`18g` of glucose `(C_(6)H_(12)O_(6))` is added to `178.2g` of water. The vapour pressure of water for this aqueous solution at `100^(@)C` is-

Options

A. `7.60` Torr
B. `76.00` Torr
C. `752.40` Torr
D. `759.00` Torr

Explanation

To calculate the vapor pressure of the aqueous solution at 100°C, we will use Raoult's Law, which states that the vapor pressure of a solvent in a solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of the solvent in the solution.

### Step-by-Step Solution:

**Step 1: Calculate the number of moles of glucose (C₆H₁₂O₆).**

The molar mass of glucose (C₆H₁₂O₆) can be calculated as follows:
- Carbon (C): 12.01 g/mol × 6 = 72.06 g/mol
- Hydrogen (H): 1.008 g/mol × 12 = 12.096 g/mol
- Oxygen (O): 16.00 g/mol × 6 = 96.00 g/mol

Total molar mass of glucose = 72.06 + 12.096 + 96.00 = 180.156 g/mol

Now, calculate the number of moles of glucose:
\[ \text{Moles of glucose} = \frac{\text{mass of glucose}}{\text{molar mass of glucose}} = \frac{18 \, \text{g}}{180.156 \, \text{g/mol}} \approx 0.0999 \, \text{mol} \]

**Step 2: Calculate the number of moles of water (H₂O).**

The molar mass of water (H₂O) is:
- Hydrogen (H): 1.008 g/mol × 2 = 2.016 g/mol
- Oxygen (O): 16.00 g/mol × 1 = 16.00 g/mol

Total molar mass of water = 2.016 + 16.00 = 18.016 g/mol

Now, calculate the number of moles of water:
\[ \text{Moles of water} = \frac{\text{mass of water}}{\text{molar mass of water}} = \frac{178.2 \, \text{g}}{18.016 \, \text{g/mol}} \approx 9.88 \, \text{mol} \]

**Step 3: Calculate the mole fraction of water in the solution.**

The mole fraction of water (X₁) is given by:
\[ X_1 = \frac{\text{moles of water}}{\text{moles of water} + \text{moles of glucose}} = \frac{9.88}{9.88 + 0.0999} \approx \frac{9.88}{9.9799} \approx 0.989 \]

**Step 4: Determine the vapor pressure of pure water at 100°C.**

The vapor pressure of pure water at 100°C is approximately 101.3 kPa.

**Step 5: Calculate the vapor pressure of the solution using Raoult's Law.**

Using Raoult's Law:
\[ P_{solution} = X_1 \times P_{pure \, water} \]
\[ P_{solution} = 0.989 \times 101.3 \, \text{kPa} \approx 100.0 \, \text{kPa} \]

### Final Answer:
The vapor pressure of the aqueous solution at 100°C is approximately **100.0 kPa**.