Chigozie's Water Potential and Molarity Mixup

Chigozie Amonu

Mr. Hammer

AP Biology

21 November 2014

Water Potential & Molarity Mixup

            Argument One

For this lab, my lab group and I had to determine the unknown molarities of the six unlabeled solutions. The molarities of each different colored solution could be 0.0M, 0.2M, 0.4M, 0.6M, 0.8M, or 1.0M. After the conduction of experiments and compilation and analysis of data from Activity One and Two, we determined that the order of molarity of the solutions, from least to greatest (0.0M to 1.0M), was orange, dark green, light green, blue, yellow, and red. To get these results, two separate experiments were done. One was done with dialysis tubing bags, and the other was done using two vegetables, squash and parsnip. My lab group and I analyzed all of the data that we got from both experiments. We eventually decided to use the data from the experiments done with squash and parsnip because these two results proved this claim while the data from the dialysis tubing did not. Simply, the majority of our evidence supported this claim.

For the vegetable experiment, my lab group and I cut six pieces from each vegetable and weighed each piece. We then measured out 75 mL of solution to put the vegetables in. The six pieces of each vegetable were put into the six different solutions, and the cup and beakers that they were in were covered with foil and left to sit overnight. The next day in class, my lab group and I weighed the vegetables and recorded the weight in data tables. The percent change for each piece of vegetable was calculated (percent change= [final mass minus initial mass] divided by [initial mass] times 100). For the parsnip, the percent changes were -25.32% for red, 47.62% for orange, -20.96% for yellow, 10.64% for light green, 32.09% for dark green, and -13.24% for blue. For the squash, the percent changes were -45.68% for red, -6.94% for orange, -43.04% for yellow, -4.55% for light green, 20.37% for dark green, and -31.43% for blue.

Parsnip Data

Color of Solution in Which Parsnip was Placed	Initial Weight of Parsnip	Weight of Parsnip After Overnight Stay	Percent Change
Red	15.8 g	11.8 g	-25.32%
Orange	12.6 g	13.2 g	47.62%
Yellow	16.7 g	13.2 g	-20.96%
Light Green	14.1 g	15.6 g	10.64%
Dark Green	13.4 g	17.7 g	32.09%
Blue	13.6 g	11.8 g	-13.24%

Squash Data

Color of Solution in Which Squash was Placed	Initial Weight of Squash	Weight of Squash After Overnight Stay	Percent Change
Red	8.1 g	4.4 g	-45.68%
Orange	7.2 g	6.7 g	-6.94%
Yellow	7.9 g	4.5 g	-43.04%
Light Green	8.8 g	8.4 g	-4.55%
Dark Green	5.4 g	6.5 g	20.37%
Blue	7.0 g	4.8 g	-31.43%

We determined that a negative percent change indicated the shrinking of the cell. This would mean that the cell was in a hypertonic solution that had a high concentration of water in the cell and a low concentration of water outside of the cell. The water would move across the concentration gradient from high concentration to low concentration through osmosis, a type of diffusion. This information led us to believe that the most negative percent change of the vegetable equaled the highest molarity of the solution in which it is placed since there would be less water and more solute, in this case sucrose, outside of the cell (in the solution). On the other hand, a positive percent change would indicate the swelling of a cell. This would mean that the cell was in a hypotonic solution that had a low concentration of water in the cell and a high concentration of water outside of the cell. The water would move into the cell through osmosis, causing it to swell, and proving that there was a less solute concentration and higher water concentration in the solution. The most positive percent change in the vegetable equaled the lowest molarity of the solution in which the vegetable is placed.

My lab group and I then arranged these percent changes from greatest to least. For the parsnip, the order was orange, dark green, light green, blue, yellow, and red. For the squash, the order was dark green, light green, orange, blue, yellow, and red. We compared these orders and found that the blue, red, and yellow were 0.6M, 0.8M, and 1.0M, respectively, for each vegetable. The dark green, light green, and orange were in different orders on each greatest to least list. For the squash, the order was dark green, light green, orange. For the parsnip, the order was orange, dark green, light green. We decided to use the parsnip order because it seemed more logical to move one color (orange) to the top since the dark and light green were ordered consecutively on each list and would move up and down the final list as one entity. After analyzing all of our data, we concluded* that the order of molarities is orange (0.0 M), dark green (0.2M), light green (0.4M), blue (0.6M), yellow (0.8M), and red (1.0M).

