Answer
$123.1\text{ kPa}$
$492.2\text{ kPa}$
Work Step by Step
According to Raoults's law, when the mole fraction of the ammonia liquid is $20 \%$,
$$
P_{\mathrm{NH} 3}=y_{f, \mathrm{NH} 3} P_{\mathrm{sat}, \mathrm{NH} 3}(T)=0.20(615.3 \mathrm{kPa})=123.1\ \mathrm{kPa}
$$ When the mole fraction of the ammonia liquid is $80 \%$, $$
P_{\mathrm{NH} 3}=y_{f, \mathrm{NH} 3} P_{\mathrm{sat}, \mathrm{NH} 3}(T)=0.80(615.3 \mathrm{kPa})=492.2\ \mathrm{kPa}
$$
