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:

The van't Hoff factor, denoted as $$i$$, measures the extent of dissociation or association of a solute in solution. For electrolytes, it is given by:

$$i = \frac{\text{observed colligative property}}{\text{calculated colligative property assuming no dissociation}}$$

NaCl is a strong electrolyte that dissociates completely in ideal conditions:

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

The theoretical van't Hoff factor for complete dissociation is 2, as one formula unit produces two ions.

However, in real solutions, interionic attractions reduce the effective number of ions, making the observed $$i$$ less than 2. This effect is more pronounced at higher concentrations due to stronger interionic interactions.

Given the concentrations:

• Solution A: 0.1 M (highest concentration)
• Solution B: 0.01 M
• Solution C: 0.001 M (lowest concentration)

As concentration decreases, interionic attractions weaken, and the van't Hoff factor increases towards the theoretical value of 2.

Therefore:

• For A (0.1 M), $$i_A$$ is the smallest due to strongest interionic attractions.
• For B (0.01 M), $$i_B$$ is greater than $$i_A$$ but less than $$i_C$$.
• For C (0.001 M), $$i_C$$ is the largest and closest to 2.

Thus, the order is $$i_A < i_B < i_C$$.

Comparing with the options:

• A. $$i_A < i_B < i_C$$ (matches)
• B. $$i_A < i_C < i_B$$ (incorrect)
• C. $$i_A = i_B = i_C$$ (incorrect, as concentration affects i)
• D. $$i_A > i_B > i_C$$ (incorrect, opposite order)

The correct answer is option A.