The freezing point of benzene decreases by 2.12 K when 2.5 g of benzoic acid `(C_6H_3COOH)` is dissolved in 25 g of benzene. If benzoic acid forms a dimer in benzene, calculate the van't Hoff factor and the percentage association of benzoic acid. [`K_f` for benzene = 5.12 K kg `"mol"^(-1)` ]

Explanation

Apply the relation : `Delta T_f = i xx K_f xx m ` where i is van.t Hoff factor.
`K_f` = 5.12, molalily (m) = ` (2.5)/(122) xx 1000/25`
Substituting the values in the equation above, we have
`2.12 = i xx 5.12 xx (2.5)/(12) xx 1000/25`
`i = (2.12 xx 122 xx 25)/(5.12 xx 2.5 xx 1000) = 0.505`
Two molecules of `C_6H_5COOH `associate to form one molecule of `(C_6H_5COOH)_2`.
For association i = `1 - alha/2` where a is degree of association.
Substituting the values, we have
`0.505 = 1 - alpha/2`
`alpha/2 = 1- 0.505 " or " alpha/2 = 0.495 " or " alpha = 0.99`
Percentage association of benzoic acid = 99