A substance ‘X’ (1.5 g) dis — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

vent ‘Y’ (molar mass = 300 g 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. (Nearest integer) [Given : Kb of the solvent = 5.0 K kg mol –1] Assume the solution to be dilute and no association or dissociation of X takes place in solution. Ans. (3) Sol. Tb = i × Kb × m 0.5 = i × m × 5 i × m = 0.5 5 = 0.1 i × a = 15 (where a = moles of solute) Now, o S o P P P  = i Xsolute = i × a a  = a 15 /1000 i 1/ 2 1/ 2   = 30 1000 = 3 × 10 –2 = 3