Which compound has the same value of Van't Hoff factor i as that of `K_4[Fe(CN)_6]`

Options

A. `NaSO_4`
B. `NaCl`
C. `Al_2(SO_4)_3`
D. `Al(NO_3)_3`

Answer

To determine which compound has the same Van't Hoff factor (i) as that of \( K_4[Fe(CN)_6] \), we first need to calculate the Van't Hoff factor for \( K_4[Fe(CN)_6] \) and then compare it with the options provided.

### Step 1: Calculate the Van't Hoff factor for \( K_4[Fe(CN)_6] \)

1. **Dissociation of \( K_4[Fe(CN)_6] \)**:
- The compound dissociates into 4 potassium ions (\( K^+ \)) and 1 complex ion (\( [Fe(CN)_6]^{4-} \)).
- The dissociation can be represented as:
\[ K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-} \]

2. **Count the total number of ions produced**:
- From the dissociation, we have:
- 4 \( K^+ \) ions
- 1 \( [Fe(CN)_6]^{4-} \) ion
- Total ions = \( 4 + 1 = 5 \)

3. **Calculate the Van't Hoff factor (i)**:
\[ i = \frac{\text{Total number of ions after dissociation}}{\text{Total number of formula units before dissociation}} = \frac{5}{1} = 5 \]

### Step 2: Analyze the options provided

Now, we need to find a compound from the options that also has a Van't Hoff factor of 5.

1. **Option 1: \( Na_2SO_4 \)**:
- Dissociation:
\[ Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-} \]
- Total ions = \( 2 + 1 = 3 \)
- \( i = \frac{3}{1} = 3 \)

2. **Option 2: \( NaCl \)**:
- Dissociation:
\[ NaCl \rightarrow Na^+ + Cl^- \]
- Total ions = \( 1 + 1 = 2 \)
- \( i = \frac{2}{1} = 2 \)

3. **Option 3: \( Al_2(SO_4)_3 \)**:
- Dissociation:
\[ Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-} \]
- Total ions = \( 2 + 3 = 5 \)
- \( i = \frac{5}{1} = 5 \)

4. **Option 4: \( Al(NO_3)_3 \)**:
- Dissociation:
\[ Al(NO_3)_3 \rightarrow Al^{3+} + 3NO_3^{-} \]
- Total ions = \( 1 + 3 = 4 \)
- \( i = \frac{4}{1} = 4 \)

### Conclusion

The compound that has the same Van't Hoff factor (i) as \( K_4[Fe(CN)_6] \) is \( Al_2(SO_4)_3 \) with \( i = 5 \).

### Final Answer
**The correct answer is \( Al_2(SO_4)_3 \)**.

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