A solution of acetone and chloroform shows a negative deviation from Raoult's law, while a solution of ethanol and acetone shows a positive deviation. Explain, in terms of the relative strength of A–B interactions compared with A–A and B–B interactions, why each mixture deviates

Question

A solution of acetone and chloroform shows a negative deviation from Raoult's law, while a solution of ethanol and acetone shows a positive deviation. Explain, in terms of the relative strength of A–B interactions compared with A–A and B–B interactions, why each mixture deviates the way it does, and state the sign of $\Delta_{mix}H$ expected in each case.

Accepted Answer

Concept understanding — Ideal and Non-Ideal Solutions

Ideal and Non-Ideal Solutions

Imagine you have two friends, A and B. When they work together, sometimes they get along perfectly — each does exactly their share, no extra effort, no friction. Other times, they either pull apart (making the job harder) or cling together (making it easier than expected). A solution of two liquids behaves the same way.

The Intuition: What "Ideal" Really Means

An ideal solution is the "perfect teamwork" case. The molecules of A and B are so similar that they don't care whether they are next to an A or a B. They interact with each other exactly as they would with their own kind. Think of mixing two grades of petrol — the molecules are nearly identical, so the mixture behaves predictably.

In an ideal solution, two things happen:

No heat is absorbed or released when you mix them (zero enthalpy change, $\Delta H_{\text{mix}} = 0$ ).
No volume change occurs — the total volume is exactly the sum of the individual volumes ( $\Delta V_{\text{mix}} = 0$ ).

Why? Because the forces between A-A, B-B, and A-B are all the same. No energy is needed to break old contacts or form new ones; no space is saved or wasted.

The Precise Statement: Raoult's Law

Raoult's law is the mathematical definition of an ideal solution. For a mixture of two volatile liquids A and B, the partial vapour pressure of each component above the solution is proportional to its mole fraction in the liquid:

p_A = x_A \, p_A^0 \quad \text{and} \quad p_B = x_B \, p_B^0

where $p_A^0$ and $p_B^0$ are the vapour pressures of pure A and pure B at that temperature. The total vapour pressure is simply the sum:

P_{\text{total}} = p_A + p_B = x_A p_A^0 + x_B p_B^0

P_{\text{total}} = x_A p_A^0 + x_B p_B^0

This is a straight line when plotted against mole fraction. Every ideal solution obeys this law at all compositions and temperatures.

Non-Ideal Solutions: When Molecules Disagree

Real solutions are rarely ideal. The molecules of A and B are different — different sizes, polarities, or hydrogen-bonding abilities. Now the A-B interaction is not the same as A-A or B-B. This causes deviations from Raoult's law.

Positive deviation occurs when A-B interactions are weaker than A-A and B-B. The molecules "prefer their own company" and escape into the vapour more easily than expected. The actual vapour pressure is higher than Raoult's law predicts. Examples: ethanol + water (initially), acetone + carbon disulphide. The mixture absorbs heat ( $\Delta H_{\text{mix}} > 0$ ) and expands slightly ( $\Delta V_{\text{mix}} > 0$ ).

Negative deviation occurs when A-B interactions are stronger than A-A and B-B. The molecules "hold on to each other" and escape less easily. The actual vapour pressure is lower than Raoult's law predicts. Examples: chloroform + acetone, nitric acid + water. The mixture releases heat ( $\Delta H_{\text{mix}} < 0$ ) and contracts ( $\Delta V_{\text{mix}} < 0$ ).

Watch out

A common mistake: thinking "positive deviation means the solution is better" or "negative means worse." The words refer only to the direction of the pressure deviation from Raoult's law — nothing about quality.

The Key Table

Property	Ideal Solution	Positive Deviation	Negative Deviation
A-B interaction	= A-A, B-B	< A-A, B-B	> A-A, B-B
Vapour pressure	Follows Raoult's law	Higher than predicted	Lower than predicted
$\Delta H_{\text{mix}}$	0	> 0 (endothermic)	< 0 (exothermic)

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In the acetone-chloroform mixture, the chloroform C–H hydrogen-bonds to the acetone carbonyl oxygen, an attractive A–B interaction that is stronger than the A–A and B–B interactions in the two pure liquids; molecules are then held together more tightly than Raoult's law assumes, so fewer molecules escape into the vapour, giving a lower-than-ideal total vapour pressure (negative deviation), and mixing releases energy as the new bond forms, so $\Delta_{mix}H < 0$ . In the ethanol-acetone mixture, mixing partially disrupts ethanol's own hydrogen-bonded network without fully replacing it with an equally strong ethanol-acetone interaction, so A–B attraction is weaker than the average of A–A and B–B attraction; molecules escape into the vapour more readily than Raoult's law predicts, giving a higher-than-i …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 24 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. Example of an ideal solution is$
(a) Mixture of n-hexane and n-heptane
(b) Mixture of acetone and chloroform
(c) Mixture of acetone and ethanol
(d) Mixture of phenol and aniline
›Reveal solutionSolution
An ideal solution obeys Raoult's law over the entire range of concentration, with ΔHmix = 0 and ΔVmix = 0; this happens when the two components are chemically very similar.

