Answer:
Freezing point of solution = - 10.3°C
Boiling point of solution = 102.8°C
Explanation:
This is about colligative properties:
Freezing point depression → ΔT = Kf . m
ΔT = Freezing point of pure solvent - Freezing point of solution
Boiling point elevation → ΔT = Kb . m
ΔT = Boiling point of solution - Boiling point of pure solvent
Let's determine m which means molality (moles of solute in 1 kg of solvent)
Solute: Ethylene glycol → Mass = 25 g
Moles = Mass / Molar mass → 25 g / 60g/mol = 0.416 moles
Solvent: Water
25 % means, 25 g of solute in 100 g of solution
Therefore, the mass of water is 75 g (100 -25) . (Solution = Solute + Solvent)
We convert the mass from g to kg → 75 g . 1kg /1000g = 0.075 kg
Molality (mol/kg) = 0.416 m / 0.075kg = 5.55 m
We replace data in the formulas:
<u>Freezing point depression</u>:0° - Freezing point of solution= 1.86°C/m. 5.55 m
Freezing point of solution = - (1.86°C/m . 5.55m) = - 10.3°C
<u>Boiling point elevation</u>:Boiling point of solution - 100°C = 0.51°C /m . 5.55 m
Boiling point of solution = 0.51°C /m . 5.55 m + 100°C = 102.8°C
