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2g of benzoic acid (C6H5COOH) is dissolv...

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?

Text Solution

Verified by Experts

The correct Answer is:
`99.16%`
`W_(B) = 2g ` Molar mass of `C_(6) H_(5) CO OH`
`M_(B) = 84 + 6 + 32 = 122 g mol^(-1)`
`W_(A) = 25 g = 0 . 0 25 kg`,
`Delta T_(f) = 1 . 62 K`
`K_(f) = 4 . 9 K kg mol^(-1)`
`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 xx 122 g mol^(-1) xx 0 . 025 kg)/( 4 . 9 kg mol^(-1) xx 2 g)`
= 0 . 05042
`2 C_(6) H_(5) CO OH hArr (C_(6) H_(5) CO OH)_(2)`
`{:("Initial Cone 1 "),("At. epn " . 1 - alpha" " alpha // 2 ):}`
`i = (1 - alpha + alpha//2)/(1) = 1 - (alpha)/(2)`
`1= (alpha)/(2) = 0 . 5042 `
`(alpha)/(2) = 1 - 0.05042`
`alpha = 2 xx 0 . 4958 = 0 . 9916`
`:.` Percent association ` = 0.9916 xx 100`
`= 99 . 16%`
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