Question

The van't Hoff factor of centi normal solution of
K3[Fe(CN), ) is 3.333. the percentage dissociation of
K3[Fe(CN)6] is
23.33
a
b
0.78
33.33
c.
d
78​

Answer 1

Answer:

Explanation:

Correct option is

C

78

K  

3

​  

[Fe(CN)  

6

​  

]→3K  

+

+[Fe(CN)  

6

​  

]  

3−

 

i=αn+(1−α)=1+α(n−1)=1+α(4−1)  

Van't Hoff factor (i) is given to be 3.333.

n= Total moles formed after dissociation.

Let degree of dissociation be α  

So,

3.333=1+α(4−1)  

α(4−1)=3.333−1  

α(4−1)=2.333

3α=2.333

α=2.333/3=0.777  

Percent dissociation =α×100=77.7=78%.