What is the freezing point of a 0.29 m solution of glucose (C₆H₁₂O₆)in water? (K_fp for water is 1.858 °C/m.) A) 0.27 °C B) 0.54 °C C) -0.54 °C D) -0.27 °C E) -1.08 °C

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Quizplus · Accepted Answer

The freezing point depression of the solution can be calculated using the formula ΔTf = i * Kf * m, where ΔTf is the change in freezing point, i is the van't Hoff factor (which is 1 for glucose, a non-electrolyte), Kf is the cryoscopic constant for water (1.858 °C/m), and m is the molality of the solution (0.29 m). Thus, ΔTf = 1 * 1.858 °C/m * 0.29 m = 0.539 °C. Since this represents a depression, the change is negative, making the correct answer -0.54 °C.

What Is the Freezing Point of a 0