Chemistry - Surface Chemistry Question with Solution | TestHub

According to the Freundlich adsorption isotherm,$\frac{x}{m}=K{p}^{\frac{1}{n}}$

Taking log on bot sides.

$\mathrm{l}\mathrm{o}\mathrm{g}\left(\frac{x}{m}\right)=\mathrm{l}\mathrm{o}\mathrm{g}K+\frac{1}{n}\log P$

Slope$=\frac{1}{n}$

Slope from graph$=\frac{y_2-y_1}{x_2-x_1}=\frac{2}{4}=\frac{1}{2}$

or,$\frac{1}{n}$=$\frac{1}{2}$

$\therefore \frac{x}{m}=K{p}^{\left(\frac{1}{2}\right)}$

or$\frac{x}{m}\propto p^{\left(\frac{1}{2}\right)}$