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

For very dilute solutions ,
`(p^(@) -p)/(p^(@)) = x_(B)(x_(B)` = mole fraction of substance).
` (100-98.88)/(100) = x_(B)`
`therefore " "x_(B) = 0.0112`
Now `" "x_(B) = (n_(B))/(n_(A)) + n_(B) = (n_(B))/(n_(A))` (for dilute solution)
` = (n_(B))/(w_(A)) xx M_(A) " "....(i)`
Molaltiy `" "m, = (n_(B) xx 1000)/(w_(A))" ".........(ii)`
Dividing equation (i) by eq.(ii)
`(x_(B))/(m) =(M_(A))/(1000)`
or `" " m = (x_(B) xx 1000)/(M_(A))`
`therefore " " m = (0.011 xx 1000)/(78) = 0.1436`
`Delta T_(f) = k_(f) xx m`
or `" " K_(f) = (Delta T_(f))/(m) = (0.73)/(0.1436)= 5.08 Km^(-1)`