The plot of $\log \frac{x}{m}$ (y-axis) and $\log p(x$-axis) is a straight line inclined at an angle of…
The plot of $\log \frac{x}{m}$ (y-axis) and $\log p(x$-axis) is a straight line inclined at an angle of $45^{\circ}$. When the intercept, $\mathrm{K}$ is 10 and pressure is $0.3 \mathrm{~atm}$ the amount of solute in grams adsorbed per gram of adsorbent $(\log 3=$ $0.4771) ?$
$30.0$
$2.0$
$3.0$
$20.0$
Solution
According to Freundlich adsorption isotherm:-
$\log \frac{x}{m}=\frac{1}{n} \log p+\log k(y=m x+c)$
We have, $\log \mathrm{k}=\log 10=1$
$\log \mathrm{p}=\log (0.3)=-0.523$
$\theta=45^{\circ} \text { so slope }=\frac{1}{n}=\tan \theta=\tan 45^{\circ}=1$
Thus, $\log \frac{\mathrm{x}}{\mathrm{m}}=(1)(-0.523)+1=0.4771$
$\Rightarrow \quad \frac{x}{m}=3.0$