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 will follow these steps:

### Step 1: Identify the given values
- Molal boiling point constant of water (Kb) = 0.53 °C
- Moles of glucose (solute) = 2 moles
- Mass of water (solvent) = 4000 g

### Step 2: Convert the mass of the solvent from grams to kilograms
To calculate molality, we need the mass of the solvent in kilograms:
\[ \text{Mass of water in kg} = \frac{4000 \text{ g}}{1000} = 4 \text{ kg} \]

### Step 3: Calculate molality (m)
Molality (m) is defined as the number of moles of solute per kilogram of solvent:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
Substituting the values:
\[ m = \frac{2 \text{ moles}}{4 \text{ kg}} = 0.5 \text{ mol/kg} \]

### Step 4: Calculate the elevation in boiling point (ΔTb)
The elevation in boiling point can be calculated using the formula:
\[ \Delta T_b = K_b \times m \]
Substituting the values:
\[ \Delta T_b = 0.53 \text{ °C} \times 0.5 \text{ mol/kg} = 0.265 \text{ °C} \]

### Step 5: Calculate the boiling point of the solution
The boiling point of the solution (Tb) can be found using the formula:
\[ T_b = T_{b0} + \Delta T_b \]
Where \( T_{b0} \) is the boiling point of the pure solvent (water), which is 100 °C:
\[ T_b = 100 \text{ °C} + 0.265 \text{ °C} = 100.265 \text{ °C} \]

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

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