Question

Volatile liquids A and B forming an azeotrope (at a particular temperature range) have vapour pressures of $180mm$ and $100mm$ respectively. If the mole fraction of A in the solution mixture is 0.5, and the mixture has a total vapour pressure of $200mm$.

Choose the correct option(s) for the azeotropic mixture.

• A. It shows positive deviation from Raoult's law
• B. It shows negative deviation from Raoult's law
• C. It forms a maximum boiling azeotrope.
• D. It forms a minimum boiling azeotrope.

Answer 1

Answer: (A), (D)

It forms a minimum boiling azeotrope.

Given,
$\text{Pure vapour pressure of A ,}{p}_A^{\circ }=180mm\text{Pure vapour pressure of B ,}{p}_B^{\circ }=100mm$

$\text{Mole fraction of A ,}{X}_A=0.5\therefore \text{Mole fraction of B,}{X}_B=1-0.5=0.5$

By Raoult's law
Total vapour pressure of solution,
$P_T=X_A.p_A^{\circ }+X_B.p_B^{\circ }$
$P_T=180\times 0.5+100\times 0.5$
$=140mm.$

$P_T>X_A.p_A^{\circ }+X_B.p_B^{\circ }$
$200mm>140mm$
Thus, the given mixture shows positive deviation from the Raoult's law and forms minimum boiling azeotrope