2g of benzoic acid `(C_(6)H_(5)COOH)` dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62K. Molal depression constant for benzene is 4.9 K kg `mol^(-1)`. What is the percentage association of acid if it forms dimer in solution?

Explanation

`W_(2)=2.0 g`,`W_(1)=25.0 g`,`DeltaT_(f)=1.62 K`,
`K_(f)=4.9 K kg^(-1)`,`Mw_(2)(C_(6)H_(5)COOH)=122 g mol^(-1)`
Substituting these values in equation
`DeltaT_(f)=ixx K_(f)xxm`
`=ixx K_(f)xx(W_(2)xx1000)/(Mw_(2)xxW_(1))`
`1.62 K=i xx4.9kg^(-1) mol^(-1) xx (2.0 gxx1000)/(122 g mol^(-1) xx 25.0 g)`
`:.i=0.504`
Benzoic acid exists as dimer, therefore

Total number of moes at equilibrium=`1-alpha+(alpha//2)`
`=1-(alpha//2)`
`:.i="Number of moles at equilibrium"/"Number of moles initially"`
`:.0.504=(1-(alpha//2))/1`
`:. alpha=0.992`
Therefore, degree of association of benzoic acid in benzene is `99.2%`.