• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Diffusion and Osmosis Lab

Extracts from this document...

Introduction

Diffusion and Osmosis Lab Hemi Ryu 11/21/05 Per. 4B Background Information: All atoms and molecules possess kinetic energy due to their constant motion. Diffusion is the movement of molecules from an area of higher concentration to an area of lower concentration and is caused by the atoms' kinetic energy. Osmosis is the diffusion of water through a selectively permeable membrane, a membrane that allows the diffusion of only selected solutes and water. During osmosis, water moves from a region of higher water potential to a region of lower water potential until equilibrium is reached. Water potential is the measure of free energy of water in a solution. Cells have ways to move ions or molecules incapable of diffusion and osmosis into and out of themselves. This process is called active transport, where carrier proteins use energy from ATP to move substances across the cell membrane against a concentration gradient, from a region of lower concentration to a region of higher concentration. Two solutions are either hypertonic, hypotonic, or isotonic to each other. Hypertonic describes a solution that has a higher solute concentration than another solution. Hypotonic describes a solution that has a lower solute concentration than another solution. Isotonic is used to describe solutions with equal solute concentrations. Experiment 1: Diffusion Hypothesis: If a semi-permeable bag filled with glucose and starch solution is placed in a beaker of distilled water, then the glucose and starch will ...read more.

Middle

3) Fill each beaker with two-thirds of distilled water 4) Record the mass of each bag. 5) Immerse each bag in a beaker of distilled water and label the beakers to indicate the molarity of the solution in the dialysis bag. 6) Wait 30 minutes. 7) Record the mass of each bag. 8) Determine the percent change in mass for each bag: % change = final mass - initial mass x 100 initial mass Data: Contents in Dialysis Tube Initial Mass (g) Final Mass (g) Percent Change in Mass a) distilled water 24.00 23.66 -1.42% b) 0.2-M sucrose 24.19 20.13 -16.78% c) 0.4-M sucrose 25.55 21.45 -16.05% d) 0.6-M sucrose 26.13 27.23 4.21% e) 0.8-M sucrose 27.19 27.53 1.25% f) 1.0-M sucrose 27.41 29.34 7.04% Analysis: The masses of the bags with the distilled water, 0.2-M sucrose, and 0.4-M sucrose solutions decreased and each gave a negative percent change in mass. These are errors and were probably caused by inexact measurements of solutes and solutions. The mass of the distilled water should have stayed the same because there would have been no net movement of water because the two liquids are both distilled water. The mass of all the sucrose-containing bags should have increased because even though sucrose comes out of the bags, the amount of water that diffuses into the bag has more mass than the sucrose. ...read more.

Conclusion

Or, the other way, when blotting the potato cells after taking them out of the solutions, too much solution may have been blotted. According to the graph, the molar concentration of sucrose at which the mass of the potato cores does not change is 0.4. To find this out, I drew a line of best fit in the graph. The point at which the line crosses the x-axis represents the molar concentration of sucrose with a water potential that is equal to the potato tissue water potential. At this concentration there is no net gain or loss of water. However, from my data, the potatoes decreased in mass in the 0.4-M solution. So, either my graph is wrong, or my data of the 0.4-M potato final mass is wrong. Using the formula, I found the osmotic potential of the 0.4-M sucrose solution to be - 9.84 bars. Because osmotic potential is equal to the water potential when the pressure potential of the solution is zero, the water potential is also - 9.84 bars. Conclusion: If potatoes are placed in distilled water, then their mass will increase because the water is hypotonic to the potato solution. If potatoes are placed in sucrose solutions of 0.2 and 0.4-M, then their mass will increase because the solutions are also hypotonic to the potato solution. Howver, if potatoes are placed in sucrose solutions of 0.6, 0.8, and 1.0-M, then they will decrease in mass because water will diffuse out of the potatoes because they are immersed in a hypertonic environment. ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our AS and A Level Exchange, Transport & Reproduction section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related AS and A Level Exchange, Transport & Reproduction essays

  1. Marked by a teacher

    Investigate the water potential of celeriac.

    5 star(s)

    Therefore, some results represented a false percentage change in mass, either losing more mass than they should have or not losing enough mass. The solution would be to have a machine that cut pieces equally, therefore eliminating any human error.

  2. Marked by a teacher

    Determining the Water Potential of Sweet Potato Tissue

    4 star(s)

    changes in mass are not greater than at 50 % the maximum is 24 % this matches my prediction. I have also found out that ?s and ?p are opposite and plant cells become flaccid or turgid to get the ?

  1. Marked by a teacher

    Osmosis. Aim: To find the molarity of potato tubers cell sap. BIOLOGICAL ...

    4 star(s)

    from the cells to maintain the concentration on either side of the membrane. 0.25 molar glucose solution however showed an unexpected result of positive percentage change. The only probable reason for this is that the molarity of the strips was more than 0.25 molar solution.

  2. Investigating osmosis on swede cells.

    1st 2nd 3rd Average 0 15.9 28 15.3 19.7 0.2 6.8 10.8 8.84 8.81 0.4 8.19 21.35 6.72 12.08 0.6 5.5 -15.4 7.69 -0.75 0.8 -15.88 -3.6 -5.55 -8.34 1 -4.95 -1.98 2.8 -1.37 Length percentage change: Concentration (M): Percentage change (%)

  1. Investigation on Osmosis using a potato.

    Final tests For these next three tests I am going to change as I have explained in all of the above bullet points to help ensure it is a fair and worthy test. Obtaining evidence After the experiment, I had to dry the potatoes first, so that the water outside

  2. To find out the factors affecting the refractive index of liquid by using different ...

    Since the density of oil (100% concentration) is the highest, I predict that the refractive index of oil is greater than sugar and salt solution. Thirdly when I compare the same concentrations of sucrose and glucose solution, the refractive index should be similar; since they are both hexose and they

  1. Design an experiment to investigate the effect of temperature on the movement of a ...

    96 90 98 93 91 96 90 94 94 45 90 92 95 93 90 91 85 93 87 82 90 82 89 89 50 82 91 85 88 81 85 69 88 78 74 84 71 81 81 55 40 57 61 45 60 73 49 75 60 66

  2. Find out the concentration of sucrose solution that is equivalent to the osmotic potential ...

    of water going into the cell is more than the flow of water leaving the cell. * If there is a lower concentration of water surrounding the planet or animal cell and a higher concentration of water in the cell, then the cell will shrink and lose weight.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work