Henry's law constant for solubility of methane in benzene is `4.2 xx 10^(-5)` mm Hg at a particular constant temperature At this temperature. Calculate the solubility of methane at i) 750 mm Hg ii) 840 mm Hg

Explanation

`(K_H)_("bonzene") = 4.2 xx 10^(-5) mm Hg`
Solubility of methane = ?
`P + 750mm Hg P = 840 mm Hg`
According to Henrys Law,
`P = K_(H)* x_("in solution")
`750mm Hg = 4.2 xx 10^(-5) mm Hg* x_("in solution")`
`implies x_("in solution") = (750)/(4.2 xx 10^(5))`
i.e, solubility `= 178.5 xx 10^(5)`
similarly at `P = 840 mm Hg`
Solubility `=(840)/(4.2xx10^(-5))= 200 xx 10^(-5)`