Question

Graph between $log\left(\frac{x}{m}\right)$ and log P is straight line at angle of $45^{\circ }$ with the intercept of 0.6020.

The extent of adsorption $\left(\frac{x}{m}\right)$ at a pressure of 1 atm is

Answer 1

Answer: 4

$log\left(\frac{x}{m}\right)=logK+\frac{1}{n}logP$
$=0.602+1log1$
= 0.602
So $\frac{x}{m}=4$ .