For a solution to be ideal, the A-A, B-B and A-B intermolecular interactions must be nearly identical, so mixing causes no enthalpy or volume change.

n-Hexane and n-heptane are both non-polar straight-chain alkanes of similar size, shape and polarity, so their mutual interactions closely match their pure-component interactions - this pair behaves ideally.

Acetone + chloroform show strong negative deviation (H-bonding between the two). …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. Assertion [A]: Nitric acid and water form maximum boiling Azeotrope. Reason [R]: Azeotropes are binary mixture showing the same composition in liquid and vapour phase.$
(a) Both [A] and [R] are true and [R] is the correct explanation of [A].
(b) Both [A] and [R] are true, but [R] is not the correct explanation of [A].
(c) [A] is true, but [R] is false.
(d) [A] is false, but [R] is true.
›Reveal solutionSolution

Nitric acid–water is a genuine example of a maximum boiling azeotrope, and the reason given (the defining property of azeotropes) correctly explains why such a mixture cannot be separated by simple fractional distillation.

Assertion: $HNO_3$ – $H_2O$ shows large negative deviation from Raoult's law (due to strong H-bonding/attraction between unlike molecules), and such systems form a maximum boiling azeotrope at a specific composition (≈68% $HNO_3$ by mass, boiling at ~120.5°C, higher than either pure component's boiling point). This is TRUE.

…

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CBSE 2026Set ANNUAL1 mark
$Q. Fill in the blank- The solutions which obey ......... law over the entire range of concentration are known as ideal solutions.$
›Reveal solutionSolution

The solutions that obey Raoult's law over the entire concentration range are called ideal solutions.

Concept. For an ideal solution, the partial vapour pressure of each component equals its mole fraction times its pure-component vapour pressure, at all compositions:

$p_A = x_A\,p_A^{0}, \qquad p_B = x_B\,p_B^{0}$

…

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CBSE 2026Set ANNUAL1 mark
$Q. When water and ethanol are mixed together the volume of the solution is not equal to the sum of the volumes of the two liquid components. Give reason.$
›Reveal solutionSolution

Ethanol + water is a non-ideal solution; the A–B (ethanol–water) interactions differ in strength from the A–A and B–B interactions of the pure liquids, so mixing changes the packing and the volume is not additive.

In an ideal solution the interactions between the two kinds of molecules are identical to those in the pure components, so volumes add up ( $V_{mix}=V_1+V_2$ ) and Raoult's law is obeyed exactly.

…

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CBSE 2025Set ANNUAL1 mark
$Q. Explain the following — Ideal solution$
›Reveal solutionSolution

Ideal solutions have identical A–A, B–B and A–B interactions.

An ideal solution is one that obeys Raoult's law over the entire range of concentration at all temperatures, i.e. $p_A = x_A p_A^0$ and $p_B = x_B p_B^0$ for every composition. In such a solution, the intermolecular forces between solute-solute (A-A), solvent-solvent (B-B), and solute-solvent (A-B) molecules are of similar magnitude, so there is no enthalpy change ( $\Delta H_{mix} = 0$ ) and no volume cha …

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CBSE 2025Set D1 markMCQ
$Q. The solution which shows positive or negative deviation from Raoult's law is called$
(a) Ideal solution
(b) True solution
(c) Non-ideal solution
(d) Colloidal solution
›Reveal solutionSolution

Deviation from Raoult's law (positive or negative) defines a non-ideal solution.

An ideal solution obeys Raoult's law over the entire range of concentration; its enthalpy of mixing and volume of mixing are zero, and A-B interactions equal A-A and B-B interactions.

When solute-solvent (A-B) interactions differ from A-A and B-B interactions, the solution deviates from Raoult's law: …

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CBSE 2025Set A1 mark
$Q. Write True or False: The solutions which obey Raoult's law over the entire range of concentrations are known as ideal solutions.$
›Reveal solutionSolution

This is the exact definition of an ideal solution, so the statement is True.

