The freezing point of a solution of 2.40 g of biphenyl (`C_(12)H_(10)`) in 75.0 g of benzene (`C_(6)H_(6)`) is `4.40^(@)C`. The normal freezing point of benzene is `5.50^(@)C`. What is the molal freezing point constant `(@C//m)` for benzene ?

Options

A. 5.3
B. 5.1
C. 4.6
D. 4.8

Answer

To solve the problem, we need to find the molal freezing point constant (Kf) for benzene using the given data. Here’s a step-by-step breakdown of the solution:

### Step 1: Determine the Depression in Freezing Point (ΔTf)
The depression in freezing point (ΔTf) can be calculated using the formula:
\[ \Delta T_f = T_f^0 - T_f \]
Where:
- \(T_f^0\) = normal freezing point of benzene = 5.50 °C
- \(T_f\) = freezing point of the solution = 4.40 °C

Calculating ΔTf:
\[ \Delta T_f = 5.50 °C - 4.40 °C = 1.10 °C \]

### Step 2: Calculate the Molality (m) of the Solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent. We need to calculate the number of moles of biphenyl (C₁₂H₁₀) and then find the molality.

1. **Find the molar mass of biphenyl (C₁₂H₁₀)**:
- Carbon (C): 12.01 g/mol × 12 = 144.12 g/mol
- Hydrogen (H): 1.008 g/mol × 10 = 10.08 g/mol
- Total molar mass = 144.12 g/mol + 10.08 g/mol = 154.20 g/mol

2. **Calculate the number of moles of biphenyl**:
\[ \text{Moles of biphenyl} = \frac{\text{mass of biphenyl}}{\text{molar mass of biphenyl}} = \frac{2.40 \, \text{g}}{154.20 \, \text{g/mol}} \approx 0.01557 \, \text{mol} \]

3. **Calculate the molality (m)**:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.01557 \, \text{mol}}{0.075 \, \text{kg}} \approx 0.2076 \, \text{mol/kg} \]

### Step 3: Use the Freezing Point Depression Formula
The freezing point depression can be expressed as:
\[ \Delta T_f = K_f \cdot m \]
Where:
- \(K_f\) = molal freezing point constant
- \(m\) = molality

Rearranging the formula to solve for \(K_f\):
\[ K_f = \frac{\Delta T_f}{m} \]

Substituting the values:
\[ K_f = \frac{1.10 °C}{0.2076 \, \text{mol/kg}} \approx 5.29 °C \cdot \text{kg/mol} \]

### Final Answer
The molal freezing point constant \(K_f\) for benzene is approximately **5.29 °C kg/mol**.

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