Question

Henry law constant for the molality of Methane in Benzene at 298 kelvin is 4.27 into 10 to the power 5 mm HGg.calculate the solubility of Methane in Benzene at 298 K under 760 mm Hg​

Answer 1

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p = k × c

760 mm = 4.25 × 10^5 mm× C

C = 760 / 4.25 × 10^5

= 178 × 10^5

= 1.78 × 10 ^-3

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Answer 2

Answer:

Given :

KH=4.27×105KH=4.27×105 mm

P=760mmP=760mm

Applying Hentry's law

P=KHxP=KHx

x=PKHx=PKH

=760m4.27×105mm=760m4.27×105mm

=1.78×10−3=1.78×10−3

(ie) Mole fraction of methane in benzene =1.78×10−3

Explanation:

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