A solution is prepared by dissolving 4.66 g of KCl in enough distilled water to give 250 mL of solution. KCl is a strong electrolyte. How will the freezing point of the solution be different from that of pure water? (Note: K_f for water is 1.86°C\M.) A)solution will be 0.930°C lower than water B)solution will be 0.475°C lower than pure water C)solution will be 0.930°C higher than pure water D)solution will be 0.475°C higher than pure water

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Quizplus · Accepted Answer

The freezing point depression can be calculated using the formula ΔTf = i * Kf * m, where i is the van't Hoff factor (2 for KCl), Kf is the freezing point depression constant (1.86°C/m for water), and m is the molality. First, calculate the molality: moles of KCl = 4.66 g / 74.55 g/mol = 0.0625 mol; molality = 0.0625 mol / 0.250 kg = 0.25 m. Then, ΔTf = 2 * 1.86°C/m * 0.25 m = 0.930°C. Thus, the solution's freezing point is 0.930°C lower than pure water.

A Solution Is Prepared by Dissolving 4