van't Hoff factor for `K_(4)[FeC(N)_(6)]` dissociated 10% is

Options

A. 1.1
B. 1.4
C. 0.86
D. 1.6

Answer

To find the van't Hoff factor (i) for the compound \( K_4[Fe(CN)_6] \) that dissociates 10%, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the dissociation of the compound**:
The compound \( K_4[Fe(CN)_6] \) dissociates in solution to form:
\[ 4K^+ + [Fe(CN)_6]^{4-} \]
This means that for every formula unit of \( K_4[Fe(CN)_6] \), we get a total of 5 ions (4 potassium ions and 1 complex ion).

2. **Determine the total number of ions formed (N)**:
From the dissociation, we can see that:
\[ N = 5 \]
(4 potassium ions + 1 complex ion)

3. **Calculate the degree of dissociation (α)**:
The problem states that the compound dissociates 10%, which can be expressed as:
\[ \alpha = \frac{10}{100} = 0.1 \]

4. **Use the formula for the van't Hoff factor (i)**:
The relationship between the van't Hoff factor (i), degree of dissociation (α), and the number of particles formed (N) is given by:
\[ \alpha = \frac{i - 1}{N - 1} \]
Rearranging this gives:
\[ i = \alpha (N - 1) + 1 \]

5. **Substitute the known values into the equation**:
Plugging in the values we have:
\[ i = 0.1 \times (5 - 1) + 1 \]
\[ i = 0.1 \times 4 + 1 \]
\[ i = 0.4 + 1 \]
\[ i = 1.4 \]

6. **Conclusion**:
The van't Hoff factor (i) for \( K_4[Fe(CN)_6] \) dissociated 10% is:
\[ i = 1.4 \]