In an experiment, `72.5 g` of `C_(6)H_(5)OH` (phenol) is dissolved in a solvent of `K_(f)=14`. If the depression in freezing point is `7 K`, find the percentage of phenol that dimerizes.

Answer

`alpha=70%`

Explanation

Phenol is dimerized as follows:
`2C_6H_5OHhArr(C_6H_5OH)_2`
`{:(At t=0,1,0),(At Eq,(1-alpha),alpha//2):}`
Total moles after dimerization
=`1-alpha+alpha//2=1-alpha//2`
Van't Hoff factor
=`"Normal molecular weight of phenol"/"Abnormal molecular weight of phenol"=94/Mw_(ab)`
`Mw_(ab)=(1000xxK_(f)xxW_(2))/(DeltaT_(f)xxW_(1))`
Given that, `K_(f)=14`
`W_(2)=72.5 g`
`W_(1)=1000 g`
`:. Mw_(ab)=(1000xx14xx72.)/(7xx1000)=145`
`:. i=94/145`
(i)=`" Number of particles after dimerizaion"/"Number of particles before dimerizaion"`
`=(1-alpha//2)/1`
From Eqs.(i) and (ii), we get
`1-alpha//2=94/145=0.65`
`alpha/2=1-0.65=0.35 :. alpha=70%`
Hence, `70%` phenol is present in dimeric form.