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?

Answer

`T_(f)=278.3K`

Explanation

`P_("benzenen")^(@)=(0.0821xx760xx293xx0.877)/(0.78xx10^(-3)xx2750)=74.75`
`ln((P)/(74.75))=(394.57)/(8.314){(1)/(293)-(1)/(300)} rArrP=75`
`(P-74.76)/(74.76)="molality"xx(78)/(1000)` and we have, `DeltaT_(f)=` molality `xxk_(f)`
so freezing point solution `=(278.5-DeltaT_(f))`