Question

Suppose you have taken a dilute solution of oleic acid in such a way that its concentration becomes 0.01 cm³ of oleic acid per cm³ of the solution. Then you make a thin film of this solution (monomolecular thickness) of area 4 cm² by considering 100 spherical drops of radius ${\left(\frac{3}{40\mathrm{\pi }}\right)}^{\frac{1}{3}}\mathrm{x}{10}^{-3}\mathrm{cm}.$ Then the thickness of the oleic acid layer will be $\mathrm{x}\times {10}^{-14}\mathrm{m}$ where x is _________.

Answer 1

Step 1: Given data

Radius r=${\left(\frac{3}{40\mathrm{\pi }}\right)}^{\frac{1}{3}}\mathrm{x}{10}^{-3}\mathrm{cm}.$

Number of droplets= 100

Area= $4{\mathrm{cm}}^2$

Step 2: Calculating the thickness

$\mathrm{Volume}=\mathrm{Area}\mathrm{of}\mathrm{film}\times \mathrm{thickness}$

Also, we know that,

Volume of film= volume of 100 spherical drops

$$\begin{gathered} \therefore 4{\mathrm{t}}_{\mathrm{T}}=100\times \left(\frac{4}{3}\right){\mathrm{\pi r}}^3 \\ \Rightarrow 4{\mathrm{t}}_{\mathrm{T}}=100\times \left(\frac{4\mathrm{\pi }}{3}\right)\mathrm{x}\left(\frac{3}{40\mathrm{\pi }}\right)\mathrm{x}{10}^{-9} \\ \Rightarrow 4{\mathrm{t}}_{\mathrm{T}}={10}^{-8}{\mathrm{cm}}^3 \end{gathered}$$

Hence thickness of the film is

$$\begin{gathered} \Rightarrow {\mathrm{t}}_{\mathrm{T}}=\frac{{10}^{-8}}{4} \\ \Rightarrow {\mathrm{t}}_{\mathrm{T}}=2.5\times {10}^{-9}\mathrm{cm}=25\times {10}^{-10}\mathrm{cm} \\ \Rightarrow {\mathrm{t}}_{\mathrm{T}}=25\times {10}^{-12}\mathrm{m} \end{gathered}$$

And for oleic acid, the thickness will be

${\mathrm{t}}_0=0.01{\mathrm{t}}_{\mathrm{T}}\Rightarrow 25\times {10}^{-14}\mathrm{m}$

Thus the value of x becomes $25$

Therefore, Then the thickness of the oleic acid layer will be $\mathrm{x}\times {10}^{-14}\mathrm{m}$ where x is $25$.