What is the freezing point (°C)of a solution prepared by dissolving 11.3 g of Ca(NO₃)₂ in 115 g of water? The molal freezing point depression constant for water is (Assume 100% ionization of Ca(NO₃)₂.) A)-3.34 B)-1.11 C)3.34 D)1.11 E)0.00

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First, we need to calculate the molality of the solution: moles of Ca(NO₃)₂ = 11.3 g / 164.3 g/mol = 0.0688 mol mass of water = 115 g molality = 0.0688 mol / 0.115 kg = 0.599 mol/kg Next, we can use the formula for freezing point depression: ΔTf = Kf * molality ΔTf = -1.86°C * 0.599 mol/kg = -1.11°C (note the negative sign, indicating a decrease in freezing point) Therefore, the freezing point of the solution is 0°C - 1.11°C = -1.11°C. The answer is B.

What Is the Freezing Point (°C)of a Solution Prepared by Dissolving