1 mole glucose is added to 1 L of water `K_(b)(H_(2)O)=0.512" K kg mole"^(-1)` boiling point of solution will be

Options

A. `373.512^@C`
B. `100.512^@C`
C. `99.488^@C`
D. `372.488^@C`

Answer

To find the boiling point of the solution when 1 mole of glucose is added to 1 L of water, we can follow these steps:

### Step 1: Understand the Concept of Boiling Point Elevation
The boiling point of a solution is elevated when a solute is added to a solvent. This phenomenon is described by the formula:
\[ \Delta T_b = K_b \times m \]
where:
- \(\Delta T_b\) = boiling point elevation
- \(K_b\) = ebullioscopic constant of the solvent (water in this case)
- \(m\) = molality of the solution

### Step 2: Calculate the Molality of the Solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent. In this case, we have:
- Moles of glucose = 1 mole
- Mass of water = 1 L of water = 1000 g = 1 kg (since the density of water is approximately 1 g/mL)

Thus, the molality \(m\) can be calculated as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{1 \text{ mole}}{1 \text{ kg}} = 1 \text{ mol/kg} \]

### Step 3: Use the Boiling Point Elevation Formula
Now, we can substitute the values into the boiling point elevation formula:
\[ \Delta T_b = K_b \times m = 0.512 \, \text{K kg mole}^{-1} \times 1 \, \text{mol/kg} = 0.512 \, \text{K} \]

### Step 4: Calculate the New Boiling Point
The normal boiling point of water is 100 °C (or 373.15 K). Therefore, the new boiling point of the solution can be calculated as:
\[ \text{New boiling point} = 100 \, °C + \Delta T_b = 100 \, °C + 0.512 \, °C = 100.512 \, °C \]

### Final Answer
The boiling point of the solution will be **100.512 °C**.

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