Calculate the solubility of a gas in water at $0.8 \ atm$ and $25^{\circ} C$. [Henry's law constant is $6.85 \times 10^{-4} \ mol \ dm^{-3} \ atm^{-1}$]
- A$5.48 \times 10^{-4} \ mol \ dm^{-3}$
- B$3.94 \times 10^{-4} \ mol \ dm^{-3}$
- C$6.858 \times 10^{-4} \ mol \ dm^{-3}$
- D$2.74 \times 10^{-4} \ mol \ dm^{-3}$
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