Using boiling point elevation to find molality:
ΔTb=Kb×m gives 0.5=5.0×m, so m=0.1 mol/kg.
Converting molality to moles of solute:
0.1=0.15moles of X, 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
= nX+nYnX=0.015+0.50.015
=0.5150.015=0.0291≈3×10−2