Van't Hoff factor of a centinormal solution of `K_(3)[Fe(CN)_(6)]` is 3.333. The precentage dissociation of `K_(3)[Fe(CN_(6))_(6)]` is :

Options

A. `33.33`
B. `0.78`
C. 78
D. `23.33`

Answer

C

Explanation

`"Given,i=3.333`
on dissociation of `K_(3)[Fe(CN)_(6)]`, we get four ions.
`K-(3)[Fe(CN)_(6)]to3K^(+)+[Fe(CN)_(6)]^(3-)`
Total number of ions (n) = 4
`:.alpha=(i-1)/(n-1)=(3.333-1)/(4-1)`
`=(2.333)/(3)=0.7776...~~0.78`
Hence, the per cent dissociation becomes 78%.