What is the freezing point of a solution of 2.86 g MgCl₂ in 100 g of water? K_f for water is 1.86°C/m. A)-0.0558°C B)-0.558°C C)-1.12°C D)-1.67°C

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Quizplus · Accepted Answer

The freezing point depression is calculated using ΔTf = i * Kf * m, where i is the van't Hoff factor (3 for MgCl2), Kf is 1.86°C/m, and m is the molality. Molality is calculated as moles of solute per kg of solvent. Moles of MgCl2 = 2.86 g / 95.21 g/mol = 0.03 mol. Molality = 0.03 mol / 0.1 kg = 0.3 m. ΔTf = 3 * 1.86°C/m * 0.3 m = 1.674°C. The freezing point is 0°C - 1.674°C = -1.67°C.

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