Question

Calculate the solubility of gas in water at 1.2 atm and 25° C, if Henry’s law constant is 0.145 mol dm⁻³ atm⁻¹ at 25° c.

पर्याय

• 0.45 mol dm⁻³

• 0.25 mol dm⁻³

• 0.174 mol dm⁻³

• 0.31 mol dm⁻³

Answer 1

0.174 mol dm⁻³ 

Explanation:

Given, partial pressure of gas (ρ) = 1.2 atm

Henry’s law constant (K_H) = 0.145 mol dm⁻³ atm⁻¹

Solubility of gas in water (S) = `"K"_"H" xx rho`

`= 0.145 xx 1.2`

= 0.174 mol dm⁻³