Question

Calculate the Henry’s law constant at 25°C if the solubility of gas in liquid is 2.1 × 10⁻² mol dm⁻³ at 0.18 bar.

पर्याय

• 0.1166 mol dm⁻³ bar⁻¹

• 0.1445 mol dm⁻³ bar⁻¹

• 0.1730 mol dm⁻³ bar⁻¹

• 0.2014 mol dm⁻³ bar⁻¹

Answer 1

0.1166 mol dm⁻³ bar⁻¹

Explanation:

`K_H = S/P`

= `(2.1 xx 10^-2  mol  dm^-3)/(0.18  "bar")`

= 0.1166 mol dm⁻³ bar⁻¹