The Van't Hoff factor for a solution of $K_2SO_4$ in water is:
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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 (\( i=4 \)).
Step 1: Understanding the Question:
The question asks for the Van't Hoff factor (\( i \)), which measures the effect of a solute on colligative properties based on its dissociation or association in a solvent. Step 2: Key Formula or Approach:
For strong electrolytes that undergo complete dissociation, the Van't Hoff factor is equal to the total number of ions produced per formula unit.
\[ i = \text{Number of cations} + \text{Number of anions} \] Step 3: Detailed Explanation:
Potassium sulfate (\( K_2SO_4 \)) is a strong electrolyte that dissociates completely in water.
The dissociation equation is:
\[ K_2SO_4(s) \rightarrow 2K^+(aq) + SO_4^{2-}(aq) \]
1. Number of Potassium ions (\( K^+ \)) = 2.
2. Number of Sulfate ions (\( SO_4^{2-} \)) = 1.
Total number of particles (\( i \)) = \( 2 + 1 = 3 \). Step 4: Final Answer:
The Van't Hoff factor for \( K_2SO_4 \) is 3.