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. `100.25^@C`
D. `100 ^@C`

Answer

To solve the problem, we need to determine the boiling point of an aqueous solution containing 3g of glucose, given that a solution with 1g of urea boils at 100.25°C.

### Step-by-Step Solution:

1. **Understanding Boiling Point Elevation**:
The boiling point elevation (\( \Delta T_b \)) is given by the formula:
\[ \Delta T_b = K_b \cdot m \]
where \( K_b \) is the ebullioscopic constant of the solvent (water in this case), and \( m \) is the molality of the solution.

2. **Identifying the Solvent**:
Since both solutions are aqueous, we will consider water as the solvent. The value of \( K_b \) for water is constant and does not change between the two solutions.

3. **Calculating Molality**:
Molality (\( m \)) is defined as:
\[ m = \frac{\text{number of moles of solute}}{\text{mass of solvent in kg}} \]
Since both solutions have the same volume and thus the same mass of water as solvent, the molality will depend only on the number of moles of solute.

4. **Finding Moles of Urea**:
The molecular weight of urea (NH₂CONH₂) is calculated as follows:
- Nitrogen (N): 14 g/mol (2 N)
- Carbon (C): 12 g/mol (1 C)
- Oxygen (O): 16 g/mol (1 O)
- Hydrogen (H): 1 g/mol (4 H)

Total molecular weight of urea:
\[ 2(14) + 12 + 16 + 4(1) = 60 \text{ g/mol} \]
The number of moles of urea in 1g is:
\[ \text{Moles of urea} = \frac{1 \text{ g}}{60 \text{ g/mol}} = \frac{1}{60} \text{ mol} \]

5. **Finding Moles of Glucose**:
The molecular weight of glucose (C₆H₁₂O₆) is calculated as follows:
- Carbon (C): 12 g/mol (6 C)
- Hydrogen (H): 1 g/mol (12 H)
- Oxygen (O): 16 g/mol (6 O)

Total molecular weight of glucose:
\[ 6(12) + 12(1) + 6(16) = 180 \text{ g/mol} \]
The number of moles of glucose in 3g is:
\[ \text{Moles of glucose} = \frac{3 \text{ g}}{180 \text{ g/mol}} = \frac{1}{60} \text{ mol} \]

6. **Comparing Moles**:
We find that the number of moles of urea and glucose are both \( \frac{1}{60} \) mol. Since the molality depends only on the number of moles of solute and the mass of solvent is the same for both solutions, the molality of both solutions is equal.

7. **Conclusion on Boiling Point**:
Since the molality is the same for both solutions, the elevation in boiling point (\( \Delta T_b \)) will also be the same. Therefore, the boiling point of the glucose solution will also be:
\[ \text{Boiling point of glucose solution} = 100.25°C \]

### Final Answer:
The boiling point of the aqueous solution containing 3g of glucose will be **100.25°C**.

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