Question

The ratio of the value of colligative properties
K [Fe(CN)g](aq) to that of Fe [Fe(CN)613(aq) is
[Assuming 100% dissociation]​

Answer 1

Answer:

It is equal to the ratio of their van't hoff factors.

K4[Fe(CN)6]−→H2O4K+(aq)+[Fe(CN)4−6](aq)

i=1

Sucrose −→H2O Neither association nor dissociation

i=1

Thus

(i)K4[Fe(CN)6](i)sucrose=5

Explanation: