Question

$C{H}_4$ is adsorbed on $1g$ charcoal at $0^{\circ }C$ following the Freundlich adsorption isotherm. $10.0mL$ of $C{H}_4$ is adsorbed at $100mm$ of $Hg$, whereas $15.0$ $mL$ is adsorbed at $200mm$ of $Hg$. The volume of $C{H}_4$ adsorbed at $300mm$ of $Hg$ is $10^{x}mL$. The value of $x$ is _______ $\times 10^{-2}$ (Nearest integer) $[$ Use ${\log }_{10}2=0.3010,{\log }_{10}3=0.4771]$

Answer 1

Answer: 128

We know
$\frac{x}{m}=K{P}^{1/n};\text{ using }(x\propto V)$
$\Rightarrow \frac{10}{1}=K\times (100{)}^{1/n}....$ (1)
$\frac{15}{1}=K\times (200{)}^{1/n}....$ (2)
$\frac{V}{1}=K\times (300{)}^{1/n}....$ (3)
Divide
(2) / (1)
$\frac{15}{10}=2^{1/n}$
$\log \left(\frac{3}{2}\right)=\frac{1}{n}\log 2$
$\frac{1}{n}=\frac{\log 3-\log 2}{\log 2}=\frac{0.4771-0.3010}{0.3010}$
$\frac{1}{n}=0.585$
Divide
(3) / (1)
$\frac{V}{10}=3^{1/n}$
$\log \left(\frac{V}{10}\right)=\frac{1}{n}\log 3$
$\log \left(\frac{V}{10}\right)=0.585\times 0.4771=0.2791$
$\frac{V}{10}=10^{0.279}\Rightarrow V=10\times 10^{0.279}$
$\Rightarrow V=10^{1.279}=10^x$
$\Rightarrow x=1.279$
$\Rightarrow x=128\times 10^{-2}$ (Nearest integer)