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In an experiment to estimate the size of a molecule of oleicacid 1 mL of oleic acid is dissolved in 19 mL of alcohol. Then 1 mL of this solution is diluted to 20 mL by adding alcohol. Now, 1 drop of this diluted solution is placedon water in a shallow trough. The solution spreads over the surface of water forming one molecule thick layer. Now, lycopodium powderis sprinkled evenly over the film and its diameter is measured. Knowing the volume ofthedrop and area of the film we can calculate the thickness of the film which will give us the size of oleic acid molecule. Read the passage carefully and answer the following questions (a) Why do we dissolve oleic acid in alcohol ? (b) What is the role of lycopodium powder? (c) What would be the volume of oleic acid in each mL of solution prepared? (d) How will you calculate the volume of n drops of this solution of oleic acid ? (e) What will be the volume of oleic acid in one drop of this solution ? |
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Answer» Solution :(a) Oleic acid does not dissolve in water hence,it is dissolved inalcohol. (b) Lycopodium powder spreads over the entire surface of water when it is SPRINKLED EVENLY. When a drop of PREPARED solution is dropped ONWATER, oleic acid does not dissolve in water. INSTEAD it spreads on the water surface pushing the lycopodium powder away to clear a circular area where the drop falls. We can therefore, measure the area over which oleic acid spreads. (c) In each mL of solution prepared volume of oleic acid =`1/20mLxx1/20=1/400mL` (d) Volume of n drops of this solution of oleic acid can be calculated by means of a burette and measuring cylinder and measuring the number of drops. (e)If n drops of thesolution make 1 mL, the volume of oleic acid in one drop will be `1/((400)n)mL.` |
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