At , the total equilibrium vapour pressure of an ideal binary solution obtained by mixing of liquid — Solutions and Colligative Properties Chemistry Question
Question
At $25^\circ\text{C}$, the total equilibrium vapour pressure of an ideal binary solution obtained by mixing $3\text{ moles}$ of liquid A and $2\text{ moles}$ of liquid B is recorded as $184\text{ torr}$. If the pure vapour pressure of liquid A at $25^\circ\text{C}$ is known to be $200\text{ torr}$, calculate the theoretical vapour pressure (in torr) of pure liquid B at the exact same temperature.
💡 Solution & Explanation
The mole fraction of A in the liquid phase is $X_A = \frac{3}{3+2} = 0.6$, and the mole fraction of B is $X_B = \frac{2}{3+2} = 0.4$. By Raoult's law, $P_{total} = P_A^0 X_A + P_B^0 X_B$. Substituting the knowns: $184 = 200(0.6) + P_B^0(0.4) \implies 184 = 120 + 0.4 P_B^0 \implies 64 = 0.4 P_B^0$. Solving for $P_B^0$ yields exactly $160\text{ torr}$.