Ideal Solutions
A liquid–liquid solution is classified using the yardstick of Raoult's law. The solutions which obey Raoult's law over the entire range of concentration are known as ideal solutions. For every composition, the partial vapour pressure of each component equals its mole fraction in the liquid times the vapour pressure of that pure component, and the total vapour pressure varies linearly between the two pure-liquid values.
For a two-component system of A and B, the defining relations are written compactly as follows, where $p_A^{\circ}$ and $p_B^{\circ}$ are the vapour pressures of the pure liquids:
$$p_A = x_A\,p_A^{\circ}, \qquad p_B = x_B\,p_B^{\circ}, \qquad p_{\text{total}} = x_A\,p_A^{\circ} + x_B\,p_B^{\circ}$$Beyond obeying this law, ideal solutions carry two further signature properties. The enthalpy of mixing of the pure components to form the solution is zero, and the volume of mixing is also zero:
$$\Delta_{\text{mix}}H = 0, \qquad \Delta_{\text{mix}}V = 0$$A perfectly ideal solution is rare, but several mixtures are nearly ideal because the two molecules are so alike. NCERT lists the solution of $\ce{n-hexane}$ and $\ce{n-heptane}$, the mixture of bromoethane $\ce{C2H5Br}$ and chloroethane $\ce{C2H5Cl}$, and the classic pairing of benzene $\ce{C6H6}$ and toluene $\ce{C6H5CH3}$ as members of this category.
Why Mixing Is Thermoneutral
The thermodynamic signatures of an ideal solution follow directly from molecular interactions. In the pure components only A–A and B–B attractions exist. On mixing, a new A–B interaction appears alongside them. Ideal behaviour is the special case where these three are essentially identical in strength.
When the intermolecular attractive forces between A–A and B–B are nearly equal to those between A–B, no extra energy is needed to pull like molecules apart and none is released when unlike molecules come together. Consequently no heat is absorbed or evolved on mixing ($\Delta_{\text{mix}}H = 0$), and because the packing is unchanged the volume of the solution is simply the sum of the volumes of the two components ($\Delta_{\text{mix}}V = 0$). The escaping tendency of each molecule in the mixture is exactly what it was in the pure liquid scaled by its mole fraction — which is precisely Raoult's law.
Deviations only make sense once the baseline law is solid. Revisit Raoult's Law for the partial-pressure relation that ideal solutions obey at every composition.
Non-Ideal Solutions and Deviation
When a solution does not obey Raoult's law over the entire range of concentration, it is called a non-ideal solution. The vapour pressure of such a solution is either higher or lower than the value predicted by Raoult's law. If it is higher, the solution exhibits positive deviation; if it is lower, it exhibits negative deviation. Most real solutions are, in fact, non-ideal to some degree.
The cause of these deviations lies entirely in the relative strengths of the molecular interactions. The whole topic reduces to comparing the A–B force against the A–A and B–B forces — a comparison that simultaneously fixes the sign of the deviation, the sign of $\Delta_{\text{mix}}H$, and the sign of $\Delta_{\text{mix}}V$.
Positive Deviation
In a solution showing positive deviation from Raoult's law, the A–B interactions are weaker than those between A–A or B–B. The intermolecular attractive forces between the solute–solvent molecules are weaker than those among the solute–solute and solvent–solvent molecules. Molecules of A (or B) therefore find it easier to escape than in the pure state, so the vapour pressure rises and the observed value lies above the Raoult prediction. Because energy must be supplied to weaken the average attraction, $\Delta_{\text{mix}}H > 0$ (endothermic), and the loosened packing gives $\Delta_{\text{mix}}V > 0$.
The textbook example is a mixture of ethanol and acetone. In pure ethanol the molecules are hydrogen bonded. On adding acetone, its molecules get in between the host molecules and break some of those hydrogen bonds, weakening the overall interaction:
$$\ce{C2H5OH \cdots HOC2H5} \;\xrightarrow{\text{add }\ce{CH3COCH3}}\; \text{H-bonds broken} \;\Rightarrow\; \text{vapour pressure rises}$$A second example NCERT gives is carbon disulphide added to acetone, $\ce{CS2}$ + $\ce{CH3COCH3}$, where the dipolar interactions between solute and solvent are weaker than the interactions among like molecules; this mixture also shows positive deviation. The familiar ethanol + water system behaves the same way.
A solution of liquids X and Y has an observed total vapour pressure of 70 torr, while the Raoult's-law prediction works out to 73 torr. What deviation does it show?
Observed (70) is less than the calculated ideal value (73), so the vapour pressure has been suppressed: the mixture shows negative deviation. Had the observed value exceeded the predicted one, it would have been positive deviation. This is exactly the reasoning NEET set in 2025 (see PYQ snapshot below).
Negative Deviation
In a solution showing negative deviation, the intermolecular attractive forces between A–A and B–B are weaker than those between A–B. The new A–B interaction is the strongest of the three, so it decreases the escaping tendency of both components, lowers the vapour pressure, and pulls the observed value below the Raoult line. Here heat is released on mixing, $\Delta_{\text{mix}}H < 0$ (exothermic), and the tighter packing gives $\Delta_{\text{mix}}V < 0$.
