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.