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's law constant
$=K_H=8.0\times 10^{4}kPa\text{ for }{O}_2$
Density of water with dissolved oxygen $=1.0kgd{m}^{-3}]$

Answer 1

Answer: 1389

$P=K_H\cdot x$

or, $20\times 10^3=\left(8\times 10^4\times 10^3\right)\times \frac{n_{O_2}}{n_{O_2}+n_{\text{water }}}$

or, $\frac{1}{4000}=\frac{n_{O_2}}{n_{O_2}+n_{\text{water }}}=\frac{n_{O_2}}{n_{\text{water }}}$

Means 1 mole water $(=18gm=18ml)$ dissolves

$\frac{1}{4000}$ moles $O_2.$ Hence, molar solubility

$=\frac{\left(\frac{1}{4000}\right)}{18}\times 1000=\frac{1}{72}mold{m}^{-3}$

$=1388.89\times 10^{-5}mold{m}^{-3}\approx 1389mold{m}^{-3}$