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For an aqueous solution freezing point i...

For an aqueous solution freezing point is `-0.186^(@)C`. The boiling point of the same solution is `(K_(f) = 1.86^(@)mol^(-1)kg)` and `(K_(b) = 0.512 mol^(-1) kg)`

A

`0.186^(@)`

B

`100.0512^(@)`

C

`1.86^(@)`

D

`5.12^(@)`

Text Solution

Verified by Experts

The correct Answer is:
B
`(DeltaT_(b))/(DeltaT_(f)) = (K_(b))/(K_(f)) rArr DeltaT_(b) = (0.512)/(1.86) xx 0.186 = 0.0512^(@)C`
`T'_(b) = T_(b) +DeltaT_(b) = 100.0.512^(@)C`
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