A solution is prepared by dissolving 215 grams of methanol, CH₃OH, in 1000.0 grams of water. What is the boiling point of the solution? (The boiling point elevation constant for water is 0.512°C/mole solute in 1000 g of water.) A) 93.0°C B) 96.5°C C) 103.5°C D) 107.0°C

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Quizplus · Accepted Answer

The boiling point elevation can be calculated using the formula ΔT = i * K_b * m. Methanol is a non-electrolyte, so i = 1. The molality (m) is calculated by dividing the moles of methanol by the mass of water in kg. The moles of methanol are 215 g / 32.04 g/mol = 6.71 mol. The molality is 6.71 mol / 1 kg = 6.71 m. ΔT = 1 * 0.512°C/m * 6.71 m = 3.44°C. The normal boiling point of water is 100°C, so the boiling point of the solution is 100°C + 3.44°C = 103.44°C, approximately 103.5°C.

A Solution Is Prepared by Dissolving 215 Grams of Methanol