The latent heat of vaporization of a liquid of molar mass, 80 g/mol and boiling point, 127°C is 8 kc — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

The thermodynamic relation for ebullioscopic constant is: $K_b$ = (R * T_b^2 * M) / (1000 * ΔH_vap). Given: M = 80 g/mol = 0.080 kg/mol, T_b = 127 + 273 = 400 K, ΔH_vap = 8 kcal/mol = 8000 cal/mol, R = 2.0 cal/K-mol (since ΔH_vap is in cal). Substituting the values: $K_b$ = (2 * 400^2 * 80) / (1000 * 8000) = (2 * 160000 * 80) / 8000000 = 25600000 / 8000000 = 3.2 K-kg/mol.