Determining the Isotonic Point of Potatoes in Sucrose Solution

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Figure 2: Illustrates the qualitative observations from the experiment.

Qualitative observation

Explanation of observation

Potato samples losing mass.

When the potatoes were taken out of the test tubes and their mass was measured, they were simultaneously losing mass, which was being absorbed by the paper towel placed underneath for cleanliness.

Texture of potatoes.

When the potatoes were first cut out using the boring tool, they were rough and hard. When taking the potatoes out from the test tubes after 24 hours, they were very soft, and not at all rigid, almost crumbling into multiple pieces upon the slightest touch.

Color of solution

When the sucrose solution was first prepared it was colorless. After 24 hours of the potato sample being immersed in it, the solution was very cloudy, probably due to the release of starch in the potato.

Size of potatoes.

After the potato samples were submerged in solution for 24 hours, the potatoes looked a lot thicker, but their height remained relatively the same.

Color of potatoes.

The potato samples looked slightly more yellow, with a stain like color on the surface, which darken with the increase in sucrose concentration.

Data Processing: (Sample calculations)

Overview: Processing raw data is essential to the evaluation of every lab as, in most labs there are multiple values that are obtained through multiple trials. Processing this raw data allows us to further analyze it and gain a better understanding of the experiment. For this lab we used the final and initial masses to determine the mean change in mass for each of the potatoes and then converting it to a percentage. This helped identify the difference in mass between the potatoes in the start compared to the end of the experiment. Calculating the mean was important because the trials have slightly different values each, but for the graph, only the set of values that represent the average of the data points is needed. Finally, standard deviation was calculated to summarize the spread of values around the mean.

Percent change in mass:

% Change in mass = End mass – Start mass /Start mass *100%

Start mass = 4.86g

End mass = 4.49g

% Change in mass = [(4.49g ± 0.01g) – (4.86g ± 0.01g)/ (4.86g ± 0.01g)] * 100%

= [( -0.37g ± 0.02g) / (4.86g ± 0.01g)] * 100%

= [(-0.37g ± (0.02g/-0.37g *100%)] / (4.86g ± (0.01g/4.86g *100%))

= [(-0.37g ± 5.40%)] / (4.86g ± 0.20%) * 100%

= [(-0.07613g ± 5.60%)] * 100%

= [(-0.07613 ± 0.00426g)] * 100%

= -7.613 ± 0.426 %

= -7.6 ± 0.4%

Therefore, there was a 7.6 ± 0.4% loss in mass; water moved out from the potato sample.

Mean percent change in mass:

Mean percent change in mass = sum of percent change / total number of trials

Trial 1percent change = 20.04%

Trial 2 percent change = 32.30%

Mean percent change = 20.04% + 32.30% / 2

= 52.34% / 2

= 26.17 %

Therefore; the mean percent change in mass for trial 1 and trial 2 of the same sucrose solution concentration is 26.17%.

Figure 3: Illustrates the percent change in mass and the mean percent change in mass.

Sucrose solution concentration (mol/L)

Trial

Percent change in mass

Mean percent change in mass

0 mol/L

1

21.30%

20.30%

2

19.30%

0.25 mol/L

1

5.25%

4.80%

2

4.35%

0.5 mol/L

1

-14.55%

-12.30%

2

-10.05%

0.75 mol/L

1

-27.44%

-28.60%

2

-29.77%

1 mol/L

1

-37.95%

-34.10%

2

-30.26%

Figure 4: Illustrates the percent uncertainty and mean percent uncertainty of the values optained in figure 3.

Sucrose solution concentration (mol/L)

Trial

Percent uncertainty

Mean percent uncertainly

0 mol/L

1

0.44%

0.44%

2