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0.90g of a non-electrolyte was dissolved...

`0.90g` of a non-electrolyte was dissolved in `90 g` of benzene. This raised the boiling point of benzene by `0.25^(@)C`. If the molecular mass of the non-electrolyte is `100.0 g mol^(-1)`, calculate the molar elevation constant for benzene.

Text Solution

Verified by Experts

The given values are:
`W_("solute")= 0.90 g`
`W_("solvent")= 90.00 g`
`DeltaT_(b) = 0.25 ^(@)C`
`Mw_("solute")= 100.0 g mol^(-1)`
`K_(b)` = ?
Using the formula
`K_(b) = (DeltaT_(b) xx W_("solvent") xx Mw_("solvent")) / (1000 xx Mw_("solute")) = 2.5 K m^(-1)`
`:. K_(b) = (0.25 xx 100 xx 90.0)/(1000 xx 0.90) = 2.5 K m^(-1)`
Thus, `K_(b)` is `2.5 K m^(-1)`.
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