Answer:
Freezing T° of solution is -142.4°C
Explanation:
This excersise is about colligative properties, in this case freezing point depression,
ΔT = Kf . m . i
Where ΔT = Freezing T° of solvent - Freezing T° of solution
Kf = Cryoscopic constant
m = mol/kg (molality)
i = Number of ions dissolved.
Water is not ionic, so i = 1
Let's find out m.
We determine mass of water, by density
498ml . 1 g/mL = 498 g
We convert the mass of water to moles → 498 g . 1mol/18g = 27.6 moles
We determine mass of solvent by density
2500 mL . 0.789 g/mL = 1972.5 g
Notice, we had to convert L to mL to cancel units.
1 cm³ = 1 mL
We convert the mass from g to kg → 1972.5 g . 1kg /1000g = 1.97kg
We determine m = mol/kg → 27.6mol / 1.97kg = 13.9 m
Kf for ethanol is: 1.99 °C/m
Freezing T° for ethanol is: -114.6°C
We replace at formula: - 114.6°C - Freezing T° solution = 1.99 °C/m . 13.9 m . 1
- 114.6°C - Freezing T° solution = 27.8 °C
- Freezing T° solution = 27.8°C + 114.6°C
Freezing T° Solution = - 142.4 °C
