An aqueous solution containing 1g of urea boils at `100.25^(@)C`. The aqueous solution containing 3g of glucose in the same volume will boil be

Options

A. `100.75^(@)C`
B. `100.5^(@)C`
C. `100^(@)C`
D. `100.25^(@)C`

Explanation

To solve the problem of determining the boiling point of an aqueous solution containing 3g of glucose, we will use the concept of boiling point elevation, which is a colligative property. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the given data
- The boiling point of pure water (solvent) = 100°C
- The boiling point of the solution with 1g of urea = 100.25°C
- The mass of glucose in the new solution = 3g
- The molecular weight of urea (NH₂CONH₂) = 60 g/mol
- The molecular weight of glucose (C₆H₁₂O₆) = 180 g/mol

### Step 2: Calculate the elevation in boiling point for urea
The elevation in boiling point (ΔT_b) can be calculated as follows:
\[ \Delta T_b = T_b - T_{b0} \]
where \( T_b \) is the boiling point of the solution and \( T_{b0} \) is the boiling point of the pure solvent.

\[ \Delta T_b = 100.25°C - 100°C = 0.25°C \]

### Step 3: Relate boiling point elevation to molality
The formula for boiling point elevation is given by:
\[ \Delta T_b = K_b \cdot m \]
where \( K_b \) is the molal boiling point elevation constant and \( m \) is the molality of the solution.

### Step 4: Calculate molality of urea solution
The molality (m) is defined as:
\[ m = \frac{\text{number of moles of solute}}{\text{mass of solvent (kg)}} \]
For 1g of urea:
\[ \text{Number of moles of urea} = \frac{1g}{60g/mol} = \frac{1}{60} \text{ mol} \]
Assuming the mass of water (solvent) is 1 kg (which is a common assumption for dilute solutions), the molality of the urea solution is:
\[ m = \frac{1/60}{1} = \frac{1}{60} \text{ mol/kg} \]

### Step 5: Calculate Kb using urea solution
Using the boiling point elevation formula:
\[ 0.25°C = K_b \cdot \frac{1}{60} \]
Solving for \( K_b \):
\[ K_b = 0.25 \cdot 60 = 15°C \cdot kg/mol \]

### Step 6: Calculate molality for glucose solution
Now, we will calculate the molality for the glucose solution with 3g of glucose:
\[ \text{Number of moles of glucose} = \frac{3g}{180g/mol} = \frac{1}{60} \text{ mol} \]
The molality of the glucose solution is:
\[ m = \frac{1/60}{1} = \frac{1}{60} \text{ mol/kg} \]

### Step 7: Calculate boiling point elevation for glucose solution
Using the same \( K_b \):
\[ \Delta T_b = K_b \cdot m = 15°C \cdot kg/mol \cdot \frac{1}{60} = 0.25°C \]

### Step 8: Calculate the boiling point of the glucose solution
The boiling point of the glucose solution is:
\[ T_b = T_{b0} + \Delta T_b = 100°C + 0.25°C = 100.25°C \]

### Conclusion
The boiling point of the aqueous solution containing 3g of glucose in the same volume will also be **100.25°C**.