`0.75 g` of a non-electrolyte was dissolved in `87.9 g` of benzene. This raised the boiling point of benzene by `0.25^(@)C`. If the molecular mass of non-electrolyte is `103`, calculate the molal elevation constant for bezene.

`0.75 g` of a non-electrolyte was dissolved in `87.9 g` of benzene. Th

1.	`0.75 g` of a non-electrolyte was dissolved in `87.9 g` of benzene. This raised the boiling point of benzene by `0.25^(@)C`. If the molecular mass of non-electrolyte is `103`, calculate the molal elevation constant for bezene.
Answer» Correct Answer - 3 `K_(b) = (Delta T_(b) xx m xx W)/(w xx 1000)` Given, `Delta T_(b) = 0.25 K, m = 103, w= 0.75g`, `W = 87.9g` `:. K_(b) = (0.25 xx 103 xx 87.9)/(0.75 xx 1000) ~~ 3.0 K log mol^(-1)`

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`0.75 g` of a non-electrolyte was dissolved in `87.9 g` of benzene. This raised the boiling point of benzene by `0.25^(@)C`. If the molecular mass of non-electrolyte is `103`, calculate the molal elevation constant for bezene.

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