45 g of ethylene glycol is mixed with 600 g of water. What is the freezing point of the solution? K_f = 1.86 K kg mol^-1.

• A. -270.90 K
• B. 270.9 K
• C. 273 K
• D. 274.15 K

To find the freezing point of the solution, we need to use the formula for freezing point depression:

Freezing Point Depression Formula

The formula is:

ΔT_f = K_f × m

Where:

• ΔT_f = change in freezing point
• K_f = freezing point depression constant (1.86 K kg/mol for water)
• m = molality of the solution

Step 1: Calculate Moles of Ethylene Glycol

The molar mass of ethylene glycol (C₂H₆O₂) is approximately 62.07 g/mol. To find the number of moles:

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

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

Step 2: Calculate Molality

Molality (m) is defined as moles of solute per kilogram of solvent:

mass of water = 600 g = 0.600 kg

molality (m) = moles of solute / mass of solvent (kg)

molality = 0.724 moles / 0.600 kg ≈ 1.207 mol/kg

Step 3: Calculate Freezing Point Depression

Now, substitute the values into the freezing point depression formula:

ΔT_f = 1.86 K kg/mol × 1.207 mol/kg ≈ 2.245 K

Step 4: Determine New Freezing Point

The normal freezing point of water is 273 K. Therefore, the new freezing point is:

Freezing point = 273 K - ΔT_f

Freezing point = 273 K - 2.245 K ≈ 270.755 K

Final Answer

Rounding to two decimal places, the freezing point of the solution is approximately 270.76 K. Among the options provided, the closest answer is:

B. 270.9 K