Question

What is the freezing point of a solution prepared from 50.0 g ethylene glycol (C2H6O2) and 85.0 g H2O? [For water, Kf = 1.86°C/m]

Answer 1

-17.6°C

Step-by-step explanation:

Let's apply the Colligative property of Freezing point depression:

ΔT = Kf . m where:

ΔT = T° freezing pure solvent - T° freezing solution

Kf = Cryoscopic constant, for water is 1.86 °C/m

m = molality (mol of solute / 1kg of solvent)

Let's determine the molality. We convert the mass of solute to moles and the grams of solvent to kg

85 g . 1 kg / 1000 = 0.085 kg

50 g / 62 g/ mol = 0.806 g/mol

Molality = 0.806 mol / 0.085 kg → 9.48 m

Let's go the formula

0° - T° freezing solution = 1.86 °C/m . 9.48 m

T° freezing solution = - 1.86 °C/m . 9.48 m → -17.6°C