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`0.48`of a substance is dissolved in `10.6`g of `C_(6)H_(6)`.The freezing point of benzene is lowered by `1.8^(@)C`what will be the mol.wt.of the substance (`K_(f)`for benzene =`5`)A. `250.2`B. `90.8`C. `125.79`D. `102.5` |
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Answer» Correct Answer - C `DeltaT_(f)=(1000.K_(f)w)/(mW)` `m=(1000K_(f)w)/(cancelOT_(f)W)=(1000xx5xx0.48)/(1.8xx10.6)=125.79` |
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