The oxygen dissolved in water exerts a partial pressure of

Question:

The oxygen dissolved in water exerts a partial pressure of $20 \mathrm{kPa}$ in the vapour above water. The molar solubility of oxygen in water is

$\times 10^{-5} \mathrm{~mol} \mathrm{dm}^{-3}$

(Round off to the Nearest Integer).

[Given : Henry's law constant

$=\mathrm{K}_{\mathrm{H}}=8.0 \times 10^{4} \mathrm{kPa}$ for $\mathrm{O}_{2}$.

Density of water with dissolved oxygen $=1.0 \mathrm{~kg} \mathrm{dm}^{-3}$ ]

Solution:

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

or, $20 \times 10^{3}=\left(8 \times 10^{4} \times 10^{3}\right) \times \frac{\mathrm{n}_{\mathrm{O}_{2}}}{\mathrm{n}_{\mathrm{O}_{2}}+\mathrm{n}_{\text {water }}}$

or, $\frac{1}{4000}=\frac{\mathrm{n}_{\mathrm{O}_{2}}}{\mathrm{n}_{\mathrm{O}_{2}}+\mathrm{n}_{\text {water }}}=\frac{\mathrm{n}_{\mathrm{O}_{2}}}{\mathrm{n}_{\text {water }}}$

Means 1 mole water $(=18 \mathrm{gm}=18 \mathrm{ml})$ dissolves

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

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

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