The freezing point (in `.^(@)C)` of a solution containing `0.1g` of `K_(3)[Fe(CN)_(6)]` (Mol. wt. `329`) in `100 g` of water `(K_(f) = 1.86 K kg mol^(-1))` is :

Options

A. `-2.3 xx 10^(-2)`
B. `-5.7 xx 10^(-2)`
C. ( c) `-5.7 xx 10^(-3)`
D. `-1.2 xx 10^(-2)`

Explanation

To solve the problem of finding the freezing point of a solution containing \(0.1 \, g\) of \(K_3[Fe(CN)_6]\) in \(100 \, g\) of water, we will follow these steps:

### Step 1: Calculate the number of moles of the solute
The number of moles (\(n\)) can be calculated using the formula:
\[ n = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)}} \]
Given:
- Mass of solute = \(0.1 \, g\)
- Molar mass of \(K_3[Fe(CN)_6] = 329 \, g/mol\)

Substituting the values:
\[ n = \frac{0.1 \, g}{329 \, g/mol} = 3.04 \times 10^{-4} \, mol \]

### Step 2: Calculate the molality of the solution
Molality (\(m\)) is defined as the number of moles of solute per kilogram of solvent:
\[ m = \frac{n}{\text{mass of solvent (kg)}} \]
Given:
- Mass of solvent (water) = \(100 \, g = 0.1 \, kg\)

Substituting the values:
\[ m = \frac{3.04 \times 10^{-4} \, mol}{0.1 \, kg} = 3.04 \times 10^{-3} \, mol/kg \]

### Step 3: Determine the van 't Hoff factor (\(i\))
The dissociation of \(K_3[Fe(CN)_6]\) in solution can be represented as:
\[ K_3[Fe(CN)_6] \rightarrow 3K^+ + [Fe(CN)_6]^{3-} \]
This means that one formula unit of \(K_3[Fe(CN)_6]\) produces \(4\) particles in solution (3 potassium ions and 1 ferricyanide ion). Thus, the van 't Hoff factor \(i = 4\).

### Step 4: Calculate the depression in freezing point (\(\Delta T_f\))
The depression in freezing point can be calculated using the formula:
\[ \Delta T_f = K_f \cdot m \cdot i \]
Where:
- \(K_f = 1.86 \, K \cdot kg/mol\)
- \(m = 3.04 \times 10^{-3} \, mol/kg\)
- \(i = 4\)

Substituting the values:
\[ \Delta T_f = 1.86 \, K \cdot kg/mol \cdot 3.04 \times 10^{-3} \, mol/kg \cdot 4 \]
\[ \Delta T_f = 1.86 \cdot 3.04 \times 10^{-3} \cdot 4 = 0.0226 \, K \approx 2.26 \times 10^{-2} \, °C \]

### Step 5: Calculate the new freezing point
The normal freezing point of water is \(0 \, °C\). The new freezing point (\(T_f\)) can be calculated as:
\[ T_f = 0 \, °C - \Delta T_f \]
Substituting the value of \(\Delta T_f\):
\[ T_f = 0 \, °C - 0.0226 \, °C = -0.0226 \, °C \]

### Final Answer
The freezing point of the solution is approximately \(-0.0226 \, °C\).

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