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

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

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