As I mentioned before, we did not use the data from the dialysis tubing to support our claim. There was more room for error in this experiment. The bags of solution may not have been properly tied and could have leaked, causing a discrepancy in the dialysis tubing bag’s final mass. Also, we did not use the same scale to weigh the bags on the second day of the lab. We also concluded to not use the data from this experiment because the evidence did not support our claim. The percent changes were -6.25% for red, 19.94% for orange, 9.94% for yellow, 0.97% for light green, 0.68% for dark green, and 30.37% for blue. For this experiment, a more negative percent change would represent the shrinking of the cell, and therefore a hypertonic solution and a lower molarity. A more positive percent change would represent the swelling of the cell and therefore a hypotonic solution and a higher molarity. Based on this information, the order of molarity of the solutions would be red (0.0M), dark green (0.2M), light green (0.4M), yellow (0.6M), orange (0.8M), and blue (1.0M). All in all, this data did not support our original claim, and we determined the squash and parsnip experiment to be more accurate.

Dialysis Tubing Data

Color of Solution in Bag	Initial Weight of a Bag	Weight After One Day	Percent Change
Red	32.0 g	30.0 g	-6.25%
Orange	32.1 g	38.5 g	19.94%
Yellow	34.2 g	37.6 g	9.94%
Light Green	30.9 g	31.2 g	0.97%
Dark Green	29.5 g	29.7 g	0.68%
Blue	32.6 g	42.5 g	30.37%

Argument Two

            The water potential for parsnip is -12.29 bars, and the water potential for squash is -3.69 bars. To find this, my lab group and I had to first determine the solute and pressure potential. To calculate the solute potential, we used the formula Ѱs= -iCRT (solute potential equals the number of particles the molecule will make in water times molar concentration times pressure constant times temperature in degrees Kelvin). We had to find C, the molar concentration. To do this, we had to graph the percent changes of each vegetable from Activity Two (squash and parsnip) compared to the molarity of each solution. The x-axis is molarity, and the y-axis is percent change. The molar concentration ended up being where the graph of the percent changes hit the x-axis. These molar concentrations were 0.5M for the parsnip and 0.15M for the squash. We used room temperature (23ºC) as the temperature for the solute potential and converted it to Kelvin (273+23ºC). We then plugged these values into Ѱs= -iCRT. The equation for parsnip was –(1)(0.5)(0.0831)(296) and it equaled -12.29 bars, the solute potential. The equation for squash was –(1)(0.15)(0.0831)(296) and it equaled -3.69 bars, the solute potential. We then determined that the pressure potential was zero, and used the equation Ѱ= Ѱs+Ѱp (water potential= solute potential plus pressure potential) to calculate the water potential. The equation for parsnip would be Ѱ= -12.29+0, and would equal -12.29 bars, the water potential. The equation for squash would be Ѱ= -3.69+0, and would equal -3.69 bars, the water potential. The squash has a higher water potential than the parsnip. Because of this, water can move more freely in and out of the squash cells during osmosis.

Title: Percent Changes of Parsnip in Comparison to Molarities of Solution; x-axis represents the molarity, and the y-axis represents the percent change

Title: Percent Changes of Squash in Comparison to Molarities of Solution; x-axis represents the molarity, and the y-axis represents the percent change

Rameia's Molarity and Water Potential Argumentation

Molarity and Water Potential Argumentation

Rameia Ramsey

   The first challenge presented to us was to figure out the molarity of six colored solutions.  The  colored solutions of red, orange, blue, dark free, dark green, and yellow could have had the molarity of  0.0M, 0.2M, 0.4M, 0.6M, 0.8M, or 1.0M.  As a result of out experiment, we have come to the conclusion that 0.0M is the orange solution, 0.2M is the dark green solution, 0.4M is the light green solution, 0.6M is the blue solution, 0.8M is the yellow solution, and 1.0M is the red solution.  

