The freezing point of pure nirobenzene is 278.8 K. When 2.5 g of an unknown substance is dissoved in 100 g of introbenzene, the freezing point of the solution is found to be 276.8 K. If the freezing point depression of nitrobenzene is `8.0 K kg mol^(-01)`, What is the molar of the unknow substance ?

Explanation

`M_(B)-(K_(f)xxW_(B))/(DeltaT_(f)xxW_(A))`
Mass of compound`(W_(B))=0.6 g`
Mass of water `(W_(A))=21.7 g= 0.0 217 kg`
`(DeltaT_(f))=(273-272.187)=0.813 K`
Molal depression constant `(K_(f))=(1.86 K kg mol^(-1))` ltbtgt `M_(B)=(1.86 "K kg Mass of solute "(W_(B))=2.5 g`
Mass of nitrobenzene `(W_(A))=100g=0.1 kg`
Depression in freezing point `(DeltaT_(f))=278.8-276.8=2 K`
Molar depression constant `(K_(f))=8.0 K kg mol^(-1)`
`M_(B)=((8.0 K kg mol^(-1))xx(2.5 g))/((2 K)xx(0.1 kg))=100 g mol^(-1)`