Two solvents X and Y have the same molar mass and the same boiling po…

The correct answer is Option (3) → K_b of X is smaller than K_b of Y.

• The molal elevation constant is a property of the solvent. It depends on the molar mass of the solvent (M₁), its boiling point (T_b) and its enthalpy of vaporisation:

      \( \displaystyle K_b = \frac{R \times M_1 \times T_b^2}{1000 \times \Delta_{vap}H} \)

• For X and Y, R, M₁ and T_b are the same, so K_b changes only with Δ_vapH, and in the opposite direction: the larger the enthalpy of vaporisation, the smaller the K_b.

• X has the larger Δ_vapH, so X has the smaller K_b.

Why the other options are incorrect

Option (1) K_b of X is greater than K_b of Y. — Incorrect. Δ_vapH is in the denominator, so a larger enthalpy of vaporisation gives a smaller K_b, not a larger one.

Option (2) K_b of X is equal to K_b of Y, since the two solvents have the same molar mass and boiling point. — Incorrect. Equal M₁ and T_b are not enough; the enthalpy of vaporisation also enters K_b, and it differs here.

Option (4) K_b of X and K_b of Y cannot be compared without knowing the solute dissolved in them. — Incorrect. K_b depends on the nature of the solvent, not on the solute; the solute only decides the molality m in ΔT_b = K_bm.

NCERT (2026-27 edition):
1.6.3 Depression of Freezing Point – “The values of K_f and K_b, which depend upon the nature of the solvent, can be ascertained from the following relations.”

    \( \displaystyle K_b = \frac{R \times M_1 \times T_b^2}{1000 \times \Delta_{vap}H} \qquad (1.38) \)

“Here the symbols R and M₁ stand for the gas constant and molar mass of the solvent, respectively and T_f and T_b denote the freezing point and the boiling point of the pure solvent respectively in kelvin. Further, Δ_fusH and Δ_vapH represent the enthalpies for the fusion and vapourisation of the solvent, respectively.”