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Properties of Water

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Question

Calculate the molal elevation constant, $k_{b}$ for water and the boiling point of 0.1 molal urea solution. Latent heat of vaporisation of water is 9.72 kcal mol $^{- 1}$ at 373.15 K.

A

K_{b} = 0. 515 K k g m o l^{- 1}, T_{b} = 373.20 K

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B

K_{b} = 0. 515 K k g m o l^{- 1}, T_{b} = 370.20 K

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C

K_{b} = 0. 565 K k g m o l^{- 1}, T_{b} = 373.20 K

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D

K_{b} = 5.15 K k g m o l^{- 1}, T_{b} = 370.20 K

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Solution

The correct option is A $K_{b} = 0. 515 K k g m o l^{- 1}, T_{b} = 373.20 K$
The expression for the molal elevation constant is $k_{b} = \frac{0.002 (T_{b}^{o})^{2}}{L_{v}}$

Here, $T_{b}^{o}$ is the boiling point of pure solvent and $L_{v}$ is the latent heat of vaporization in cal/g of the pure solvent. It is equal to $9.72 k c a l m o l^{- 1} o r 540 c a l / g$ .

$∴ k_{b} = \frac{0.002 \times (373.15)^{2}}{540} = 0.515 K k g m o l^{- 1}$
Th depression in freezing point is $Δ T_{b} = k_{b} \times m = 0.515 \times 0.1 = 0.0515$
The boiling point is $T_{b} = 373.15 + 0.0515 \approx 373.20 K$

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