$\mathrm{CH}_{4}$ is adsorbed on $1 \mathrm{~g}$ charcoal at $0^{\circ} \mathrm{C}$ following the Freundlich adsorption isotherm. $10.0 \mathrm{~mL}$ of $\mathrm{CH}_{4}$ is adsorbed at $100 \mathrm{~mm}$ of $\mathrm{Hg}$, whereas $15.0$ $\mathrm{mL}$ is adsorbed at $200 \mathrm{~mm}$ of $\mathrm{Hg}$. The volume of $\mathrm{CH}_{4}$ adsorbed at $300 \mathrm{~mm}$ of $\mathrm{Hg}$ is $10^{x} \mathrm{~mL}$. The value of $\mathrm{x}$ is $\times 10^{-2}$.
(Nearest integer)
$\left[\right.$ Use $\left.\log _{10} 2=0.3010, \log _{10} 3=0.4771\right]$
We know
$\frac{x}{m}=K P^{1 / n} ;$ using $(x \propto V)$
$\Rightarrow \quad \frac{10}{1}=\mathrm{K} \times(100)^{1 / \mathrm{n}}$ ..................(1)
$\frac{15}{1}=\mathrm{K} \times(200)^{1 / \mathrm{n}}$ .......................(2)
$\frac{\mathrm{V}}{1}=\mathrm{K} \times(300)^{1 / \mathrm{n}}$ ....................(3)
Divide
(2) / (1)
$\frac{15}{10}=2^{1 / n}$
$\log \left(\frac{3}{2}\right)=\frac{1}{n} \log 2$
$\frac{1}{\mathrm{n}}=\frac{\log 3-\log 2}{\log 2}=\frac{0.4771-0.3010}{0.3010}$
$\frac{1}{n}=0.585$
Divide
(3) / (1)
$\frac{\mathrm{V}}{10}=3^{1 / \mathrm{n}}$
$\log \left(\frac{\mathrm{V}}{10}\right)=\frac{1}{\mathrm{n}} \log 3$
$\log \left(\frac{\mathrm{V}}{10}\right)=0.585 \times 0.4771=0.2791$
$\frac{\mathrm{V}}{10}=10^{0.279} \Rightarrow \mathrm{V}=10 \times 10^{0.279}$
$\Rightarrow \mathrm{V}=10^{1.279}=10^{\mathrm{x}}$
$\Rightarrow \mathrm{x}=1.279$
$\Rightarrow \mathrm{x}=128 \times 10^{-2}$ (Nearest integer)
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