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`1.0 g` of non-electrolyte solute dissolved in `50.0 g` of benzene lowered the freezing point of benzene by `0.40 K`. The freezing point depression constant of benzene is `5.12 kg mol^(-1)`. Find the molecular mass of the solute.

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`M_(B)-(K_(f)xxW_(B))/(DeltaT_(f)xxW_(A))`
Mass of non-electrolyte`(W_(B))=10 g`
Mass of benzene `(W_(A))=50 g=0.005 Kg`.
Depression in freezing point `(DeltaT_(P))=0.40 K`
Molal depresion constant`(K_(f))=5.12 K kg mo^(-1)`
`T_(f)=((5.12 K kg mol^(-1))xx(1.0 g))/((0.40 K)xx(0.005 kg))=256 g mol^(-1)`
by (22 points)
It is Wb = 1.0 not 10
Correct it

Regards,
Suraj.

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