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

Calculation of observed milar mass benzoic acid
`M_(B)=(K_(f)xxW_(B))/(W_(A)xxDeltaT_(f))`
`W_(B)=2.0g, W_(A)=0.025 kg, K_(f)=4.9 K kg mol^(-1),DeltaT_(f)=1.62 K`
`M_(B)=((4.9K kg mol^(-1))xx(2.0))/((0.025 kg)xx(1.62K))=242 g mol^(-1)`
Calculation of Van't Hoff factor(i)
Normal molar mass of `C_(6)H_(5)COOH=6xx12+5xx1+12+2xx16+1`
=72+5+12+32+1=122 g `mol^(-1)`
`"Van't Hoff factor (i)"=("Normal molar mass")/(" Observed molar mass")=((122g mol^(-1)))/((424g mol^(-1)))=0.504`
Calculation of percentage association of acid (alpha)
Benzonic acid associates in benzne as:
`22C_(6)H_(5)COOHhArr(C_(6)H_(5)COOH)_(2)`
Value of n=2, `alpha=(i-1)/(1//n-1)=(0.504-1)/(1//2-1)=((-0.496))/((-0.50))=0.992`
Percentage association of acid=0.992xx100=99.2%