Question

The oxygen dissolved in water exerts a partial pressure of $20\, kPa$ in the vapour above water. The molar solubility of oxygen in water is _________. $\times 10^{-5}\,\,mol\,\, dm ^{-3}$
(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 $\left.=1.0\,kg\,dm ^{-3}\right]$

Answer 1

$P = K _{ H } \cdot x$

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 $(=18\,gm =18\,ml )$ dissolves

$\frac{1}{4000}$ moles $O _{2} .$ Hence, molar solubility

$=\frac{\left(\frac{1}{4000}\right)}{18} \times 1000=\frac{1}{72} mol\, dm ^{-3}$

$=1388.89 \times 10^{-5}\,mol \,dm ^{-3} \approx 1389\,mol\,dm ^{-3}$