Suppose you have taken a dilute solution of oleic acid in such a way that its concentration becomes $0 .01 {cm}^{3}$ of oleic acid per cm³ of the solution. Then you make a thin film of this solution (monomolecular thickness) of area $4 {cm}^{2}$ by considering 100 spherical drops of radius ${(\frac{3}{40 π})}^{\frac{1}{3}} \times 10^{- 3}$ cm. then the thick ness of oleic acid layer will be $x \times 10^{- 14} m .$ where x is___

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answer is 25.

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Detailed Solution

Volume of oleic acid = $\begin{array}{l} (100 \times \frac{4}{3} {πr}^{3}) \times \frac{1}{100} = A \times t \\ [∵ V_{oleic acid} = V_{thin flim}] \end{array}$
$r = {(\frac{3}{40 π})}^{\frac{1}{3}} \times 10^{- 3} cm = {(\frac{3}{40 π})}^{\frac{1}{3}} \times 10^{- 5} m$
And A=4cm²=4 10^-4m²
$\begin{array}{l} \frac{4}{3} {πr}^{3} = A \times t \\ t = \frac{4}{3} \frac{{πr}^{3}}{A} = \frac{4}{3} π {[{(\frac{3}{40 π})}^{\frac{1}{3}}]}^{3} \times {(10^{- 5})}^{3} \times \frac{1}{4 \times 10^{- 4}} \\ \Rightarrow t = 25 \times 10^{- 14} m \\ ∴ x = 25 . \end{array}$