Question

If the initial pressure of a gas is $0.03 \,atm$, the mass of the gas adsorbed per gram of the adsorbent is _____$\times 10^{-2} g$

Answer 1

$\frac{x}{ m }= kP ^{\frac{1}{ n }} $
$\log \frac{ x }{ m }=\log k +\frac{1}{ n } \log P$
From graph
Slope $=\frac{1}{n}=1 \Rightarrow n=1$
Intercept $=\log k =0.602$
$k =4$
$\frac{x}{ m }=4 \times(0.03)^{\frac{1}{1}}$
$\frac{ x }{ m }=12 \times 10^{-2}$