Henry’s law constant for the solubility of methane in benzene is 4.27 × 10^(-5) mm Hg mol dm^(-3) at constant temperature.

Question

Solve the following :

Henry’s law constant for the solubility of methane in benzene is 4.27 × 10^-5 mm Hg mol dm^-3 at constant temperature. Calculate the solubility of methane at 760 mm Hg pressure at same temperature.

Answer 1

Given :

Henry’s law constant = K 

= 4.27 × 10^-5 mm Hg mol dm^-3 

Pressure of the gas = P = 760 mm Hg

K_H = 4.27 × 10^-5 mm^-1 mol dm^-3

= 4.27 × 10^-5 × 760 atm^-1 mol dm^-3

= 3245 × 10^-5 atm^-1 mol dm^-3

P = 760 mm = \(\frac{760}{760}\) = 1 atm

By Henry’s law,

S = K_H x P = 3.245 x 10^-2atm^-1 mol dm^-3 x 1 atm

= 3.245 × 10^-2 mol dm^-3

∴ Solubility of methane = 3.245 × 10^-2 mol dm^-3