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Bio Potato Core Osmosis Essay Sample

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Bio Potato Core Osmosis Essay Sample

Table 1: Weights of potato pieces
Solutions for the potato pieces to be dipped in(M) | Mass of potato pieces in Grams (+/-0.01)(initial mass)| Time when the potato pieces were dipped in the solution in seconds (+/- 0.01)| Time when the potato pieces were taken out of the solution in seconds(+/- 0.01)| Mass of potato after a hour in Grams (+/- 0.01) (final mass)| Water | 2.62| 0| 0| 3.19|

0.1| 2.54| 60| 180| 2.99|
0.2| 2.44| 60| 120| 2.76|
0.3| 2.56| 12| 150| 2.76|
0.4| 2.61| 60| 138| 2.67|
0.5| 2.83| 24| 180| 2.84|
0.6| 2.61| 42| 60| 2.40|

Qualitative Data:
Sucrose concentrations appeared to be colorless. Osmosis is is the net movement of solvent molecules through a partially permeable membrane into a region of higher solute concentration. Potato is made up of tiny cells and is protected by a cell membrane. Six solutions and water was taken to obtain the results of Osmosis. The potato pieces sank into the solutions. The change in the masses shows that osmosis occurred. Mostly there is a change in the mass of the potato when sank in the solution for about an hour.

Processed data:
Table 2: Difference in the initial and final masses of the potato pieces

#3 Madan

Solutions for the potato pieces to be dipped in(M)| Mass difference in the Potato pieces (Grams)| Water| 0.57|
0.1| 0.45|
0.2| 0.32|
0.3| 0.20|
0.4| 0.06|
0.5| 0.01|
0.6| -0.21|

Table 3: % change in the mass of the potato pieces
Solutions for the potato pieces to be dipped in(M)| % change in the mass of the potato pieces| Water| 21.76|
0.1| 17.71|
0.2| 13.11|
0.3| 7.81|
0.4| 0.29|
0.5| 0.35|
0.6| -8.04|

Presented Data:
Graph 1: it shows the difference in the mass of the potato pieces

Graph 2: It shows the % change in the mass of the potato pieces

Example calculations:
#4 Madan

1. To obtain the difference in the masses of the potato pieces after they were taken out of the solutions was as following

Initial mass – final mass
3.19 – 2.62 = 0.57

2. In order to obtain the % change in the mass of the potato pieces we do the following:

Difference in the masses of the potato X 100 Initial mass

0.57/2.62 *100 = 21.76%

Conclusion:
Osmosis is is the net movement of solvent molecules through a partially permeable membrane into a region of higher solute concentration. The sucrose solutions act as solvents here and the potato is the solute. The solvent moves through the semi-permeable membrane of potato. Hypotonic solution is the one with lower osmotic pressure, so 0.6m surcose solution was a hypotonic solution as the mass of the potato decreased whereas the hypertonic solution is one with a higher osmotic pressure, so 0.1m was the highest. 0.5m sucrose solution is an isotonic solution as they are the solutions that have no net movement across the membrane and the osmotic pressure is equal.

#5 Madan

Evaluation:
There are certain limitations in the following experiment that can affect the results of the experiment. Everything should be done neatly and accurately so we can obtain the correct results. The following are the limitations and improvements for the experiment that should be taken care of.

Limitations| Improvements|
The size of the potato might be different. Some could be smaller than the others, some could be bigger than the others| The size of the potato should be equal.| Noting the time might mess up| The time noted should be precise and taken care as it will tell us how much it took for the potato to absorb the solutions| Uncertainties| Uncertainties should be taken care or else they might affect the results in calculations.| Skin of the potato could be left or might not be removed properly| The skin of the potato should be removed or else it will act as a barrier and lower the rate of osmosis| Potato pieces could be wet before soaked in the solutions| Potato pieces should be dried by taping tissue over it as if they are wet the might affect in the rate for Osmosis.

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