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ChemistryLiquid SolutionHenry's LawMedium2 minPYQ_2009

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The Henry's law constant for the solubility of $N_{2}$ gas in water at $298 K$ is $1.0 \times 105 atm$ . 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 of $298 K$ and $5 atm$ pressure is

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Answer:

Solution:

$P_{N_{2}} = K_{H} \times$ mole-fraction $(N_{2})$ mole-fraction $N_{2} \frac{1}{1 0 ^{5}} \times 0.8 \times 5 = 4 \times 1 0^{- 5} mol^{- 1}$ In 10 mole solubility in $4 \times 1 0^{- 4}$ .

Subject:ChemistryTopic:Liquid SolutionSubtopic:Henry's Law

⏱ 2mℹ️ Source: PYQ_2009

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