An aqueous solution of glucose boils at `100.01^(@)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 (ΔTb)
The boiling point of the solution is given as 100.01°C. The boiling point of pure water is 100.00°C.

\[ \Delta T_b = T_{solution} - T_{pure \, water} = 100.01°C - 100.00°C = 0.01°C \]

### Step 2: Convert the elevation in boiling point to Kelvin
Since the boiling point elevation is already in Celsius and the change in temperature is the same in Kelvin, we can keep it as:

\[ \Delta T_b = 0.01 \, K \]

### 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 kmol⁻¹kg)
- \( m \) is the molality of the solution

### Step 4: Rearrange the formula to find molality (m)
Rearranging the formula gives us:

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

Substituting the known values:

\[ m = \frac{0.01 \, K}{0.5 \, kmol^{-1}kg} = 0.02 \, kmol/kg \]

### Step 5: Convert molality to moles of solute
Molality (m) is defined as the number of moles of solute per kilogram of solvent. We have 100 g of water, which is 0.1 kg.

\[ m = \frac{n}{W_1} \]

Where:
- \( n \) is the number of moles of solute (glucose)
- \( W_1 \) is the mass of the solvent (0.1 kg)

Rearranging gives:

\[ n = m \cdot W_1 = 0.02 \, kmol/kg \cdot 0.1 \, kg = 0.002 \, kmol \]

Converting kmol to moles:

\[ n = 0.002 \, kmol \times 1000 \, mol/kmol = 2 \, mol \]

### Step 6: Calculate the number of molecules of glucose
To find the number of molecules, we use Avogadro's number (\( N_A = 6.022 \times 10^{23} \, molecules/mol \)):

\[ \text{Number of molecules} = n \cdot N_A = 2 \, mol \cdot 6.022 \times 10^{23} \, molecules/mol \]

Calculating this gives:

\[ \text{Number of molecules} = 12.044 \times 10^{23} \, molecules \]

### Final Answer
The number of molecules of glucose in the solution containing 100 g of water is approximately:

\[ \text{Number of molecules} \approx 12.044 \times 10^{23} \]

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