72 g of glucose is dissolved in 1 kg of water in a sauce pan. At what temperature will the solution — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

Moles of glucose $= 72 \text{ g} / 180 \text{ g/mol} = 0.4$ mol. Molality $= 0.4 \text{ mol} / 1 \text{ kg} = 0.4$ m. Elevation in boiling point $\Delta T_b = K_b \times m = 0.52 \times 0.4 = 0.208$ K. At 1.013 bar (1 atm), the normal boiling point of pure water is 373.0 K (as per options scale context). Thus, the boiling point of the solution is $373.0 + 0.208 = 373.208$ K. Therefore, correct answer is C.