At certain Hill-station pure water boils at `99.725^(@)C`. If `K_(b)` for water is `0.513^(@)C kg mol^(-1)`, the boiling point of `0.69m` solution of urea will be:

Options

A. `100.079^(@)C`
B. `103^(@)C`
C. `100.359^(@)C`
D. un predictable

Explanation

To solve the problem, we will use the formula for boiling point elevation, which 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 of the solvent (water in this case),
- \(m\) is the molality of the solution.

### Step 1: Identify the given values
- The boiling point of pure water at the hill station, \(T_b^0 = 99.725^\circ C\)
- The ebullioscopic constant for water, \(K_b = 0.513^\circ C \cdot kg \cdot mol^{-1}\)
- The molality of the urea solution, \(m = 0.69 \, mol/kg\)

### Step 2: Calculate the elevation in boiling point (\(\Delta T_b\))
Using the formula:

\[ \Delta T_b = K_b \times m \]

Substituting the values:

\[ \Delta T_b = 0.513 \, ^\circ C \cdot kg \cdot mol^{-1} \times 0.69 \, mol/kg \]

Calculating this:

\[ \Delta T_b = 0.513 \times 0.69 = 0.35397 \, ^\circ C \approx 0.354 \, ^\circ C \]

### Step 3: Calculate the new boiling point (\(T_b\))
The new boiling point of the solution can be calculated using:

\[ T_b = T_b^0 + \Delta T_b \]

Substituting the values:

\[ T_b = 99.725 \, ^\circ C + 0.354 \, ^\circ C \]

Calculating this:

\[ T_b = 100.079 \, ^\circ C \]

### Final Answer
The boiling point of the 0.69m solution of urea is approximately:

\[ \boxed{100.079 \, ^\circ C} \]

---