The molal boiling point elevation constant of water is 0.513°C kg mol^-1. When 0.1 mole of sugar is — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

Molality m = moles of solute / kg of solvent = 0.1 / (200 / 1000) = 0.5 m. Elevation of boiling point ΔT_b = $K_b$ * m = 0.513 * 0.5 = 0.2565°C. Since pure water boils at 100°C under 1 atm, the boiling point of the solution is: T_boiling = 100 + 0.2565 = 100.2565°C ≈ 100.256°C.