1.

When 1.5 g of a non-volatile solute was dissoved in 90 g of benzene, the boiling point of benzene was raised from 353.23^(-1) K to 353.93 K. Calculate the molar mas of solute (K_(b) for benzene=2.52 K kg mol^(-1)).

Answer»


Solution :`M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)XX W_(A))`
`W_(B)=1.5g , W_(A)=90.0 g=0.09" kg", DeltaT_(b)=0.7" K"`
`K_(b)=2.52" k kg MOL"^(-1)`
`M_(B)=((2.52" K kg mol"^(-1))xx(1.5" g"))/((0.7" K")xx(0.09" kg"))=60.0" g mol"^(-1)`.


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