The ratio of the value of any collogative property for `K_(4)[Fe(CN)_(6)]` solution to that of `Fe_(4)[Fe(CN)_(6)]_(3)`(prussian blue,)solution is nearly

Options

A. `1`
B. `0.71`
C. `1.4`
D. Less than `1`

Explanation

To find the ratio of the value of any colligative property for \( K_4[Fe(CN)_6] \) solution to that of \( Fe_4[Fe(CN)_6]_3 \) (Prussian blue) solution, we will follow these steps:

### Step 1: Determine the dissociation of \( K_4[Fe(CN)_6] \)

1. **Identify the formula**: \( K_4[Fe(CN)_6] \) is a salt that dissociates in solution.
2. **Dissociation equation**:
\[ K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-} \]
3. **Total particles produced**: From this dissociation, we see that 1 formula unit produces 5 particles (4 \( K^+ \) ions and 1 \( [Fe(CN)_6]^{4-} \) ion).

### Step 2: Determine the dissociation of \( Fe_4[Fe(CN)_6]_3 \)

1. **Identify the formula**: \( Fe_4[Fe(CN)_6]_3 \) is a complex that also dissociates in solution.
2. **Dissociation equation**:
\[ Fe_4[Fe(CN)_6]_3 \rightarrow 4Fe^{3+} + 3[Fe(CN)_6]^{3-} \]
3. **Total particles produced**: From this dissociation, we see that 1 formula unit produces 7 particles (4 \( Fe^{3+} \) ions and 3 \( [Fe(CN)_6]^{3-} \) ions).

### Step 3: Calculate the van 't Hoff factor (i)

1. **For \( K_4[Fe(CN)_6] \)**:
\[ i = 5 \]
2. **For \( Fe_4[Fe(CN)_6]_3 \)**:
\[ i = 7 \]

### Step 4: Calculate the ratio of colligative properties

1. **Ratio of colligative properties**: The ratio of any colligative property (like boiling point elevation, freezing point depression, etc.) is directly proportional to the van 't Hoff factor. Therefore, the ratio of the values of any colligative property for the two solutions is:
\[ \text{Ratio} = \frac{i_{K_4[Fe(CN)_6]}}{i_{Fe_4[Fe(CN)_6]_3}} = \frac{5}{7} \]

### Final Answer

The ratio of the value of any colligative property for \( K_4[Fe(CN)_6] \) solution to that of \( Fe_4[Fe(CN)_6]_3 \) solution is nearly \( \frac{5}{7} \).

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