Question

The boling point of benzone is 353.23K. When 1.80 g of a non-valatioe solute is mixed in 90 g of benzene, the boling point id raised to 354.11 Calculate the molar mass of the solute. Given that `K_(b)` benzene is 2.53 K kg `mol^(-1)`

Answer 1

`M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A))`
`W_(B)=1.80 kg,k_(b)=2.53K Kg mol^(-1)`
`DeltaT_(b)=(354.11-353.23)=0.88 K`
`M_(B)=((1.80)xx(2.53 kg mol^(-1)))/((0.88)xx(0.090 kg))=9.5 g mol^(-1)`