What is the freezing point (°C)of a solution prepared by dissolving 11.3 g of Ca(NO₃)₂ (formula weight = 164 g/mol)in 115 g of water? The molal freezing point depression constant for water is 1.86 °C/m. A)-3.34 B)-1.11 C)3.34 D)1.11 E)0.00

Question

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First, we need to calculate the molality of the solution: moles of Ca(NO₃)₂ = 11.3 g / 164 g/mol = 0.0689 mol mass of water = 115 g molality = 0.0689 mol / 0.115 kg = 0.599 m Next, we need to calculate the freezing point depression: ΔTf = Kf * molality ΔTf = 1.86 °C/m * 0.599 m = 1.11514 °C Finally, we can calculate the freezing point of the solution: freezing point = 0 °C - 1.11514 °C = -1.11514 °C Rounding to two decimal places, we get -1.11 °C. Therefore, the correct answer is (A).

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