Question

The vapor pressures of two liquids \(P\) and \(Q\) are \(80\) torr and \(60\) torr respectively. The total vapor pressure of the solution obtained by mixing \(3\) mol of \(P\) and \(2\) mol of \(Q\) would be:

• A. \(68\) torr
• B. \(140\) torr
• C. \(72\) torr
• D. \(20\) torr

Hint:

For binary ideal solutions: \[ P_{total}=X_AP_A^\circ + X_BP_B^\circ \] Always calculate mole fractions first before applying Raoult's law.

Answer 1

The Correct Option is C

Concept: For an ideal liquid solution, Raoult's law states: \[ P_{total}=P_P+P_Q \] where: \[ P_P = X_P P_P^\circ \] \[ P_Q = X_Q P_Q^\circ \] Here:
• \(X_P,\;X_Q\) are mole fractions
• \(P_P^\circ,\;P_Q^\circ\) are vapor pressures of pure liquids Thus, \[ P_{total}=X_P P_P^\circ + X_Q P_Q^\circ \]

Step 1: Calculating mole fractions.
Number of moles: \[ n_P=3 \] \[ n_Q=2 \] Total moles: \[ n_{total}=3+2=5 \] Mole fraction of \(P\): \[ X_P=\frac{3}{5} \] Mole fraction of \(Q\): \[ X_Q=\frac{2}{5} \]

Step 2: Applying Raoult's law.
Given: \[ P_P^\circ = 80\;\text{torr} \] \[ P_Q^\circ = 60\;\text{torr} \] Using: \[ P_{total}=X_P P_P^\circ + X_Q P_Q^\circ \] \[ P_{total}=\frac{3}{5}\times80+\frac{2}{5}\times60 \] \[ P_{total}=48+24 \] \[ P_{total}=72\;\text{torr} \] Therefore, the total vapor pressure of the solution is: \[ 72\;\text{torr} \]