Henry's law constant for the molality of methane in benzene at 298 K is `4.27 xx 10^5` mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

Explanation

`K_H = 4.27 xx 10^5 mm, p = 760 mm`
Applying Henry.s law
`p = K_H x " or " x = (p)/(K_H)`
Substituting the values, we have
` x=(760mm)/(4.27 xx 10^5 mm) =- 1.78 xx 10^(-3) `
i.e., Mole fraction of methane in benzene = `1.78 xx 10^(-3)`