Question #44528

A volatile gas \ce{G} is dissolved in water at $298\text{ K}$ under a constant partial pressure $P_G = 760\text{ Torr}$. The ideal Henry's law constant for \ce{G} in pure water is $K_H^\circ = 7.6 \times 10^5\text{ Torr}$. To this solution, a non-volatile, fully dissociating strong electrolyte \ce{Ca(NO3)2} is added such that its molality is $2.0576\text{ mol kg}^{-1}$. Due to the 'salting-out' effect, the chemical potential of the solvent decreases, which increases the activity coefficient ($\gamma_G$) of the dissolved gas. The apparent Henry's law constant becomes $K_H = K_H^\circ \cdot \gamma_G$. If $\gamma_G$ is thermodynamically linked to the vapor pressure lowering of the solvent such that $\gamma_G = \frac{P^\circ_{\ce{H2O}}}{P_{\ce{H2O}}}$, calculate the new equilibrium mole fraction of gas \ce{G} ($x_G'$) in the solution. (Assume the amount of dissolved gas is negligible compared to the solvent and \ce{Ca(NO3)2}, and the molar mass of water is $18\text{ g mol}^{-1}$).