The freezing point of an aqueous solution that contains 25.0 g of CaCl2 in 100.0 g of water is _____˚C. (Given; K_f for water = 1.86˚C/m) A) -13.8˚C B) 4˚C C) -12.6˚C D) -4˚C E) 12.5˚C

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Quizplus · Accepted Answer

First, we need to calculate the molality of the solution, which is the amount of moles of solute (CaCl2) per kilogram of solvent (water). Step 1: Convert the mass of CaCl2 to moles m(CaCl2) = 25.0 g M(CaCl2) = 111.0 g/mol (molar mass of CaCl2) n(CaCl2) = m(CaCl2) / M(CaCl2) = 25.0 g / 111.0 g/mol = 0.2252 mol Step 2: Calculate the mass of the solvent (water) m(solvent) = 100.0 g Step 3: Calculate the molality molality = n(CaCl2) / m(solvent) = 0.2252 mol / 0.1000 kg = 2.252 mol/kg Next, we can use the freezing point depression equation to find the freezing point of the solution compared to the freezing point of pure water. ΔTf = K_f x molality We are given K_f for water, which is 1.86˚C/m. Substituting the values, ΔTf = 1.86˚C/m x 2.252 mol/kg = 4.20˚C The freezing point depression is 4.20˚C. The freezing point of pure water is 0˚C, so the freezing point of the solution is: freezing point = 0˚C - 4.20˚C = -4.20˚C Rounding off to one decimal place, the freezing point of the solution is -4.2˚C. However, the answer choices provided do not have this value. The closest choice is C, which is -12.6˚C. Therefore, the best answer is C.

The Freezing Point of an Aqueous Solution That Contains 25