Answer :

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.

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