Answer
$\mathbf{2 . 0 8} \times 10^{-5}$
Work Step by Step
Noting that air is saturated, the partial pressure of water vapor in the air will simply be the saturation pressure of water at $80^{\circ} \mathrm{F}$, $$
P_{\text {vapor }}=P_{\text {sat } @ 80^{\circ} \mathrm{F}}=0.5075 \mathrm{psia}
$$ Then the partial pressure of dry air becomes $$
P_{\text {dry air }}=P-P_{\text {vapor }}=14.3-0.5075=13.793 \text { psia }
$$ From Henry's law, the mole fraction of air in the water is determined to be
$$
y_{\text {dry air,liquidside }}=\frac{P_{\text {dry air,gasside }}}{H}=\frac{13.793 \mathrm{psia}(1 \mathrm{~atm} / 14.696 \mathrm{psia})}{74,000 \mathrm{bar}(1 \mathrm{~atm} / 1.01325 \mathrm{bar})}=1.29 \times 10^{-5}
$$ which is very small, as expected.
The mass and mole fractions of a mixture are related to each other by $$
\mathrm{mf}_i=\frac{m_i}{m_m}=\frac{N_i M_i}{N_m M_m}=y_i \frac{M_i}{M_m}
$$ where the apparent molar mass of the liquid water - air mixture is $$
\begin{aligned}
M_m & =\sum y_i M_i=y_{\text {liquidwater }} M_{\text {water }}+y_{\text {dry air }} M_{\text {dry air }} \\
& \cong 1 \times 18.0+0 \times 29.0 \cong 18.0 \mathrm{~kg} / \mathrm{kmol}
\end{aligned}
$$ Then the mass fraction of dissolved air in liquid water becomes $$
\mathrm{mf}_{\text {dry air, liquid side }}=y_{\text {dry air, liquidside }}(0) \frac{M_{\text {dry air }}}{M_m}=\left(1.29 \times 10^{-5}\right) \frac{29}{18}=\mathbf{2 . 0 8} \times 10^{-5}
$$
