If 1 mole of a non-volatile, non-electrolyte solute in 1000 g of water depresses the freezing point — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

First solution: 1 mole of solute in 1000 g of water has molality m_1 = 1.0 m. ΔT_f = $K_f$ * m_1 = 1.86°C => $K_f$ = 1.86 K-kg/mol. Second solution: 1 mole of solute in 500 g of water has molality m_2 = 1.0 / 0.500 kg = 2.0 m. ΔT_f = $K_f$ * m_2 = 1.86 * 2.0 = 3.72°C. Since pure water freezes at 0°C, the freezing point of the second solution is -3.72°C.