1.

A solution containing 0.90 g of a non-volatile solute in 100 mL of benzene having density 0.88 g mL^(-1) at 298 K, boils at a temperature 0.25^(@) higher than benzne. Calculate the molecular mass of solute. Molal boiling point elevation constant for benzenene is 2.52 K kg mol^(-1).

Answer»


Solution :`T_(b)^(@)=78^(@)C,T_(b)=78.28^(@)C,DeltaT_(b)=T_(b)-T_(b)^(@)=78.28-780.28^(@)C=0.28 K`
`K_(b)=1.15" K KG mol"^(-1), W_(B)=1.12 g, W_(A)=0.032 kg, M_(B)= ?`
`M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A))=((1.5" K kg mol"^(-1))xx(1.12 g))/((0.28 K)xx(0.032 Kg ))=143.75" g mol"^(-1)`.


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