Question

What is the Van’t Hoff factor (\(i\)) for a completely dissociated solution of \(K_2SO_4\) in water?

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

Van’t Hoff factor equals the number of ions produced after complete dissociation. Examples: \(NaCl \rightarrow Na^+ + Cl^-\) → \(i = 2\) \(CaCl_2 \rightarrow Ca^{2+} + 2Cl^-\) → \(i = 3\) \(K_2SO_4 \rightarrow 2K^+ + SO_4^{2-}\) → \(i = 3\)

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The Van’t Hoff factor (\(i\)) measures the effect of solutes on colligative properties. For electrolytes, it represents the number of particles into which a substance dissociates.
Step 2: Key Formula or Approach:
For 100% dissociation:
\[ i = \text{number of ions produced per formula unit} \]
Step 3: Detailed Explanation:
1. Potassium sulfate (\(K_2SO_4\)) is a strong electrolyte.
2. In water, it dissociates according to the equation:
\[ K_2SO_4 \rightarrow 2K^+ + SO_4^{2-} \]
3. Counting the ions:
- 2 Potassium ions (\(K^+\)).
- 1 Sulfate ion (\(SO_4^{2-}\)).
4. Total particles = \(2 + 1 = 3\).
5. Since dissociation is complete (100%), \(i = 3\).
Step 4: Final Answer:
The Van’t Hoff factor is 3 (C).