The rise in the boiling point of a solution containing 1.8 g of glucose in 100 g of a solvent is 0.1 — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

Molar mass of glucose (C6H12O6) = 180 g/mol. Moles of glucose = 1.8 / 180 = 0.01 mol. Molality of glucose m = 0.01 / 0.1 kg = 0.1 m. Elevation in boiling point ΔT_b = $K_b$ * m => 0.1°C = $K_b$ * 0.1 m => $K_b$ = 1.0 K-kg/mol = 1 K/m.