Answer:
Freezing T° of solution = - 3.72°C
Boiling T° of solution = 101.02°C
Explanation:
To solve this we apply colligative properties. Firstly, freezing point depression:
ΔT = Kf . m . i
ΔT = Freezing T° of pure solvent - Freezing T° of solution
Kf = Cryoscopic constant, for water is 1.86 °C/m
m = molality (moles of solute in 1kg of solvent)
i = Ions dissolved in solution
Our solute is sucrose, an organic compound so no ions are defined. i = 1.
Let's determine the moles: 68.4 g . 1mol/ 342g = 0.2 moles
molality = 0.2 mol / 0.1kg of water = 2 m
We replace data: ΔT = 1.86°C/m . 2m . 1
Freezing T° of solution = - 3.72°C
Now, we apply elevation of boiling point: ΔT = Kb . m . i
ΔT = Boiling T° of solution - Boiling T° of pure solvent
Kf = Ebulloscopic constant, for water is 0.512 °C/m
We replace:
Boiling T° of solution - Boiling T° of pure solvent = 0.512 °C/m . 2 . 1
Boiling T° of solution = 0.512 °C/m . 2 . 1 + 100°C → 101.02°C
