`0.5 m` solution of acetic acid `(Mw=60)` in benzene `(Mw=78)` boils at `80.80^(@)C`. The normal boiling point of benzene is `80.10^(@)C`. And `Delta_(vap)H=30.775 kJ mol^(-1)`. Calculate the percent of association of acetic acid in benzene.

Explanation

Given,
`T_(0)("boiling point of benzene")=80.10^(@)C=353.1 K`
`Delta_(vap)H=30.775 kJ mol^(-1)`
`Mw_(1)("benzene")=78`
`K_(b)=(RT_(0)^(2)Mw_(1))/(1000 Delta_(vap)H) =(8.314 xx (353.1)^(2) xx 78)/(1000 xx 30.775 xx 10^(3))=2.63`
`DeltaT_(b)=iK_(b)m rArr i=(DeltaT_b)/(K_(b)m)=0.70/(2.65 xx 0.5) =0.532`
Considering association of acetic acid

Total number of moles =`1-alpha+(alpha/2)=1- (alpha/2)`
`i="Total number of moles"/"Initial moles" =1-(alpha/2)`
`0.532 =1-alpha/2`
`alpha=0.936`
% of association =93.6