42. In an experiment to estimate the size of a molecule of oleic acid 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 placed on water in a shallow through. The solution spreads over the surface of water forming one molecule thick layer. Now, lycopodium powder is sprinkled evenly over the film and its diameter is measured. Knowing the volume of the drop and area of the film we can calculate the thickness of the film which will give us the size of the 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 the 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?
(a) Oleic acid does not dissolve in water hence, it is dissolved in alcohol.
(b) Lycopodium powder spreads over the entire surface of the water when it is sprinkled evenly. When a drop of the prepared solution is dropped on water, 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
(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 the solution make 1 mL, the volume of oleic acid in one drop will be .
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