Question

If O₂ gas is bubbled through water at 303 K, the number of millimoles of O₂ gas that dissolve in 1 litre of water is_______. (Nearest integer)
(Given : Henry’s Law constant for O₂ at 303 K is 46.82 k bar and partial pressure of O₂ = 0.920 bar)
(Assume solubility of O₂ in water is too small, nearly negligible)

Answer 1

Correct Answer: 1

According to Henry’s law,
\(X(\text{oxygen}) = \frac{p(\text{oxygen})}{K_H} = \frac{0.920}{46.82 \times 10^3} = 1.96 \times 10^{-5}\)

Since, 1 litre of water contains 55.5 mol of it,
therefore,\(→ n \) represents moles of O₂ in solution.
\(X(\text{oxygen}) = \frac{n}{n + 55.5} \approx \frac{n}{55.5}\)

\(\frac{n}{55.5} = 1.96 \times 10^{-5}\)

\(n = 108.8 \times 10^{-5} = 1.08 \times 10^{-3} \, \text{moles}\)
m moles of oxygen = 1.08 × 10^–3 × 10³= 1 m mole
So, the answer is 1.