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Calculate the solubility of gas in water at $260 \mathrm{~mm} \mathrm{Hg}$ and $25^{\circ} \mathrm{C}$ if Henry's law constant of gas is $0.159 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~atm}^{-1}$ at $25^{\circ} \mathrm{C}$
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The correct answer is:
$5.4 \times 10^{-2} \mathrm{~mol} \mathrm{dm}^{-3}$
$\begin{aligned} & S=p K \\ & =\frac{260}{760} \times 0.159 \\ & =5.4 \times 10^{-2} \frac{\mathrm{mol}}{\mathrm{dm}^3}\end{aligned}$
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