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The boiling point of ethanol is 78^(@)C and its nolal boiling point elevation constant per 1000 g is 1.15 K. A solution of 1.12 g of camphor in 32 g of ethanol has a boiling point of 78.28^(@)C. Cakculate the molecular mass of camphor.

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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