Question

The Van't Hoff factor for a solution of $K_2SO_4$ in water is:

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

For salts, just count the subscripts in the chemical formula to find the number of ions. For example, $NaCl$ has 2 ions (\( i=2 \)), $K_2SO_4$ has 3 ions (\( i=3 \)), and $AlCl_3$ has 4 ions (\( i=4 \)).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks for the Van't Hoff factor (\( i \)) for the electrolyte $K_2SO_4$ (assuming "KSO" is a typo for Potassium Sulfate, $K_2SO_4$, as it is a common exam example) when dissolved in water. The Van't Hoff factor represents the degree of dissociation of a solute.

Step 2: Key Formula or Approach:
For strong electrolytes that dissociate completely in dilute solutions:
\[ i = \text{total number of ions produced per formula unit} \] For an electrolyte \( A_xB_y \) that dissociates into \( xA^{y+} \) and \( yB^{x-} \), the number of ions is \( n = x + y \).

Step 3: Detailed Explanation:
When Potassium Sulfate ($K_2SO_4$) dissolves in water, it dissociates into its constituent ions as follows:
\[ K_2SO_4(s) \rightarrow 2K^+(aq) + SO_4^{2-}(aq) \] - Number of potassium ions (\( K^+ \)) = 2
- Number of sulfate ions (\( SO_4^{2-} \)) = 1
Total number of ions produced = \( 2 + 1 = 3 \).
Assuming complete dissociation (ideal solution), the Van't Hoff factor \( i \) is equal to the number of ions produced.
Therefore, \( i = 3 \).

Step 4: Final Answer:
The Van't Hoff factor for a solution of $K_2SO_4$ in water is 3.