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`

Options

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

Explanation

To solve the problem of finding the vapor pressure of water for the aqueous solution of glucose at 100°C, we will follow these steps:

### Step 1: Determine the number of moles of glucose and water.
1. **Calculate moles of glucose (C₆H₁₂O₆)**:
- Molar mass of glucose = 6(12) + 12(1) + 6(16) = 180 g/mol
- Moles of glucose = mass / molar mass = 18 g / 180 g/mol = 0.1 moles

2. **Calculate moles of water (H₂O)**:
- Molar mass of water = 2(1) + 16 = 18 g/mol
- Moles of water = mass / molar mass = 178.2 g / 18 g/mol = 9.9 moles

### Step 2: Calculate the mole fraction of glucose.
- Mole fraction of glucose (X_glucose) = moles of glucose / (moles of glucose + moles of water)

\[ X_{\text{glucose}} = \frac{0.1}{0.1 + 9.9} = \frac{0.1}{10} = 0.01 \]

### Step 3: Use the relative lowering of vapor pressure formula.
- The formula for relative lowering of vapor pressure is given by:

\[ \frac{P_0 - P_s}{P_0} = X_{\text{solute}} \]

Where:
- \(P_0\) = vapor pressure of pure solvent (water at 100°C = 760 torr)
- \(P_s\) = vapor pressure of the solution
- \(X_{\text{solute}} = X_{\text{glucose}} = 0.01\)

### Step 4: Substitute the values into the equation.
- Rearranging the formula gives us:

\[ P_0 - P_s = P_0 \cdot X_{\text{solute}} \]

Substituting the known values:

\[ P_0 - P_s = 760 \cdot 0.01 \]

\[ P_0 - P_s = 7.6 \text{ torr} \]

### Step 5: Solve for \(P_s\).
- Now, we can find \(P_s\):

\[ P_s = P_0 - 7.6 = 760 - 7.6 = 752.4 \text{ torr} \]

### Final Answer:
The vapor pressure of the aqueous solution at 100°C is **752.4 torr**.

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