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Explore how the concentration of a sucrose solution affects the rate of osmosis.

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Introduction

Investigate the factors affecting osmosis in plant tissue Aim: To explore how the concentration of a sucrose solution affects the rate of osmosis Introduction: Diffusion is the movement of particles from a high concentration to a low concentration until they are spread out evenly. An example of diffusion is when an aerosol is sprayed. The particles spread out from the high concentration at the nozzle into the rest of the room and that is how the smell moves. Osmosis is the passage of water molecules from a weaker solution to a stronger solution through a partially permeable membrane. Osmosis is a type of diffusion involving water - the water molecules move from a weak solution (with a high concentration of water) into a strong solution (with a low concentration of water). This can be shown in a diagram The cell membrane in a plant cell is partially permeable - it has small holes that can let in small molecules but not large ones. This allows water through and therefore allows osmosis. When the cell has all the water it can take inside of it the osmosis process stops. The water pushes up against the cell wall which is strong enough to stop it bursting. The cell is turgid and the plant needs turgid cells to give it rigidity and allow it to stand upright. ...read more.

Middle

Weigh each potato chip and record it's new mass. Subtract the original mass from the new mass to determine the change in mass. A positive number denotes an increase in mass while a negative number denotes a loss of mass. For each potato chip be sure to keep a record of the concentration of solution it was in. Clean out all of the boiling tubes and the measuring cylinder and repeat this experiment twice more in order to have three sets of results and a more accurate set of averages. Results: These are my raw results. Concentration (molars) 0.0 0.1 0.2 0.3 0.4 Original Mass (g) 4.39 5.05 4.78 6.16 4.71 New Mass (g) 5.37 5.43 4.88 5.77 4.26 Change in Mass (g) 0.98 0.38 -0.1 -0.39 -0.45 Concentration (molars) 0.0 0.1 0.2 0.3 0.4 Original Mass (g) 5.46 4.69 4.96 5.12 5.75 New Mass (g) 6.14 4.95 4.98 4.70 5.10 Change in Mass (g) 0.68 0.26 0.02 -0.42 -0.65 Concentration (molars) 0.0 0.1 0.2 0.3 0.4 Original Mass (g) 4.92 5.67 4.81 4.71 4.65 New Mass (g) 5.53 5.88 4.76 4.26 3.88 Change in Mass (g) 0.61 0.21 -0.05 -0.45 -0.77 In order to account for the differences in mass of the original potato chips I have decided to display my results for the final mass as a percentage change from the original mass. ...read more.

Conclusion

I would also like to do an experiment using the same concentrations as I did in this one, but repeating it about 5 times, each time leaving the potato chip in the solution for different time periods. I could then compare the gradients of the lines of best fit for the 5 different times, and also draw graphs for each molarity across the 5 time periods. I could also do an experiment using the same concentrations as I did in this experiment, but measuring the mass of the potato chips after every 3 or 4 hours until the mass stays the same, and see how long potato chips in different solutions took to reach a final mass and to see how large it's mass would get. Finally I would like to do the same experiment as I did here, but try it out on different types of plants and compare the rates of osmosis of the different plants. This would give an idea of which plants were more efficient at taking up water and I could see what types of plants had the fastest rate of osmosis, and whether there was a link between the rate of osmosis in a plant and the habitat it exists in. For example I might find that plants that live in hot, dry conditions have a faster rate of osmosis than plants which live in cold, wet environments. These experiments would help give a better idea of how the rate of osmosis is affected by the concentration of a solution. ...read more.

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