The molal boiling point constant of water is `0.53^(@)C`. When 2 mole of glucose are dissolved in 4000 gm of water, the solution will boil at:

Options

A. `100.53^(@)C`
B. `101.06^(@)C`
C. `100.265^(@)C`
D. `99.47^(@)C`

Answer

To solve the problem, we need to calculate the boiling point of the solution when 2 moles of glucose are dissolved in 4000 grams of water. We will use the formula for boiling point elevation.

### Step-by-Step Solution:

1. **Identify the given data:**
- Molal boiling point constant of water (K_b) = 0.53 °C
- Number of moles of glucose (solute) = 2 moles
- Mass of water (solvent) = 4000 grams

2. **Convert the mass of the solvent from grams to kilograms:**
\[ \text{Mass of water in kg} = \frac{4000 \text{ grams}}{1000} = 4 \text{ kg} \]

3. **Calculate the molality (m) of the solution:**
\[ \text{Molality (m)} = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} = \frac{2 \text{ moles}}{4 \text{ kg}} = 0.5 \text{ mol/kg} \]

4. **Calculate the elevation in boiling point (ΔT_b):**
\[ \Delta T_b = K_b \times m = 0.53 \, °C \times 0.5 \, \text{mol/kg} = 0.265 \, °C \]

5. **Determine the boiling point of the solution:**
- The boiling point of pure water = 100 °C
- Therefore, the boiling point of the solution = Boiling point of water + ΔT_b
\[ \text{Boiling point of solution} = 100 \, °C + 0.265 \, °C = 100.265 \, °C \]

### Final Answer:
The solution will boil at **100.265 °C**.