1.

Henry's law constant for the molality of methane in benzene at 298 K is 4.27xx10^(5). Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

Answer»

Solution :Here, p = 760 mm Hg
`K_(H)=4.27xx10^(5)` mm Hg
According to Henry.s law,
`p=K_(H).x`
`x=(p)/(K_(H))`
`= (760 " mm Hg")/(4.27xx10^(5)" mm Hg")`
`=1.77.99xx10^(-5)`
`= 1.78xx10^(-5)` (approximately)
Hence, the MOLE FRACTION of methane in benzene is `1.78xx10^(-5)`.


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