Answer
$0.24\%$
$99.76\%$
Work Step by Step
The vapor pressures are $$
\begin{aligned}
& P_{\mathrm{H}_2 \mathrm{O}}=y_{f, \mathrm{H}_2 \mathrm{O}} P_{\text {sat }, \mathrm{H}_2 \mathrm{O}}(T)=0.40(4.247 \mathrm{kPa})=1.70\ \mathrm{kPa} \\
& P_{\mathrm{NH}_3}=y_{f, \mathrm{NH}_3} P_{\text {sat }, \mathrm{NH}_3}(T)=0.60(1167.4 \mathrm{kPa})=700.44\ \mathrm{kPa}
\end{aligned}
$$ Thus the total pressure of the mixture is $$
P_{\text {total }}=P_{\mathrm{H}_2 \mathrm{O}}+P_{\mathrm{NH}_3}=(1.70+700.44) \mathrm{kPa}=702.1\ \mathrm{kPa}
$$ Then the mole fractions in the vapor phase become $$
\begin{aligned}
& y_{g, \mathrm{H}_2 \mathrm{O}}=\frac{P_{\mathrm{H}_2 \mathrm{O}}}{P_{\text {total }}}=\frac{1.70 \mathrm{kPa}}{702.1 \mathrm{kPa}}=\mathbf{0 . 0 0 2 4} \text { or } 0.24 \% \\
& y_{\mathrm{g}, \mathrm{NH}_3}=\frac{P_{\mathrm{NH}_3}}{P_{\text {total }}}=\frac{700.44 \mathrm{kPa}}{702.1 \mathrm{kPa}}=\mathbf{0 . 9 9 7 6} \text { or } 99.76 \%
\end{aligned}
$$
