The van’t Hoff factor for strong electrolyte is:
i = Total number of ions after dissociation/association/Total number of ions before dissociation/association.
For NaCl, the van’t Hoff factor is 2 as it dissociates into two ions i.e.
Van’t Hoff factor is independent of the molality of solute added.
Thus, iA = iB = iC .
Thus, Option (iii) is the correct answer.
Rate this question :