Question

Vant hoff factor (i) for 30% dissociation of k4[fe(cn)6] is

Answer 1

Answer:

K4[Fe(CN)6] 4 K+ + [Fe(CN)6] 4- Since one molecule of K4[Fe(CN)6] dissociates to produce 5 ions, the value of van't Hofff factor is 5.

Answer 2

i=2.2

Explanation:

α=

n−1

i−1

;

n=5, since K

4

[Fe(CN)

6

] gives 5 ions in the solution.

K

4

[Fe(CN)

6

]→4K

+

+[Fe(CN)

6

]

4−

Now,

∴0.3=

5−1

i−1

i=2.2