The freezing point of benzene decreases by `0.45^(@)C` when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, then percentage association of acetic acid in benzene will be ?
(`K_(f)` for benzene = 5.12 K kg `mol^(-1)`)

Explanation

Let the degree of association of acetic acid `(CH_(3)COOH)` in benzene is `alpha`, then
`2CH_(2)COOH harr (CH_(3)COOH)_(2)`
`{:("Initial moles",1,0),("Moles at equilibrium",1-alpha,(alpha)/(2)):}`
`therefore` Total moles `= 1 - alpha + (alpha)/(2) = 1 - (alpha)/(2) or i = 1 - (alpha)/(2)`
Now, depression in freezing point `(Delta T_(f))` is given as :
`Delta T_(f) = i K_(f)m" "...(i)`
where, `K_(f)` = molal depression constant or cryoscopic constant, m = molality
Molality `= ("Number of moles of solute")/("Weight of solvent (in kg)")`
`= (0.2)/(60) xx (1000)/(20)`
Putting the values in Eq. (i)
`therefore" "0.45 = [1-(alpha)/(2)](5.12) [(0.2)/(60) xx (1000)/(20)]`
`1 - (alpha)/(2) = (0.45 xx 60 xx 20)/(5.12 xx 0.2 xx 1000)`
`rArr" "1 - (alpha)/(2) = 0.527 rArr (alpha)/(2) = 1 - 0.527`
`therefore " "alpha = 0.946`
Thus, percentage of association = 94.6%