45 g of ethylene glycol (C2H6O2) is mixed with 600 g of water. Calculate (a) the freezing point depression and (b) the freezing point of the solution.

To tackle the problem of calculating the freezing point depression of a solution made from ethylene glycol and water, we can break it down into a few logical steps. First, we need to understand the concept of freezing point depression, which is a colligative property. This means that the freezing point of a solvent decreases when a solute is added, and the extent of this depression depends on the number of solute particles in the solution rather than their identity.

Step 1: Gather Necessary Information

We have the following data:

• Mass of ethylene glycol (C2H6O2): 45 g
• Mass of water: 600 g
• Molar mass of ethylene glycol: approximately 62.07 g/mol
• Molar mass of water: approximately 18.02 g/mol
• Freezing point depression constant (Kf) for water: 1.86 °C kg/mol

Step 2: Calculate Moles of Ethylene Glycol

To find the number of moles of ethylene glycol, we use the formula:

moles = mass (g) / molar mass (g/mol)

Substituting the values:

moles of C2H6O2 = 45 g / 62.07 g/mol ≈ 0.724 moles

Step 3: Calculate the Mass of the Solvent in Kilograms

Since the freezing point depression constant (Kf) is given in °C kg/mol, we need the mass of water in kilograms:

mass of water = 600 g = 0.600 kg

Step 4: Calculate the Freezing Point Depression

The formula for freezing point depression (ΔTf) is:

ΔTf = i × Kf × m

Where:

• i = van 't Hoff factor (for ethylene glycol, i = 1 since it does not dissociate)
• Kf = freezing point depression constant (1.86 °C kg/mol for water)
• m = molality of the solution (moles of solute per kg of solvent)

First, we calculate the molality (m):

m = moles of solute / mass of solvent (kg) = 0.724 moles / 0.600 kg ≈ 1.207 mol/kg

Now we can calculate ΔTf:

ΔTf = 1 × 1.86 °C kg/mol × 1.207 mol/kg ≈ 2.25 °C

Step 5: Determine the Freezing Point of the Solution

The normal freezing point of pure water is 0 °C. To find the freezing point of the solution, we subtract the freezing point depression from the freezing point of pure water:

Freezing point of solution = 0 °C - ΔTf = 0 °C - 2.25 °C = -2.25 °C

Final Results

In summary, the calculations yield:

• Freezing Point Depression (ΔTf): 2.25 °C
• Freezing Point of the Solution: -2.25 °C

This means that the addition of ethylene glycol to water lowers the freezing point of the solution to -2.25 °C, which is an important consideration in applications like antifreeze in vehicles. Understanding these principles helps in various fields, including chemistry, environmental science, and engineering.