JEE Mains Chemistry · Solutions

Raoult's Law for Volatile Liquids

In an ideal mixture of two volatile liquids each exerts x·p° and the total is their sum; the vapour is richer in the more volatile liquid, and real mixtures deviate above or below the ideal line.

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Why this matters

Seventeen PYQs, twelve of them multiple choice, and five from 2026. Four find the total vapour pressure or a pure-liquid vapour pressure from two measured totals; seven move between the liquid and the vapour mole fractions; six ask which mixtures deviate from Raoult's law, in which direction and with which azeotrope.

Concept 1 of 3: Total vapour pressure of an ideal mixture

Each liquid escapes in proportion to its share of the surface, so liquid A contributes

x_A p^\circ_A

and liquid B contributes

x_B p^\circ_B

. The total is a straight line from

p^\circ_B

(pure B) to

p^\circ_A

(pure A). Two measured totals at two compositions give two linear equations for the two unknown pure vapour pressures.

Definition

Partial pressures: $p_A = x_A p^\circ_A$ , $p_B = x_B p^\circ_B$ .
Total: $P = x_A p^\circ_A + x_B p^\circ_B = p^\circ_B + (p^\circ_A - p^\circ_B)x_A$ .
Mole fractions come from moles: adding moles of one liquid changes BOTH mole fractions.
The liquid with the higher $p^\circ$ is the MORE volatile one.

Raoult's law for two volatile liquids

P = x_A p^\circ_A + x_B p^\circ_B = p^\circ_B + \left(p^\circ_A - p^\circ_B\right)x_A

Worked example

Liquids P and Q form an ideal solution. With

x_P = 0.4

the vapour pressure is 360 mmHg; with

x_P = 0.8

it is 440 mmHg. Find

p^\circ_P

and

p^\circ_Q

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 2 · Q47Moderate

Example 1 · Solutions · Raoult's Law for Volatile Liquids

Two liquids A and B form an ideal solution. At 320 K, the vapour pressure of the solution, containing 3 mol of A and 1 mol of B is 500 mm Hg. At the same temperature, if 1 mol of A is further added to this solution, vapour pressure of the solution increases by 20 mm Hg. Vapour pressure (in mm Hg ) of B in the pure state is

\_\_\_\_

. (Nearest integer)

Attach each mole fraction to its own liquid

With 1 mol A and 3 mol B,

x_A = 0.25

multiplies

p^\circ_A

, not

p^\circ_B

. Swapping them gives a pure vapour pressure several times too large, and that value is always among the options.

Higher pure vapour pressure means more volatile

The liquid that escapes more easily has the larger

p^\circ

. A question that asks for the LEAST volatile component wants the one with the smaller

p^\circ

Drill 3 more on total vapour pressure of an ideal mixture

Concept 2 of 3: Composition of the vapour over an ideal mixture

Raoult's law gives the partial pressures; Dalton's law turns them into vapour mole fractions,

y_A = p_A/P

. The more volatile liquid pushes more molecules into the vapour, so the vapour is richer in it than the liquid is.

Definition

$y_A = \dfrac{x_A p^\circ_A}{P}$ , $y_B = 1 - y_A$ .
From the vapour side: $\dfrac{1}{P} = \dfrac{y_A}{p^\circ_A} + \dfrac{y_B}{p^\circ_B}$ , then $x_A = \dfrac{y_A P}{p^\circ_A}$ .
Dividing: $\dfrac{y_A}{y_B} = \dfrac{x_A}{x_B}\cdot\dfrac{p^\circ_A}{p^\circ_B}$ . If $p^\circ_A < p^\circ_B$ , then $\dfrac{x_A}{x_B} > \dfrac{y_A}{y_B}$ .
Given $y_A$ and $P$ : $p_A = y_A P$ (Dalton) and $p^\circ_A = p_A/x_A$ (Raoult).
Straight-line form: $\dfrac{1}{x_1} = \dfrac{p^\circ_1}{p^\circ_2}\cdot\dfrac{1}{y_1} + \dfrac{p^\circ_2 - p^\circ_1}{p^\circ_2}$ .

