Diffusion and Osmosis

Topics: Osmosis, Solution, Diffusion Pages: 5 (1606 words) Published: March 31, 2013
Kristen Demaline
Bio 1113, Lab 3: Diffusion and Osmosis

Osmolarity of Plant Cells
In this class, we learned about hypertonic, hypotonic, and isotonic solutions. Hypertonic solutions have a higher concentration of solutes outside of the membrane, hypotonic solutions have a lower concentration of solutes outside the membrane, and isotonic solutions have an equal amount of solutes inside and outside of the membrane (Morgan & Carter, 66). When the solute concentration is not equal, the water concentration is not equal, so water will move from a higher concentration to a lower concentration in a process called osmosis. In this experiment, we cut 4 pieces of potato, weighed them, and let each soak in a different sucrose solution for about an hour and a half. Our solutions consisted of distilled water (.0 sucrose molarity), .1 sucrose molarity, .3 sucrose molarity, and .6 sucrose molarity. Our question was “which solutions are hypertonic, which are hypotonic, and which are isotonic?”. This can all be determined through weight change. We hypothesized that distilled water would be a hypotonic solution, the .1M would be a hypotonic solution, the .3M would be an isotonic solution, and the .6M would be a hypertonic solution. We thought that .3M would be the isotonic solution because its molarity is in the middle. If .3M is in fact an isotonic solution, then the water concentration is the same inside and outside of the membrane and there should be no water movement resulting in no weight change. If distilled water and .1M are hypotonic solutions, then the concentration of water is higher on the outside, so water will move into the potato where water concentration is lower, causing a weight gain. Finally if .6M is hypertonic, then water concentration is lower on the outside, so water will move from the inside of the potato to the solution, causing the potato to lose weight. After about an hour and a half we took the potato pieces out of the solutions they were soaking in, patted the water off of them, and weighed them for a second time. The initial weight and final weight was recorded, which can be seen in Table 1. The potato piece that was soaking in the distilled water had a 3.1% weight gain, and the potato piece that was soaking in .1M sucrose had a 2.1% weight gain. The potato piece had no weight change in the .3M sucrose solution. And the potato piece that was soaking in .6M sucrose solution had a 5.7% weight loss. The weight changes can be easily seen in Graph 1.

Table 1: Change in Weight
|Sucrose Molarity: |0M |0.1M |0.3M |0.6M | |final weight (g) |16.4 |14.7 |17.7 |13.2 | |initial weight (g) |15.9 |14.4 |17.7 |14 | |weight change (g) |0.5 |0.3 |0 |0.8 | |%change in weight |3.10% |2.10% |0% |5.70% |

Graph 1:
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As you can see, the results supported our hypothesis. Distilled water is a hypotonic solution, which makes sense because there is no concentration of solute in it. The water moved to the potato because the potato has more sucrose concentration, meaning a lower water concentration. The potato that was soaking in .1M sucrose solution also gained weight as an effect of having a lower water concentration inside, but its weight gain percentage was lower because the solution had more solute than the distilled water. The potato soaking in .3M sucrose solution had no change because the concentration of sucrose was the same in the potato as it was in the solution, as we predicted. The potato lost weight in the .6M sucrose solution because the amount of sucrose inside the potato was less than the solution causing water movement from the potato to...

References: Morgan, J.G. and M.E.B. Carter. 2013. Energy Transfer and Development Lab Manual. Pearson Learning Solutions, Boston, MA.
|  |Points |Self-Assessment |Total Earned |
|Introduction |2 | 2 |  |
|Results |2 | 2 |  |
|Figures/Tables |3 | 3 |  |
|Discussion |3 | 3 |  |
|Total |10 | 10 |  |