A very small amount of a non-volatile solute (that does not dissociate) is dissolved in `56.8 cm^(3)` of benzene (density `0.889 g cm^(3))`. At room temperature, vapour pressure of this solution is `98.88 mm Hg` while that of benzene is `100 mm Hg` . Find the molality of this solution. If the freezing temperature of this solution is `0.73` degree lower than that of benzene, what is the value of molal the freezing point depression constant of benzene?

Explanation

`(Deltap)/(p_(0))=x_(B)`
`(100-98.8)/(100)=x_(B)`
`x_(B)=0.012`
Molality `= (x_(B)xx1000)/((1-x_(B))m_(A))=(0.012xx1000)/(0.988xx78)=0.1557`
`DeltaT=K_(f)xx` Molality
`0.73=K_(f)xx0.1557`
`K_(f)=4.688`.