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0.48of a substance is dissolved in 10.6g...

`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`

Text Solution

Verified by Experts

The correct Answer is:
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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