Ap Bio- Lab 1: Osmosis and Diffusion
AP Biology 16 April 2013 Lab 1: Osmosis and Diffusion The purpose of this experiment is to decide which flasks correspond to the solute concentration of sucrose, which are 0. 8M, 0. 6M, 0. 4M, 0. 8M. This can be determined by the usage of potato cores. If a potato core’s mass decreases after a day soaked in the sucrose solution, then that flask will have a high concentration of sucrose. A) The first step in this lab would be cut out four cores of a potato and measure the masses of each individual potato piece. The mass of the potato core would be a variable factor in this experiment.
A way that this variable can be manipulated would be each group would be assigned to a specific potato core mass. A controlled variable in this lab would be to pour a certain amount of sucrose solution from a flask into a breaker. An example would be pour 200ML from flask A into a beaker labeled A, 200ML from flask B into a beaker labeled B, and etc. Next, put 1 potato core in each beaker: A, B, C, and D. Leave the cores in the flasks for a day. After leaving the potato cores to soak in the solutions for a whole day, take out the potato cores.
Measure the final mass of each potato core and record this data. Afterwards, calculate the percent change in mass. To calculate the percent change in mass, use this formula: [(mass initial – mass final) / (mass initial) ] x 100%. Measure the data results of the percent change of mass in grams. To show a visual picture of the percent change in mass, graph it. This will help in seeing the sucrose molarity among the solutions. To finish, share data with other groups and compare results among one another to obtain an average of all of the results.
This will verify that the results acquired were incorrect or correct. B) My hypothesis was that if a potato core’s mass decreases after a day soaked in the sucrose solution, then that flask will have a high concentration of sucrose. Due to this, I would expect that the potato cores in the 0. 4M, 0. 6M, and 0. 8M sucrose solutions will lose mass. Considering this, the percent change in mass will be negative. In contrast, I would expect that the potato cores in the 0. 2M sucrose solution will gain mass. As a result of that happening, their percent change in mass will be positive.
C) Considering the principles of osmosis and diffusion, these would be reasons on why I have these high expectations of my results. In osmosis, water moves through a selectively permeable membrane from a region of higher concentration to a region of lower concentration. This selective membrane only allows certain types of molecules to flow through while other molecules are not able to. In diffusion, molecules are in constant motion and move from an area of higher concentration to an area of lower concentration. In this experiment, the potato core will act like a selectively permeable membrane.
In the 0. 2M sucrose solution, the sucrose will not be able to go through the selectively permeable membrane because the sucrose molecules are too big to go through the passages of the membrane. It will, however, allow water molecules to go through the passages of the selectively permeable membrane due to diffusion. Water molecules will be able to pass the selectively permeable membrane because there will be a lower concentration of water in potato then the outside, making the water flowed into the potato core.
By this happening, it would make the potato core gain mass. In the high sucrose solutions, such as the 0. 8 solution, the sucrose will still not be able to go through the selectively permeable membrane because of the size of the sucrose molecule. The factor that makes these high sucrose solutions different is the result of having a high sucrose molarity. Having a high sucrose molarity will make the water from the potato cores move out. This would ultimately make the potato cores in high sucrose solutions lose mass.