Calculate the solubility of a gas in water at 0.8 atm and 2 5 ∘ C [Henry's law constant is 6.85 × 1 0 − 4 mol dm − 3 atm − 1 ]

S = P k = 0.8 × 6.85 × 1 0 − 4 = 5.48 × 1 0 − 4 mol / dm − 3

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ChemistryLiquid SolutionHenry's LawEasy2 minPYQ_2022

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Calculate the solubility of a gas in water at $0.8 atm$ and $2 5^{\circ} C$ [Henry's law constant is $6.85 \times 1 0^{- 4} mol dm^{- 3} atm^{- 1}$ ]

Answer:

Solution:

$S = P k = 0.8 \times 6.85 \times 1 0^{- 4} = 5.48 \times 1 0^{- 4} mol / dm^{- 3}$

Stream:MHT_CETSubject:ChemistryTopic:Liquid SolutionSubtopic:Henry's Law

⏱ 2mℹ️ Source: PYQ_2022

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