Rana's Water Potential and Molarity Mixup Arguments

Rana Srouji

Mr. Hammer

AP biology

November 21, 2014

Argument 1

Using our understanding of water potential and tonicity, my lab partners and I were able to determine the molarities of the sucrose solutions given.  The order of the solutions from the greatest molarity to the least is yellow with 1.0M, red with 0.8M, orange with 0.6M, blue with 0.4M, light green with 0.2M, and lastly dark green with 0.0M.  

In order to obtain these molarities experiments needed to be performed and data had to be collected.  The first experiment we created to test the solutions required my lab partners and I to create cells by measuring out 30 mL of each solution and putting them into 15 cm of dialysis tubing.  We measured the initial weight of the “cells” and then placed them into a plastic container of water for 24 hours.  Once they were taken out, we measured the final weight of each of them.  Then we calculated the percent change of mass. All of the data can be seen in the data table below.  We calculated the percent change in mass by subtracting the final mass by the initial mass, and dividing the difference by the initial mass.  The red solution had a percent change of 63.7%, the orange solution had a percent change of 26.8%, the yellow solution had a percent change of 47.8%, the dark green solution had a percent change of 10.1%, the light green solution had a percent change of 18.2%, and lastly the blue solution had a percent change of 27.2%.  However, this data was not used to find the molarity of each solution.  This is because we had accidentally made an error in our procedure.  Instead of using one dark green solution cell and one light green solution cell, we used two light green cells.  So we based the molarity off of how firm each cell was, considering the more firm the cell is, the more molar it is. The most firm to least firm cells went in order from yellow, red, orange, blue, and light green. In order to get factual evidence to determine the molarity of each solution, we compromised and used the data that was collected from the second experiment that we created, using pieces of two vegetables per solution.  The vegetables used were turnip and artichoke.  After which we calculated the percent change in mass per vegetable.  The vegetables with the least percent change determined that the solutions had the most molarity.  The solutions with turnip had percent changes of -9.8% for red, 0.06% for orange, -6.8% for yellow, 28.7% for dark green, 14.7% for light green, and 11.5% for blue.  The solutions with artichoke had percent changes of 11.11% for red, 10% for orange, -22.2% for yellow, 25% for dark green, 20% for light green, and 11.1% for blue. Thus, the most negative overall was the yellow solution, then the red, orange, blue, light green, and lastly dark green.  Their molarity went from greatest to least 1.0M, 0.8M, 0.6M, 0.4M, 0.2M, and 0.0M.  

Molarity is an amount of solute in a solution.  This is the reasoning for why the more firm the cell is, the greater molarity it has as well. Since hypertonic solutions have more solute in them than the solution around them, this means they have greater molarity.  These hypertonic solutions in the dialysis tubing cells would then swell because of osmosis.  Osmosis is when the water goes from a low solute concentration to a high solute concentration.  The dialysis tubing cells were filled with a solution that would have caused water to move into the cell because it has more solute than water.  So the only cell that would have 0.0M is the one with the least firmness, because it would have no solute, and the surrounding water would not have stayed inside the cell.  This is also known as an isotonic solution, which we concluded was the dark green solution because it was the only one that we had not tested, and the other solutions in the cells were greater than 0.0M.  Also, we calculated the percent change of the dialysis tubings because the greater the percent change in mass meant the greater molarity of the solution.  This is because the percent change in mass would determine the percentage of how much the cell either gained in mass or lost in mass.  This would show if water entered or left the cell.  If water left the cell, then the percent change in mass would be negative.  If water entered the cell, the percent would be positive.  Since all of the percent change in mass increased for each solution that we used, that meant that each of them gained water and had solute in them.  Which would once again, leave the dark green to be the only one we did not use, and must have been the solution with 0.0M.  So we determined that the solutions with the largest percentage would have the least molarity, and the smallest percentage would have the most molarity.  However, we still decided to use the vegetable percentages for more accurate results since we used all of the solutions in that experiment.  The vegetable percentages were the opposite of the dialysis tubing cells percentages because the vegetable was put into the sucrose solutions, while the tubing was placed into water.  So instead of the water entering the cell, it would have left the cell and gone into the solution.  Meaning that the vegetables were hypotonic and would have most likely become smaller after being placed into the solutions.  Thus, the vegetables with the least percent change in mass would mean the solution it was placed in had the most molarity or solute.  

Argument 2

In our second experiment, the water potential of the turnip is -21.2 bars, and the water potential of the artichoke is -15.0 bars.  Using the data we collected, we calculated the percent change in mass of each piece of vegetable that was placed in each of the sucrose solutions.  Using the percent change we determined which of the solutions had the different molarities.  Once they were found, we plotted the percentages on a graph which can be seen below.  On the y-axis is the percent, and on the x-axis is the molarity.  Both the turnip values and artichoke values were plotted on the graph.  Each line was a different color so that it is easily shown which of the lines represented each vegetable.  After the points were plotted, we calculated the points where each line crosses the x-axis.  The turnip line crossed the x-axis at about 0.86M.  The artichoke line crossed the x-axis at about 0.61M.  We used this information to calculate the solute potential (Ψs) of the vegetables.  This is because, the equation for water potential (Ψ) is the pressure potential (Ψp) plus the solute potential.  The solute potential equation is Ψs=-iCRT.  The i is the number of particles the molecule will make in water, in this case the i = 1.  The C is the molar concentration, which is the point that was found on the graph.  The R is the pressure constant which is 0.0831 liter bar/ mole K.  The T is the temperature in degrees kelvin of the solution, which was 296 degrees.  When the math was calculated, which can be seen below, the number for turnips came out to be -21.2 bars, and the number for artichokes came out to be -15.0 bars.  The pressure potential was 0 because water has no pressure potential.  Thus, the pressure potential added to the solute potential equals the water potential which is -21.2 bars for turnips and -15.0 bars for artichokes.