Question

At $25{}^{\circ }\text{C}$, the vapour pressure of pure liquid A (mol. wt. $=40$) is $100\text{torr}$, while that of pure liquid B is $40\text{torr}$ (mol. wt. $=80$). The vapour pressure at $25{}^{\circ }\text{C}$ of a solution containing $20\text{g}$ of each A and B is:

• A. $80\text{torr}$
• B. $59.8\text{torr}$
• C. $68\text{torr}$
• D. $48\text{torr}$

Answer 1

The correct option is A $80\text{torr}$

Moles of A $=\frac{20}{40}=0.5$
Moles of B $=\frac{20}{80}=0.25$
${\chi }_A=\frac{0.5}{0.5+0.25}=0.67$
${\chi }_B=1-{\chi }_A=0.33$
The total vapour pressure is
$P_{total}=P_A^{\circ }{\chi }_A+P_B^{\circ }{\chi }_B=100\times 0.67+40\times 0.33=80\text{torr}$