2.5 g of a non-volatile, non-electrolyte is dissolved in 100 g of water at 25°C. The solution showed a boiling point elevation by 2°C. Assuming the solute concentration is negligible with respect to the solvent concentration, the vapour pressure of the re

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Solution

Q.
2.5 g of a non-volatile, non-electrolyte is dissolved in 100 g of water at 25°C. The solution showed a boiling point elevation by 2°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 ${mol}^{- 1}$ ] [2024]

Ans.
(711)

$Δ T_{b} = m \times k_{b}$

$2 = m \times 0.52$

   $m = \frac{2}{0.52} m$

$\frac{x_{solute} \times 1000}{x_{solvent} \times M_{solvent}} = \frac{2}{0.52}$

   $\frac{x_{solute} \times 1000}{x_{solvent} \times 18} = \frac{2}{0.52}$

$\frac{x_{solute}}{x_{solvent}} = \frac{2}{0.52} \times \frac{18}{1000}$

   $\frac{x_{solvent}}{x_{solute}} = \frac{130}{9}$

$x_{solvent} = \frac{130}{9 + 130} = \frac{130}{139}$

By Raoult's law:

   $Δ T_{b} = m \times k_{b}$

   $P_{solution} = P_{solvent}^{\circ} x_{solvent}$

   $= 760 \times \frac{130}{139} mmHg$

   $= 710.8 mmHg \approx 711 mmHg$

Note: $P_{solvent}^{\circ}$ is not given in question, question is solved by taking it as 760 mmHg which is not correct as 760 mmHg is vapor pressure of water at $100 ° C$ but temperature in question is $25 ° C$ .

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