Question

What is the depression of freezing point, when mole fraction of non-electrolyte solute in aqueous solution is 0.01? ($ K_f $ of $ \text{H}_2\text{O} = 1.86 \, \text{K kg mol}^{-1} $)

• A. 1.246 K
• B. 1.380 K
• C. 1.528 K
• D. 1.043 K

Hint:

Molality relates moles of solute to mass of solvent. Mole fraction relates moles of solute to total moles.

Answer 1

The Correct Option is D

To determine the depression of the freezing point when the mole fraction of a non-electrolyte solute is 0.01, we use the formula for depression in freezing point:

ΔT_f = K_f * molality

Where:

• ΔT_f is the depression in freezing point.
• K_f is the cryoscopic constant of the solvent (for water, K_f = 1.86 K kg mol^-1).
• Molality (m) is moles of solute per kilogram of solvent.

The relationship between mole fraction and molality is given by:

molality = (mole fraction of solute / mole fraction of solvent) * (1000 / Molar mass of solvent)

In our case, the solvent is water with a molar mass of approximately 18 g/mol (or 0.018 kg/mol). Given that the mole fraction of the solute is 0.01, the mole fraction of the solvent is:

Mole fraction of solvent = 1 - mole fraction of solute = 1 - 0.01 = 0.99

Now, calculate the molality:

molality = (0.01 / 0.99) * (1000 / 18)

Calculating the above:

molality ≈ 0.0101 * 55.56 ≈ 0.561

Finally, calculate the depression in freezing point:

ΔT_f = 1.86 * 0.561 ≈ 1.043 K

Thus, the depression of the freezing point is 1.043 K.