Two grams of benzoic acid `(C_(6)H_(5)COOH)` dissolved in `25.0 g` of benzene shows a depression in freezing point equal to `1.62 K`. Molal depression constant for benzene is `4.9 K kg^(-1)"mol^-1`. What is the percentage association of acid if it forms dimer in solution?

Answer

`alpha=99.2%`

Explanation

Benzonic acid-B, Benzene-A
Benzonic acid dimerises ,therefore `"n=2"`
`"i"=1+((1)/(n)-1)alpha`
`"I"=1-(alpha)/(2)`
`Delta"T"_("f")="i"xx("K"_("f")xx"W"_("B")xx1000)/("m"_("B")xx"W"_("A"))`
`1.62=(1-(alpha)/2)(4.9xx2xx1000)/(122xx25)`
`(1-(alpha)/2)=(1.62xx25xx122)/(4.9xx2xx1000)`
`alpha=0.992`
Percentage association=99.2%