Question

Calculate Henry’s law constant if solubility of nitrogen gas in water at 25°C is 6.85 × 10^–4 mol dm^–3.

[Partial pressure of nitrogen gas = 0.9 bar]

Options

• 6.85 × 10^–4 mol dm^–3 bar^–1

• 4.71 × 10^–4 mol dm^–3 bar^–1

• 3.43 × 10^–4 mol dm^–3 bar^–1

• 7.6 × 10^–4 mol dm^–3 bar^–1

Answer 1

7.6 × 10^–4 mol dm^–3 bar^–1

Explanation:

S = K_H × P

`K_H = S/P`

`= (6.85 xx 10^-4  "mol dm"^-3)/(0.9  "bar"^-1)`

= 7.6 × 10^–4 mol dm^–3 bar^–1