Raoult's law states that the partial vapour pressure of each volatile component in a solution is proportional to its mole fraction: $p_i = x_i p_i^{\circ}$ . A solution is called an ideal solution precisely when it obeys Raoult's law at every composition, from pure solvent to pure solute, over the entire concentration range, at all temperatures. Ideal solutions also have zero enthalpy of mixing ( $\Delta H_{mix}=0$ ) …

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CBSE 2025Set ANNUAL1 mark
$Q. Under what condition do non-ideal solutions show positive deviation from Raoult's law?$
›Reveal solutionSolution
Positive deviation occurs when mixing weakens the average intermolecular forces compared to the pure components, letting molecules escape into the vapour phase more easily than Raoult's law predicts.

Condition for positive deviation

In an ideal solution, A–B interactions are comparable in strength to A–A and B–B interactions. A solution shows positive deviation from Raoult's law when the solute–solvent (A–B) interactions are weaker than the solute–solute (A–A) and solvent–solvent (B–B) interactions present in the pure liquids.

When this happens, the molecules in the mixture are held less tightly than in the pure liquids, so they escape into the vapour phase more readily than expected. As a result:

the observed vapour pressure of the solution is higher than the Raoult's-law-predicted value,

mixing is typically accompanied by $\Delta_{mix}H > 0$ (endothermic) and $\Delta_{mix}V > 0$ (volume expansion). …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following pairs will not form an ideal solution?$
(a) Benzene and Toluene
(b) Ethanol and Acetone
(c) n-hexane and n-heptane
(d) Bromoethane and Chloroethane
›Reveal solutionSolution

An ideal pair needs closely similar molecular size/polarity/intermolecular forces; ethanol's hydrogen bonding is disrupted when mixed with acetone, breaking ideality.

An ideal solution requires A–A, B–B and A–B intermolecular forces to be essentially equal (Raoult's law holds exactly). Benzene–toluene, n-hexane–n-heptane, and bromoethane–chloroethane are each pairs of structurally very similar molecules (comparable size, shape and polarity), so they form essentially ideal solutions. Ethanol, however, is extensively hydrogen-bonded to itself in the pure liquid; mixing it wit …

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CBSE 2024Set D1 markMCQ
$Q. An azeotropic mixture of HCl and H2O has$
(a) 48% HCl
(b) 36% HCl
(c) 22.2% HCl
(d) 20.2% HCl
›Reveal solutionSolution

Hydrochloric acid forms a maximum-boiling azeotrope with water at about 20.2% HCl by mass (b.p. ~110 deg C), so it cannot be concentrated further by distillation.

HCl-water is a negative-deviation system that forms a maximum-boiling azeotrope. This constant-boiling mixture contains about 20.2% HCl by mass and boils around 110 deg C. Because it …

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CBSE 2024Set D1 markMCQ
$Q. Which of the following show positive deviation from Raoult's law?$
(a) C6H6 and C6H5CH3
(b) C6H6 and CCl4
(c) CHCl3 and C2H5OH
(d) CHCl3 and CH3COCH3
›Reveal solutionSolution
Chloroform + ethanol shows positive deviation from Raoult's law.

Positive deviation occurs when A–B interactions are weaker than A–A and B–B interactions, so the observed vapour pressure is higher than predicted.

In ethanol, molecules are strongly hydrogen-bonded. When chloroform is added it disrupts the ethanol H-bond network without forming equally strong new interactions, so escaping tendency increases → positive deviation. …
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CBSE 2024Set ANNUAL1 mark
$Q. What type of deviation from Raoult's law is observed when chloroform and acetone are mixed?$
›Reveal solutionSolution
Chloroform and acetone form a hydrogen bond with each other that is stronger than their own like-like interactions, so the mixture shows negative deviation from Raoult's law.

Deviation from Raoult's law depends on how the A–B (unlike-molecule) intermolecular forces compare to the A–A and B–B (like-molecule) forces:

If A–B interactions are weaker than A–A/B–B, molecules escape more easily than expected → positive deviation (higher vapour pressure than ideal).

If A–B interactions are stronger than A–A/B–B, molecules are held back more than expected → negative deviation (lower vapour pressure than ideal).

In a chloroform–acetone mixture, the hydrogen atom of chloroform ( $CHCl_3$ , made acidic by the three electronegative Cl atoms) forms a hydrogen bond with the lone pair on acetone's carbonyl oxygen: $Cl_3C-H\cdots O=C(CH_3)_2$ . This new intermolecular hydrogen bond is stronger than the (comparatively weak, dipole-dipole-only) interactions each pure liquid has with itself.

…
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