`0.1M` `K_(4) [Fe (CN)_(6)]` is `60%` ionized.What will be its Van't Hoff factor?

The electrolyte solutions show abnormal colligative porperties.To account for this effect we define a quantity called the Van't Hoff factor given by i=("Actual number of particles in solution after dissociation")/("Number of formula units initially dissolved in solution") i=1 ("for non-electrolytes") igt1 ("for electrolytes, undergoing dissociation") ilt1 ("for solutes, undergoing association") Answer the following questions: 0.1 M K_(4)[Fe(CN)_(6)] is 60% ionized. What will be its Van't Hoff factor?

What is Van't Hoff factor ?

0.1 M K_(4)[Fe(CN)_(6)] is 50% ionised in aqueous medium. What will be its van't Hoff factor ?

If van't Hoff factor i=1,then

The degree of dissociation for K_(4)[Fe(CN)_(6)] is 60% . Calculate the Van't Hoff factor.

If Hg_(2)Cl_(2) is 100% ionised, then find out its van't Hoff factor.

An electrolyte XY_(2) is 40% ionized. Calculate van't Hoff factor