JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Solutions

2.5 g of a non-volatile, non-electrolyte is dissolved in 100 g of water at

2 5^{\circ} C

. The solution showed a boiling point elevation by

2^{\circ} C

. Assuming the solute concentration in negligible with respect to the solvent concentration, the vapour pressure of the resulting aqueous solution is _____ mm of Hg (nearest integer)
[Given : Molal boiling point elevation constant of water

(K_{b}) = 0.52 K . kg mol^{- 1}

,
1 atm pressure

= 760 mm

of Hg , molar mass of water

= 18 g mol^{- 1}

]

Answer: 707

Solution

Boiling-point elevation gives the molality:

Δ T_{b} = K_{b} m \Rightarrow m = \frac{2}{0.52} = 3.846 mol/kg .

For a dilute solution the relative lowering of vapour pressure is

\frac{Δ P}{P ^{\circ}} \approx \frac{m M _{solvent}}{1000} = \frac{3.846 \times 18}{1000} = 0.0692.

Δ P = 760 \times 0.0692 \approx 52.6 mm \Rightarrow P = 760 - 52.6 \approx 707 mm Hg .

Answer: 707.

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Integer answer

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JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Solutions

Solution

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