A substance ' X ' (1.5 g) dissolved in 150 g of a solvent ' 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 : K _ b 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.

Step-by-step solution

Using boiling point elevation to find molality: T_b = K_b m gives 0.5 = 5.0 m , so m = 0.1 mol/kg. Converting molality to moles of solute: 0.1 = moles of X 0.15 , giving moles of X = 0.015 mol. Molar mass of X = 1.5 g / 0.015 mol = 100 g/mol. For vapor pressure lowering, the relative lowering equals the mole fraction of solute. Moles of solvent Y = 150 g / 300 g/mol = 0.5 mol. Relative lowering of vapor pressure = n_X n_X + n_Y = 0.015 0.015 + 0.5 = 0.015 0.515 = 0.0291 3 10⁻²