Pure benzene freezes at `5.45^(@)C` at a certain place but a 0.374m solution of tetrachloroethane in benzene freezes at `3.55^(@)C`. The `K_(f)` for benzene is

Options

A. 5.08 K kg `mol^(-1)`
B. 508 kg `mol^(-1)`
C. 0.508 k kg `mol^(-1)`
D. `50.8^(@)C kg mol^(-1)`

Answer

To find the freezing point depression constant \( K_f \) for benzene, we will follow these steps:

### Step 1: Identify the freezing points
- The freezing point of pure benzene (\( T_f^0 \)) is \( 5.45^\circ C \).
- The freezing point of the solution (\( T_f \)) is \( 3.55^\circ C \).

### Step 2: Calculate the depression in freezing point (\( \Delta T_f \))
The depression in freezing point is calculated using the formula:
\[ \Delta T_f = T_f^0 - T_f \]
Substituting the values:
\[ \Delta T_f = 5.45^\circ C - 3.55^\circ C = 1.90^\circ C \]

### Step 3: Use the formula for freezing point depression
The formula for freezing point depression is:
\[ \Delta T_f = K_f \cdot m \]
where:
- \( K_f \) is the freezing point depression constant for the solvent (benzene in this case),
- \( m \) is the molality of the solution.

### Step 4: Substitute the known values
From the problem, we know:
- \( \Delta T_f = 1.90^\circ C \)
- The molality (\( m \)) of the solution is \( 0.374 \, \text{m} \).

Substituting these values into the equation:
\[ 1.90 = K_f \cdot 0.374 \]

### Step 5: Solve for \( K_f \)
Rearranging the equation to solve for \( K_f \):
\[ K_f = \frac{1.90}{0.374} \]
Calculating this gives:
\[ K_f \approx 5.08 \, \text{K kg}^{-1} \text{mol}^{-1} \]

### Conclusion
The value of \( K_f \) for benzene is approximately \( 5.08 \, \text{K kg}^{-1} \text{mol}^{-1} \).