Answer:
T° freezing solution = -22.8°C
Explanation:
To solve this problem we apply, the Freezing Point Depression. This is a colligative property which its formula is:
T° freezing pure solvent - T° freezing solution = Kf . molality . i
i = Van't Hoff factor.
We have been informed is a nonionizing solute, so i = 1
Our solute is ethylene glycol, so le'ts determine the moles to get molality
388 g . 1 mol / 62.07 g = 6.25 moles
molality (m) = moles of solute /kg of solvent
We convert mass of solvent, water, to kg → 510 g . 1kg /1000g = 0.510 kg
6.25 mol /0.510kg = 12.25 m
We replace at formula → 0°C - T° freezing solution = 1.86°C/m . 12.25 m . 1
T° freezing solution = -22.8°C
