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 of finding the boiling point of a solution when 2 moles of glucose are dissolved in 4000 grams of water, we can follow these steps:

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

### Step 2: Convert the mass of water to kilograms
Since molality is defined as moles of solute per kilogram of solvent, we need to convert the mass of water from grams to kilograms:
\[ \text{Mass of water in kg} = \frac{4000 \text{ g}}{1000} = 4 \text{ kg} \]

### Step 3: Calculate the molality of the solution
Molality (m) is calculated using the formula:
\[ m = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} \]
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:
\[ \Delta T_b = 0.53 \text{ °C} \times 0.5 = 0.265 \text{ °C} \]

### Step 5: Calculate the final boiling point (Tf)
The final boiling point can be found using the equation:
\[ T_f = T_i + \Delta T_b \]
Where \(T_i\) is the initial boiling point of water, which is 100 °C. Thus:
\[ T_f = 100 \text{ °C} + 0.265 \text{ °C} = 100.265 \text{ °C} \]

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

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