For a gas, the Henry’s law constant is 1.25 × 10^(-3) mol dm^(-3) atm^(-1) at 25 °C. Calculate the solubility

Question

Solve the following :

For a gas, the Henry’s law constant is 1.25 × 10^-3 mol dm^-3 atm^-1 at 25 °C. Calculate the solubility of the given gas at 2.5 atm and 25 °C.

Answer 1

Given : 

Henry’s law constant = K_H

= 1.25 x 10^-3 mol dm^-3 atm^-1

Pressure of the gas = P = 2.5 atm

Solubility of the gas = S = ? 

By Henry’s law,

S = K_H × P

= 1.25 × 10^-3 mol dm^-3 atm^-1 x 2.5 atm

= 3.125 × 10^-3 mol dm^-3

∴ Solubility of gas = 3.125 × 10^-3 mol dm^-3