Two grams of benzoic acid `(C_(6)H_(5)COOH)` dissolved in `25.0 g` of benzene shows a depression in freezing point equal to `1.62 K`. Molal depression constant for benzene is `4.9 K kg^(-1)"mol^-1`. What is the percentage association of acid if it forms dimer in solution?

Explanation

Given, `w= 2g, W= 25 g, DeltaT=1.62, K'_(f)=4.9`
`because DeltaT=(1000xxK'_(f)xxw)/(mxxW)`
`1.62=(1000xx4.9xx2)/(25xxm)`
`:. M_(exp.)=241.98`
`{:(,nC_(6)H_(5)COOH,hArr,(C_(6)H_(5)COOH)_(n),),("Before dissociation",1,,0,),("After dissociation",1-alpha,,(alpha)/n,):}`
`:.` Total number of mole at equilibrium `=1-alpha+(alpha)/(n)`
`(m_(N))/(m_(exp.))=1-alpha+(alpha)/(n)`
For dimer formation, `n=2 (122.0)/(241.98)=1-alpha+(alpha)/(2)`
`(m_(N)=122.0 "for" C_(6)H_(5)COOH)`
or `1-(alpha)/(2)=0.504`
`:. alpha=0.992 or 99.2%`