d
\(P = K _{ H } \cdot x\)

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

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

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