1.

Calculate the boiling point of a solution containing 1.8 g of a non-volatile solute dissolved in 90 g of benzene. The boiling point of pure benzene is 353.23 K, (K_(b)=2.53 K kg mol^(-1) , density of water= 1 g mol^(-1)).

Answer»


Solution :`W_(B)=1.8g, W_(A)=90g=0.09 kg, T_(b)^(@)=353.23 K`
`K_(b)=2.053" K kg mol"^(-1), M_(B)=58"g mol"^(-1),T_(b)=?`
`DeltaT_(b)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A))=((2.53" K kg mol"^(-1))XX(1.8g))/((58" g mol"^(-1))xx(0.09 kg))=0.872 K`
`T_(b)=T_(b)^(@)+DeltaT_(b)=353.23+0.872=354.102 K.`


Discussion

No Comment Found

Related InterviewSolutions