Osmosis Lab Report

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Osmosis Lab Report I. Introduction Diffusion is the random movement of molecules from an area of high concentration to one of lower concentration. Osmosis is the diffusion of water molecules from an area of high concentration to one of low concentration through a selectively permeable membrane. In this experiment, we will be looking at the osmosis of cells within a potato, which allows water to pass through its membrane, but not sucrose. The objective of this experiment is to discover how a change in solvent concentration affects the net magnitude and direction of water through a selectively permeable membrane. II. Hypothesis Potato Core Method: If the sucrose concentration of a solution is increased then there will be an increase of water diffusing out of the potato cells within the solution. Dialysis Tubing Method: If the sucrose concentration of a solution within a dialysis tube surrounded by water is increased

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Osmosis Lab for AP biology from DeBakey High School

Transcript of Osmosis Lab Report

Page 1: Osmosis Lab Report

Osmosis Lab Report

I. Introduction

Diffusion is the random movement of molecules from an area of high concentration to one of

lower concentration. Osmosis is the diffusion of water molecules from an area of high

concentration to one of low concentration through a selectively permeable membrane. In this

experiment, we will be looking at the osmosis of cells within a potato, which allows water to

pass through its membrane, but not sucrose. The objective of this experiment is to discover how

a change in solvent concentration affects the net magnitude and direction of water through a

selectively permeable membrane.

II. Hypothesis

Potato Core Method: If the sucrose concentration of a solution is increased then there will be an

increase of water diffusing out of the potato cells within the solution.

Dialysis Tubing Method: If the sucrose concentration of a solution within a dialysis tube

surrounded by water is increased then there will be an increase of water diffusing into the

dialysis tubing.

III. Materials:

The materials required for the two methods are as follows. In the Potato Core method a potato, a

potato core cutter, 5 beakers, 500 mL of water, 500 mL of sucrose, and a sharpie. In the Dialysis

Tubing method 5 30-cm piece of 2.5cm dialysis tubing, 5 pipettes, 5 water filled beakers, and 5

prepared solutions with varying sucrose concentration: one with 0M, one with .25M, one

with .50M, one with .75M, and one with 1M concentration.

IV. Procedures:

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In the Potato Core method, start by cutting out 5 potato cores using the potato core cutter,

then measure and record the mass in grams of each core. Next fill up one of the beakers with 200

mL of water and 0 mL of sucrose, creating a 0M sucrose solution, one with 150 mL of water and

50 mL of sucrose, creating a 0.25M sucrose solution, one with 100mL of water and 100 mL of

sucrose, creating a 0.50M sucrose solution, one with 50mL of water and 150 mL of sucrose,

creating a 0.75M sucrose solution, and one with 0mL of water and 200mL of sucrose, creating a

1M sucrose solution. Then place each core into one of the beakers, taking care to label each

beaker with its sucrose concentration and the weight of the potato core in it, in order to avoid

confusion. After approximately 48 hours, remove the potato cores from the beakers and take the

final mass of each core. Record the measurements and change in mass.

In the Dialysis Tubing method, obtain five 30-cm piece of 2.5cm dialysis tubing that has been

soaking in water and tie off one end of the tubing to form a bag. Then, using a pipette, fill each

of the dialysis bags with one of the sucrose solutions prepared. make sure each bag receives a

different solution concentration. Record the mass of the bags and then put each one in a separate

water-filled beaker. After approximately 48 hours, remove the bags from the beakers and take

the final mass. Record the measurements and change in mass.

Zucchini Method: N/A

V. Data:

Sucrose Concentration

(M)

Initial Mass of Potato (g)

Final Mass of Potato (g)

Change in Mass(g)

Percent Change in Mass

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0 2.15 2.74 +.59 +27.4

.25 1.8 1.72 -.08 -4.44

.5 2.0 1.61 -.39 -19.5

.75 1.93 1.21 -.72 -37.3

1 1.93 1.4 -.53 -27.46

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Sucrose Concentration

(M)

Initial Mass of tubing (g)

Final Mass of tubing (g)

Change in Mass(g)

Percent Change in Mass

0 9.51 10.2 +.69 7.3

.25 10.2 14.53 +4.33 42.45

.50 5.05 9.26 +4.21 83.36

.75 7.53 14.32 +6.79 90.01

1 8.87 16.35 +7.48 84.3

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Sucrose Concentration (M) % Change in Sucrose Concentration

0 20

0.2 10

0.4 -3

0.6 -17

0.8 -25

1 -30

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VI. Conclusion

The purpose of this experiment was to determine how changes in solvent concentration affect the

direction and magnitude of the diffusion of water. In the potato core method, as the concentration

of sucrose increased in the beaker solution the percent change in mass lowered, because the

water in the potato core moved through the membrane to the area of lower water concentration.

In the dialysis Tubing method, as the concentration of sucrose increased within the dialysis bag

the percent change in mass increased, because the water within the beaker diffused into the bag

of lower water concentration. The Zucchini method behaved identical to the potato core method,

where the percent change in mass decreased as the sucrose concentration in the beaker increased,

because as the concentration rose to that above the concentration within the zucchini the water

within the zucchini diffused to the lower water concentration(remember that higher sucrose

concentration means lower water concentration). In all three methods, water diffused towards the

area of higher sucrose concentration. As the concentration of sucrose rose, either inside the

membrane or outside the membrane, so did the magnitude of the water that diffused towards it.

VII. Questions

1. Explain the relationship between the change in mass and the molarity of sucrose within

the dialysis bags.

The percent change in mass and the molarity of the sucrose within the dialysis bag are directly

proportionate, because as the concentration of sucrose in the bag increases more water is needed

to diffuse into to the bag, so that the concentration gradient both inside and outside the bag

become close to equal.

2. Predict what would happen to the mass of each bag in this experiment if all the bags were

placed in a 0.4M sucrose solution instead of distilled water. Explain your response.

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The masses of the bags that had a sucrose concentration of 0M and 0.25M would decrease,

because water would diffuse out of the higher water concentrated bags into the lower water

concentrated solution(remember that higher sucrose concentration means lower water

concentration). The masses of the bags that had a sucrose concentration of 0.5M, 0.75M, and 1M

would decrease, because water would diffuse out of the lower water concentrated bags into the

higher water concentrated solution.

3. Why did you calculate the percent change in mass rather than simply using the change in

mass?

Because the percent change in mass creates a standard for all the masses out of a hundred, so that

the change can be compared regardless of the core’s initial mass. For example, a core with initial

mass 10g would lose twice as much water as a core with mass 5g in the same solution

concentration, but the percent change would be the same.

4. A dialysis bag is filled with distilled water and then placed in a sucrose solution. The

bag’s initial mass is 20g, and its final mass is 18g. Calculate the percent change of mass,

showing your calculations in the space below.

Percent Change of mass= (change in mass)/(initial mass)X100 ;

change in mass of mass= (final mass)-(initial mass)

Percent Change of mass= [(final mass)-(initial mass)]/(initial mass)X100

Percent Change of mass= [(18-20)/20]X100

Percent Change of mass= (-2/20)X100

Percent Change of mass= -10

7. b) What is the molar concentration of solutes within the zucchini cells?

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The molar concentration is the x-axis intercept, because it is the point where there is no net gain

or loss due to osmosis, meaning that the concentration inside and outside are equal.