The normal freezing point of nitrobenzne `(C_(6)H_(5)NO_(2))` is 278.82 K.A 0.25 m solution of solute A in nitrobenzene decreases the freezing point to 276.82 K. The value of `K_(f)` for nitrobenzene is

Options

A. 2K kg `mol^(-1)`
B. 4K kg `mol^(-1)`
C. 6K kg `mol^(-1)`
D. 8.0K kg `mol^(-1)`

Answer

To find the value of \( K_f \) for nitrobenzene, we will use the formula for depression in freezing point, which is given by:

\[ \Delta T_f = K_f \times m \]

Where:
- \( \Delta T_f \) is the depression in freezing point,
- \( K_f \) is the freezing point depression constant,
- \( m \) is the molality of the solution.

### Step 1: Calculate the depression in freezing point (\( \Delta T_f \))

The normal freezing point of nitrobenzene is given as \( 278.82 \, K \) and the freezing point of the solution is \( 276.82 \, K \).

\[ \Delta T_f = T_{0f} - T_f = 278.82 \, K - 276.82 \, K = 2.00 \, K \]

### Step 2: Use the molality of the solution

The molality \( m \) of the solution is given as \( 0.25 \, m \) (molal).

### Step 3: Substitute values into the depression in freezing point formula

Now, we can substitute \( \Delta T_f \) and \( m \) into the formula to find \( K_f \):

\[ \Delta T_f = K_f \times m \]
\[ 2.00 \, K = K_f \times 0.25 \, m \]

### Step 4: Solve for \( K_f \)

Rearranging the equation to solve for \( K_f \):

\[ K_f = \frac{\Delta T_f}{m} = \frac{2.00 \, K}{0.25 \, m} = 8 \, K \cdot kg^{-1} \cdot mol^{-1} \]

### Final Answer

Thus, the value of \( K_f \) for nitrobenzene is:

\[ K_f = 8 \, K \cdot kg^{-1} \cdot mol^{-1} \]