0.01 m aqueous solution of `K_(3)[Fe(CN)_(6)]` freezes at `-0.062^(@)C`. What is the apparent percentage of dissociation ? (`K_(f)` for water `= 1.86" K kg mol"^(-1)`)

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the change in freezing point (ΔTf)
The freezing point depression (ΔTf) is given by the formula:
\[ \Delta T_f = T_f^0 - T_f \]
Where:
- \(T_f^0\) is the freezing point of pure solvent (water), which is \(0^\circ C\).
- \(T_f\) is the freezing point of the solution, which is given as \(-0.062^\circ C\).

Calculating ΔTf:
\[ \Delta T_f = 0 - (-0.062) = 0.062^\circ C \]

### Step 2: Use the freezing point depression formula
The formula for freezing point depression is:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \(i\) is the van 't Hoff factor (number of particles the solute dissociates into).
- \(K_f\) is the cryoscopic constant, given as \(1.86 \, K \, kg \, mol^{-1}\).
- \(m\) is the molality of the solution, given as \(0.01 \, mol/kg\).

### Step 3: Rearranging the formula to find \(i\)
Rearranging the formula gives:
\[ i = \frac{\Delta T_f}{K_f \cdot m} \]

### Step 4: Substitute the values
Substituting the known values:
\[ i = \frac{0.062}{1.86 \cdot 0.01} \]
Calculating:
\[ i = \frac{0.062}{0.0186} \approx 3.33 \]

### Step 5: Determine the number of particles (n)
The dissociation of \(K_3[Fe(CN)_6]\) can be represented as:
\[ K_3[Fe(CN)_6] \rightarrow 3K^+ + [Fe(CN)_6]^{3-} \]
This results in a total of 4 ions (3 potassium ions and 1 complex ion).

Thus, \(n = 4\).

### Step 6: Calculate the percentage of dissociation (α)
The formula for percentage of dissociation is:
\[ \alpha = \frac{i - 1}{n - 1} \]
Substituting the values:
\[ \alpha = \frac{3.33 - 1}{4 - 1} = \frac{2.33}{3} \approx 0.7766 \]

### Step 7: Convert to percentage
To express α as a percentage:
\[ \text{Percentage of dissociation} = \alpha \times 100 \approx 0.7766 \times 100 \approx 77.66\% \]

### Final Answer
The apparent percentage of dissociation of \(K_3[Fe(CN)_6]\) is approximately **77.66%**.

---