Question
Which of the following pairs shows positive deviation from Raoult’s Law?
(A) Acetone + Chloroform (B) Water + Nitric Acid (C) Acetone + Carbon Disulphide (CS₂) (D) Water + Hydrochloric Acid
(JEE Main 2024 — this exact pair appears frequently in both JEE Main and CBSE board exams.)
Solution — Step by Step
For an ideal solution, the partial vapour pressure of each component equals its mole fraction times its pure vapour pressure: P_A = x_A · P°_A. The total vapour pressure is just the sum of both partial pressures.
Deviation happens when intermolecular forces in the mixture are weaker or stronger than in the pure components.
Pure acetone has dipole–dipole interactions between its molecules. Pure CS₂ is a non-polar molecule — it has only weak van der Waals (London dispersion) forces.
When we mix the two, acetone’s polar dipoles cannot interact as strongly with non-polar CS₂. The A–B interactions are weaker than the average of A–A and B–B interactions.
Because the mixture holds molecules less tightly than the pure liquids did, vapour pressure goes up — above what Raoult’s Law predicts.
This is the definition of positive deviation: P_observed > P_Raoult. The answer is (C).
Acetone (C=O group) and chloroform (CHCl₃) form a hydrogen bond between them — the H of CHCl₃ and the lone pair on the C=O oxygen. This A–B interaction is stronger than A–A or B–B.
Stronger interactions → molecules escape less easily → vapour pressure drops below Raoult’s prediction → negative deviation.
Water + HNO₃ and Water + HCl both involve acid dissociation and strong ion–dipole interactions in solution. Both show negative deviation. Eliminate both.
Final Answer: (C) Acetone + CS₂ — Positive Deviation
Why This Works
The core idea is simple: Raoult’s Law assumes A–B interactions are identical to A–A and B–B interactions. Real solutions almost never satisfy this.
When you mix two substances that don’t “like” each other (one polar, one non-polar), the molecules at the surface feel less attraction holding them back. They escape into the vapour phase more readily — vapour pressure rises. Positive deviation is essentially a “mismatch” situation.
Negative deviation is the opposite: when A and B form stronger bonds with each other than with their own kind (like H-bonding between CHCl₃ and acetone), molecules are harder to pull apart. Vapour pressure drops, and in some cases these mixtures form minimum-boiling-point azeotropes (positive) or maximum-boiling-point azeotropes (negative).
Quick shortcut for the exam: “Polar + Non-polar = Positive deviation.” Whenever you see a polar–nonpolar pair (acetone + benzene, acetone + CS₂, ethanol + cyclohexane), go with positive deviation. For pairs that can H-bond with each other, think negative.
Alternative Method — The ΔH_mix Approach
Thermodynamically, you can identify the deviation type from the enthalpy of mixing:
ΔH_mix | ΔV_mix | Deviation | |
|---|---|---|---|
| Ideal | 0 | 0 | None |
| Acetone + CS₂ | +ve (endothermic) | +ve | Positive |
| CHCl₃ + Acetone | −ve (exothermic) | −ve | Negative |
When mixing is endothermic (ΔH_mix > 0), energy is absorbed to break A–A and B–B bonds, and the weaker A–B bonds release less energy. Net result: higher vapour pressure, positive deviation.
This approach is more rigorous — and ΔH_mix data sometimes appears directly in JEE problems where they ask you to identify deviation type from a given enthalpy value.
Common Mistake
Many students confuse which component breaks which rule. A common wrong answer here is (A) acetone + chloroform, because both involve acetone and both are “organic solvents.” The trap is assuming polarity alone determines deviation. What matters is whether A–B interactions are stronger or weaker than A–A and B–B combined. Chloroform actively H-bonds with acetone (Cl₃C–H···O=C), making the mixture more stable — negative deviation, not positive.
Also watch out: students sometimes write that acetone + CS₂ shows negative deviation because “CS₂ is more volatile, so vapour pressure increases.” That’s a completely wrong reasoning chain. Vapour pressure increases because of weaker intermolecular forces in the mixture, not because of the pure component’s boiling point.