Chapter 13 - The Chemistry of Solutes and Solutions - Questions for Review and Thought - Topical Questions - Page 605b: 30

$2.4\times10^{-3}\,g/L$

Given: $P_{O_{2}}= 100.\,mmHg$ and $k_{H}=1.5\times10^{-6} mol \,L^{-1} mmHg^{-1}$ According to Henry's law, $S_{O_{2}}=k_{H} \times P_{O_{2}}$ $=1.5\times10^{-6} mol \,L^{-1} mmHg^{-1}\times100\,mmHg=1.5\times10^{-4}mol/L$ Concentration of $O_{2}$ in g/L= $1.5\times10^{-4}mol/L\times\frac{16\,g}{1\,mol}=2.4\times10^{-3}\,g/L$