Measure The Mass Of Sucrose Solution Lab Report

Procedure
Activity 1:
Fill 6 large beaker halfway with distilled water, making sure all beakers have equal amounts of water.
Cut 6 30 cm dialysis bags and label each bag with a letter, A through F.
Fill each dialysis bag with 15 mL of solution A through F that corresponds with the lab on each bag. For example, bag A is filled with solution A.
Measure the mass of each dialysis bag and record masses of each bag in BILL.
Cover the beakers with paper towel and leave the bags in the beakers overnight.
Remove the dialysis bags from the beakers and let dry.
Measure the mass of each dialysis bag and record it in BILL
Order dialysis bags A through F from highest to lowest mass, which becomes 1.0, 0.8, 0.6, 0.4, 0.2, 0.0 molar solutions respectively.

The molarity of sucrose is the amount of sucrose, ergo, each solution had different amounts of sucrose, the solute to this solution. As the amount of sucrose increased, the amount of water decreased. To have more of a solute, there must be less of a solvent, as the milliliters of the solutions tested remained the same. An increased molarity of sucrose led to an increased net diffusion of water into the cell, a greater mass change, because water potential of the cell decreased as sucrose increased. From these masses, a percent of mass change was calculated: final minus the initial, all divided by the initial. When the percents were ordered from least to greatest, the molarity of each solution can be deduced, as those with the least change had a molarity closest to zero and those with the most change had a molarity that was the highest. This is due to the fact that if they had 0.0M of solute, they are isotonic and should have no net diffusion of water in a single direction. If they have higher concentrations of sucrose, the water potentials are no longer equal and to decrease entropy of the system, water must diffuse to achieve equilibrium. In this way, it was deduced that Solution B is 0.0M, Solution D is 0.2M, Solution F is 0.4M, Solution A is 0.6M, Solution E is 0.8M, and Solution C is 1.0M. The graph seconded this data by portraying an almost constant

As the molarity of the solutions as been solved, they serve as indicators of water potential. This is because when the potato slices are placed in each of the six solutions, some experience net diffusion of water into the potato cells, out of the cell, or no det diffusion at all. This is due to water potential moving down its concentration gradient:free water flows from high concentration to low concentration, to reach equilibrium. In the lab, the potato slices with sucrose solutions less than 0.28M increased in mass because their water potential was lower than the solution around them: water diffused into the potato samples to reach equilibrium. These were hypotonic sucrose solutions. Potato slices that were in solutions greater than 0.28M decreased in mass because the water potential of the potato was high than that of the solution around them: water diffused out of the potato cells and into the sucrose solution around them. These sucrose solutions were hypertonic to the slices. As the graph indicates, the potato is isotonic to the sucrose molarity at 0.28M. That indicates that the water potential reaches equilibrium at 0.28M of sucrose. That isotonic value provides the molar concentration of the solute in the potato, which can then be plugged into the equation for solute potential: multiply the negative ionization constant, the pressure constant, the temperature,