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Calculate the solubility of a gas in water at $0.8 \mathrm{~atm}$ and $25^{\circ} \mathrm{C}$
[Henry's law constant is $6.85 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~atm}^{-1}$ ]
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[Henry's law constant is $6.85 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~atm}^{-1}$ ]
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The correct answer is:
$5.48 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3}$
$\begin{aligned} & S=P k \\ & =0.8 \times 6.85 \times 10^{-4} \\ & =5.48 \times 10^{-4} \mathrm{~mol} / \mathrm{dm}^{-3}\end{aligned}$
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