Question

What is the freezing point of a solution of 498mL of water (solute) dissolved in 2.50 L of ethanol (solvent), C2H5OH? The density of C2H5OH is 0.789g/cm3. (Remember that water has a density of 1.0 g/cm3.)

Answer 1

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