To solve the problem of finding the vapor pressure of water for an aqueous solution of glucose at 100°C, we will follow these steps:
### Step 1: Calculate the moles of glucose
The formula for glucose is \( C_6H_{12}O_6 \). The molar mass of glucose can be calculated as follows:
- Carbon (C): \( 6 \times 12 = 72 \, \text{g/mol} \)
- Hydrogen (H): \( 12 \times 1 = 12 \, \text{g/mol} \)
- Oxygen (O): \( 6 \times 16 = 96 \, \text{g/mol} \)
Adding these together:
\[
\text{Molar mass of glucose} = 72 + 12 + 96 = 180 \, \text{g/mol}
\]
Now, we can calculate the moles of glucose:
\[
\text{Moles of glucose} = \frac{\text{mass}}{\text{molar mass}} = \frac{18 \, \text{g}}{180 \, \text{g/mol}} = 0.1 \, \text{mol}
\]
### Step 2: Calculate the moles of water
The molar mass of water \( H_2O \) is:
\[
\text{Molar mass of water} = 2 \times 1 + 16 = 18 \, \text{g/mol}
\]
Now, we can calculate the moles of water:
\[
\text{Moles of water} = \frac{178.2 \, \text{g}}{18 \, \text{g/mol}} = 9.9 \, \text{mol}
\]
### Step 3: Calculate the mole fraction of glucose
The mole fraction \( X \) of glucose is calculated using the formula:
\[
X_{\text{glucose}} = \frac{\text{moles of glucose}}{\text{moles of glucose} + \text{moles of water}} = \frac{0.1}{0.1 + 9.9} = \frac{0.1}{10} = 0.01
\]
### Step 4: Calculate the vapor pressure of the solution
The vapor pressure lowering can be calculated using Raoult's Law:
\[
\frac{P^0 - P}{P^0} = X_{\text{solute}}
\]
Where:
- \( P^0 \) is the vapor pressure of pure water at 100°C, which is approximately \( 760 \, \text{mmHg} \).
- \( P \) is the vapor pressure of the solution.
Rearranging the equation gives:
\[
P = P^0 \times (1 - X_{\text{solute}})
\]
Substituting the values:
\[
P = 760 \, \text{mmHg} \times (1 - 0.01) = 760 \, \text{mmHg} \times 0.99 = 752.4 \, \text{mmHg}
\]
### Final Answer
The vapor pressure of the water for this aqueous solution at 100°C is approximately **752.4 mmHg**.
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