Pure water boils at `99.725^(@)C` at Shimla. If `K_(b)` for water is `0.51 K mol^(-1) kg` the boiling point of `0.69` molal urea solution will be:

Options

A. `100.35^(@)C`
B. `100.08^(@)C`
C. `99.37^(@)C`
D. none of these

Answer

To find the boiling point of a 0.69 molal urea solution at Shimla, we can follow these steps:

### Step 1: Understand the Concept of Boiling Point Elevation
The boiling point elevation (\( \Delta T_b \)) is the increase in the boiling point of a solvent when a solute is added. It can be calculated using the formula:
\[ \Delta T_b = K_b \times m \]
where:
- \( K_b \) is the ebullioscopic constant of the solvent (water in this case).
- \( m \) is the molality of the solution.

### Step 2: Identify the Given Values
From the problem, we have:
- \( K_b \) for water = 0.51 °C kg/mol
- Molality of urea solution = 0.69 molal
- Boiling point of pure water at Shimla = 99.725 °C

### Step 3: Calculate the Elevation in Boiling Point
Using the formula for boiling point elevation:
\[ \Delta T_b = K_b \times m \]
Substituting the values:
\[ \Delta T_b = 0.51 \, \text{°C kg/mol} \times 0.69 \, \text{mol/kg} \]
Calculating this gives:
\[ \Delta T_b = 0.3519 \, \text{°C} \]

### Step 4: Calculate 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:
\[ T_b = \Delta T_b + T_{b0} \]
Substituting the values:
\[ T_b = 0.3519 \, \text{°C} + 99.725 \, \text{°C} \]
Calculating this gives:
\[ T_b = 100.0769 \, \text{°C} \]
Rounding this to two decimal places, we get:
\[ T_b \approx 100.08 \, \text{°C} \]

### Final Answer
The boiling point of the 0.69 molal urea solution is approximately **100.08 °C**.

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