JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Solutions

A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass

= 300 g mol^{- 1}

) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is

____

\times 1 0^{- 2}

. (Nearest integer)
[Given:

K_{b}

of the solvent

= 5.0 K kg mol^{- 1}

]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.

Answer: 3

Solution

Boiling point elevation fixes the molality:

0.5 = 5.0 \times m

, so

m = 0.1 mol k g^{- 1}

, and the

0.150 kg

of solvent then holds

0.1 \times 0.150 = 0.015 mol

of X. The solvent itself amounts to

150/300 = 0.5 mol

. For a dilute solution the relative lowering of vapour pressure equals the mole fraction of solute,

\frac{0.015}{0.015 + 0.5} \approx 0.03

. Expressed in the stated units the relative lowering is

3 \times 1 0^{- 2}

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Integer answer