0.5 molal aqueous solution of a weak acid (HX) is 20% ionized. If K_f…

The correct answer is Option (2) → 1.12 K

A weak acid ionises only partly, so the number of solute particles is more than the number of formula units dissolved. The van’t Hoff factor i accounts for this:

    \( \ce{HX <=> H^+ + X^-} \)

With degree of ionisation α = 0.20, one mole of HX gives (1 − α) mol HX + α mol H⁺ + α mol X⁻:

    i = (1 − α) + α + α = 1 + α = 1 + 0.20 = 1.2

The depression of freezing point is:

    ΔT_f = i K_f m

    ΔT_f = 1.2 × 1.86 K kg mol⁻¹ × 0.5 mol kg⁻¹

    ΔT_f = 1.116 K ≈ 1.12 K

The lowering ΔT_f = T_f° − T_f is a positive quantity.

NCERT (2026-27 edition):
Abnormal Molar Masses – “Inclusion of van’t Hoff factor modifies the equations for colligative properties as follows: … Depression of Freezing point, ΔT_f = i K_f m”