The Henry's law constant for the dissolution | vedclass

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(C) Given: Henry's law constant $K_H = 3 	imes 10^2 \ atm$,molality $m = \frac{5}{9} \ mol/kg$.First,calculate the mole fraction of the gas $(X_A)$ i

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$3$

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(C) Given: Henry's law constant $K_H = 3 	imes 10^2 \ atm$,molality $m = \frac{5}{9} \ mol/kg$.First,calculate the mole fraction of the gas $(X_A)$ in the solution.$X_A = \frac{n_{gas}}{n_{gas} + n_{water}} = \frac{m}{m + \frac{1000}{18}} = \frac{5/9}{5/9 + 55.55} = \frac{0.555}{0.555 + 55.55} = \frac{0.555}{56.105} \approx 0.01$.Using Henry's law: $P = K_H 	imes X_A$.$P = (3 	imes 10^2 \ atm) 	imes 0.01 = 3 \ atm$.

The Henry's law constant for the dissolution of a gas in an aqueous medium is $3 \times 10^2 \ atm$. At what partial pressure of the gas will the molality of the gas in the aqueous solution be $\frac{5}{9} \ m$?

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