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

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.