w2 g of a non-volatile solute of molar mass M2 is dissolved in w1 g of a solvent of molar mass M1, giving a dilute solution of molality m. Match List-I with List-II.
| List-I | List-II |
|---|---|
| (A) Depression of freezing point, ΔTf | (I) \( \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1} \) |
| (B) Molar mass of the solute, M2 | (II) \( \frac{R \times M_1 \times T_b^2}{1000 \times \Delta_{vap}H} \) |
| (C) Molal depression constant, Kf | (III) Kf m |
| (D) Molal elevation constant, Kb | (IV) \( \frac{R \times M_1 \times T_f^2}{1000 \times \Delta_{fus}H} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-III, B-I, C-IV, D-II
For a dilute solution the depression of freezing point is proportional to the molality. The constants Kf and Kb belong to the solvent: Kf uses its freezing point and enthalpy of fusion, Kb its boiling point and enthalpy of vaporisation.
(A) Depression of freezing point, ΔTf — This matches (III) Kf m.
(B) Molar mass of the solute, M2 — This matches (I) \( \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1} \). Putting \( m = \frac{w_2 \times 1000}{M_2 \times w_1} \) into ΔTf = Kf m and solving for M2:
\( \displaystyle M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1} \)
(C) Molal depression constant, Kf — This matches (IV) \( \frac{R \times M_1 \times T_f^2}{1000 \times \Delta_{fus}H} \), the relation with Tf and ΔfusH.
(D) Molal elevation constant, Kb — This matches (II) \( \frac{R \times M_1 \times T_b^2}{1000 \times \Delta_{vap}H} \), the relation with Tb and ΔvapH.
NCERT (2026-27 edition):
1.6.3 Depression of Freezing Point – “Similar to elevation of boiling point, depression of freezing point (ΔTf) for dilute solution (ideal solution) is directly proportional to molality, m of the solution. Thus,”
\( \displaystyle \Delta T_f = K_f\,m \qquad (1.34) \)
\( \displaystyle M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1} \qquad (1.36) \)
“The values of Kf and Kb, which depend upon the nature of the solvent, can be ascertained from the following relations.”
\( \displaystyle K_f = \frac{R \times M_1 \times T_f^2}{1000 \times \Delta_{fus}H} \qquad (1.37) \)
\( \displaystyle K_b = \frac{R \times M_1 \times T_b^2}{1000 \times \Delta_{vap}H} \qquad (1.38) \)
“Here the symbols R and M1 stand for the gas constant and molar mass of the solvent, respectively and Tf and Tb denote the freezing point and the boiling point of the pure solvent respectively in kelvin. Further, ΔfusH and ΔvapH represent the enthalpies for the fusion and vapourisation of the solvent, respectively.”