The standard example is a mixture of chloroform and acetone. The chloroform molecule forms a hydrogen bond with the acetone molecule, an attraction absent in either pure liquid:
$$\ce{Cl3C-H \cdots O=C(CH3)2}$$A second NCERT example is the mixture of phenol $\ce{C6H5OH}$ and aniline $\ce{C6H5NH2}$, where the hydrogen bond between the phenolic proton and the lone pair on the nitrogen of aniline is stronger than the hydrogen bonding between like molecules. The widely used nitric-acid–water mixture is another negative-deviation system.
Master Comparison Table
The three cases line up cleanly when read against one another. Memorise this grid as a single block — the sign of the deviation, the sign of $\Delta_{\text{mix}}H$ and $\Delta_{\text{mix}}V$, and the azeotrope type all move together.
| Property | Ideal solution | Positive deviation | Negative deviation |
|---|---|---|---|
| Raoult's law | Obeyed at all compositions | Not obeyed; $p_{\text{obs}} > p_{\text{Raoult}}$ | Not obeyed; $p_{\text{obs}} < p_{\text{Raoult}}$ |
| A–B vs A–A, B–B forces | $A\text{–}B \approx A\text{–}A \approx B\text{–}B$ | $A\text{–}B$ weaker than $A\text{–}A$, $B\text{–}B$ | $A\text{–}B$ stronger than $A\text{–}A$, $B\text{–}B$ |
| $\Delta_{\text{mix}}H$ | = 0 |
> 0 (endothermic) |
< 0 (exothermic) |
| $\Delta_{\text{mix}}V$ | = 0 |
> 0 |
< 0 |
| Escaping tendency | Unchanged | Increased (escapes more easily) | Decreased (escapes less easily) |
| NCERT examples | $\ce{C6H6}$ + $\ce{C6H5CH3}$; $\ce{n-hexane}$ + $\ce{n-heptane}$ | $\ce{C2H5OH}$ + $\ce{CH3COCH3}$; $\ce{CS2}$ + $\ce{CH3COCH3}$; ethanol + water | $\ce{CHCl3}$ + $\ce{CH3COCH3}$; phenol + aniline; $\ce{HNO3}$ + water |
| Azeotrope formed | None | Minimum-boiling azeotrope | Maximum-boiling azeotrope |
Azeotropes
Some liquids on mixing form azeotropes, which are binary mixtures having the same composition in the liquid and the vapour phase and which boil at a constant temperature. Because the vapour leaving the boiling liquid has exactly the same composition as the liquid, distilling it does not change the proportions — so it is not possible to separate the components of an azeotrope by fractional distillation. There are two kinds.
Solutions that show a large positive deviation from Raoult's law form a minimum-boiling azeotrope at a specific composition. The ethanol–water mixture obtained by fermentation of sugars is the classic case: on fractional distillation it yields a solution of roughly 95% ethanol by volume, and once this azeotropic composition is reached the liquid and vapour have the same composition so no further separation occurs.
Solutions that show a large negative deviation form a maximum-boiling azeotrope at a specific composition. Nitric acid and water is the standard example, with an azeotrope of approximately 68% nitric acid and 32% water by mass that boils at 393.5 K.
Match the deviation to the right azeotrope
The most frequent slip is pairing the deviations with the wrong azeotrope. High vapour pressure means the mixture boils more easily (lower boiling point), so positive deviation gives a minimum-boiling azeotrope. Low vapour pressure means it boils with difficulty (higher boiling point), so negative deviation gives a maximum-boiling azeotrope.
Positive deviation ($\Delta_{\text{mix}}H > 0$) → minimum-boiling azeotrope (e.g. ethanol–water, ~95% ethanol). Negative deviation ($\Delta_{\text{mix}}H < 0$) → maximum-boiling azeotrope (e.g. $\ce{HNO3}$–water, ~68% acid, 393.5 K).
Six lines before the exam
- Ideal solution: obeys Raoult's law at all compositions, with $\Delta_{\text{mix}}H = 0$ and $\Delta_{\text{mix}}V = 0$; A–B forces ≈ A–A ≈ B–B.
- Near-ideal examples: $\ce{C6H6}$ + $\ce{C6H5CH3}$, $\ce{n-hexane}$ + $\ce{n-heptane}$, bromoethane + chloroethane.
- Positive deviation: A–B weaker → $p_{\text{obs}} > p_{\text{Raoult}}$, $\Delta_{\text{mix}}H > 0$, $\Delta_{\text{mix}}V > 0$; e.g. ethanol + acetone, ethanol + water.
- Negative deviation: A–B stronger → $p_{\text{obs}} < p_{\text{Raoult}}$, $\Delta_{\text{mix}}H < 0$, $\Delta_{\text{mix}}V < 0$; e.g. acetone + chloroform, phenol + aniline.
- Azeotrope: same composition in liquid and vapour, boils at constant T, cannot be separated by fractional distillation.
- Positive → minimum-boiling azeotrope; negative → maximum-boiling azeotrope.