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

Mass of solute (benzoic acid), `w_(2)=2.0 g", Mass of solvent (benzene), "w_(1)=25.0g`
Observed `DeltaT_(f)=1.62K", "K_(f)="4.9 K kg mol"^(-1)`
`therefore" Observed molar mass of benzoic acid"`
`M_(2)=(1000xxK_(f)xxw_(2))/(DeltaT_(f)xxw_(1))=("1000 g kg"^(-1)xx"4.9 K kg mol"^(-1)xx"2.0 g")/("1.62 K"xx"25.0 g")="242 g mol"^(-1)`
Calcualted molar mass of benzoic acid `(C_(6)H_(5)COOH)=72+5+12+32+1="122 g mol"^(-1)`
`"van't Hoff factor, i"=("Calculated mol. mass")/("Observed mol. mass")=(122)/(242)=0.504`
If `alpha` is the degree of association of benzoic acid, then we have
`{:(,2C_(6)H_(5)COOH,hArr,(C_(6)H_(5)COOH)_(2)),("Initial moles"," 1",," 0"),("After association"," "1-alpha,," "alpha//2):}`
`therefore" Total number of moles after association "=1-alpha+(alpha)/(2)=1-(alpha)/(2) therefore i=(1-alpha//2)/(1)=0.504 or 1-(alpha)/(2)=0.504`
`alpha=(1-0.504)xx2=0.496xx2=0.992" or % age association = 99.2%".`