Graph between $\log \left(\frac{\mathrm{x}}{\mathrm{m}}\right)\mathrm{v}\mathrm{s}\log \mathrm{P}$ is provided for adsorption of $\mathrm{N}{\mathrm{H}}_3$ gas on metal surface. Calculate weight of $\mathrm{N}{\mathrm{H}}_3$ gas (in gm) absorbed by $24\mathrm{ }\mathrm{g}$ of metal surface at 2 atm pressure. (Take $\log 2=0.3$ )

Solution

Slope $=\frac{1}{\mathrm{n}}=\tan {45}^{\mathrm{o}}=1$
$\frac{\mathrm{x}}{\mathrm{m}}=\mathrm{K}{\mathrm{P}}^{\frac{1}{\mathrm{n}}}$
$\log \left(\frac{\mathrm{x}}{\mathrm{m}}\right)=\log \mathrm{K}+\frac{1}{\mathrm{n}}\log \mathrm{P}$
$=\log 2+\frac{1}{\mathrm{n}}\log 2$
$\log \left(\frac{\mathrm{x}}{\mathrm{m}}\right)=\log 2+\left(1\right)\log \left(2\right)$
$\frac{\mathrm{x}}{\mathrm{m}}=2\times 2$
$\frac{\mathrm{x}}{24}=4$
$\text{x}=4\times 24=96\text{g}$

Hence, the correct answer is 96.