For Freundlich adsorption isotherm, a plot of log (x/m) (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 $\_\_\_\_$ $\times 10^{-4}$ g. $(\log 3=0.4771)$
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Solution
Freundlich adsorption isotherm:
$$
\begin{aligned}
& \frac{x}{m}=K p^{1 / n} \\
\Rightarrow \quad & \log \frac{x}{m}=\log K+\frac{1}{n} \log p \\
& \frac{1}{n}=2 \text { and } \log K=0.4771 \\
& =\log 3 \\
\therefore \quad & K=3 \\
& \frac{x}{m}=3 \cdot p^2
\end{aligned}
$$
mass of gas adsorbed per gram of adsorbent
$$
\begin{aligned}
& =3 \times(0.04)^2 \\
& =48 \times 10^{-4}
\end{aligned}
$$
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