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

As `(P^(@)-P_(s))/(P_(s))=(n_(1))/(n_(2))=(w_(1)xxm_(2))/(m_(1)xxw_(2))`
`rArr (P^(0)-P_(s))/(P_(s))=((1000xxw_(1))/(m_(1)xxw_(2)))xx(m_(2))/(1000)=mxx(m_(2))/(1000)`
`{:(m="molality"),(m_(2)="molecular wt of solvent"):}`
`:.(100-98.88)/(98.88)=mxx(78)/(1000)`
`:.m=(1.12xx1000)/(78xx98.88)=0.1452`
Also, `DeltaT_(f)=K_(t)xx` molality
`:.0.73=K_(f)xx0.1452`
`K_(i)= 6.028 K` molality