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The freezing point of a solution contani...

The freezing point of a solution contaning `0.3 g` of acetic acid in `43 g` of benzene reduces by `0.3^(@)`. Calculate the Van's Hoff factor
"(`K_(f)` for benzene =`5.12 K kg mol^(-1)`)"

Text Solution

Verified by Experts

The given values are
`W_(B)=0.3 g`, `DeltaT_(f)=0.30`
`W_(A)=43 g`, `DeltaK_(f)=5.12 K kg mol^(-1)`
Now, using formula
`Mw_(B)=(K_(f)xxW_(B)xx1000)/(DeltaT_(f)xxW_(A))`
`:.Mw_(B)=(5.12xx0.3xx1000)/(0.3xx43)=120 g mol^(-1)`
Normal molar mass =60
`:.i="Normal molar mass"/"Observed molar mass"=60/120=0.5`
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