The molal boiling-point elevation constant of water is $K_b = 0.52$ K kg/mol. What is the boiling point of a solution of $0.5$ mol of glucose in $500$ g of water? (Boiling point of pure water = $373$ K)

Molality $m = \dfrac{\text{moles of solute}}{\text{kg of solvent}} = \dfrac{0.5}{0.5} = 1$ mol/kg. Elevation: $\Delta T_b = K_b \cdot m = 0.52 \times 1 = 0.52$ K. New boiling point: $373 + 0.52 = 373.52$ K. Glucose is a non-electrolyte (van't Hoff factor $i \approx 1$), so no correction is needed. For NaCl with the same molality, $\Delta T_b$ would roughly double.