At 100°C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If K_b = 0.52, the boiling point of this solution will be :

1. 101°C

2. 100°C

3. 102°C

4. 103°C

Doubtly Asked 30 Oct 2026 NEET UG 2016 Phase 1 Practice test neet Save WhatsApp Report

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Best answer

Answer: 1. 101°C

Key idea: use the lowering of vapour pressure to get the mole fraction, and then the boiling point elevation.

At 100°C pure water has $P^0=760$ mm.

$$\frac{P^0-P}{P^0}=\frac{28}{760}=0.0368=x_{solute}$$

This gives $n_2/n_1\approx0.0383$ with $n_1=100/18=5.56$, so $n_2\approx0.213$ mol. Molality $=0.213/0.1=2.13\ \mathrm{m}$.

$$\Delta T_b=K_bm=0.52\times2.13\approx1.1\approx1^\circ\mathrm{C}$$

Boiling point $\approx101^\circ\mathrm{C}$, option 1.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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