Question

Henry's law constant for the molality of methane in benzene at 298 is 4.27x10^5mm Hg
Calculate the solubility of methane in benzene at 298K under 760 mm Hg​

Answer 1

K  H =4.27×10  ^5  mm Hg (at 298 K)

p=760mm

Applying Henry's law,

p=K  H x,       where x= Mole fraction or solubility of methane.

x=p/KH=(760mm)/(4.27*10^5mm)

==1.78×10  ^−3