The Henry's law constant for the solubility of N2 gas in water at 298 – InterviewSolution

1.	The Henry's law constant for the solubility of N2 gas in water at 298 K is 1.0×105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298 K and 5 atmospheric pressure is x×10−y. Find the value of xy.
Answer» The Henry's law constant for the solubility of $N_{2}$ gas in water at $298 K$ is $1.0 \times 10^{5} a t m$ . The mole fraction of $N_{2}$ in air is $0.8$ . The number of moles of $N_{2}$ from air dissolved in $10$ moles of water at $298 K$ and $5$ atmospheric pressure is $x \times 10^{- y}$ . Find the value of $\frac{x}{y}$ .

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