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

Using Raoult's law `P_(1) =chi_(1)P_(1)^(@)`
We get `(98.88 mm Hg)=chi_(1)(100 mmHg)`
or `chi_(1)=0.9888`
Let `1 mol` be the total amount of solvent and solute. We will have
`n_(1)=0.9888 mol n_(2)=0.0112 mol`
`W_(1)= n_(1)Mw_(1) =(0.9888 mol)(78 g mol^(-1)) = 77.126 g`
The molality of the solution is
`m= n_(1)/W_(1)=(0.0112 mol)/(77.126xx10^(-3) kg) = 0.1452 mol kg^(-1)`
Molal freezing point depression constant of benzene would be
`K_(f) =(-DeltaT_(f))/m=(0.73 K)/0.1452 mol Kg ^(-1) =5.028 K kg mol^(-1)`