The values of the Henry's law constant of Ar, | vedclass

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(C) According to Henry's law,$p = K_{H} 	imes \chi$,where $p$ is the partial pressure of the gas,$K_{H}$ is the Henry's law constant,and $\chi$ is th

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$CH_{4} > CO_{2} > O_{2} > Ar$

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(C) According to Henry's law,$p = K_{H} 	imes \chi$,where $p$ is the partial pressure of the gas,$K_{H}$ is the Henry's law constant,and $\chi$ is the mole fraction of the gas in the solution.This implies $\chi = \frac{p}{K_{H}}$.For a given partial pressure $p$,the solubility (represented by mole fraction $\chi$) is inversely proportional to the Henry's law constant $K_{H}$ (i.e.,$\chi \propto \frac{1}{K_{H}}$).Given $K_{H}$ values: $Ar (40.30) > O_{2} (34.86) > CO_{2} (1.67) > CH_{4} (0.41) \ kbar$.Since solubility is inversely proportional to $K_{H}$,the order of solubility is $CH_{4} > CO_{2} > O_{2} > Ar$.

The values of the Henry's law constant of $Ar$,$CO_{2}$,$CH_{4}$ and $O_{2}$ in water at $25^{\circ} C$ are $40.30$,$1.67$,$0.41$ and $34.86 \ kbar$,respectively. The order of their solubility in water at the same temperature and pressure is

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