For Freundlich adsorption isotherm, a plot of $\log \frac{x}{m}$ at (y-axis) and $\log P$ (x-axis) gives a straight line. The intercept and slope for the line is $0.4771$ and 2, respectively. The mass of gas adsorbed per gram of adsorbent if the initial pressure is $0.04 atm$, is $\mathrm{x}\times {10}^{-4} \mathrm{g}$. $(\log 3=0.4771)$.

Determine the value of x.

According to Freundlich Isotherm it can be written as,

$\log \frac{x}{m}=\log K+\frac{1}{n}\log P$

According to the above Freundlich Isotherm equation, slope and intercept are $\frac{1}{n} and \log k$ respectively, hence 

$$\begin{gathered} \frac{1}{n}=2 \\ \log k=0.4771 \\ \mathrm{k}=3 \end{gathered}$$

Determining the value of $\frac{x}{m}$ at 0.04 atm pressure, using another equation of Freundlich Isotherm:

$\frac{x}{m}=K{P}^{\frac{1}{n}}$

Substitute the values to know the value of x.

$\frac{x}{m}=3\times (0.04{)}^2$

$\frac{\mathrm{x}}{\mathrm{m}}=\underset{\mathrm{x}}{\underbrace{48}}\times {10}^{-4}$