In an adsorption experiment a graph between $\log \frac{\mathrm{x}}{\mathrm{m}} \mathrm{vs} \log \mathrm{P}$ is found to be linear with a slope of $45°$. The intercept on the $Y$-axis was found to be $0.3010$. What is $\frac{\mathrm{x}}{\mathrm{m}}$ if pressure is $0.6$ $bar\left(\tan 45°=1\right)?$

According to Freundlich adsorption isotherms,

$\log \frac{\mathrm{x}}{\mathrm{m}}=\log \mathrm{k} +\frac{1}{\mathrm{n}}\log \mathrm{P}... \left(1\right)$ or $\frac{\mathrm{x}}{\mathrm{m}}={\mathrm{kP}}^{\frac{1}{\mathrm{n}}}...\left(2\right)$

$\tan {45}^{\circ }=1, \mathrm{hence} \frac{1}{\mathrm{n}}=1 \mathrm{or} \mathrm{n}=1$

Value of k is calculated using intercept value.

$$\begin{gathered} \log \mathrm{k}=0.3010 \\ \mathrm{k}=2 \end{gathered}$$

Using equation (2)

$$\begin{gathered} \frac{\mathrm{x}}{\mathrm{m}}={\mathrm{kP}}^1 \\ \frac{\mathrm{x}}{\mathrm{m}}=2\times 0.6=1.2 \end{gathered}$$

Therefore, option (B) is correct.