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 `K_f` for benzene?

Explanation

We know that `(p^@-P_s)/(p^@) =(wxxM)/(mxxW)`
Substituting values,
`(100-98.88)/100 = w/m xx 78/W xx 1000/1000 = " molality " xx78/100`
(molecular weight of benzene = 78)
or, Molality `= (1.12xx1000)/(78xx100)=0.1436`.
We know, `De,taT_f=K_f xx` molality
`:. K_f = (DeltaT_f)/("molality") = (0.73)/(0.1436) = 5.08 K "molality"^(-1)`