An aqueous solution of glucose boils at `100.02^(@)C`.The molal elevation constant for water is `0.5 kmol^(-1)kg`. The number of molecules of glucose in the solution containing `100g` of water is

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the elevation in boiling point (ΔT_b)
The boiling point of the solution is given as 100.02 °C. The boiling point of pure water is 100 °C. Therefore, the elevation in boiling point (ΔT_b) can be calculated as follows:

\[ \Delta T_b = \text{Boiling point of solution} - \text{Boiling point of pure water} \]

\[ \Delta T_b = 100.02 °C - 100 °C = 0.02 °C \]

### Step 2: Convert ΔT_b to Kelvin
Since the molal elevation constant (K_b) is in Kelvin, we convert the elevation in boiling point from Celsius to Kelvin. The conversion is straightforward because a change of 1 °C is equivalent to a change of 1 K.

\[ \Delta T_b = 0.02 K \]

### Step 3: Use the formula for boiling point elevation
The formula for boiling point elevation is given by:

\[ \Delta T_b = K_b \cdot m \]

Where:
- \( K_b \) is the molal elevation constant (0.5 K kg/mol),
- \( m \) is the molality of the solution.

### Step 4: Calculate the mass of the solvent in kg
The mass of water is given as 100 g. We need to convert this to kg:

\[ \text{Mass of water (W)} = 100 \, \text{g} = 0.1 \, \text{kg} \]

### Step 5: Rearrange the formula to find molality (m)
We can rearrange the formula to find molality:

\[ m = \frac{\Delta T_b}{K_b} \]

Substituting the known values:

\[ m = \frac{0.02 \, K}{0.5 \, K \cdot \text{kg/mol}} = 0.04 \, \text{mol/kg} \]

### Step 6: Calculate the number of moles of glucose
Molality (m) is defined as the number of moles of solute per kg of solvent. Therefore, we can calculate the number of moles of glucose (n) using the formula:

\[ m = \frac{n}{W} \]

Rearranging gives:

\[ n = m \cdot W \]

Substituting the values:

\[ n = 0.04 \, \text{mol/kg} \cdot 0.1 \, \text{kg} = 0.004 \, \text{mol} \]

### Step 7: Calculate the number of molecules of glucose
To find the number of molecules, we use Avogadro's number (\( N_A \)), which is approximately \( 6.022 \times 10^{23} \) molecules/mol:

\[ \text{Number of molecules} = n \cdot N_A \]

Substituting the values:

\[ \text{Number of molecules} = 0.004 \, \text{mol} \cdot 6.022 \times 10^{23} \, \text{molecules/mol} \]

Calculating this gives:

\[ \text{Number of molecules} \approx 2.41 \times 10^{21} \, \text{molecules} \]

### Final Answer
The number of molecules of glucose in the solution containing 100 g of water is approximately \( 2.41 \times 10^{21} \) molecules.

---