Question

Volatile liquids A and B have volumes of $100mL$ and $40mL$ respectively at a given temperature . They are mixed to form a solution which has volume of $125mL$.
Choose the correct option:

• A. Mixture boils at a temperature as same as the boiling point of its components
• B. Mixture boils at a temperature lower than the boiling point of its components
• C. It is an ideal mixture
• D. Mixture boils at a temperature higher than the boiling point of its components

Answer 1

The correct option is D Mixture boils at a temperature higher than the boiling point of its components

Volume of 'A' = $100mL$
Volume of 'B' = $40mL$
For ideal condition,
$\text{Volume of solution of A + B}=140mL$
$\text{Original volume of solution}=125mL$
Thus, the given solution is a non-ideal solution which does not obey Raoult's law.

Here,
$\Delta V_{mix}=125-140=-15<0$
This condition occurs whe solvent-solute interaction is more than the solvent-solvent and the solute-solute interactions.
Thus, the total vapour pressure of the solution will be less than the ideal vapour pressure.
$P_T<X_A.p_A+X_B.p_B$

Hence, a non-ideal solution showing large negative deviation from Raoult’s law and solution mixture boils at a temperature higher than the boiling point of it components.
Thus, it is a maximum boiling azeotrope.