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

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

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

Intercept $\mathrm{logK}=0.602=\log 4$

$\mathrm{K}=4$

Slope $\Rightarrow \frac{1}{\mathrm{n}}=1$ so $\mathrm{n}=1$

so $\left(\frac{\mathrm{x}}{\mathrm{m}}\right)=4\left(0.03\right)=0.12=12\times {10}^{-2}$g

Hence, the mass of the gas adsorbed per gram of the adsorbent is $12\times {10}^{-2}\mathrm{g}$