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. Unpredictable

Answer

To find the boiling point of a 0.69 molal solution of urea at a hill station where pure water boils at 99.725°C, we can follow these steps:

### Step 1: Understand the formula for boiling point elevation
The elevation in boiling point (\( \Delta T_b \)) can be calculated using the formula:
\[ \Delta T_b = K_b \times m \]
where:
- \( K_b \) is the ebullioscopic constant (0.513°C kg/mol for water),
- \( m \) is the molality of the solution (0.69 mol/kg).

### Step 2: Calculate the elevation in boiling point
Substituting the given values into the formula:
\[ \Delta T_b = 0.513 \, \text{°C kg/mol} \times 0.69 \, \text{mol/kg} \]
Calculating this gives:
\[ \Delta T_b = 0.35397 \, \text{°C} \]

### Step 3: Determine the boiling point of the solution
The boiling point of the solution (\( T_b \)) can be found by adding the elevation in boiling point to the boiling point of pure water:
\[ T_b = T_{b0} + \Delta T_b \]
where \( T_{b0} \) is the boiling point of pure water (99.725°C). Thus:
\[ T_b = 99.725 \, \text{°C} + 0.35397 \, \text{°C} \]
Calculating this gives:
\[ T_b = 100.07897 \, \text{°C} \]

### Step 4: Round off the boiling point
Rounding off to three decimal places, the boiling point of the solution is:
\[ T_b \approx 100.079 \, \text{°C} \]

### Final Answer
The boiling point of the 0.69 molal solution of urea is approximately **100.079°C**.

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