then the others, because it has a larger mass of water. However, using percentage change in mass makes the graphs fair, because it uses the mass change over the original, giving us a percentage of the original. The equation used for this was: Percentage Change = Change x 100 Original ...read more.

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