Question

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 1

Correct Answer - 60.0 g `mol^(-1)`
`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)`.