Question:
In Freundlich adsorption isotherm, slope of AB line is :
Correct Option: , 4
Solution:
$\frac{x}{m}=K(P)^{1 / n}$
$\log \left(\frac{x}{m}\right)=\log K+\frac{1}{n} \log P$
$y=c+m x$
$\mathrm{m}=1 / \mathrm{n}$ so slope will be equal to $1 / \mathrm{n}$.
