Pure benzene freeze at `5.45^(@)"C"`. A solution containing 7.24g of `"C"_(2)"H"_(2)"Cl_(4)` in 115.3 g of benzene was observed to freeze at `3.55^(@)"C".` What is the molal freezing point constant of benzene?

Explanation

To find the molal freezing point constant (Kf) of benzene, we can follow these steps:

### Step 1: Calculate the depression in freezing point (ΔTf)
The depression in freezing point is calculated as:
\[ \Delta T_f = T_f^{\text{pure}} - T_f^{\text{solution}} \]
Where:
- \(T_f^{\text{pure}} = 5.45^\circ C\) (freezing point of pure benzene)
- \(T_f^{\text{solution}} = 3.55^\circ C\) (freezing point of the solution)

Calculating:
\[ \Delta T_f = 5.45^\circ C - 3.55^\circ C = 1.90^\circ C \]

### Step 2: Calculate the molality (m) of the solution
Molality is defined as the number of moles of solute per kilogram of solvent. First, we need to find the number of moles of the solute (C2H2Cl4).

1. **Calculate the molar mass of C2H2Cl4**:
- Carbon (C): 12.01 g/mol × 2 = 24.02 g/mol
- Hydrogen (H): 1.008 g/mol × 2 = 2.016 g/mol
- Chlorine (Cl): 35.453 g/mol × 4 = 141.812 g/mol
- Total molar mass = 24.02 + 2.016 + 141.812 = 167.848 g/mol (approximately 168 g/mol)

2. **Calculate the number of moles of C2H2Cl4**:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{7.24 \text{ g}}{168 \text{ g/mol}} \approx 0.0431 \text{ moles} \]

3. **Convert the mass of benzene to kilograms**:
\[ \text{Mass of benzene} = 115.3 \text{ g} = 0.1153 \text{ kg} \]

4. **Calculate molality (m)**:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.0431 \text{ moles}}{0.1153 \text{ kg}} \approx 0.373 \text{ mol/kg} \]

### Step 3: Use the formula for freezing point depression
The freezing point depression is related to the molality and the freezing point constant (Kf) by the formula:
\[ \Delta T_f = K_f \cdot m \]

Rearranging for Kf:
\[ K_f = \frac{\Delta T_f}{m} \]

### Step 4: Substitute the values into the equation
Substituting the values we calculated:
\[ K_f = \frac{1.90^\circ C}{0.373 \text{ mol/kg}} \approx 5.08^\circ C \cdot \text{kg/mol} \]

### Final Answer
The molal freezing point constant of benzene (Kf) is approximately:
\[ K_f \approx 5.08^\circ C \cdot \text{kg/mol} \]

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