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

Options

A. `6.023xx10^(23)`
B. `6.023xx10^(22)`
C. `12.046xx10^(20)`
D. `12.046xx10^(23)`

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Identify the given data
- Boiling point of the solution, \( T_b = 100.01^\circ C \)
- Boiling point of pure water, \( T_b^0 = 100^\circ C \)
- Molal elevation constant for water, \( K_b = 0.5 \, \text{kmol}^{-1}\text{kg} \)
- Mass of water, \( m_{water} = 100 \, g = 0.1 \, kg \)

### Step 2: Calculate the elevation in boiling point
The elevation in boiling point, \( \Delta T_b \), is given by:
\[ \Delta T_b = T_b - T_b^0 = 100.01^\circ C - 100^\circ C = 0.01^\circ C \]

### Step 3: Use the formula for boiling point elevation
The formula for boiling point elevation is:
\[ \Delta T_b = K_b \cdot m \]
where \( m \) is the molality of the solution.

### Step 4: Rearrange the formula to find molality
Rearranging the formula to solve for molality \( m \):
\[ m = \frac{\Delta T_b}{K_b} \]
Substituting the known values:
\[ m = \frac{0.01}{0.5} = 0.02 \, \text{kmol/kg} \]

### Step 5: Convert molality to moles of solute
Molality \( m \) is defined as the number of moles of solute per kilogram of solvent. Therefore, we can find the number of moles of glucose in the solution:
\[ \text{Number of moles of glucose} = m \times \text{mass of solvent (in kg)} = 0.02 \, \text{kmol/kg} \times 0.1 \, kg = 0.002 \, \text{kmol} \]

### Step 6: Convert moles to molecules
To find the number of molecules, we use Avogadro's number, \( N_A = 6.022 \times 10^{23} \, \text{molecules/mol} \):
\[ \text{Number of molecules} = \text{Number of moles} \times N_A = 0.002 \, \text{kmol} \times 1000 \, \text{mol/kmol} \times 6.022 \times 10^{23} \, \text{molecules/mol} \]
\[ = 0.002 \times 1000 \times 6.022 \times 10^{23} = 1.2044 \times 10^{21} \, \text{molecules} \]

### Final Answer
The number of molecules of glucose in the solution containing 100 g of water is approximately \( 1.2044 \times 10^{21} \) molecules.

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