A solution of `122gm` of benzoic acid is `1000gm` of benzene shows a `b.p.` elevation of `1.4^(@)`. Assuming that solute is dimerized to the extent of `80` percent `(80^(@)C)` calculate normal `b.p.` of benzene given molar enthalpy of vap. Of benzine`=7.8 Kcal//mol`.

Explanation

`T_(b)-T_(i)=(122xx1000)/(122xx1000)xxK_(b)(1+((1)/(n)-1))0.8`
`rArrK_(b)[(1)/(2)xx0.8]`
`T_(b)-T_(i)=K_(b)xx0.6`
`K_(b)=(RT_(b)^(2))/(1000xxL_(vap.))`
`K_(b)=(2xxT_(b)^(2)xx78)/(1000xx7.8xx1000)`
`1.4=(0.4xx2)/(10^(5))T_(b)^(2)`
`T_(b)=418.33K`