What is the freezing point of a solution prepared from 50.0 g ethylene glycol (C₂H₆O₂) and 85.0 g H₂O? (For water, K_f = 1.86°C/m) A) 17.6°C B) -176°C C) -1.50°C D) 1.50°C E) -17.6°C

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The freezing point depression of a solution can be calculated using the formula ΔTf = i * Kf * m, where ΔTf is the freezing point depression, i is the van't Hoff factor (which is 1 for ethylene glycol since it does not ionize in solution), Kf is the freezing point depression constant for the solvent (1.86°C/m for water), and m is the molality of the solution. Ethylene glycol (C2H6O2) has a molar mass of approximately 62.07 g/mol. The molality (m) of the solution is the moles of solute per kilogram of solvent. Moles of ethylene glycol = 50.0 g / 62.07 g/mol = 0.805 mol. Mass of water in kilograms = 85.0 g / 1000 = 0.085 kg. Thus, molality (m) = 0.805 mol / 0.085 kg = 9.47 m. ΔTf = 1 * 1.86°C/m * 9.47 m = 17.61°C. Since this is a depression, the freezing point is lowered by this amount from 0°C (the freezing point of pure water), resulting in a new freezing point of -17.6°C.

What Is the Freezing Point of a Solution Prepared from 50.0