# The oxygen dissolved in water exerts a partial pressure of 20 kPa in the vapour above water.

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 __________ $\times 10^{-5} \mathrm{moldm}^{-3}$ (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:

(25)

$\mathrm{P}_{-}\{(\mathrm{g})\}=\backslash$ left $\left[\mathrm{K}_{-}\{\mathrm{H}\} \backslash\right.$ right $] \mathrm{X} 20 \backslash$ times $10^{\wedge}\{3\}=\backslash$ left $\left[8 \backslash\right.$ times $10^{\wedge}\{4\}$

times $10^{\wedge}\{3\}$ |right $]$ SolubilitySolubility $=\left\langle\right.$ frac $\{20\}\left\{8 \backslash\right.$ times $\left.10^{\wedge}\{4\}\right\}$

$=2.5 \mid$ times $10^{\wedge}\{-4\}=25 \mid$ times $10^{\wedge}\{-5\}$ S