1.

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) :

Answer»

`0.186^(@)C`
`0.0512^(@)C`
`0.092^(@)C`
`0.2372^(@)C`

SOLUTION :`(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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