Question

We have three aqueous solutions of NaCl labelled as ‘A’, ‘B’ and ‘C’ with concentration 0.1 M, 0.01M & 0.001 M, respectively. The value of vant’ Haft factor (i) for these solutions will be in the order.

• A. \( i_A<i_B<i_C \)
• B. \( i_A<i_C<i_B \)
• C. \( i_A = i_B = i_C \)
• D. \( i_A>i_B>i_C \)

Answer 1

The Correct Option is A

The problem asks to determine the order of the van't Hoff factor (i) for three aqueous solutions of NaCl with different concentrations: 0.1 M (A), 0.01 M (B), and 0.001 M (C).

Concept Used:

The van't Hoff factor (\(i\)) is a measure of the extent of dissociation or association of a solute in a solution. For an electrolyte, it is defined as:

\[ i = \frac{\text{Actual number of particles in solution after dissociation}}{\text{Number of formula units initially dissolved in solution}} \]

For an electrolyte that dissociates, the van't Hoff factor is related to the degree of dissociation (\(\alpha\)) by the formula:

\[ i = 1 + \alpha(n - 1) \]

where \(n\) is the number of ions produced from one formula unit of the solute.

For strong electrolytes like NaCl, while dissociation is considered almost complete, it is not 100% in practice, especially at higher concentrations. This is due to inter-ionic attractions between the cations and anions in the solution. At higher concentrations, ions are closer together, leading to stronger attractions and the formation of "ion pairs," which reduces the effective number of free-moving particles. As the solution is diluted (concentration decreases), the average distance between ions increases, reducing inter-ionic forces. This leads to a higher degree of dissociation (\(\alpha\)). Consequently, the van't Hoff factor (\(i\)) increases with dilution and approaches the theoretical value (\(n\)) at infinite dilution.

Step-by-Step Solution:

Step 1: Analyze the dissociation of NaCl.

Sodium chloride (NaCl) is a strong electrolyte that dissociates in water into one sodium ion (\(\text{Na}^+\)) and one chloride ion (\(\text{Cl}^-\)).

\[ \text{NaCl}_{(aq)} \rightarrow \text{Na}^+_{(aq)} + \text{Cl}^-_{(aq)} \]

From one formula unit of NaCl, two ions are produced. Therefore, the theoretical maximum value of the van't Hoff factor is \(n=2\).

Step 2: Relate van't Hoff factor to the degree of dissociation for NaCl.

Using the formula \(i = 1 + \alpha(n - 1)\) with \(n=2\), we get:

\[ i = 1 + \alpha(2 - 1) = 1 + \alpha \]

This equation shows that the van't Hoff factor (\(i\)) for NaCl is directly proportional to its degree of dissociation (\(\alpha\)).

Step 3: Analyze the effect of concentration on the degree of dissociation.

The three solutions have the following concentrations:

• Solution A: 0.1 M
• Solution B: 0.01 M
• Solution C: 0.001 M

The order of concentrations is: \( \text{Concentration (A)} > \text{Concentration (B)} > \text{Concentration (C)} \).

As the concentration of the electrolyte solution decreases, the solution becomes more dilute. In a more dilute solution, the ions are farther apart, and the inter-ionic forces of attraction are weaker. This leads to a greater degree of dissociation (\(\alpha\)).

Therefore, the order of the degree of dissociation for the three solutions will be the reverse of the order of their concentrations:

\[ \alpha_C > \alpha_B > \alpha_A \]

Step 4: Determine the order of the van't Hoff factor.

Since \(i = 1 + \alpha\), the van't Hoff factor (\(i\)) will follow the same order as the degree of dissociation (\(\alpha\)).

Let \(i_A\), \(i_B\), and \(i_C\) be the van't Hoff factors for solutions A, B, and C, respectively. Because \( \alpha_C > \alpha_B > \alpha_A \), it follows that:

\[ i_C > i_B > i_A \]

Final Result:

The most dilute solution (C) will have the highest degree of dissociation and thus the largest van't Hoff factor, while the most concentrated solution (A) will have the lowest degree of dissociation and the smallest van't Hoff factor. The value of \(i\) for all three solutions will be less than the theoretical value of 2, but will approach 2 as the concentration decreases.

The order of the van't Hoff factor for these solutions will be \(i_A < i_B < i_C\).