Calculate the freezing point of a solution when 370 grams of Ca₃(PO₄)is added to 3000mL of water.(density of water is 1.00 g/mL).The K_f for water is 1.86°C/m and normal freezing point of water is 0.00°C. A)-0.74 B)-0.0030 C)-3.70 D)-1.27

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Quizplus · Accepted Answer

First, we need to calculate the molality of the solution: molality = moles of solute / mass of solvent in kg The molar mass of Ca₃(PO₄) is 40.078 g/mol + 32.066 g/mol + 13(15.999 g/mol) + 1(30.974 g/mol) = 407.642 g/mol moles of Ca₃(PO₄) = 370 g / 407.642 g/mol = 0.907 mol mass of solvent (water) = 3000 mL x 1.00 g/mL = 3000 g = 3 kg molality = 0.907 mol / 3 kg = 0.302 mol/kg Using the freezing point depression equation: ΔTf = Kf x molality where ΔTf = change in freezing point and Kf = freezing point depression constant. We can calculate the freezing point depression: ΔTf = 1.86°C/m x 0.302 mol/kg = 0.56172°C Finally, we can calculate the freezing point of the solution: freezing point = normal freezing point - ΔTf = 0.00°C - 0.56172°C = -0.56172°C Therefore, the answer is C) -3.70 (rounded to two decimal places).

Calculate the Freezing Point of a Solution When 370 Grams