Calculate the solubility of gas in water at 1.2 atm and 2 5 ∘ C if Henry's law constant is 0.45 mol dm − 3 atm − 1 at 2 5 ∘ C .

Solubility ( s ) = k × p = 0.45 mol dm − 3 atm − 1 × 1.2 atm = 0.54 mol dm − 3

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Chemistry - Liquid Solution Question with Solution | TestHub

ChemistryLiquid SolutionHenry's LawEasy2 minPYQ_2022

ChemistryEasysingle choice

Calculate the solubility of gas in water at $1.2 atm$ and $2 5^{\circ} C$ if Henry's law constant is $0.45 mol dm^{- 3} atm^{- 1}$ at $2 5^{\circ} C$ .

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Answer:

Solution:

$Solubility (s) = k \times p = 0.45 mol dm^{- 3} atm^{- 1} \times 1.2 atm = 0.54 mol dm^{- 3}$

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

⏱ 2mℹ️ Source: PYQ_2022

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