Vapour mole fraction

y_A = \frac{x_A p^\circ_A}{P} \qquad \frac{1}{P} = \frac{y_A}{p^\circ_A} + \frac{y_B}{p^\circ_B}

Worked example

An ideal mixture has

x_A = 0.25

, with

p^\circ_A = 90

and

p^\circ_B = 30

torr. Find the total pressure and the mole fraction of A in the vapour.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 2 · Q29Moderate

Example 2 · Solutions · Raoult's Law for Volatile Liquids

Two liquids A and B form an ideal solution at temperature T K. At T K, the vapour pressures of pure A and B are 55 and

15kNm^{- 2}

respectively. What is the mole fraction of A in solution of A and

B

in equilibrium with a vapour in which the mole fraction of A is 0.8 ?

The vapour mole fraction needs the total pressure

y_A

x_A p^\circ_A

divided by the TOTAL vapour pressure, not by

p^\circ_A

. Stopping at

x_A p^\circ_A

gives a partial pressure, not a mole fraction.

The vapour is richer in the more volatile liquid

The liquid with the larger

p^\circ

always has a bigger share in the vapour than in the liquid. An answer with the vapour poorer in that liquid has a slip in it.

Drill 6 more on composition of the vapour over an ideal mixture

Concept 3 of 3: Positive and negative deviations from Raoult's law

An ideal mixture has A–B attractions equal to the average of A–A and B–B. If the new A–B attractions are weaker, molecules escape more easily: the vapour pressure rises above the Raoult line (positive deviation). If they are stronger, often a new hydrogen bond, the vapour pressure falls below it (negative deviation).

Definition

Ideal: $\Delta H_{\text{mix}} = 0$ , $\Delta V_{\text{mix}} = 0$ ; obeys Raoult's law at every composition.
Positive deviation: A–B weaker; $\Delta H_{\text{mix}} > 0$ , $\Delta V_{\text{mix}} > 0$ ; higher vapour pressure, lower boiling point; forms a MINIMUM-boiling azeotrope.
Negative deviation: A–B stronger; $\Delta H_{\text{mix}} < 0$ , $\Delta V_{\text{mix}} < 0$ ; lower vapour pressure, higher boiling point; forms a MAXIMUM-boiling azeotrope.
An azeotrope boils at constant composition: its vapour has the same composition as the liquid, so fractional distillation cannot separate it.

Mixture	Deviation	Reason	Vapour pressure and boiling point
Benzene + toluene	None (ideal)	Similar molecules, similar attractions	On the Raoult line; $\Delta V_{\text{mix}} = 0$
n-Hexane + n-heptane	None (ideal)	Two similar non-polar chains	On the Raoult line; $\Delta H_{\text{mix}} = 0$
Acetone + $\mathrm{CS_2}$	Positive	$\mathrm{CS_2}$ breaks the dipole attraction between acetone molecules	Vapour pressure above the line; boils lower
Ethanol + water	Positive	Ethanol breaks some of water's hydrogen bonds	Minimum-boiling azeotrope, about 95% ethanol by volume
Methanol + $\mathrm{CCl_4}$	Positive	$\mathrm{CCl_4}$ breaks the hydrogen bonds of methanol	Vapour pressure above the line; boils lower
Chloroform + acetone	Negative	The C–H of chloroform hydrogen-bonds to the C=O of acetone	Maximum-boiling azeotrope The standard negative-deviation pair; the answer to 'maximum-boiling azeotrope'.
Acetone + aniline	Negative	The N–H of aniline hydrogen-bonds to the C=O of acetone	Vapour pressure below the line; boils higher
Nitric acid + water	Negative	Strong attraction between the acid and water	Maximum-boiling azeotrope, about 68% nitric acid by mass

