Question

The van’t Hoff Factor (i) for a dilute aqueous solution of \( \text{Na}_2\text{SO}_4 \) is

• A. \( 1 - \alpha \)
• B. \( 1 - 2\alpha \)
• C. \( 1 + \alpha \)
• D. \( 1 + 2\alpha \)

Hint:

To verify the formula, consider complete dissociation (\( \alpha = 1 \)). For \( \text{Na}_2\text{SO}_4 \), complete dissociation yields 3 ions, so \( i \) should equal 3. Substituting \( \alpha = 1 \) into the correct option (D) gives \( i = 1 + 2(1) = 3 \), confirming the formula.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to determine the expression for the van't Hoff factor (\( i \)) of a dilute sodium sulfate (\( \text{Na}_2\text{SO}_4 \)) solution in terms of its degree of dissociation (\( \alpha \)).

Step 2: Key Formula or Approach:
The relation between the van't Hoff factor (\( i \)) and the degree of dissociation (\( \alpha \)) for an electrolyte that dissociates into \( n \) ions is:
\[ i = 1 + (n - 1)\alpha \]

Step 3: Detailed Explanation:
1. Sodium sulfate dissociates in aqueous solution according to the equation:
\[ \text{Na}_2\text{SO}_4\text{(aq)} \rightleftharpoons 2\text{Na}^+\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \]
2. Here, one formula unit of \( \text{Na}_2\text{SO}_4 \) dissociates to produce a total of \( n = 2 + 1 = 3 \) ions.
3. Substituting \( n = 3 \) into the relation:
\[ i = 1 + (3 - 1)\alpha \]
\[ i = 1 + 2\alpha \]

Step 4: Final Answer:
The van’t Hoff Factor (\( i \)) is \( 1 + 2\alpha \), which corresponds to option (D).