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`.

Options

A. `1.78 xx 10^(-4)`
B. `1.78 xx 10^(-5)`
C. `17.8 xx 10^(-4)`
D. `1.78 xx 10^(-6)`

Answer

C

Explanation

`p = K_(H) X x`
Where x =mole fraction or solubility of methane
`x=p/K_(H)=760/(4.27 xx 10^(5))=1.78 xx 10^(-3) = 17.8 xx 10^(-4)`