The boiling point of 0.1 molal aqueous solution of urea is `100.18^(@)C` at 1 atm. The molal elevation constant of water is

Options

A. `1.8`
B. `0.18`
C. 18
D. `18.6`

Explanation

To find the molal elevation constant (Kb) of water using the given data, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We are given the boiling point of a 0.1 molal aqueous solution of urea, which is 100.18°C at 1 atm. We need to find the molal elevation constant (Kb) of water.

2. **Identify the Formula**: The formula for boiling point elevation is:
\[ \Delta T_b = K_b \times m \]
where:
- \(\Delta T_b\) = elevation in boiling point
- \(K_b\) = molal elevation constant
- \(m\) = molality of the solution

3. **Calculate \(\Delta T_b\)**:
- The boiling point of pure water (\(T_0^b\)) is 100°C.
- The boiling point of the solution (\(T_s\)) is 100.18°C.
- Therefore, the elevation in boiling point is:
\[ \Delta T_b = T_s - T_0^b = 100.18°C - 100°C = 0.18°C \]

4. **Substitute Values into the Formula**:
- We know \(\Delta T_b = 0.18°C\) and \(m = 0.1\) molal.
- Substitute these values into the formula:
\[ 0.18 = K_b \times 0.1 \]

5. **Solve for \(K_b\)**:
- Rearranging the equation gives:
\[ K_b = \frac{0.18}{0.1} = 1.8 \]

6. **Conclusion**: The molal elevation constant of water is \(K_b = 1.8\).

### Final Answer:
The molal elevation constant of water is **1.8**.