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

Give I =3.333
on dissociation of `K_(3)[Fe(CN)_(6)]` we get four ions
`K_(3)[Fe(CN)_(6)] rarr 3k^(+)+[Fe(cN)_(6)]^(3-)`
total number of ions (n)=4
`therefore alpha=(i-1)/(n-1)=(3.333)/(4-1)=(2.333)/(3)=0.7776` ltbrge0.78
hence the per cent dissociation becomes 78%