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) ]

The freezing point of benzene decreases by 2.12 K when 2.5 g of benzoi

1.	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) ]
Answer» Solution :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