The mass of gas adsorbed, x , per unit mass of adsorbate, m , was...

$$\begin{array}{ll}\left(\dfrac{x}{m}\right) = k ( P) \dfrac{1}{n}\end{array}$$

$\log \left(\dfrac{x}{m}\right) = logk + \dfrac{1}{n}logP$

Slope $= \dfrac{1}{n} = 2$

so $n = \dfrac{1}{2}$

Intercept , $logk = 0 . 477$ So $k = Antilog ( 0 . 477 ) = 3$

so \left(\dfrac{x}{m}\right) = k ( P) \dfrac{1}{n}$</p><p>$= 3 [ 4{]}^{2} = 48