Surface area / Volume ratio Experiment
The surface area to volume ratio in living organisms is very important. Nutrients and oxygen need to diffuse through the cell membrane and into the cells. Most cells are no longer than 1mm in diameter because small cells enable nutrients and oxygen to diffuse into the cell quickly and allow waste to diffuse out of the cell quickly. If thecells were any bigger than this then it would take too long for the nutrients and oxygen to diffuse into the cell so the cell would probably not survive. Single celled organisms can survive as they have a large enough surface area to allow all the oxygen and nutrients they need to diffuse through. Larger multi celled organisms need specialist organs to respire such as lungs or gills.
ApparatusNeeded For the Experiments:
2. Gelatin blocks mixed containing universal indicator
3. 0.1M Hydrochloric acid
4. Stop Watch
7. Safety glasses
1. A block of gelatin which has been dyed with universal indicator should be cut into blocks of the following sizes (mm).
5 x 5 x 5
10 x 10 x 10
15 x 15 x 15
20 x 20 x 20
10 x 10x 2
10 x 10 x 10 (Triangle)
10 x 15 x 5
20 x 5 x 5
The triangle is of the following dimensions.
The rest of the blocks are just plain cubes or rectangular blocks.
Universal indicator is a neutral indicator. In the alkali conditions of the gelatin it is blue and when it gets exposed to acid it turns red.
Gelatin is used for these tests as it is permeable and so it acts like a cell. It iseasy to cut into the required sizes and the hydrochloric acid can diffuse at an even rate through it.
2. A small beaker was filled with 100cm³ of 0.1 molar Hydrochloric acid. This is to ensure that all the block sizes are fully covered in acid when dropped into the beaker.
3. One of the blocks is dropped into this beaker and the time for the entire universal indicator to disappear is noted in atable such as the one below.
Dimensions (mm) |Surface Area |Volume (mm³) |Surface Area / Volume Ratio |Test 1 |Test 2 |Test 3 |Average Time | | | | | | | | | | | | | | | | | | | |4. This test should be repeated for all the sizes of blocks three times to ensure a fair test. Fresh acid should be used for each block to ensure that this does not affect the experiment’s results.
5. The surface area /volume ratio and an average of the results can then be worked out. A graph of Time against Surface Area to Volume Ratio can then be plotted. From this graph we will be able to see how the surface area affects the time taken for the hydrochloric acid to penetrate to the centre of the cube.
I predict that as the Surface Area / Volume Ratio increases the time taken for thehydrochloric acid to penetrate to the centre of the cube will go down. This is because a small block has a large amount of surface area compared to it’s volume so the hydrochloric acid will have a large surface area to diffuse through. A larger block has a smaller amount of surface area in relation to it’s size so it should take longer for the hydrochloric acid to diffuse into the centre of the cube. Theactual rate of the hydrochloric acid diffusing through the gelatin should be the same for all the blocks but when the surface area / volume ratio goes up it will take less time for the hydrochloric acid to reach the centre of the cube.
Dimensions (mm) |Surface Area |Volume (mm³) |Surface Area / Volume Ratio |Test 1 |Test 2 |Test 3 |Average Time | |5 x 5 x 5 |150 |125 |1.2:1 | || | | |10 x 10 x 10 |600 |1,000 |0.6:1 | | | | | |15 x 15 x 15 |1,350 |3,375 |0.4:1 | | | | | |20 x 20 x 20 |2,400 |8,000 |0.3:1 | | | | | |10 x 10 x 2 |280 |200 |1.4:1 | | | | | |10 x 15 x 5 |550 |750 |0.73:1 | | | | | |20 x 5 x 5 |450 |500 |0.9:1 | | | | | |10 x 10 x 10
(Triangle) |441.42 |500 |0.88:1 | | | | | |The Surface area to Volume ratio is calculated:
Surface Area to Volume Ratio =...
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