The ratio of the value of any colligative property for `K_4[Fe(CN)_6]` to that of `Fe_4[Fe(CN)_6]_3` solution is nearly

Options

A. 0.62
B. 0.71
C. 1.4
D. 1.2

Answer

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

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

The compound \( K_4[Fe(CN)_6] \) dissociates in solution as follows:
\[ K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-} \]
From this dissociation, we can see that it produces a total of 5 ions (4 potassium ions and 1 hexacyanoferrate ion).

### Step 2: Calculate the van 't Hoff factor \( i_1 \) for \( K_4[Fe(CN)_6] \)

The van 't Hoff factor \( i \) is the number of particles the solute dissociates into. For \( K_4[Fe(CN)_6] \):
\[ i_1 = 5 \]

### Step 3: Identify the dissociation of \( Fe_4[Fe(CN)_6]_3 \)

The compound \( Fe_4[Fe(CN)_6]_3 \) dissociates in solution as follows:
\[ Fe_4[Fe(CN)_6]_3 \rightarrow 4Fe^{3+} + 3[Fe(CN)_6]^{3-} \]
From this dissociation, we can see that it produces a total of 7 ions (4 iron ions and 3 hexacyanoferrate ions).

### Step 4: Calculate the van 't Hoff factor \( i_2 \) for \( Fe_4[Fe(CN)_6]_3 \)

For \( Fe_4[Fe(CN)_6]_3 \):
\[ i_2 = 7 \]

### Step 5: Calculate the ratio of the van 't Hoff factors

Now, we can find the ratio of the van 't Hoff factors:
\[ \text{Ratio} = \frac{i_1}{i_2} = \frac{5}{7} \]

### Step 6: Calculate the numerical value of the ratio

Calculating the numerical value:
\[ \frac{5}{7} \approx 0.714 \]

### Conclusion

Thus, the ratio of the value of any colligative property for \( K_4[Fe(CN)_6] \) to that of \( Fe_4[Fe(CN)_6]_3 \) is approximately \( 0.71 \).

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