A new hydrogen bond between the two liquids means a negative deviation; broken attractions within one liquid mean a positive one.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 8 Apr 2026 Shift 2 · Q27Moderate

Example 3 · Solutions · Raoult's Law for Volatile Liquids

Given below are two statements : Given : Molar mass of

C,H,O,Cl

are

12,1,16

and

35.5\text{ }g{\text{ }mol}^{- 1}

, respectively. Statement I : In

30\%(w/w)

solution of methanol in

{CCl}_{4}

(at T K ), the mole fraction of

{CCl}_{4}

is equal to 0.33 . Statement II : Mixture of methanol and

{CCl}_{4}

shows positive deviation from Raoult's law. In the light of the above statements, choose the correct answer from the option given below :

Positive deviation gives the MINIMUM-boiling azeotrope

A positive deviation raises the vapour pressure, so the mixture boils at a lower temperature than either liquid near the azeotrope. Ethanol + water is minimum-boiling; chloroform + acetone is maximum-boiling. Swapping the two is the match-list trap.

A new hydrogen bond means a negative deviation

When mixing makes a hydrogen bond that neither pure liquid had (chloroform with acetone, aniline with acetone), the liquids hold each other more tightly and the vapour pressure falls. When mixing breaks hydrogen bonds (methanol or ethanol diluted by a non-polar liquid), the deviation is positive.

Drill 5 more on positive and negative deviations from raoult's law

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Total vapour pressure of an ideal mixture
Raoult's law for two volatile liquids
$P = x_A p^\circ_A + x_B p^\circ_B = p^\circ_B + \left(p^\circ_A - p^\circ_B\right)x_A$
Composition of the vapour over an ideal mixture
Vapour mole fraction
$y_A = \frac{x_A p^\circ_A}{P} \qquad \frac{1}{P} = \frac{y_A}{p^\circ_A} + \frac{y_B}{p^\circ_B}$

Reference tables (1)

Positive and negative deviations from Raoult's law8 rows

Mixture	Deviation	Reason	Vapour pressure and boiling point
Benzene + toluene	None (ideal)	Similar molecules, similar attractions	On the Raoult line; $\Delta V_{\text{mix}} = 0$
n-Hexane + n-heptane	None (ideal)	Two similar non-polar chains	On the Raoult line; $\Delta H_{\text{mix}} = 0$
Acetone + $\mathrm{CS_2}$	Positive	$\mathrm{CS_2}$ breaks the dipole attraction between acetone molecules	Vapour pressure above the line; boils lower
Ethanol + water	Positive	Ethanol breaks some of water's hydrogen bonds	Minimum-boiling azeotrope, about 95% ethanol by volume
Methanol + $\mathrm{CCl_4}$	Positive	$\mathrm{CCl_4}$ breaks the hydrogen bonds of methanol	Vapour pressure above the line; boils lower
Chloroform + acetone	Negative	The C–H of chloroform hydrogen-bonds to the C=O of acetone	Maximum-boiling azeotrope The standard negative-deviation pair; the answer to 'maximum-boiling azeotrope'.
Acetone + aniline	Negative	The N–H of aniline hydrogen-bonds to the C=O of acetone	Vapour pressure below the line; boils higher
Nitric acid + water	Negative	Strong attraction between the acid and water	Maximum-boiling azeotrope, about 68% nitric acid by mass

A new hydrogen bond between the two liquids means a negative deviation; broken attractions within one liquid mean a positive one.

Watch out for (6)

Attach each mole fraction to its own liquid
→ Total vapour pressure of an ideal mixture
Higher pure vapour pressure means more volatile
→ Total vapour pressure of an ideal mixture
The vapour mole fraction needs the total pressure
→ Composition of the vapour over an ideal mixture
The vapour is richer in the more volatile liquid
→ Composition of the vapour over an ideal mixture
Positive deviation gives the MINIMUM-boiling azeotrope
→ Positive and negative deviations from Raoult's law
A new hydrogen bond means a negative deviation
→ Positive and negative deviations from Raoult's law

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