Question

We are given with three NaCl samples and their Van't Hoff factors. Which is correct option ?

Sample	Van't Haff Factor
Sample - 1 (0.1 M)	\(i_1\)
Sample - 2 (0.01 M)	\(i_2\)
Sample - 3 (0.001 M)	\(i_2\)

• A. \(i_1=i_2=i_3\)
• B. \(i_1>i_2>i_3\)
• C. \(i_1>i_2>i_3\)
• D. \(i_1>i_3>i_2\)

Answer 1

The Correct Option is A

To determine the correct option regarding the Van't Hoff factors for the given NaCl samples, we need to consider the concept of the Van't Hoff factor in solutions, particularly those involving ionic compounds like NaCl.

NaCl dissociates into its ions in solution as follows:

NaCl \rightarrow Na^+ + Cl^−

For an ideal scenario where there is complete dissociation, the theoretical Van't Hoff factor, i, for NaCl is 2, because each formula unit of NaCl gives 2 ions.

The Van't Hoff factor is influenced by the concentration of the solute. At higher concentrations, there might be ion pairing, resulting in a Van't Hoff factor less than the theoretical value. However, as the concentration decreases, the solution approaches ideal behavior and the Van't Hoff factor approaches the theoretical value.

Analyzing the samples:

• Sample 1: 0.1 M NaCl has a certain degree of ion pairing, so the observed Van't Hoff factor i_1 might be slightly less than 2.
• Sample 2: 0.01 M NaCl is more dilute, resulting in less ion pairing, making i_2 closer to the theoretical value.
• Sample 3: 0.001 M NaCl is even more dilute, leading i_3 to be almost the theoretical value of 2.

Given these insights, at sufficiently dilute concentrations as mentioned above, the Van't Hoff factors for NaCl solutions tend to align, particularly at the lower concentrations where the interactions become negligible.

Therefore, the answer is:

i_1 = i_2 = i_3

This means in very dilute solutions, the degree of dissociation is nearly complete, and the Van't Hoff factors of these samples are approximately equal. This accounts for ideal solution behavior at progressively lower concentrations.