The Henry law constant for dissolution of a gas in aqueous medium is 3 × 1 0 2 atm . At what partial pressure of the gas (in atm), the molality of gas in aqueous solution will be 9 5 m

p = K H × X ; X = n 1 + n 2 n 1 X = 5/9 + 1000/18 5/9 = 9.89 × 1 0 − 3 ; p = 300 × 9.89 × 1 0 − 3 ≈ 3

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The Henry law constant for dissolution of a gas in aqueous medium is $3 \times 1 0^{2} atm$ . At what partial pressure of the gas (in atm), the molality of gas in aqueous solution will be $\frac{5}{9} m$

Answer:

Solution:

$p = K_{H} \times X$ ; $X = \frac{n _{1}}{n _{1} + n _{2}}$

$X = \frac{5/9}{5/9 + 1000/18} = 9.89 \times 1 0^{- 3}$ ; $p = 300 \times 9.89 \times 1 0^{- 3} \approx 3$

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

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Chemistry - Liquid Solution Question with Solution | TestHub

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JEE Main 2027 (Class 12+)

JEE Main 2027 (Class 12)