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A drop of olive oil of radius 0.20mm spreads as a film on the water surface. Calculate the molecular size of the olive oil ifthe film is (a) circular, with a radius size of 10 cm (b) retangular, with size 20cm xx 10 cm. |
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Answer» Solution :Radius of oil drop `=0.20mm` (a) THICKNESS of olive oil film `=("Volume of oil drop")/("Area of the film")` `=((4)/(3)pi xx(0.020)^(3) cm^(3))/(pi xx (10)^(2) cm^(2))` `=(4)/(3)xx(20)^(3)xx10^(-11) cm` `=1.067xx10^(-7)cm` ASSUMING that the film has one molecular thickness, molecular size of olive oil `=1.067xx10^(-7)cm`. (b) Thickness of olive oil film `=("Volume of oil drop")/("Area of the film")` `=((4)/(3)pixx(0.020)^(3)cm^(3))/((20xx10)cm^(2))` `=(4)/(3)pi xx ((20)^(3))/(2)xx10^(-11)cm` `=1.674xx10^(-7)cm` Molecular size of olive oil `=1.674xx10^(-7)cm`. |
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