5 . 0 g of phenol dissolved in 250 g of benzene shows a depression in freezing point equal to ` 0 . 70^(@) C.` What is the percentage association of phenol ? (`K_(f)` for benzene = 5.12 K kg.`MOL^(-1)` )

Answer

`71 . 6%`

Explanation

`W_(B) = 5 . 0 g, W_(A) = 250 g = 0 . 25 kg`
`K_(f)` (benzene )= 5 . 12 K kg ` mol^(-1)`
Molar mass of phenol, `(C_(6) H_(5) OH)`
`M_(B) = 72 + 6 + 16 = 94 g mol^(-1)`
`Delta T_(f) = 0 . 70^(@) C = 0 . 70 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))/( W_(B) xx K_(f))`
`= ((0 . 70 K) xx (94 g mol^(-1)) xx (0 . 25 kg)) /( 5 . 0 g xx (5 . 12 K kg mol^(-1)))`
`= (0 . 70 xx 94 xx 0 . 25)/( 5 0 xx 5 . 12) = 0 . 6426`
`2 C_(6) H_(5) OH overset("benzene")hArr (C_(6) H_(5) OH)_(2)`
`{:("Initial Cone 1" ),("At . eqn. "1 - alpha " " (alpha)/(2)):}`
`i = (1 - alpha + alpha//2)/( 1) = 1 - (alpha)/( 2)`
`:. 0 . 6426 = 1 - (alpha)/( 2)`
`(alpha)/(2) = 1 - 0.6426 = 0 . 3574`
`alpha = 0 . 3574 xx 2 = 0 . 7148`
Percentage association ` = 100 xx alpha `
`= 100 xx 0 . 7148 = 71 . 48%`