1. Concept of van ’t Hoff factor

For an electrolyte like $\mathrm{NaCl}$ in water,

$\mathrm{NaCl} \rightarrow \mathrm{Na^+} + \mathrm{Cl^-}$

If dissociation were complete and ideal, then

$i = 2$

But in real solutions, due to inter-ionic attraction, the effective number of particles is slightly less than ideal, especially at higher concentration.

2. How concentration affects $i$

As the solution becomes more dilute:

• ions are farther apart,
• inter-ionic interactions decrease,
• dissociation/effective particle behavior becomes closer to ideal,
• hence van ’t Hoff factor $i$ increases.

Therefore, for the given solutions:

• $A = 0.1\,\mathrm{M}$
• $B = 0.01\,\mathrm{M}$
• $C = 0.001\,\mathrm{M}$

we expect

$i_A < i_B < i_C$

3. Check options

• A: $i_A < i_C < i_B$ ❌
• B: $i_A < i_B < i_C$ ✅
• C: $i_A > i_B > i_C$ ❌
• D: $i_A = i_B = i_C$ ❌

4. Final answer

$\boxed{i_A < i_B < i_C}$

So, the correct option is B.