Identify the correct relation between the molar mass of solute and Eb…

The correct answer is Option (3) → M₂ = (1000 × w₂ × K_b) / (ΔT_b × w₁)

The elevation of boiling point is proportional to the molality: ΔT_b = K_b m, where K_b is the ebullioscopic constant. For w₂ g of solute of molar mass M₂ in w₁ g of solvent:

    \( \displaystyle m = \frac{w_2/M_2}{w_1/1000} = \frac{1000 \times w_2}{M_2 \times w_1} \)

    \( \displaystyle \Delta T_b = \frac{K_b \times 1000 \times w_2}{M_2 \times w_1} \)

    \( \displaystyle M_2 = \frac{1000 \times w_2 \times K_b}{\Delta T_b \times w_1} \)

So the solute mass and K_b are in the numerator, and ΔT_b and the solvent mass in the denominator.

Why the other options are incorrect

• Option (1) M₂ = (1000 × w₂ × w₁)/(ΔT_b × K_b) — Incorrect. It has w₁ in the numerator and K_b in the denominator; the solvent mass must divide and K_b must multiply.

• Option (2) ΔT_b = (1000 × w₂ × K_b × M₂)/w₁ — Incorrect. It puts M₂ in the numerator of ΔT_b; a larger molar mass gives fewer moles and a smaller ΔT_b, so M₂ belongs in the denominator.

• Option (4) M₂ = (1000 × w₁ × K_b)/(ΔT_b × w₂) — Incorrect. It interchanges w₁ and w₂; the mass of solute, w₂, goes in the numerator.

NCERT (2026-27 edition):
Elevation of Boiling Point – “… Experiments have shown that for dilute solutions the elevation of boiling point (ΔT_b) is directly proportional to the molal concentration of the solute in a solution. Thus
ΔT_b ∝ m   (1.29)
or  ΔT_b = K_b m   (1.30)
Here m (molality) is the number of moles of solute dissolved in 1 kg of solvent and the constant of proportionality, K_b is called Boiling Point Elevation Constant or Molal Elevation Constant (Ebullioscopic Constant).”