# IB Biology Potato Lab

Table 1:

Trial Number| Concentration of Sucrose Solution (M) ±0.2 ml| Initial Mass of Potato Core Slice(g) ±0.1 | Final Mass of Potato Core Slices (g) ±0.1| 1| 0.0| 7.7| 9.3|

2| | 6.0| 8.1|

3| | 6.2| 7.4|

4| | 10.2| 13.2|

5| | 8.7| 10.3|

6| | 4.9| 6.0|

7| | 9.2| 10.4|

1| 0.2| 5.8| 6.0|

2| | 11.6| 12.1|

3| | 2.5| 3.1|

1| 0.4| 14.4| 13.9|

2| | 2.6| 2.8|

3| | 8| 6.5|

1| 0.6| 7.3| 5.3|

2| | 10.7| 7.3|

3| | 9.6| 7.4|

4| | 2.9| 2.8|

1| 0.8| 5.6| 3.6|

2| | 16.0| 13.1|

3| | 11.5| 5.9|

1| 1.0| 7.9| 5.4|

2| | 10.0| 6.7|

3| | 9.6| 6.2|

4| | 4.7| 3.2|

Table 2: Calculations of Averages

Concentration of Sucrose Solution (M) ±0.2 ml| Average Initial Mass of Potato Core Slice(g) ±0.1 | Average Final Mass of Potato Core Slices (g) ±0.1| Change in Mass (g) ±0.1| Percentage Change in Mass (%)| Standard Deviation of Initial Mass| Standard Deviation of Final Mass| 0.0| 7.6| 9.2| 1.6| 22.3| 1.9| 2.4|

0.2| 6.6| 7.1| 0.5| 6.5| 4.6| 4.6|

0.4| 8.3| 7.7| -0.6| -7.2| 5.9| 5.7|

0.6| 7.6| 5.7| -1.9| -25.2| 3.5| 2.2|

0.8| 11.0| 7.5| -3.5| -31.7| 5.2| 5.0|

1.0| 8.1| 5.4| -2.7| -33.2| 2.4| 1.5|

Sample Calculations:

To find average of initial mass of potato core with 0.2M solution: # 1 + # 2 + # 3 / 3

5.8 + 11.6 +2.5 / 3 = 6.6

To find average of final mass of potato core with 1.0M solution: # 1 + # 2 + # 3 + # 4 / 4

5.4 + 6.7 + 6.2 + 3.2 / 4 = 5.4

To find change in mass in 0.4M solution:

Final mass – initial mass = change in mass

7.7 – 8.3 = -0.6

To find percentage change in mass in 0.6M solution:

Final – initial / initial x 100

5.7 – 7.6 / 7.6 x 100 = -25.2

Figure 1: The effect of sucrose solution on the mass of potato cores

Figure 1: In the above graph, it’s visible that with an increasing concentration of sucrose solution there is also a decrease in the percent change in mass. The r² value of 0.9416 represents that there is a good relationship in the data between the concentration of sucrose solution and the percent change in mass. The relationship between the data can also be proven by the error bars, representing the standard deviation from the data points and the amount of uncertainty. With the small error bars we know that the data is reliable; although as the last few points overlap this indicates that the data is similar.

Conclusion:

I found that as the concentration of sucrose increased, the change in mass and percentage of the change in mass decreased. This data did support the hypothesis, as we knew from the term osmosis. Osmosis is the process of diffusion of water molecules from an area of higher concentration to lower concentration. The concentration gradient between the potato and the sucrose solution lead to the amount of molecules coming in and out of the potato. Therefore, the data supports the hypothesis as when the water concentration was lower in the potato than in the sucrose solution, the water molecules moved through the semi-permeable membrane into the potato which caused it to gain weight. With a higher concentration of water in the potato, the result would be the opposite. Hence the prediction that the lower the concentration of sucrose, the higher the final weight of the potato was reinforced. The hypothesis is supported by the evidence of the graph. With a concentration of 0.0M sucrose solution the increase in the change of mass was 1.6g and a percentage change of 22.3%. In comparison, the 0.8M sucrose solution had a change in mass of -3.5g and a loss of 31.7%. Although in Figure 1 the r² value gives the impression that the data is very reliable, some of the error bars do overlap. While observing the data there are no outliers present, although when inspecting the change in mass the development between the 0.8M concentration of sucrose and the 1.0M numbers slightly...

Bibliography: Diffusion and osmosis. (n.d.). Retrieved from http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/diffus.html

McGraw-Hill. (2006). How osmosis works. Retrieved from http://highered.mcgraw-hill.com/sites/0072495855/student_view0/chapter2/animation__how_osmosis_works.html

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