1.

A solution of 1.25 g of a certain non-electrolyte in 20.0 g of water freezes at 271.94 K. Calculate the molecular mass of the solute, K_(f)for water is 1.86 K/m.

Answer»


Solution :`W_(B)=1.25g, W_(A)=0.02 KG, K_(f)-=1.86" K kg mol"^(-1),T_(f)^(@)=273 K`
`T_(f)=271.94 K, DeltaT_(f)=273-271.94=1.06 K, M_(B)=?`
`M_(B)=(K_(f)xxW_(B))/(DeltaT_(f)xxW_(A))=((1.86" K kg mol"^(-1))XX(1.25 G))/((1.06 K)xx(0.02 kg))=109.67" g mol"^(-1).`


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