MHT-CET Chemistry · Solutions and Colligative Properties

Vapour Pressure and Raoult's Law

Each volatile component contributes its pure vapour pressure times its mole fraction; a non-volatile solute lowers the solvent's vapour pressure by the solute's mole fraction — the relative lowering (P° − P)/P° = x₂.

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Why this matters

33 PYQs, one HARD. Two-thirds are the relative lowering of vapour pressure — from the two pressures, from moles, or in the dilute form W₂M₁/(M₂W₁) to recover a molar mass — and the rest are Raoult's law for two volatile liquids solved for a mole fraction or a pure vapour pressure, plus the recall of which mixtures are ideal, positive-deviation or negative-deviation. The only real trap is dividing by the wrong pressure.

Concept 1 of 3: Raoult's Law for Two Volatile Liquids: P = x_A P_A° + x_B P_B°

Each liquid evaporates as if it alone were present, scaled down by its mole fraction. The total vapour pressure is the sum, and the same equation solved backwards gives an unknown mole fraction or an unknown pure-component pressure.

Definition

$P_A = x_A P_A^\circ$ , $P_B = x_B P_B^\circ$ , $P_{\text{total}} = x_A P_A^\circ + x_B P_B^\circ$ with $x_A + x_B = 1$ .
$2$ mol A ( $420$ ) and $3$ mol B ( $610$ ): $0.4 \times 420 + 0.6 \times 610 = 534$ mm Hg.
Unknown mole fraction: $500 = 400(1 - x_B) + 575x_B \Rightarrow x_B = 0.57$ . Unknown pure pressure: $600 = 0.6P_A^\circ + 0.4 \times 900 \Rightarrow P_A^\circ = 400$ ; $600 = 0.6 \times 400 + 0.4P_B^\circ \Rightarrow P_B^\circ = 900$ .
Statement form: 'the partial vapour pressure of any volatile component equals the vapour pressure of the pure component multiplied by its mole fraction' — Raoult's law.
The mole fraction of a component in the VAPOUR is $y_A = \dfrac{P_A}{P_{\text{total}}}$ (Dalton), richer in the more volatile component.

Raoult's law

P_{\text{total}} = x_A P_A^\circ + x_B P_B^\circ \qquad y_A = \frac{x_A P_A^\circ}{P_{\text{total}}}

Worked example

Liquids A (

P^\circ = 300

mm Hg) and B (

P^\circ = 500

mm Hg) are mixed in the mole ratio

1 : 3

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

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 4th May Shift 1 · Q54Easy

Example 1 · Solutions and Colligative Properties · Vapour Pressure and Raoult's Law

Calculate vapour pressure of a solution containing mixture of 2 moles of volatile liquid A and 3 moles of volatile liquid B at room temperature.

P_A^\circ = 420

P_B^\circ = 610

mmHg

Using the given mole fraction for the wrong component

x_B = 0.4

is given,

x_A = 0.6

multiplies

P_A^\circ

. Swapping them turns

400

into a value that is also on the list.

Drill 5 more on raoult's law for two volatile liquids: p = x_a p_a° + x_b p_b°

Concept 2 of 3: Relative Lowering of Vapour Pressure = Mole Fraction of the Solute

A non-volatile solute contributes nothing to the vapour but takes up a share of the surface, so

P = x_1 P^\circ

and the fractional drop

\dfrac{P^\circ - P}{P^\circ} = x_2

. For a dilute solution

x_2 \approx \dfrac{n_2}{n_1} = \dfrac{W_2 M_1}{M_2 W_1}

, which turns a measured lowering into a molar mass.

Definition

$\dfrac{P^\circ - P}{P^\circ} = x_2 = \dfrac{n_2}{n_1 + n_2}$ ; e.g. $\dfrac{32 - 30}{32} = 0.0625$ , $\dfrac{640 - 590}{640} = 0.078$ , $\dfrac{40}{550} = 0.072$ .
$1$ mol solute in $36$ g water ( $2$ mol): $x_2 = \tfrac13$ , $P = 32 \times \tfrac23 = 21.44$ mm Hg. $0.1$ mol in $16.2$ g water: $x_1 = 0.9$ , $P = 21.6$ mm Hg.
Dilute form $\dfrac{\Delta P}{P^\circ} = \dfrac{W_2 M_1}{M_2 W_1}$ : $3$ g urea in $50$ g water gives $\dfrac{3 \times 18}{60 \times 50} = 0.018$ ; $20$ g solute in $200$ g water with lowering $0.02$ gives $M_2 = \dfrac{20 \times 18}{200 \times 0.02} = 90$ .
Lowering is proportional to $x_2$ : doubling the lowering ( $10 \to 20$ mm Hg) doubles the solute mole fraction ( $0.2 \to 0.4$ ). Given the relative lowering, $P = P^\circ(1 - 0.018) = 17.68$ mm Hg for $P^\circ = 18$ .
Identify the solute: $100$ g water, $17.53 \to 17.22$ mm Hg, $17.10$ g of X: $x_2 = 0.0177$ , $n_2 \approx 0.1$ , $M \approx 171 \approx 180$ — glucose.

