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Emission and Absorption Spectra

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Question

Calculate the molal elevation constant, $K_{b}$ for water and the boiling of 0.1 molal urea solution. Latent heat of vaporisation of water 9.72 kcal $m o l^{1}$ at 373.15K.

A

K_{b} = 0.512 k g m o l K^{- 1}, T_{b} = 373.20 K

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B

K_{b} = 05.12 k g m o l K^{- 1}, T_{b} = 378.20 K

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C

K_{b} = 1.02 k g m o l K^{- 1}, T_{b} = 383.20 K

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D

K_{b} = 0.512 k g m o l K^{- 1}, T_{b} = 385.70 K

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Solution

The correct option is A $K_{b} = 0.512 k g m o l K^{- 1}, T_{b} = 373.20 K$
$K_{b} = \frac{R \times M_{1} \times T_{b}^{2}}{1000 \times Δ_{v a p} H}$

$M_{1} = 18 g, T_{b} = 373.15 K$

$Δ_{v a p} H = 9.72 k c a l / m o l = 40629.6 J / m o l$

$K_{b} = 0.512 k g m o l K^{- 1}$

$Δ T_{b} = K_{b} \times m = 0.512 \times 0.1 = 0.0512 K$

Hence, $T_{b} = 373.15 + 0.0512 = 373.20 K$

Hence, option A is correct

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