The vapour pressure of water at `293K` is `17.535mm Hg`. Calculate the vapour pressure of water at `293K` when `25g` of glucose is dissolved in `450g` of water.

Explanation

To calculate the vapor pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water, we will use Raoult's Law. Here’s the step-by-step solution:

### Step 1: Identify the given data
- Vapor pressure of pure water at 293 K, \( P_0 = 17.535 \, \text{mm Hg} \)
- Mass of glucose, \( W_2 = 25 \, \text{g} \)
- Mass of water, \( W_1 = 450 \, \text{g} \)

### Step 2: Calculate the number of moles of glucose and water
- **Molecular weight of glucose (C₆H₁₂O₆)**: \( 180 \, \text{g/mol} \)
- **Molecular weight of water (H₂O)**: \( 18 \, \text{g/mol} \)

**Number of moles of glucose (\( n_2 \))**:
\[ n_2 = \frac{W_2}{\text{Molecular weight of glucose}} = \frac{25 \, \text{g}}{180 \, \text{g/mol}} \approx 0.1389 \, \text{mol} \]

**Number of moles of water (\( n_1 \))**:
\[ n_1 = \frac{W_1}{\text{Molecular weight of water}} = \frac{450 \, \text{g}}{18 \, \text{g/mol}} = 25 \, \text{mol} \]

### Step 3: Calculate the mole fraction of water
The mole fraction of water (\( X_1 \)) is given by:
\[ X_1 = \frac{n_1}{n_1 + n_2} = \frac{25}{25 + 0.1389} \approx \frac{25}{25.1389} \approx 0.993 \]

### Step 4: Apply Raoult's Law
Raoult's Law states that the vapor pressure of the solution (\( P_s \)) is given by:
\[ P_s = P_0 \cdot X_1 \]
Substituting the values:
\[ P_s = 17.535 \, \text{mm Hg} \cdot 0.993 \approx 17.42 \, \text{mm Hg} \]

### Step 5: Final Result
The vapor pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water is approximately:
\[ \boxed{17.42 \, \text{mm Hg}} \]