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The boiling point elevation contant for ...

The boiling point elevation contant for benzene is `2.57^(@)C//m`. The boiling water of benzene is `81.0^(@)C`. Determine the boiling point of a solution formed when `5 g` of `C_(14)H_(12)` is dissolved I `15 g` fo benzene.

Text Solution

Verified by Experts

Given weight of `(C_(14)H_(12)(W_(B)) = 5 g`
Weight of benzene `(W_(A)) = 15 g`
`DeltaT_(b) = K_(b) [(W_(B) /(Mw_(B)) xx 1000] /W_(A)]`
` = 2.57 [[5 /180 xx 1000] /15] = 4.76^(@)C`
Now,`DeltaT_(b)=T_(b)-T_(b)^(@) [{:(T_(b)^(@)="boiling point of pure solvent"),(T_(b)="boiling point of solution"):}]`
`:. T_(b) = DeltaT_(b) + T_(b)^(@)`
`= 4.76 + 81.0`
`=85.76^(@)C`
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