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

We have ,
`(p^(0)-p)/(p^(0))=(n)/(N)`
`(100-98.88)/(100)=(n)/(0.6474)` `{N=(56.8xx0.889)/(78)=0.6474}`
`n=0.00725`
Molality `=(0.00725xx1000)/(56.8xx0.889)=0.1435m`
Further,
`K_(f)=(DeltaT_(f))/(m)=(0.73)/(0.1435)=5.087`