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Effect of different sucrose concentration solutions on potato samples.

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Introduction

Effect of different sucrose concentration solutions on potato samples. Aim In order to find out how the mass of potato samples are affected by various concentrations of sucrose solution, I will use six concentrations ranging from a 1-mol dm�� solution to distilled water (0 mol dm��). The effects upon the potato samples will be concluded by comparing the mass of the sample before and after being placed in their solution. I predict that a unique type of diffusion called osmosis will affect the mass of the potato samples in their sucrose solutions. Background Diffusion is defined as the net movement of molecules (or ions) from a region of their higher concentration to lower concentration: down a concentration gradient. However, osmosis is a unique type of diffusion as it only involves water molecules. During osmosis, the energy of the water molecules tendency to move from one region to another, by exerting pressure on the membrane of the plant cell, is known as water potential. Therefore, osmosis is defined as the movement of water molecules from a region of higher water potential to a region of lower water potential through a partially permeable membrane. The water potential of pure distilled water is set at zero, but this is lowered (negative value) if a solute is present in the solution. Consequently, the value of water potential becomes increasingly negative as the amount of solute increases. The value the water potential is lowered by the amount of solute present in the solution is the solute potential. This means that the solute potential values of all solutions are negative. When osmosis occurs in plant cells, the following features of the cell are affected: - * 'The cell wall - allows all molecules through; it is fully permeable. It is strong enough to resist the tension the cell can develop due to high internal pressure. * Vacuole - contains a watery sap solution of salts and other dissolved substances. ...read more.

Middle

* Amount of total solution/concentration. This must be maintained seeing that if there were differences in the total solution present, i.e. use of too much sucrose solution or distilled water, the percentage change in mass may occur to an extent it shouldn't. These factors were noted during my preliminary experiment and could affect the results of my final experiment. By carrying out the preliminary experiment, was made aware of these variables and therefore, I will be able to control when the experiment is repeated. Method To understand how osmosis occurs across a range of concentrations, I will use the following in this investigation: 0 mol dm�� (distilled water), 0.2 mol dm��, 0.4 mol dm��, 0.6 mol dm��, 0.8 mol dm� and a 1 mol dm� sucrose solution. The 1 mol dm�� solution is one in which 1 dm� (a litre) of solution contains the molecular mass (one mole) of the substance in grams. In this case, the sucrose 1 mol dm��solution, 342g of sucrose will be dissolved in 1000 cm� of distilled water. For this investigation, such a large quantity of concentration is not required and will instead be made up of one quarter of the stated amounts, which will still make up a 1 concentration; 85.5g of sucrose in 250cm� of distilled water. Below is the procedure I will follow: - 1) Collect apparatus and other requirements needed to carry out the experiment stated. 2) Label boiling tubes with stickers and the concentration in mol dm� � to be contained within them. 3) Make up solutions by following ratio figures outlined in the following table: Solution (mol dm��) Ratio of Pure Distilled Water: 1M Solution (mls) 0 10:0 0.2 8:2 0.4 6:4 0.6 4:6 0.8 2:8 1 0:10 Using a pipette and placing it halfway in the sucrose solution can carry this out. Draw up required ratio amount of sucrose solution (none is required for the 0 mol dm��) ...read more.

Conclusion

Initially, the potato samples gain mass in the 0 mol dm�� concentration and then as the solute potential decreases (more negative) so does the mass they gain decrease. Nonetheless, eventually the samples will begin to gain in mass in solutions of higher solute potentials and the highest being the 1 mol dm�� solution. This relationship between molartiy and solute potential is shown in graph 3. The process of osmosis causes this to occur since as we add solute to water, the water molecules from a shell around each solute molecule. So this decreases the number of free water molecules that are able to exert a pressure on the membrane. Therefore, the water potential of the cells becomes lower. The process of plasmolysis may have occurred in such conditions because the cells of the potato sample lose water due to osmosis, the protoplast shrinks away from the cell wall. The solute potential in the potato samples is higher than that of the concentrations, which causes them to gain mass as water molecules diffuse through the cells of the sample due to the difference of the concentration gradient. The cells of the sample may have become turgid due to the build up of internal pressure (pressure potential). Using these findings, regarding how specific concentrations of solutions affected the mass of the potato samples, I can determine the concentration at which the potato samples are neither lose or gain in mass due to osmosis. This means, I can use my results to work out the solute potential of the potato samples. This is determined by the point at which the average percentage change line equals zero (osmosis does not occur in this concentration due to equal concentration gradient) and crosses the x-axis. For that reason, the solution at which the sample would not gain or decrease in mass is approximately 0.280 mol dm��, which according to graph 3 indicates the solute potential of the potato samples is around -850 kPa. ...read more.

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