Question

The vapour pressure of two pure liquids $\text{A}$ and $\text{B}$, that forms an ideal solution, are 100 and $900\text{torr}$ respectively at temperature $\text{T}$. This liquid solution of $\text{A}$ and $\text{B}$ is composed of 1 mole of $\text{A}$ and 1 mole of $\text{B}$. What will be the pressure, when 1 mole of this mixture has been vaporized?

• A. $800\text{torr}$
• B. $500\text{torr}$
• C. $300\text{torr}$
• D. None of these

Answer 1

The correct option is C $300\text{torr}$

After 1 mole of liquid is vapourised lets say mole fraction as well as the number of moles of A in the liquid is $x_A$ so for B will $(1-X_A)$ respectively
In the vapour phase, the mole fraction as well as the number of moles of A and B will be $(1-X_A)$ and $x_A$ respectively
From the ideal gas equation
$P=\frac{nRT}{V}$ hence
$P_A=P_A^0{x}_A$
$\frac{(1-x_A)RT}{V}=100x_A$ (i) similarly for B
$\frac{x_{A}RT}{V}=900(1-x_A)$ (ii)
from equation (i) and (ii) we will get
$x_A=3/4$
The total vapour pressure for this concentration will be
$P_T=100\times 3/4+900\times 1/4$
$=300torr$