Question

$100\text{mL}$ of liquid 'A' and $50\text{mL}$ of liquid 'B' are mixed to form $138\text{mL}$ of solution. The given solution is a/an:

• A. None of the above
• B. Ideal solution
• C. Minimum boiling Azeotrope
• D. Maximum boiling Azeotrope

Answer 1

The correct option is D Maximum boiling Azeotrope

Volume of 'A' = $100\text{mL}$
Volume of 'B' = $50\text{mL}$
For ideal condition,
$\text{Volume of solution of A and B}=150\text{mL}$
$\text{Original volume of solution}=138\text{mL}$
Thus, the given solution is a non-ideal solution which does not obey Raoult's law.

Here,
$\Delta V_{mix}=138-150=-12<0$
This condition occurs whe solvent-solute interaction is more than the solvent-solvent and the solute-solute interactions.
Thus, the total vapour presure 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.