Relative lowering

\frac{P^\circ - P}{P^\circ} = x_2 = \frac{n_2}{n_1 + n_2} \approx \frac{W_2\,M_1}{M_2\,W_1}

Worked example

9

g of a non-volatile solute in

90

g of water lowers the vapour pressure from

30

29.4

mm Hg. Find the solute's molar mass.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 2 · Q71Easy

Example 2 · Solutions and Colligative Properties · Vapour Pressure and Raoult's Law

Calculate the relative lowering of vapour pressure if vapour pressure of pure solvent and vapour pressure of solution at

25^\circ\text{C}

are 32 and 30 mm Hg respectively.

Dividing by the solution's pressure, or reporting the solvent's mole fraction

\dfrac{2}{30}

gives

0.067

, and

0.9375

x_1

, the SOLVENT's mole fraction — both are options on the

32/30

stem. The relative lowering divides by

P^\circ

and equals the SOLUTE's mole fraction.

Drill 12 more on relative lowering of vapour pressure = mole fraction of the solute

Concept 3 of 3: Ideal and Non-Ideal Solutions: Which Way a Mixture Deviates

An ideal solution obeys Raoult's law at every composition because A–B attractions match A–A and B–B, so

\Delta H_{\text{mix}} = 0

and

\Delta V_{\text{mix}} = 0

. Weaker A–B attractions push the vapour pressure ABOVE Raoult (positive deviation); stronger ones, usually new hydrogen bonds, pull it below (negative deviation).

Definition

Ideal: $\Delta H_{\text{mix}} = 0$ , $\Delta V_{\text{mix}} = 0$ , obeys Raoult over the whole range — benzene + toluene, n-hexane + n-heptane.
Positive deviation: $P > P_{\text{Raoult}}$ , $\Delta H_{\text{mix}} > 0$ , $\Delta V_{\text{mix}} > 0$ — ethanol + acetone, carbon disulphide + acetone, ethanol + water.
Negative deviation: $P < P_{\text{Raoult}}$ , $\Delta H_{\text{mix}} < 0$ , $\Delta V_{\text{mix}} < 0$ — chloroform + acetone, phenol + aniline, nitric acid + water.
The vapour pressure of a non-ideal solution can lie OUTSIDE the range of the pure components' pressures (a maximum or a minimum), which is what makes azeotropes; 'always lies between' is the false statement.

Type	Raoult's law	ΔH mix, ΔV mix	Examples
Ideal	Obeyed at every composition	Both zero	Benzene + toluene; hexane + heptane The exam's default 'obeys Raoult's law' answer is benzene + toluene.
Positive deviation	P above Raoult	Both positive	Ethanol + acetone; CS₂ + acetone; ethanol + water Acetone breaks ethanol's hydrogen bonds — weaker A–B attraction, higher vapour pressure.
Negative deviation	P below Raoult	Both negative	Chloroform + acetone; phenol + aniline; HNO₃ + water Chloroform's H bonds to acetone's oxygen — a NEW attraction, lower vapour pressure.

Deviation follows the strength of the A–B attraction relative to A–A and B–B.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 22 April Shift I · Q56Moderate

Example 3 · Solutions and Colligative Properties · Vapour Pressure and Raoult's Law

Which from following mixtures exhibits positive deviation from Raoult's law?

Calling chloroform + acetone positive

Chloroform's C–H hydrogen-bonds to acetone's C=O, a new attraction that LOWERS the vapour pressure: negative deviation. Ethanol + acetone is the positive one.

Drill 3 more on ideal and non-ideal solutions: which way a mixture deviates

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)

Raoult's Law for Two Volatile Liquids: P = x_A P_A° + x_B P_B°
Raoult's law
$P_{\text{total}} = x_A P_A^\circ + x_B P_B^\circ \qquad y_A = \frac{x_A P_A^\circ}{P_{\text{total}}}$
Relative Lowering of Vapour Pressure = Mole Fraction of the Solute
Relative lowering
$\frac{P^\circ - P}{P^\circ} = x_2 = \frac{n_2}{n_1 + n_2} \approx \frac{W_2\,M_1}{M_2\,W_1}$

Reference tables (1)

Ideal and Non-Ideal Solutions: Which Way a Mixture Deviates3 rows

Type	Raoult's law	ΔH mix, ΔV mix	Examples
Ideal	Obeyed at every composition	Both zero	Benzene + toluene; hexane + heptane The exam's default 'obeys Raoult's law' answer is benzene + toluene.
Positive deviation	P above Raoult	Both positive	Ethanol + acetone; CS₂ + acetone; ethanol + water Acetone breaks ethanol's hydrogen bonds — weaker A–B attraction, higher vapour pressure.
Negative deviation	P below Raoult	Both negative	Chloroform + acetone; phenol + aniline; HNO₃ + water Chloroform's H bonds to acetone's oxygen — a NEW attraction, lower vapour pressure.

Deviation follows the strength of the A–B attraction relative to A–A and B–B.

Watch out for (3)

Using the given mole fraction for the wrong component
→ Raoult's Law for Two Volatile Liquids: P = x_A P_A° + x_B P_B°
Dividing by the solution's pressure, or reporting the solvent's mole fraction
→ Relative Lowering of Vapour Pressure = Mole Fraction of the Solute
Calling chloroform + acetone positive
→ Ideal and Non-Ideal Solutions: Which Way a Mixture Deviates

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