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Question
Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 × 105 mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.
Solution
Here,
p = 760 mm Hg
KH = 4.27 × 105 mm Hg
According to Henry’s law,
p = KHx
`=> "x" = "p"/"K"_"H"`
= `(760 "mm Hg")/(4.27xx10^(5) "mm Hg")`
= 1.78 × 10-3
i.e., solubility in terms of mole fraction of methane in benzene = 1.78 × 10-3
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