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Boiling point of one molal NaCl is 373.936 K. Degree of ionization of NaCl is $(K_{b} of water = 0.52 K . Kg . {mol}^{- 1})$

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0.75

0.92

0.66

0.8

answer is B.

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Detailed Solution

$∆ T_{b} = (373.936 - 373) K = 0.936 K;$

$∆ T_{b} = i . K_{b} . m;$

$i = \frac{0.936}{0.52 \times 1} = 1.8;$

$for NaCl, n = 2$

$i = 1 - α + 2 α = 1 + α;$

$α = 0.8;$

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