The value of Van't Hoff factors for `K_(2)SO_(4),NaCl` and KCL is:

Options

A. 2, 2,2
B. 3, 2,2
C. 2,2,3
D. 3, 3,2

Explanation

To find the Van't Hoff factors (i) for the compounds \( K_2SO_4 \), \( NaCl \), and \( KCl \), we need to determine how many particles each compound dissociates into when dissolved in water.

### Step-by-Step Solution:

1. **Identify the Dissociation of \( K_2SO_4 \)**:
- When \( K_2SO_4 \) dissolves in water, it dissociates into:
\[ K_2SO_4 \rightarrow 2K^+ + SO_4^{2-} \]
- Here, \( K_2SO_4 \) produces 2 potassium ions (\( K^+ \)) and 1 sulfate ion (\( SO_4^{2-} \)).
- Therefore, the total number of ions produced is:
\[ 2 + 1 = 3 \]
- Thus, the Van't Hoff factor \( i \) for \( K_2SO_4 \) is **3**.

2. **Identify the Dissociation of \( NaCl \)**:
- When \( NaCl \) dissolves in water, it dissociates into:
\[ NaCl \rightarrow Na^+ + Cl^- \]
- Here, \( NaCl \) produces 1 sodium ion (\( Na^+ \)) and 1 chloride ion (\( Cl^- \)).
- Therefore, the total number of ions produced is:
\[ 1 + 1 = 2 \]
- Thus, the Van't Hoff factor \( i \) for \( NaCl \) is **2**.

3. **Identify the Dissociation of \( KCl \)**:
- When \( KCl \) dissolves in water, it dissociates into:
\[ KCl \rightarrow K^+ + Cl^- \]
- Here, \( KCl \) produces 1 potassium ion (\( K^+ \)) and 1 chloride ion (\( Cl^- \)).
- Therefore, the total number of ions produced is:
\[ 1 + 1 = 2 \]
- Thus, the Van't Hoff factor \( i \) for \( KCl \) is **2**.

### Summary of Van't Hoff Factors:
- For \( K_2SO_4 \): \( i = 3 \)
- For \( NaCl \): \( i = 2 \)
- For \( KCl \): \( i = 2 \)

### Final Answer:
The values of Van't Hoff factors are:
- \( K_2SO_4 \): 3
- \( NaCl \): 2
- \( KCl \): 2

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