What is Henry's law constant of a gas if solubility of gas in water at 2 5 ∘ C is 0.028 mol dm − 3 ? [Partial pressure of gas = 0.346 bar ]

K H = P S = 0.346 0.028 = 0.081 mol dm − 3 bar − 1

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ChemistryLiquid SolutionHenry's LawMedium2 minPYQ_2023

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What is Henry's law constant of a gas if solubility of gas in water at $2 5^{\circ} C$ is $0.028 mol dm^{- 3}$ ? [Partial pressure of gas $= 0.346 bar]$

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Answer:

Solution:

$K_{H} = \frac{S}{P} = \frac{0.028}{0.346} = 0.081 mol dm^{- 3} bar^{- 1}$

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

⏱ 2mℹ️ Source: PYQ_2023

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