Q. 20

We have three aqueous solutions of NaCl labelled as ‘A’, ‘B’ and ‘C’ with concentrations 0.1M, 0.01M and 0.001M, respectively. The value of van’t Hoff factor for these solutions will be in the order______.
A. iA < iB < iC

B. iA > iB > iC

C. iA = iB = iC

D. iA < iB > iC

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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