We have three aqueous solutions of NaCl labelled as ' A ', ' B ' and ' C ' with concentration 0.1 M, 0.01 M and 0.001 M, respectively. The value of van 't Hoff factor(i) for these solutions will be in the order :

Solution

Chemistry ma’am · 4 steps

NaCl is a strong electrolyte, but its ions cling together more when crowded, so dilute means i closer to 2.

1. Ideal value. Each NaCl gives one Na⁺ and one Cl⁻, so the ideal van 't Hoff factor is i = 2.
2. Why i falls short. Oppositely charged ions attract each other. In a concentrated solution they are close together, so some move about as ion pairs and count as one particle. That lowers the number of free particles, so i drops below 2.
3. Effect of dilution. As the solution gets more dilute, the ions sit farther apart and ion pairing weakens. So i rises towards 2 as concentration falls.

   Solution	Concentration	Ion pairing	Van 't Hoff factor
   A	0.1 M	most	smallest, furthest below 2
   B	0.01 M	less	in between
   C	0.001 M	least	largest, closest to 2

4. So. The order is i_A < i_B < i_C, which is option (b).

The idea to keep. For a strong electrolyte, i reaches its ideal value only at infinite dilution. The more concentrated the solution, the more ion pairing, and the smaller i becomes.

Answer from the NTA final answer key. Solution by PrepWiser.