• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  14. 14
    14
  15. 15
    15
  16. 16
    16
  17. 17
    17
  18. 18
    18
  19. 19
    19
  20. 20
    20
  21. 21
    21
  22. 22
    22
  23. 23
    23

Osmosis Investigation

Extracts from this document...

Introduction

Osmosis Investigation To understand my task, I will first understand the concept of osmosis. Osmosis is the process of water movement from a region of low concentration to a region of high concentration across a semi permeable membrane, which is simply a thin membrane allowing the passage of small molecules. Ex: water molecules. The solutions of high concentration and low concentration are generally referred to as hypertonic and hypotonic solutions respectively. Hypertonic solutions have less water molecules and thus are said to have lower water potential than hypotonic solutions; in hypertonic solutions, solute molecules (ex: sugar, salts) are dissolved and when this occurs, some water molecules form a cluster around them. Thus, there are less free moving water molecules. Therefore, water potential is simply the measure of whether the solution is likely to gain or lose water; it is also the pressure exerted by the freely moving water molecules. It is measured in kilopascals. (kPa) Therefore, osmosis may also be defined as the water movement from a region of high water potential to a region of low water potential across a semi permeable membrane. Osmosis is a very important biological process as it involves the transfer of water in and out of cells. Animal and plant cells both have the presence of the jelly-like substance known as the cytoplasm, made up of about 90 percent water with dissolved sugars and salts. This forms a weak solution. Plant cells also have a large space known as the vacuole containing a weak solution of sugars, salts and water known as the cell sap. If the concentration of the solution is more concentrated in the cell than its external solution, then water moves into the cell. This is particularly important in plant cells, as they will become turgid and strong. If too much water enters animal cells, they will burst and this is damaging. However, plant cells can limit the amount of water that enters the cell due to the cell wall; when much water enters ...read more.

Middle

Furthermore, when inserting the cork borer into the potato, I have made sure to place the potato on a raised platform and made sure the potato was not cupped in my hand to prevent any injury to my hand. Analyzing Evidence With the data obtained from my experiment, it is now necessary to fully understand them and analyze the results. More than that, it is important to prove why the results are given as so. To do so, it may help to construct a bar graph to clearly present the data obtained from the experiment. The bar graph on the next page clearly reveals the change in length for the potato tubors in the different concentrations of solution; it presents the initial length of the tubors and its final length. From this graph, it is obvious that with a less concentrated solution, the potato tubor actually gains length. For example, when the potato tubors are placed in distilled water, from four centimeters, they increase in length to 4.43 centimeters, an increase of 43 millimeters. This relies on the fact of osmosis; if a cell is placed in a less concentrated solution, then in order for equalization to occur, that is for the concentration to be equal on both sides, then water must move into the cell so that its cell sap becomes diluted and so that the concentration gradient is eliminated. In this case, since the potato cells inside the tubor obviously have cell sap, a mixture of salts and sugars with water, it forms a weak solution. Even though it is weak, it is still more concentrated than the distilled water. Thus, in order for equalization to occur, some of the distilled water must move across the cell membrane of the potato cell and enter it. In this case, the cells will become bloated or turgid; they become firm and full. Due to this, the potato tubor will increase in length as the potato cells have become firm and turgid; they occupy more space and so the tubor increases in length. ...read more.

