JEE Main 2026 Chemistry question
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, Kb of the solvent = 0.3 K kg mol⁻¹]
Answer: 5
Solution
ΔTb = 0.5 K gives molality = 0.5/0.3 = 1.667 mol/kg, so = 1.667 × 0.040 = 0.0667
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