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Addition of 0.643 g of a compound to 50 ...

Addition of 0.643 g of a compound to 50 mL of benzene (density 0.679 h/mL) lower the freezing point from `5.51^(@)C` to `5.03^(@)C`. If `K_(f)=5.12 km^(-1)`, calculate the molecular mass of the compound.

Text Solution

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The correct Answer is:
156 g `mol^(-1)`
`W_(B)=0.643g, W_(A)=(50 mL)xx(0.879" g mL"^(-1))=43.95g=0.04395 kg`
`Deltat_(f)=5.51-5.03=0.48^(@)C=0.48 K, K_(f)=5.12" K kg mol"^(-1)`
`M_(B)=(W_(B)xxK_(f))/(DeltaT_(f)xxW_(A))=((5.12" g kg mol"^(-1))xx(0.643g))/((0.48 K)xx(0.04395 kg))=156" g mol"^(-1)`
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