Question

A solution of two miscible liquids showing negative deviation from Raoult's law will have:

• A. Increased vapour pressure, increased boiling point
• B. Increased vapour pressure, decreased boiling point
• C. Decreased vapour pressure, decreased boiling point
• D. Decreased vapour pressure, increased boiling point

Answer 1

The Correct Option is D

A solution showing negative deviation from Raoult’s law has lower vapour pressure than expected because the interactions between the components are stronger than in the pure liquids. This results in a decreased vapour pressure, which leads to an increase in boiling point.

For solutions showing negative deviation:

$$ P_T < P_A^0 X_A + P_B^0 X_B $$

where \( P_T \) is the total vapour pressure, \( P_A^0 \) and \( P_B^0 \) are the vapour pressures of pure components, and \( X_A \) and \( X_B \) are their mole fractions.