Answer
$m_{a}=95.8\text{ kg}$
$m_{v}=1.10\text{ kg}$
Work Step by Step
The partial pressure of water vapor and dry air are determined to be
$$
\begin{aligned}
& P_v=\phi P_g=\phi P_{\text {sat } @ 26 \mathrm{C}}=(0.50)(3.364 \mathrm{kPa})=1.682 \mathrm{kPa} \\
& P_a=P-P_v=93-1.682=91.32\ \mathrm{kPa}
\end{aligned}
$$ The masses are determined to be $$
\begin{aligned}
& m_a=\frac{P_a V}{R_a T}=\frac{(91.32 \mathrm{kPa})\left(90 \mathrm{~m}^3\right)}{\left(0.287 \mathrm{kPa} \cdot \mathrm{m}^3 / \mathrm{kg} \cdot \mathrm{K}\right)(299 \mathrm{~K})}=95.8 \mathrm{~kg} \\
& m_v=\frac{P_v V}{R_v T}=\frac{(1.682 \mathrm{kPa})\left(90 \mathrm{~m}^3\right)}{\left(0.4615 \mathrm{kPa} \cdot \mathrm{m}^3 / \mathrm{kg} \cdot \mathrm{K}\right)(299 \mathrm{~K})}=1.10 \mathrm{~kg}
\end{aligned}
$$
