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

Explanation

Here,
`p = 760 mm Hg`
`K_(H) = 4.27 xx 10^(-5) mm Hg`
According to Henry's law,
`p = k_(H)x`
`rArr x =(p)/(k_(H)) = (760 mm Hg)/(4.27 xx 10^(5) mm Hg) = 177.99 xx 10^(-5) = 178 xx 10^(-5)` (approximately)
Hence, the mole fraction of methane in benzene is `178 xx 10^(-5)`