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Diffusion and
Osmosis
By Lexi Parker and Tana-Rae Halfkenny
Exercise 1A: Diffusion
Evidence:
Table 1: Diffusion Results
Initial
Contents
Initial
Solution
Colour
After Lugol’s
and
Benedict’s
Final Solution
Colour
Presence of
Glucose
Positive
Positive
Bag:
15% Glucose
and 1%
Starch
Milky
White
Light Blue
Separated
(clear on
bottom; and
milky on top)
Beaker:
H₂O and IKI
Clear
Light Yellow
Foggy White
Discussion: Diffusion is the movement of molecules from high concentration to low concentration. As an
example, if Mr. Trentham walked into the classroom with some really stinky cologne, it would diffuse
into the air and all of his students would be able to smell it because the fragrance particles dispersed
into the air because the molecules will move to open spaces (low concentration). If we place dialysis
tubing filled with a solution of 15% glucose and 1% starch, then the tubing will expand because there is a
high concentration of water outside of the tubing (hypotonic solution) and the particles inside of the bag
will diffuse out (if the selectively permeable membrane allows it). The purpose of this lab was to
measure the effects of diffusion through dialysis with a solution of glucose and starch. The manipulated
variable in this test is the addition of Benedict’s reagent and Lugol’s reagent, the responding variable in
this lab is the colour of the substance, and the controlled variable in this test is the substances used,
temperature, and the amount of reagent used. Water is the control group in this test. Water is used as
the control group to show a test with no changes because it is a universal solvent. The contents in the
bag reacted with the Benedict’s reagent because it has glucose in it. The bag’s final color was a result of
diffusion. During the waiting period, the glucose in the bag transferred through the semi-permeable
membrane into the solution. The solution was then polluted with glucose molecules causing it to be a
foggy white color. The evidence that supports our answer includes the knowledge of hypotonic and
hypertonic solutions. The solution in the beaker is hypotonic and the solution in the bag is hypertonic.
The concentration of glucose and starch inside the bag was higher that outside the bag. This means that
the glucose and starch would have passed through the semi-permeable membrane. The pores in the
dialysis tubing allowed the glucose and starch to pass through, but not the Benedict’s and Lugol’s
reagent. This is due to the fact that the pores are not big enough to fit the reagents. We could have
measured the weight of the beaker and the bag before and after the test to get quantitative data. We
could have used this data to further prove the diffusion of water in the dialysis bag. Based on our
observations, the following are ranked by size: (Smallest-Largest) water molecules, IKI molecules,
glucose molecules, membrane pores and starch molecules. These were determined by the reactions in
the lab and information we already know. Water will obviously have the smallest pores because the
atom size is very small due to the oxygen and hydrogen that make it up. Iodine and glucose both passed
through the membrane so we can infer they are smaller than the membrane pores. Starch is the largest
because it didn’t pass through the pores. If the results started with a glucose and IKI solution inside the
bag and starch and water outside, the glucose and starch would pass through the membrane, while the
starch wouldn't pass through, due to the relative sizing of the atoms.
Exercise 1B: Osmosis
Purpose: The purpose of this lab is to use dialysis tubing to determine solute concentration and water
movement through a selectively permeable membrane by the process of osmosis.
Hypothesis: If we place a dialysis bag of distilled water in a beaker of distilled water, there will be no
change because there will be no net movement since this is an isotonic solution. If we place a dialysis
bag of 0.8 M sucrose in a beaker of distilled water, the mass will increase because the net movement
will cause the water to move into the bag, causing expansion. This will occur because the water is a
hypotonic solution while the sucrose is a hypertonic solution.
Variables: The manipulated variable in this lab is the addition of the dialysis bag into the distilled water.
The responding variable is the mass of the bag after osmosis. The controlled variables are the amount of
distilled water in the beaker, the amount of solution and the time waited for osmosis to occur.
Evidence:
Table 2 : Dialysis Bag Results : Individual Data
Contents in
Bag
Initial Mass
Final Mass
Mass Difference
Percent Change in
Mass
Distilled
Water
23.55
22.49
1.06
4.50%
0.8 M
sucrose
20.05
23.45
-3.4
16.96%
Table 3 : Dialysis Bag Results : Class Data
Grp. 1
Distilled
Water
Grp. 2
0%
Grp. 4
Grp. 5
1.4%
0.2 M
0.4 M
Grp. 3
5.9%
Grp. 7
Total
Class
Avrg.
4
1.3%
4.6%
12.6
4.2%
5.4%
22.3
7.4%
Grp. 8
2.6%
4%
11%
Grp. 6
4%
0.6 M
10%
11%
0.8 M
13.6%
1.0 M
14.7%
𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑀𝑎𝑠𝑠 =
=
16%
23%
25.4%
21
10.5%
52.6
17.5%
40.1
20.1%
𝐹𝑖𝑛𝑎𝑙 𝑀𝑎𝑠𝑠 − 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑀𝑎𝑠𝑠
∗ 100
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑀𝑎𝑠𝑠
18𝑔 − 20𝑔
∗ 100
20𝑔
= −0.1 ∗ 100
= −10
The sucrose solution in the beaker would have been hypertonic to the distilled water in the bag.
Graph 1: Dialysis Bag Results
Exercise 1C: Water Potential
Purpose: To determine what will happen when a potato is placed in solutions with different solute
potential.
Hypothesis: If we place a potato in distilled water, it will cause it to expand because the water potential
is lower in the potato which causes the water to flow into it because water will always move to an area
of lower water potential. If we place a potato in 0.8 M sucrose, it will cause it to shrivel up because the
water potential in the sucrose is lower which causes the water to flow into it because water will always
move to an area of lower water potential.
Variables: The manipulated variable in this lab is the placement of the potatoes in the solutions. The
responding variable is the reaction of the potato in the solutions. The controlled variables are the
amount of potatoes used in each solution, the amount of solution, and the amount of time waited for
the reaction.
Evidence:
Table 4: Potato Core: Class Results
Contents Initial
in
Mass
Beaker
Distilled 3.32
Water
Final
Mass
3.88
Mass
Percent
Difference Change
in Mass
-0.56
16.87%
0.2 M
5.00
5.87
0.87
17.00%
0.4 M
2.20
2.26
-0.60
-2.70%
0.6 M
2.10
1.87
0.23
10.90%
0.8 M
2.67
2.06
0.61
-0.23%
1.0 M
5.27
6.45
1.18
22.40%
Exercise 1D: Calculations of Water Potential from Experimental Data