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ChemistryInteger+4 / -1Question 75 of 75

JEE Main 2026 · 4 Apr · Shift 1, Chemistry question 75

A non-volatile, non-electrolyte solid solute when dissolved in 40 g of a solvent, the vapour pressure of the solvent decreased from 760 mm Hg to 750 mm Hg. If the same solution boils at 320 K, then the number of moles of the solvent present in the solution is ________. (Nearest integer)
[Given: boiling point of the pure solvent = 319.5 K, KbK_b of the solvent = 0.3 K kg mol−1^{-1}]

Answer and worked solution

Official answer: 5

Elevation. 0.5=0.3m0.5=0.3m, so m=53m=\frac53 mol kg⁻¹ and nsolute=53×0.040=0.0667n_\text{solute}=\frac53\times0.040=0.0667 mol.

Raoult's law. 10760=nsolutensolvent+nsolute\dfrac{10}{760}=\dfrac{n_\text{solute}}{n_\text{solvent}+n_\text{solute}}, so nsolvent+nsolute=76×0.0667=5.07n_\text{solvent}+n_\text{solute}=76\times0.0667=5.07.

Solvent ≈5\approx5 mol.

Answer: 5

Topics:Colligative propertiesRaoult's law and vapour pressure
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