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Q.

Molar heat of vapourization for water is 9.72 $kcal {mol}^{- 1} at 373 K .$ The boiling point of water is 373 K the molal elevation constant for water (K_b) is

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a

$0 .51 {Kkgmol}^{- 1}$

b

$9 .72 {Kkgmol}^{- 1}$

c

$0 .59 Kkgmol = 1$

d

$5.4 Kkg {mol}^{- 1}$

answer is A.

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Detailed Solution

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$K_{b} = \frac{{RT}_{0}^{2} M}{1000 {ΔH}_{V}} = \frac{2 \times 10^{- 3} \times (373)^{2} \times 18}{1000 \times 9.72} = 0.51 Kkg {mol}^{- 1}$

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Molar heat of vapourization for water is 9.72 kcal mol−1 at 373K. The boiling point of water is 373 K the molal elevation constant for water (Kb) is