Freezing point depression of millimolal `K_nFe(CN)_(6)` is `7.1 xx 10^(-3) K.` If `K_f = 1.86 k Kg mol^(-1)` . Calculate the value of n.

Answer

To solve the problem of calculating the value of \( n \) in the freezing point depression of millimolal \( K_nFe(CN)_6 \), we will follow these steps:

### Step 1: Understand the relationship between freezing point depression and molality
The freezing point depression (\( \Delta T_f \)) is given by the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- \( \Delta T_f \) is the freezing point depression,
- \( i \) is the van 't Hoff factor (number of particles the solute breaks into),
- \( K_f \) is the cryoscopic constant,
- \( m \) is the molality of the solution.

### Step 2: Identify the components of the formula
From the problem:
- \( \Delta T_f = 7.1 \times 10^{-3} \, K \)
- \( K_f = 1.86 \, K \cdot kg \cdot mol^{-1} \)
- The solution is millimolal, which means \( m = \frac{1}{1000} \, mol/kg \).

### Step 3: Determine the van 't Hoff factor \( i \)
For the complex \( K_nFe(CN)_6 \), it dissociates into \( n \) \( K^+ \) ions and \( 1 \) \( Fe(CN)_6^{3-} \) ion. Therefore, the total number of ions produced is:
\[ i = n + 1 \]

### Step 4: Substitute the known values into the freezing point depression formula
Substituting the known values into the formula:
\[ 7.1 \times 10^{-3} = (n + 1) \cdot 1.86 \cdot \frac{1}{1000} \]

### Step 5: Simplify the equation
Rearranging the equation gives:
\[ 7.1 \times 10^{-3} = (n + 1) \cdot 1.86 \times 0.001 \]
\[ 7.1 \times 10^{-3} = (n + 1) \cdot 0.00186 \]

### Step 6: Solve for \( n + 1 \)
Dividing both sides by \( 0.00186 \):
\[ n + 1 = \frac{7.1 \times 10^{-3}}{0.00186} \]
Calculating the right side:
\[ n + 1 \approx 3.82 \]

### Step 7: Solve for \( n \)
Now, solving for \( n \):
\[ n = 3.82 - 1 \approx 2.82 \]
Rounding to the nearest whole number gives:
\[ n \approx 3 \]

### Final Answer
Thus, the value of \( n \) is approximately \( 3 \).

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