Which of the following is incorrect for the given graph?

The correct answer is Option (2) → \(\Delta V_{mix}\) = positive

For an ideal solution the total vapour pressure lies on the straight line joining p_A° and p_B°. In the graph the solid curves of the partial and total vapour pressures sag below the ideal values, so the solution shows negative deviation from Raoult’s law.

In such a solution the A–B attractions are stronger than the A–A and B–B attractions (as in chloroform + acetone, which form a hydrogen bond). Fewer molecules escape, so the vapour pressure falls; heat is released on mixing, and the molecules pack more closely, so the volume shrinks:

    p_A < p_A°x_A,   Δ_mixH < 0,   Δ_mixV < 0

• Option (1) ΔV_mix = negative — Correct for this graph. Stronger A–B attraction pulls the molecules closer, so the volume decreases on mixing.

• Option (2) ΔV_mix = positive — Incorrect for this graph. A positive volume change belongs to positive deviation; here the volume shrinks.

• Option (3) ΔH_mix = negative — Correct for this graph. Forming the stronger A–B interactions releases heat (exothermic mixing).

• Option (4) P_A < P_A°X_A — Correct for this graph. The partial vapour pressure is lower than Raoult’s law predicts; that is what negative deviation means.

NCERT (2026-27 edition):
Non-ideal Solutions – “… In case of negative deviations from Raoult’s law, the intermolecular attractive forces between A-A and B-B are weaker than those between A-B and leads to decrease in vapour pressure resulting in negative deviations. An example of this type is a mixture of phenol and aniline.”