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

To investigate the relationship between the time taken for diffusion and cell dimension.

Extracts from this document...


Banu Thuraisingam Aim: To investigate the relationship between the time taken for diffusion and cell dimension. Procedure: 1. Place the gelatin block on a tile or Petri dish and use a scalpel or razor blade to cut it in half, producing two cubes of 10mm sides 2. Keep on of these cubes intact and cut the other in half 3. Repeat this cutting operation until you have 4 more cubes. 4. Fill a test-tube to within 10ml of the top with dilute HCl 5. Note the time: starting with the largest block drop all the blocks into the acid in the test-tube and close it securely with a rubber bung or cork. 6. Tilt the tube to spread the gelatin blocks along its length. Hold the tube horizontally and rotate it so that you can see each block clearly and from all sides. Try not to warm the tube too much with your hands or the gelatin may dissolve. 7. Note the time taken for the acid to penetrate to the center of the block as indicated by the disappearance of the orange color. Data Collection: This is a table of results showing the time for hydrochloric acid to penetrate a gelatin block with the variance of size which includes surface area, volume and surface area to volume ratio. ...read more.


Both holding the test tube with my warm hands and holding the test tube to the sunlight could have increased the temperature of the Hydrochloric acid and thus increasing the rate of penetration. The results could have been affected due to the following reasons of human error. One place where human error takes place is when cutting the gelatin into blocks as we can never be sure if the sizes we have cut are exactly the sizes that are wanted. There are such problems as cutting in a straight line as well as measuring and handling a jelly-like substance. Another human error would be making the distinction between when the orange gelatin has become fully pink. By doing the experiment I can honestly say that it is very difficult to decide when the orange has become pink as it is a gradual change and not an immediate one. The results are based on what people perceive the color to be. There is also human error in stopping the stop watch. When you decide that the color has fully changed and you want to stop the stopwatch it takes at least a few seconds for your brain to process that and for your finger to hit the stop button. ...read more.


Therefore, since a cell must maintain a certain surface area to volume ratio, its size is limited. Cells make up organisms so if we think of this lab on a larger scale we can deduce more information. Organisms have to exchange substances like food, waste, gases and heat with their surroundings. These substances diffuse between the organism and the surroundings. The rate at which a substance diffuses is given by Fick's Law: Rate of Diffusion a surface area x concentration difference distance The organism's surface area that is in contact with the surroundings determines the rate of exchange of substances. The volume of the organism determines the requirement for materials. Thus the ability to meet requirements depends on the surface area to volume ratio. As organisms get bigger their volume and surface are both get bigger, however volume increases more rapidly than surface area and we can also see this in our results. There is a problem with size however. When the organisms increase in size it becomes harder for them to exchange materials with their surroundings. Thus, as mentioned earlier, this problem sets a limit on the maximum size for a single cell of about 100mm. If it is any bigger than this the materials just can't diffuse fast enough to support the reactions need to sustain life. ...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 Physical Chemistry 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 Physical Chemistry essays

  1. Marked by a teacher

    Electrochemical Cells

    5 star(s)

    Each of the compounds have different molar masses, so I have to use equations to work out how to make 1mol dm-3 of them in 250ml of water. Copper (II) Sulphate: 1mol dm-3 for 250ml Moles= 1 x 250 x 10-3 = 0.25mol Molar Mass= 64 + 32 + (4 x 16)

  2. Investigating the Rate of the Reaction between Bromide and Bromate Ions in Acid Solution

    To test this, I will look at the correlation between and the values for concentration squared. If the reaction is second order with respect to hydrogen ions, then this graph will be linear. [H+]2 /10-6 mol2 dm-6 Average time (t)

  1. Double Displacement Reactions

    K2CrO4(aq) + Cu(NO3)2(aq) --> 2KNO3(aq) + Cu(CrO4)2(s) K2CrO4(aq) + Pb(NO3)2(aq) --> 2KNO3(aq) + Pb(CrO4)2(s) 3K2CrO4(aq) + 2Al(NO3)3(aq) --> 6KNO3(aq) + Al2(CrO4)3(s) K2CrO4(aq) + (NH4)2SO4(aq) --> K2SO4(aq) + (NH4)2CrO4(aq) K2CrO4(aq) + 2NaC2H3O2(aq) --> 2KC2H3O2(aq) + NaCrO4(aq) K2CrO4(aq) + MgCl2(aq) --> 2KCl(aq) + MgCrO4(s) K2CrO4(aq) + 2NaOH(aq) --> 2KOH(aq) + Na2CrO4(aq)

  2. Investigating the Volume of a Drop

    ml 0.19 ml 0.019 ml 2.80 ml 2.98 ml 0.18 ml 0.018 ml Average volume per drop: (0.011+0.018+0.020+0.018) / 5 = 0.0134ml Rounded: 0.013 ml Uncertainty: + 0.02 / 10 = + 0.002ml Volume per drop = 0.011ml to 0.015ml Cold saturated salt water at 0.5 + 0.02 degrees Celsius

  1. Electrochemistry - Inventing Better Batteries

    According to the Standard Reduction Potentials for Half-Reductions (SRPHR), Copper and Zinc is a good pair since Copper is a fairly strong oxidizing agent and Zinc is a very good reducing agent which gives a high voltage between these two half cells.

  2. Investigating how concentration affects rate of reaction

    e = mathematical value similar to pi (has a value of 2.71828) Ea = the activation enthalpy R = the gas constant (has a value of 8.31 J K-1 mol-1 T = temperature (in Kelvin) k = Ae-Ea/RT This equation can be rearranged by taking the log (to the base e)

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