An aqueous solution of glucose boils at `100.05^(@)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. 3.01 x `10^21`
B. 6.02 x `10^21`
C. 5.14 x `10^21`
D. 7.21 x `10^21`

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Determine the elevation in boiling point (ΔTv)
The boiling point of pure water is 100°C, and the boiling point of the glucose solution is 100.05°C.

\[ \Delta T_v = \text{Boiling point of solution} - \text{Boiling point of pure solvent} = 100.05°C - 100°C = 0.05°C \]

### Step 2: Convert ΔTv to Kelvin
Since the molal elevation constant is given in Kelvin, we can directly use the value in Celsius as it is equivalent for small temperature changes.

\[ \Delta T_v = 0.05 \text{ K} \]

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

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

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

### Step 4: Rearranging the formula to find molality (m)
We can rearrange the formula to find molality:

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

Substituting the known values:

\[ m = \frac{0.05 \text{ K}}{0.5 \text{ K kg/mol}} = 0.1 \text{ mol/kg} \]

### Step 5: Calculate the number of moles of glucose
Molality (m) is defined as the number of moles of solute per kilogram of solvent. Given that we have 100 g of water, we convert this to kg:

\[ \text{Mass of water} = 100 \text{ g} = 0.1 \text{ kg} \]

Now, we can calculate the number of moles of glucose:

\[ \text{Number of moles of glucose} = m \times \text{mass of solvent (kg)} = 0.1 \text{ mol/kg} \times 0.1 \text{ kg} = 0.01 \text{ moles} \]

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

\[ \text{Number of molecules} = \text{Number of moles} \times \text{Avogadro's number} = 0.01 \text{ moles} \times 6.022 \times 10^{23} \text{ molecules/mol} \]

Calculating this gives:

\[ \text{Number of molecules} = 6.022 \times 10^{21} \text{ molecules} \]

### Final Answer
The number of molecules of glucose in the solution containing 100 g of water is approximately \(6.022 \times 10^{21}\) molecules.

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