2g of benzoic acid (`C_6H_5COOH`) is dissolved in 25g of benzene show depression in freezing point equal to 1.62K. Molar depression constant for benzene, `K_f`=4.9K `kg mol^-1`. What is percentage association of acid if it forms a dimer in solution?

Answer

`99.2%`

Explanation

`W_(B) = 2g, W_(A) = 25 . 0 g = 0 .025 kg`
Molar mass of benzoic acid , `C_(6) H_(5) CO OH`
`M_(B) = 84 + 6 + 32 `
`= 122 g mol^(-1)` `K_(f)` (benzene) = 49 K kg ` mol^(-1)`
`Delta T_(f) = 1 . 62^(@) C = 1 . 62 K`
`Delta T_(f) = i xx K_(f) xx (W_(B))/(M_(B)) xx (1)/(W_(A) (kg))`
`i = (Delta T_(f) xx M_(B) xx W_(A) (kg))/( K_(f) xx W_(B))`
`= ((1 . 62 K) xx (122 g mol^(-1)) xx (0 . 025 kg))/(( 4 . 9 K kg mol^(-1)) xx (2 g))`
= 0 . 5042
`2 C_(6) H_(5) CO OH overset ("benzene ") (hArr) (C_(6) H_(5) CO OH)_(2)`
`{:("Initial Cone. 1 0" ),("At. eqn " 1 - alpha " " (alpha )/(2)):}`
`i = (1 - alpha + alpha//2)/( 1) = 1 - (alpha)/(2)`
` 1 - (alpha )/(2) = 0 . 5042`
`(alpha)/( 2) = 1 - 0 . 5042`
`(alpha)/( 2) = 0. 4658`
= 0 . 4958
` 0 . 4858 xx 2 = 0 .9916`
`:.` Percentage association `= 0 . 9916 xx 100`
` = 99 . 16 %`