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(C) According to Henry's law,the solubility $(S)$ of a gas in a liquid is given by the formula: $S = K_H 	imes P$ Where: $K_H = 6.8 	imes 10^{-4} \

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$5.44 	imes 10^{-4} \ M$

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(C) According to Henry's law,the solubility $(S)$ of a gas in a liquid is given by the formula: $S = K_H 	imes P$ Where: $K_H = 6.8 	imes 10^{-4} \ mol \ dm^{-3} \ atm^{-1}$ $P = 0.8 \ atm$ Substituting the values: $S = (6.8 	imes 10^{-4} \ mol \ dm^{-3} \ atm^{-1}) 	imes (0.8 \ atm)$ $S = 5.44 	imes 10^{-4} \ mol \ dm^{-3}$ Since $1 \ mol \ dm^{-3} = 1 \ M$,the solubility is $5.44 	imes 10^{-4} \ M$.

Calculate the solubility of a gas in a solvent at $25^{\circ} C$ and $0.8 \ atm$ if Henry's law constant for the solvent is $6.8 \times 10^{-4} \ mol \ dm^{-3} \ atm^{-1}$.

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