Answer:
Freezing T° of solution = - 7.35 °C
Explanation:
This is about the freezing point depression, a colligative property which depends on solute.
The formula is: Freezing T° pure solvent - Freezing T° solution = m . Kf . i
Freezing T° of pure solvent is -3.1°C
At this case, i = 1. As an organic compound the urea does not ionize.
We determine the molality (mol/kg of solvent)
We convert mass to moles:
12.3 g . 1mol / 60.06 g = 0.205 moles
0.205 mol / 0.3 kg = 0.682 mol/kg
We replace data in the formula:
-3.1°C - Freezing T° of solution = 0.682 mol/kg . 6.23 kg°C/mol . 1
Freezing T° of solution = 0.682 mol/kg . 6.23 kg°C/mol . 1 + 3.1°C
Freezing T° of solution = - 7.35 °C
