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1.0 gof a non-electrolyte solute (molar ...

`1.0 g`of a non-electrolyte solute (molar mass`250gmol^(-1)` was dissolved in 51.2 g of benzene. If the freezing point depression constant of benzene is `5.12K kgmol^(-1)` the lowering in freezing point will be

Text Solution

Verified by Experts

Given values are:
`W_(solute)=1.0 g`
`W_(solvent)=50.0 g`
`DeltaT_(f) =0.40 K`
`K_(f)=5.12 K kg mol^(-1)`
Substituting this value in equation
`Mw_(solute)=(K_(f)xxW_(solute)xx1000)/(DeltaT_(f)xxW_(solvent)`
=`(5.12 K kg mol^(-1)xx1.0xx1000)/(0.40 Kxx50.0 g)`
=`256.0 g mol^(-1)`
Thus, molecular mass of the solute =`256.0^(-1)`
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1.00 g of a non-electrolyte solute (molar mass 250g mol^(–1) ) was dissolved in 51.2 g of benzene. If the freezing point depression constant K_(f) of benzene is 5.12 K kg mol^(–1) , the freezing point of benzene will be lowered by:-

1.0 g of a non-electrolyte solute( mol. Mass 250.0 g mol^(-1) ) was dissolved in 5.12 g benzene. If the freezing point depression constant, K_(f) of benzene is 5.12 K kg mol^(-1) , the freezing point of benzene will be lowered by: