Question

The oxygen dissolved in water exerts a partial pressure of $20kPa$ in the vapour above water. The molar solubility of oxygen in water is ________$\times {10}^{-5}mold{m}^{-3}$
(Round off to the Nearest Integer).
[Given: Henry'law constant, $K_H=8.0\times {10}^{4}kPa$ for $O_2$. Density of water with dissolved oxygen = $1.0kgd{m}^{-3}$]

Answer 1

By Henry law,
$P_{C{O}_2}=K_H\times S$
Where,
S is solubility of gas
$K_H$ is Henry's constant
$P_{C{O}_2}$ is partial pressure of the gas

$20\times {10}^3=[8.0\times {10}^3\times {10}^4]\times \text{Solubility}$

$\text{Solubility}=\frac{20\times {10}^3}{8.0\times {10}^7}$

$\text{Solubility}=25\times {10}^{-5}mol/d{m}^3$