Conclusion

Furthermore, in order to prevent the tubors from being irregular, I will use only those potato tubors that are almost a perfect cylinder. In this case, the readings in mass and length are more accurate. Another common error to overcome is that of parallax; when using the measuring jar to confirm 30mL of solution to be used, I will make sure to take this reading at eye level so that I am actually using an exact volume of 30mL exactly. Also the parallax error can be avoided when cutting the potato tubors to an exact four centimeters; its length should be noted at eye level to confirm the length of the tubor is four centimeters. And finally, in order to prevent the concentration of potato cell sap from altering, I will use the same potato. Though my experiment is suitable and appropriate, it can be improved further more my using a different range of concentrations of sucrose solution to better reveal the change in mass and length of the potato tubors when placed in these solutions but more importantly to reveal the exact concentration of the potato cell sap. Thus, in this case, it may perhaps be better to use a range of concentrations between 0.25M and 0.5M. Thus, this will help to narrow down the exact concentration of the cell sap. The experiment may also been carried out using an egg rather than potato tubors or perhaps another substance could have been used to more appropriately show what is happening when the substance is placed in different sucrose solutions. Even though, in conclusion, I have found out that when a substance is placed in a more concentrated solution, it loses water in order for equalization to occur and thus in the case of the potato tubors, they will decrease in mass and length. Furthermore, when placed in a less concentration solution, the tubors will gain water in order for equalization to occur and so they will increase in mass and length. And also from my experiment, I have determined the potato cell sap concentration to be 0.46M. ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our GCSE Life Processes & Cells 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 GCSE Life Processes & Cells essays

  1. Marked by a teacher

    Osmosis in Potatoes Lab. At which concentration of sucrose in water (% mass ...

    5 star(s)

    > Step 1: Find the differences in the masses of the potato cell by subtracting the final masses at each time from the initial mass (Difference= final mass- initial mass) 0% Mass by Sucrose in Solution o At .5 Hours: 1.619g- 1.506= 0.113g o At 24 Hours: 1.553g- 1.506= 0.047g

  2. Marked by a teacher

    An experiment to show osmosis in potato chips in different concentrations of sucrose solution ...

    3 star(s)

    leave another two-minute interval and repeat again with potato chips C1 and C2 and beaker C. * Cover the beakers with cling film preventing the solution form evaporating * Leave the potato chips in the solution for a minimum of half a day.

  1. Marked by a teacher

    An Experiment to find the concentration of a potato cell cytoplasm compared to sucrose.

    3 star(s)

    * These results were noted on the table and a graph was created. Results Table: Distilled Water 0.2 Molar (sucrose) 0.4 Molar 0.6 Molar 0.8 Molar 1.0 Molar Start Weight (grams) 1.8 1.8 1.85 1.8 1.8 1.9 Final Weight 2.0 1.9 1.8 1.7 1.7 1.7 Change in Weight +0.2 +0.1

  2. Aim To determine the water potential of a potato tuber cell

    Initial mass (g) 1st try of the mass after solution 2nd try of the mass after solution Average mass Change in mass Percentage of mass loss 1st try 2nd try 0.0 0.2 0.4 0.6 0.8 1.0 Fair Test The experiment must be a fair test, if it is not we will be obtaining the wrong results.

  1. To investigate the effect of different concentration of sucrose on osmosis in potato chips

    4.40 4.21 4.28 4.30 0.50M 4.31 4.13 4.13 4.19 3.70 4.63 3.71 4.01 0.75M 4.24 4.25 4.24 4.24 3.51 3.47 3.54 3.50 1.00M 4.27 4.26 4.10 4.21 3.41 3.27 3.10 3.26 Concentration/M % Change in mass 0.00M 9.15 0.25M 0.70 0.50M -4.30 0.75M -1.75 1.00M -22.56 Analysis of data From

  2. The effect of sucrose concentration on osmosis in potato chips.

    and measured at 60mm. The same balance is used to weigh my potato chips. This is because the measurements may slightly vary between scales. The volume of the solution that the potato chips are kept in must be fair. This must be 10 ml.

  1. Osmosis. An experiment to find the concentrations of salt solution which is the ...

    before and after the experiment Apparatus: 6 boiling tubes, test tube rack, potato, white tile, stanley knife, 10% salt solution, stopwatch, cork borer, syringe. Method: Concentration % of salt solution Volume of 10% salt solution/ cm3 Volume of water/ cm3 Total Volume 10 20 0 20 8 16 4 20

  2. To find out the internal concentration of a potato cell.

    Preliminary Experiment An experiment was conducted to test the effects of differing concentrations on osmosis in visking tubing. Six pieces of visking tubing were filled with 0.6M sugar solution and tied at both ends with cotton. They were dried to remove excess liquid and the mass of each was taken.

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