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On dissolving 0.25 g of a non-volatile s...

On dissolving 0.25 g of a non-volatile substance in 30 mL of benzene (density 0.8 g `mL^(-1)`), its freezing point decreases by `0.40^(@)C`. Calculate the molecular mass of the non-volatile substance. `K_(f)` for benzene is 5.12 `Km^(-1)` .

Text Solution

Verified by Experts

The correct Answer is:
133.33 g `mol^(-1)`
`W_(B)=0.25g, W_(A)=30xx0.8=24g, 0.024 kg , DeltaT_(f)=0.40^(@)C=0.40 K`
`K_(f)=5.12" K kg mol"^(-1), M_(B)=?`
`M_(B)=(K_(f)xxW_(B))/(DeltaT_(f)xxW_(A))=((5.12" K kg mol"^(-1))xx(0.25 g))/((0.40 K)xx(0.024 kg ))=133.33" g mol"^(-1)`.
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