Water Potential and Molarity Mixup:
Activity 1: Experimental Determination of Unknown Molarities
Claim:
In this experiment, we were tasked with figuring out the molarity of six unknown sucrose solutions that were identified by different colors: red, orange, yellow, green, blue, and purple. The goal of the experiment was to figure out the molarity of the solutions within the range of 0M to 1M with increments of 0.2M (0M, 0.2M, 0.4M, 0.6M, 0.8M, and 1M). After conducting the experiment, my group concluded that the 0M solution was green, the 0.2M solution was orange, the 0.4M solution was blue, the 0.6M solution was purple, the 0.8M solution was red, and the 1M solution was yellow.
Evidence:
In order to conduct our experiment we placed varying amounts of each solution in 75 mL of another solutions using dialysis tubing and string. Bag 1 contained the purple solution and was placed in 75 mL of the purple solution. Bag 2 contained the yellow solution and was placed in 75 mL of the green solution. Bag 3 contained the orange solution and was placed in 75 mL of the blue solution. Bag 4 contained the green solution and was placed in 75 mL of the orange solution. Bag 5 contained the red solution and was placed in 75 mL of the purple solution. Bag 6 contained the blue solution and was placed in 75 mL of the red solution. Bag 7 contained the purple solution and was placed in 75 mL of the orange solution. Bag 8 contained the yellow solution and was placed in 75 mL of the blue solution. Bag 9 contained the red solution and was placed in 75 mL of the yellow solution. Bag 10 contained the blue solution and was placed in 75 mL of the green solution. Lastly, bag 11 contained the green solution and was placed in 75 mL of the red solution. We then weighed each of the bags and let them soak in their respective solutions overnight, and then weighed them again the next day. The data from this part of the experiment is below.
Activity 1 Data: Experimental Determination of Unknown Molarities
Bag Numer
|
Bag Color
|
Solution Placed In
|
Bag Weight Start (grams)
|
Bag Weight - End (grams)
|
1
|
Purple
|
Yellow
|
2.99
|
2.7
|
2
|
Yellow
|
Green
|
1.99
|
3.9
|
3
|
Orange
|
Blue
|
3.2
|
3.2
|
4
|
Green
|
Orange
|
2
|
1.7
|
5
|
Red
|
Purple
|
2.73
|
3
|
6
|
Blue
|
Red
|
3.3
|
3.5
|
7
|
Purple
|
Orange
|
4.9
|
5.85
|
8
|
Yellow
|
Blue
|
4.7
|
6.3
|
9
|
Red
|
Yellow
|
4.2
|
4.1
|
10
|
Blue
|
Green
|
5.1
|
7.1
|
11
|
Green
|
Red
|
2.9
|
2.4
|
The above data shows the results of soaking the dialysis tubing bags in their surrounding solutions overnight. Bag 1 lost weight, bag 2 gained weight, bag 3 stayed the same, bag 4 lost weight, bag 5 gained weight, bag 6 gained weight, bag 7 gained weight, bag 8 gained weight, bag 9 lost weight, bag 10 gained weight, and bag 11 lost weight. We then used the change in weight of each bag to determine which solution matched up with which molarity.
Reasoning:
After weighing the bags on the second day of the experiment we determined the relationship between each bag and its surrounding solution. For example, bag 1, being purple and being placed in the yellow solution, lost weight, which helped us conclude that the yellow solution was more concentrated than the purple solution and it was a hypertonic environment. We used this similar reasoning for each of the eleven situations. After analyzing each situation we found that the yellow solution was more concentrated than the purple, green, blue, and red solutions. We were also able to determine that the purple solution was more concentrated than the orange solution, that the green solution was the least concentrated solution, that the blue solution was more concentrated than the green solution, that the red solution was more concentrated than the green and purple solutions, and that the orange solution was more concentrated than the orange solution. We then applied the logic that if “a” is more concentrated than “b”, and “b” is more concentrated than “c”, then “a” is more concentrated than”c”. By the end of our process, we were able to conclude that the yellow solution was 1M, the red solution was 0.8M, the purple solution was 0.6M, and the green solution was 0M. This left the orange and blue solutions to be labeled, and our setup did not account for that direct relationship. In order to distinguish these two final labels, we compared the percent change in mass of the bag 4 environment and the bag 10 environment since both the orange and blue solutions were compared to the green solution. We were able to determine that blue was more concentrated than orange because the environment in bag 10 showed the greatest percent change in weight. Therefore, we were able to identify the blue solution as 0.4M and the orange solution as 0.2M. The combination of our deductive reasoning and comparisons allowed us to ultimately conclude that the 0M solution was green, the 0.2M solution was orange, the 0.4M solution was blue, the 0.6M solution was purple, the 0.8M solution was red, and the 1M solution was yellow.
Activity 2: Experimental Determination of the Water Potential of a Parsnip
Claim:
In this part of the experiment we were tasked with finding the water potential of a vegetable, which our group chose a parsnip. After conducting our experiment we were able to conclude that the water potential of a parsnip is -43.64 bars.
Evidence:
In order to conduct this part of the experiment we cut pieces off of a single parsnip using a scalpel. We then placed one of those pieces in each of the six colored solutions (one in the red, one in the orange, one in the yellow, one in the green, one in the blue, and then one in the purple). After we placed the parsnips in their respective solutions, we let it soak overnight in the attempt to determine the water potential of the vegetable. We originally did the experiment, but found that each of the parsnips gained weight overnight, so we redid the experiment and are using the data from the second trial of our experiment. After soaking the parsnips overnight, we found that, once again, each piece of parsnip gained weight, so we knew that there was no fault in the method of our experiment. We used the weight of the parsnips before and after the soaking to determine the percent change in mass of the parsnips overnight. The data from our experiment is below.
Molarity of Solution
|
Start Weight (grams)
|
End Weight (grams)
|
Percent Change in Mass
|
0.2
|
1.5
|
1.7
|
13.33333333
|
0.4
|
2.3
|
2.7
|
17.39130435
|
0.6
|
1.3
|
1.5
|
15.38461538
|
0
|
2.1
|
2.5
|
19.04761905
|
0.8
|
2.2
|
2.5
|
13.63636364
|
1
|
2.4
|
2.5
|
4.166666667
|
The above data shows that each parsnip piece gained weight. We were able to use the percent change in mass in each of the parsnips in a graph that would allow us to find the molarity of the parsnip, and thus the water potential of the parsnip.
Reasoning:
In order to figure out the water potential of the parsnip we first had to figure out concentration of the parsnip (in Moles). However, our data showed that each of the parsnips in our solution gained weight, which showed us that we were not going to be able to find a zero point on a graph because the concentration of the parsnip was going to be greater than 1M. In order to figure out the concentration of the parsnip, I made a graph using excel that placed a line of best fit on the graph of our data. The graph is included below.
The equation of the line of best fit of our data is useful because it can be used to find the point on the graph where y=0 (the concentration point of the parsnip). To calculate the concentration of the parsnip I set the equation equal to zero and found that the concentration of the parsnip was 1.78 bars. From there, I used the equation for water potential to find the water potential of the parsnip Ψ=Ψs+Ψp. Since the system was open, I knew that the pressure potential, Ψp, of the solution was zero, so all I had to figure out was the solute potential of the solution which can be found through the equation Ψs=-iCRT. I calculated the solute potential of the equation in the following way: -1*1.78*0.0831*295=-43.64 bars (assuming room temperature to be 22°C). Since the equation for water potential in an open system is Ψ=Ψs I had my answer. The water potential of our parsnip was -43.64 bars.
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