If the initial pressure of a gas is 0 . 03atm , the mass of the gas...

Equation for Freundlich's adsorption isotherm is $\left(\dfrac{x}{m}\right) = KP\dfrac{1}{n}$

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

Intercept $logK = 0 . 602 = \log 4$

$K = 4$

Slope $⇒ \dfrac{1}{n} = 1$ so $n = 1$

so \left(\dfrac{x}{m}\right) = 4\left(0 . 03\right) = 0 . 12 = 12 \times {10}^{- 2}$g</p><p>Hence, the mass of the gas adsorbed per gram of the adsorbent is$12 \times {10}^{- 2}g