The K_H values (in ktext{bar}) of argon (I),c | vedclass

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(C) According to Henry's law,the solubility of a gas in a liquid is inversely proportional to its Henry's law constant $(K_H)$.Mathematically,$	ext{S

Vedclass · Accepted Answer

$I < II < IV < III$

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(C) According to Henry's law,the solubility of a gas in a liquid is inversely proportional to its Henry's law constant $(K_H)$.Mathematically,$	ext{Solubility} \propto \frac{1}{K_H}$.Given $K_H$ values at $298 \ K$:Argon $(I)$: $40.3 \ k	ext{bar}$Carbon dioxide $(II)$: $1.67 \ k	ext{bar}$Methane $(IV)$: $0.413 \ k	ext{bar}$Formaldehyde $(III)$: $1.83 	imes 10^{-5} \ k	ext{bar}$Comparing the $K_H$ values: $1.83 	imes 10^{-5} Therefore,the order of $K_H$ is $III Since solubility is inversely proportional to $K_H$,the increasing order of solubility is $I < II < IV < III$.

The $K_H$ values (in $k\text{bar}$) of argon $(I)$,carbon dioxide $(II)$,formaldehyde $(III)$,and methane $(IV)$ are $40.3$,$1.67$,$1.83 \times 10^{-5}$,and $0.413$ respectively at $298 \ K$. The increasing order of solubility of these gases in liquid is:

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