What is the normal boiling point of a solution that contains 30.0 g of glucose (C₆H₁₂O₆)in 100.0 g of H₂O? K_b for water is 0.512°C/m.

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Quizplus · Accepted Answer

The boiling point elevation can be calculated using the formula ΔT = i * Kb * m, where i is the van't Hoff factor (1 for glucose), Kb is the ebullioscopic constant, and m is the molality. The molality is calculated as moles of solute per kg of solvent. The moles of glucose is 30.0 g / 180.18 g/mol = 0.1665 mol. The molality is 0.1665 mol / 0.1 kg = 1.665 m. ΔT = 1 * 0.512°C/m * 1.665 m = 0.852°C. The normal boiling point of water is 100°C, so the boiling point of the solution is 100°C + 0.852°C = 100.85°C.

What Is the Normal Boiling Point of a Solution That