An aqueous solution freezes at – 0.372°C. If K_fand K_b for water are 1.86K kg mol^–1 and 0.53 K kg mol^–1 respectively, the elevation in boiling point of same solution in K is :

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0.72

0.46

4.6

0.106

answer is D.

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Detailed Solution

$For a same solution, \frac{∆ T_{b}}{∆ T_{f}} = \frac{K_{b}}{K_{f}} ∆ T_{f} = 0 - (- 0.372) = 0.372 \frac{∆ T_{b}}{0.372} = \frac{0.53}{1.86} ∆ T_{b} = \frac{0.53}{1.86} \times 0.372 = 0.106 ° c$

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