The ebullioscopic constant for benzene is `2.52 K Kg mol^(-1)`. A solution of an organic substance in benzene boils at `0.125^(@)C` higher than benzene. Calculate the molality of solution?

Options

A. 10
B. 1
C. 2
D. 0.05

Explanation

To solve the problem, we need to calculate the molality of a solution given the elevation in boiling point and the ebullioscopic constant of benzene.

### Step-by-Step Solution:

1. **Identify the given values:**
- Ebullioscopic constant (Kb) for benzene = 2.52 K kg mol^(-1)
- Elevation in boiling point (ΔTb) = 0.125 °C

2. **Use the formula for elevation in boiling point:**
The formula relating the elevation in boiling point to molality is given by:
\[ \Delta T_b = K_b \cdot m \]
where:
- ΔTb = elevation in boiling point
- Kb = ebullioscopic constant
- m = molality of the solution

3. **Rearrange the formula to solve for molality (m):**
\[ m = \frac{\Delta T_b}{K_b} \]

4. **Substitute the known values into the equation:**
\[ m = \frac{0.125 \, \text{°C}}{2.52 \, \text{K kg mol}^{-1}} \]

5. **Calculate the molality:**
\[ m = \frac{0.125}{2.52} \approx 0.0496 \, \text{mol/kg} \]

6. **Round the answer to two decimal places:**
\[ m \approx 0.05 \, \text{mol/kg} \]

### Final Answer:
The molality of the solution is approximately **0.05 mol/kg**.