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Freezing point of an aqueous solution is...

Freezing point of an aqueous solution is `-0.186^(@)C`. Elevation of boiling point of the same solution is ……..if `K_(b)=0.512 K "molality"^(-1)`and `K_(f)=1.86K "molality"^(-1)` :

A

`0.186^(@)C`

B

`0.0512^(@)C`

C

`0.092^(@)C`

D

`0.2372^(@)C`

Text Solution

Verified by Experts

The correct Answer is:
b
`(DeltaT_(b))/(DeltaT_(f))=K_(b)/K_(f)orDeltaT_(b)=K_(b)/K_(f)xxDeltaT_(f)`.
`DeltaT_(b)=((0.512^(@)C))/((1.86^(@)C))xx(0.186^(@)C)`
`=0.0512^(@)C`
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