JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Solutions

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A non-volatile, non-electrolyte solid solute when dissolved in 40 g of a solvent, the vapour pressure of the solvent decreased from 760 mm Hg to 750 mm Hg. If the same solution boils at 320 K, then the number of moles of the solvent present in the solution is ______. (Nearest integer)
[Given: boiling point of the pure solvent

= 319.5 K

K_{b}

of the solvent

= 0.3 K kg mol^{- 1}

]

Answer: 5

Solution

Boiling-point elevation gives the molality:

Δ T_{b} = 320 - 319.5 = 0.5 K

, so molality

= \frac{Δ T _{b}}{K _{b}} = \frac{0.5}{0.3} = \frac{5}{3} mol kg^{- 1}

. With

40 g = 0.04 kg

of solvent, the solute amount is

n_{solute} = \frac{5}{3} \times 0.04 = \frac{1}{15} mol

. The relative lowering of vapour pressure equals the solute mole fraction:

\frac{760 - 750}{760} = \frac{1}{76} = \frac{n _{solute}}{n _{solute} + n _{solvent}}

, so

n_{solute} + n_{solvent} = 76 n_{solute}

and

n_{solvent} = 75 n_{solute} = 75 \times \frac{1}{15} = 5

. The solution contains 5 moles of solvent.

JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Integer answer

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JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Solutions

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