Which of the following is the correct formula of a colligative property?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\Delta T_f=\frac{K_f\times 1000\times w_2}{M_2\times w_1}\)
The depression of freezing point is proportional to the molality of the solute:
\( \Delta T_f = K_f \, m, \qquad m = \displaystyle\frac{w_2/M_2}{w_1/1000} \)
where w2 and M2 are the mass and molar mass of the solute and w1 is the mass of the solvent in grams. Substituting:
\( \Delta T_f = \displaystyle\frac{K_f \times 1000 \times w_2}{M_2 \times w_1} \)
This is Option (3).
Why the other options are incorrect
Option (1) \( \displaystyle \frac{P_1^\circ - P_1}{P_1^\circ} = x_1 \) — Incorrect. The relative lowering of vapour pressure equals the mole fraction of the solute, x2, not x1 (the solvent).
Option (2) \( \displaystyle \Delta T_b = \frac{K_b \times 1000 \times w_2}{M_1 \times w_1} \) — Incorrect. The elevation of boiling point uses the molar mass of the solute, M2; the printed M1 is the molar mass of the solvent.
Option (4) \( \pi = \Delta T_f R T \) — Incorrect. Osmotic pressure is π = CRT (C = molar concentration of the solute); ΔTf has no place in it.
NCERT (2026-27 edition):
Depression of Freezing Point – …