    In order to determine the molarities of the solutions we used the data collected from the dialysis tubing as well as the data collected from the vegetable we used, which were squash and parsnip.  With the dialysis tubing, we filled each "cell" with each of the solutions.  We weighed each cell and then placed them into individual cups with the different solutions.  The cells were left to soak in the solutions for 24 hours.  After the cells had soaked, we took the cells out of the cups and measured them again to check for a weight difference.  After weighing each cell, we then calculated the percent change in mass of the cells.  To calculate this we subtracted the ending mass from the initial mass and then divided by the initial mass and multiplied by 100.  The blue solution had a percent change of 30.379%, light green of 0.097%, yellow of 9.94%, orange of 19.94%, dark green of 0.68%, and red of         -6.3%.  The negative percent values indicate the cell had decreased in size meaning that water has left the cell and the cell has shrunk because the solution is hypertonic.  Positive percent values indicate that water has moved into the cell causing it to swell, symbolizing that the solution is hypnotic.  With the vegetables, we cut each type of vegetable into six pieces each.  We weighed each piece of vegetable before putting them into individual cups of the different solutions.   Again, we let the vegetables soak for 24 hours before reweighing them again.  After reweighing the vegetables, we calculated the percent change in mass.  The percent change for the squash in a red solution was -45.68%, for orange -6.94%, for yellow 43.04%, for dark green -4.35%, for light green 20.37%, and for blue -31.43%.  The percent change for the parsnips in red solution was -25.32%, for orange 47.62%, for dark green 10.64%, for light green 32.09%, and for blue -13.24%.  We decided to use the vegetables to determine the molarity because there were similarities between the two vegetables whereas the dialysis tubing was all over the place.  In order to better compare the percent change of each piece of vegetable we ordered them from greatest percent change to least. Positive percent change again means that water has moved into the vegetable causing it to swell because it was in a hypnotic solution with more solute on the inside of the cell than the outside and more water on the outside of the cell than the inside.  Negative percent changes mean that the vegetable shrank in size because water had moved out of the cell meaning the solution was hypertonic where there is more solute on the outside of the cell than the inside, and more water inside the cell than outside.  If there is more sucrose on the inside of the cell and more water outside of the cell this would mean that the vegetables that had the most positive percent change have the least amount of molarity.  The vegetables with more sucrose on the outside of the cell and more water on the inside of the cell will have a more negative percent change which means the more negative the percent change, the higher the molarity of the solution.  Based on this from most positive percent change to least positive percent change, the order of least molarity to highest molarity for parsnips would be orange, dark green, light green, blue, yellow, and red.  For the squash the order was dark green, light green, orange, blue, yellow, and red.  Since the orange and dark green switch, in order to find out which solution color had the least molarity we used the percent change of the cell from the orange solution in the dialysis tubing.  Since the orange in the dialysis tubing and the orange for the squash had the most positive percent change we decided to keep orange as the solution with the lowest molarity.  Based on this my group has come to the conclusion that orange is 0.0M, 0.2M is dark green, 0.4 is light green, 0.6 is blue, 0.8 is yellow, and 1.0 is the red solution.

In the second experiment we were challenged with finding out the water potential of our two vegetables.  The water potential of the parsnips was -12.29 bars,  the water potential of the squash was -3.69 bars.  In order to calculate the water potential we had to figure out what the solute potential(().  The equation for solute potential is () = –iCRT.  -i stands for the amount of particles the molecule, in this case sucrose, will make in the water.  C stands for the molar concentration.  We found the molar concentration by creating graphs for our two vegetables.  On the y-axis was the percent change in mass of the vegetable, on the x-axis was the solute concentration.  To figure out the molar concentration you must find where your data line hits the x-axis and that will be your molar concentration. R stands for the pressure constant which is 0.0831 liter bar/mole K   Finally, T stands for the temperature in degrees Kelvin of the solution.  When all of these numbers are found the equation will read that the solute potential of parsnips =(-1)(.5)(.5)(0.0831)(296) which equals -12.29 bars.  For squash the equation would read that solute potential is equal to (-1)(.15)(0.0831)(296) which equals -3.69. After finding the solute potential, we then had to find the pressure potential.  In this case we always used 0 as the pressure potential because this is the point at which there would be equilibrium, where there would be no pressure exerted amongst the water.  With all of this information, we then were able to calculate the water potential of out two vegetables.  The equation for water potential is Water potential () = 

pressure potential () + solute potential ().  For parsnips our equation would then look like 0+-3.69 which equals -3.69.   For squash our equation would be 0+-12.29 which equals -12.29.  This is how the water potential of the two vegetables was determined.  Also, this signifies that the water potential and solute potential are the same in the two vegetables.  

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.