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(D) According to Henry's law,the solubility $(S)$ of a gas is directly proportional to the partial pressure $(P)$ of the gas above the solution: $S =

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$1$

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(D) According to Henry's law,the solubility $(S)$ of a gas is directly proportional to the partial pressure $(P)$ of the gas above the solution: $S = K_H 	imes P$ Given: Solubility $(S)$ = $6.85 	imes 10^{-4} \ mol \ dm^{-3}$ Henry's law constant $(K_H)$ = $6.85 	imes 10^{-4} \ mol \ dm^{-3} \ bar^{-1}$ Rearranging the formula to solve for pressure $(P)$: $P = S / K_H$ $P = (6.85 	imes 10^{-4} \ mol \ dm^{-3}) / (6.85 	imes 10^{-4} \ mol \ dm^{-3} \ bar^{-1})$ $P = 1.0 \ bar$

Calculate the pressure of the gas if the solubility of the gas in water at $25^{\circ} C$ is $6.85 \times 10^{-4} \ mol \ dm^{-3}$. (Henry's law constant $K_H$ is $6.85 \times 10^{-4} \ mol \ dm^{-3} \ bar^{-1}$) (in $bar$)

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