Chemistry - Liquid Solution Question with Solution | TestHub

Given: $P_A^0=300$ torr, $P_B^0=800$ torr, and vapor mole fraction of A: $y_A=0.25$.

Total pressure of vapor mixture:

$$P=\frac{P_A^0}{y_A}=\frac{300}{0.25}=1200torr$$

For B:

$$P_B=y_{B}P=0.75\times 1200=900torr$$

Liquid mole fraction of A in condensate:

$$x_A=\frac{P_A}{P_A^0}=\frac{300}{300+900/800\times 300}$$

Using the equilibrium relation:

$$x_A=\frac{y_A{P}_B^0}{y_A{P}_B^0+(1-y_A)P_A^0}$$

$$x_A=\frac{0.25(800)}{0.25(800)+0.75(300)}=\frac{200}{425}=0.47$$

Therefore, the mole fraction of A in the first drop of condensate = 0.47 (Option C).