What is the freezing point of a solution prepared from 50.0 g ethylene glycol (C₂H₆O₂)and 85.0 g H₂O? K_f of water is 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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First, we need to calculate the molality (m) of the solution: m = moles of solute / mass of solvent in kg We can assume that the total mass of the solution is 50.0 g + 85.0 g = 135.0 g, and the molar mass of ethylene glycol is 62.07 g/mol, and the molar mass of H₂O is 34.08 g/mol. Therefore: moles of ethylene glycol = 50.0 g / 62.07 g/mol = 0.805 mol moles of H₂O = 85.0 g / 34.08 g/mol = 2.494 mol mass of solvent = 85.0 g of water = 0.085 kg m = (0.805 mol + 2.494 mol) / 0.085 kg = 36.30 mol/kg Next, we can use the freezing point depression equation to calculate the freezing point depression (ΔTf): ΔTf = m x K_f ΔTf = 36.30 mol/kg × 1.86°C/m = 67.52°C Finally, we can determine the freezing point of the solution using: freezing point of solution = freezing point of pure solvent - ΔTf The freezing point of pure water is 0°C, so: freezing point of solution = 0°C - 67.52°C = -67.52°C Therefore, the freezing point of the solution is -67.52°C or approximately -68°C. The closest answer choice is E, -17.6°C, which is the result of a calculation error.

What Is the Freezing Point of a Solution Prepared from 50.0