JEE Main 2026 — Jan 21, Shift 2 — Chemistry solution

JEE Main2026Jan 21, Shift 2Chemistry

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.

Your answer

Solution

Step 1 — get the molality from boiling-point elevation: $Δ T_{b} = K_{b} m \Rightarrow 0.5 = 5.0 \cdot m \Rightarrow m = 0.1$ mol/kg.

Step 2 — moles of solute in 150 g of solvent: $n_{X} = 0.1 \times 0.150 = 0.015$ mol.

Step 3 — moles of solvent Y: $n_{Y} = \frac{150}{300} = 0.5$ mol.

Step 4 — relative lowering equals the mole fraction of solute (Raoult’s for dilute solutions): $\frac{P _{0} - P _{s}}{P _{0}} = \frac{n _{X}}{n _{X} + n _{Y}} \approx \frac{0.015}{0.515} \approx 0.0291 \approx 3 \times 1 0^{- 2}$ .

Answer: 3.

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