`18g` glucose `(C_(6)H_(12)O_(6))` is added to `178.2g` water. The vapour pressure of water (in torr) for this aqueous solution is:

Options

A. `759.0`
B. `7.6`
C. `76.0`
D. `752.4`

Explanation

To find the vapor pressure of water in an aqueous solution containing 18 g of glucose (C₆H₁₂O₆) added to 178.2 g of water, we will follow these steps:

### Step 1: Calculate the number of moles of water.
The molecular mass of water (H₂O) is calculated as follows:
- H: 1 g/mol × 2 = 2 g/mol
- O: 16 g/mol × 1 = 16 g/mol
- Total: 2 + 16 = 18 g/mol

Now, we can calculate the number of moles of water:
\[ \text{Number of moles of water} = \frac{\text{mass of water}}{\text{molecular mass of water}} = \frac{178.2 \text{ g}}{18 \text{ g/mol}} = 9.9 \text{ moles} \]

### Step 2: Calculate the number of moles of glucose.
The molecular mass of glucose (C₆H₁₂O₆) is calculated as follows:
- C: 12 g/mol × 6 = 72 g/mol
- H: 1 g/mol × 12 = 12 g/mol
- O: 16 g/mol × 6 = 96 g/mol
- Total: 72 + 12 + 96 = 180 g/mol

Now, we can calculate the number of moles of glucose:
\[ \text{Number of moles of glucose} = \frac{\text{mass of glucose}}{\text{molecular mass of glucose}} = \frac{18 \text{ g}}{180 \text{ g/mol}} = 0.1 \text{ moles} \]

### Step 3: Calculate the total number of moles in the solution.
\[ \text{Total moles} = \text{moles of water} + \text{moles of glucose} = 9.9 + 0.1 = 10.0 \text{ moles} \]

### Step 4: Calculate the mole fraction of water.
\[ \text{Mole fraction of water} (X_{water}) = \frac{\text{moles of water}}{\text{total moles}} = \frac{9.9}{10.0} = 0.99 \]

### Step 5: Calculate the vapor pressure of the solution.
The vapor pressure of pure water (P₀) is given as 760 torr. The vapor pressure of the solution (P) can be calculated using Raoult's Law:
\[ P = P₀ \times X_{water} = 760 \text{ torr} \times 0.99 = 752.4 \text{ torr} \]

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

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