When 1 mole of a solute is dissolved in 1 kg of `H_(2)O `, boiling point of solution was found to be `100.5^(@)`C. `K_(b)` for `H_(2)O` is :

Options

A. 0.5
B. 100
C. 100.5
D. 95.5

Answer

To solve the problem, we need to calculate the ebullioscopic constant (K_b) for water based on the given information about the boiling point elevation when a solute is dissolved in it.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Boiling point of the solution (T_solution) = 100.5 °C
- Boiling point of pure water (T_water) = 100 °C
- Number of moles of solute (n) = 1 mole
- Mass of solvent (water) = 1 kg

2. **Calculate the Elevation in Boiling Point (ΔT_b):**
\[ \Delta T_b = T_{\text{solution}} - T_{\text{water}} = 100.5 °C - 100 °C = 0.5 °C \]

3. **Use the Formula for Boiling Point Elevation:**
The formula for boiling point elevation is given by:
\[ \Delta T_b = K_b \times m \]
where:
- \( \Delta T_b \) is the elevation in boiling point,
- \( K_b \) is the ebullioscopic constant,
- \( m \) is the molality of the solution.

4. **Calculate the Molality (m):**
Molality (m) is defined as the number of moles of solute per kilogram of solvent:
\[ m = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} = \frac{1 \text{ mole}}{1 \text{ kg}} = 1 \text{ molal} \]

5. **Substitute the Values into the Boiling Point Elevation Formula:**
Now, substituting the values we have into the boiling point elevation formula:
\[ 0.5 °C = K_b \times 1 \text{ molal} \]

6. **Solve for K_b:**
Rearranging the equation to solve for \( K_b \):
\[ K_b = \frac{0.5 °C}{1 \text{ molal}} = 0.5 \text{ °C kg/mol} \]

### Final Answer:
The value of \( K_b \) for water is **0.5 °C kg/mol**.

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