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JEE Mains Chemistry · Solutions

Relative Lowering of Vapour Pressure

A non-volatile solute lowers the solvent's vapour pressure, and the fractional lowering (p° − p)/p° equals the mole fraction of the solute.

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Why this matters

Eleven PYQs, seven of them numeric, and four from 2026. Seven apply the relative lowering directly to find a vapour pressure, a mass of solute or a number of moles; four reach it through a boiling-point elevation, because both depend on the same moles of solute. The skill that decides these is counting the solvent's moles correctly.

Concept 1 of 2: Relative lowering of vapour pressure equals the solute's mole fraction

Only solvent molecules can escape from the surface, and a non-volatile solute takes up part of it. The solution's vapour pressure is the solvent's share times p∘p^\circ, so the fraction lost is the solute's share.

Definition

  • p=x1p∘p = x_1 p^\circ, so p∘−pp∘=x2=n2n1+n2\dfrac{p^\circ - p}{p^\circ} = x_2 = \dfrac{n_2}{n_1 + n_2}.
  • Dilute form: p∘−pp∘≈n2n1\dfrac{p^\circ - p}{p^\circ} \approx \dfrac{n_2}{n_1}. Use it when the stem says the solution is dilute or the solute amount is negligible.
  • Electrolyte: replace n2n_2 by i n2i\,n_2, where ii, the van 't Hoff factor, is the number of particles each formula unit gives (NaCl fully dissociated: i=2i = 2; for example MgCl2\mathrm{MgCl_2} with 80% dissociation has i=2.6i = 2.6).
  • Percent w/v: grams of solute per 100 mL of SOLUTION. Mass of solvent == density ×\times volume −- mass of solute.

Raoult's law for a non-volatile solute

p∘−pp∘=x2=n2n1+n2≈n2n1\frac{p^\circ - p}{p^\circ} = x_2 = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}

Worked example

15 g of urea (M = 60 g/mol) is dissolved in 85.5 g of water. The vapour pressure of pure water is 32 mmHg. Find the vapour pressure of the solution.
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The same idea in a real exam question:

JEE Mains · 2023 · 15 Apr 2023 · Q56Moderate

Example 1 · Solutions · Relative Lowering of Vapour Pressure

The vapour pressure of 30%(w/v)30\%(w/v) aqueous solution of glucose is _____mmHg\_\_\_\_\_mmHg at 25∘C25^{\circ}C. [Given: The density of 30%(w/v)30\%(w/v), aqueous solution of glucose is 1.2 g cm−31.2\text{ }g{\text{ }cm}^{- 3} and vapour pressure of pure water is 24 mmHg24\text{ }mmHg.] (Molar mass of glucose is 180 g mol−1180\text{ }g{\text{ }mol}^{- 1}.)

Solute's mole fraction, or solvent's?

The relative lowering is the SOLUTE's mole fraction; the solution's vapour pressure divided by p∘p^\circ is the SOLVENT's. If the solute's mole fraction is 0.4, the solvent's is 0.6. Read which one the question asks for.

Mass of solution is not mass of solvent

In a w/v solution, 100 mL of solution weighs density times 100 mL, and the solute's mass must be taken out to get the solvent's. Using the solution's full mass overcounts the water.

An electrolyte multiplies the solute's moles

A salt that dissociates gives ii particles per formula unit. Leaving out ii for MgCl2\mathrm{MgCl_2} or NaCl gives a lowering that is too small.
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Concept 2 of 2: Relative lowering of vapour pressure from a boiling-point elevation

Both effects count the same solute particles. The elevation gives the molality, the molality gives the solute's moles in the solvent, and the solvent's own moles come from its molar mass. From there the relative lowering is one division.

Definition

  • Molality from the elevation: m=ΔTbKbm = \dfrac{\Delta T_b}{K_b}.
  • Moles of solute =m×= m \times kg of solvent; moles of solvent == grams of solvent /M1/M_1.
  • Dilute form in one line: p∘−pp∘≈n2n1=mM11000\dfrac{p^\circ - p}{p^\circ} \approx \dfrac{n_2}{n_1} = \dfrac{m M_1}{1000}, with M1M_1 in g/mol.
  • The exact form n2n1+n2\dfrac{n_2}{n_1 + n_2} differs only in the third significant figure for a dilute solution.

Linking the two colligative effects

p∘−pp∘≈m M11000=ΔTb M11000 Kb\frac{p^\circ - p}{p^\circ} \approx \frac{m\,M_1}{1000} = \frac{\Delta T_b\,M_1}{1000\,K_b}

Worked example

A non-volatile solute raises the boiling point of benzene (Kb=2.53K_b = 2.53 K kg/mol, M = 78 g/mol) by 0.506 K. Find the relative lowering of the vapour pressure of benzene.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 21 Jan 2026 Shift 2 · Q48Moderate

Example 2 · Solutions · Relative Lowering of Vapour Pressure

A substance ' X ' ( 1.5 g ) dissolved in 150 g of a solvent ' Y ' (molar mass =300 g mol−1= 300\text{ }g{\text{ }mol}^{- 1} ) led to an elevation of the boiling point by 0.5 K . The relative lowering in the vapour pressure of the solvent ' Y ' is ____\_\_\_\_ ×10−2\times 10^{- 2}. (Nearest integer) [Given : KbK_{b} of the solvent =5.0 K kg mol−1= 5.0\text{ }K\text{ }kg{\text{ }mol}^{- 1} ] Assume the solution to be dilute and no association or dissociation of X takes place in solution.

Moles of solute come from the elevation, not from a molar mass

When the solute's molar mass is not given, the only route to its moles is m=ΔTb/Kbm = \Delta T_b/K_b times the kilograms of solvent. Look for the molar mass of the SOLVENT instead: it is what turns grams of solvent into moles.

Kilograms for molality, grams for moles of solvent

Molality uses kilograms of solvent; moles of solvent use grams divided by its molar mass. Mixing the two gives an answer off by a factor of 1000.
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Summary — formulas & gotchas at a glance

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Formulas (2)

Watch out for (5)

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