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, percentage association of acetic acid in benzene will be
`(K_(f) "for benzene" = 5.12 K kg mol^(-1))`

Options

A. 0.646
B. 0.804
C. 0.746
D. 0.946

Answer

D

Explanation

`Delta T_(f) = ( ixxk_(f) xx 1000 xx w_(b))/(w_(A) xx M_(B))`
`0.45 = ( ixx5.12 xx 1000 xx 0.2)/(5.12 xx 1000 xx 0.2) = 0.527`
`i=(0.45 xx 20 xx 60)/(5.12 xx 1000 xx 0.2) = 0.527`
`2CH_(3) COOH hArr (CH_(3) COOH)_(2)`
`1-alpha" "alpha//2`
Total number of moles ` = 1-alpha+alpha//2 = 1- alpha//2`
` i = (1-alpha//2)/(1) = 0.527`
` - alpha//2 = 0.527 - 1 = -0.473`
` alpha = 0.473 xx 2 =0.946`
`therefore ` Percentage assoication = 94.6%.