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The problem in the image asks to calculate the vapor pressure of a solution at 293K, given the vapor pressure of pure water and the mass of glucose dissolved.

Here’s the detailed approach to solve it:

### Problem:
The vapor pressure of water at 293K is 17.535 mm Hg. When 25 g of glucose (C₆H₁₂O₆) is dissolved in 450 g of water, calculate the vapor pressure of the solution at 293K.

### Solution:

To calculate the vapor pressure of the solution, we use **Raoult's Law**:

\[
P_{\text{solution}} = P_{\text{water}} \times X_{\text{water}}
\]

Where:
- \(P_{\text{solution}}\) is the vapor pressure of the solution.
- \(P_{\text{water}}\) is the vapor pressure of pure water (17.535 mm Hg).
- \(X_{\text{water}}\) is the mole fraction of water in the solution.

### Steps:

1. **Calculate the moles of water:**

   The molar mass of water (H₂O) is:
   \[
   M_{\text{water}} = 18 \text{ g/mol}
   \]

   Given mass of water is 450 g, so the number of moles of water is:
   \[
   n_{\text{water}} = \frac{\text{mass of water}}{\text{molar mass of water}} = \frac{450 \text{ g}}{18 \text{ g/mol}} = 25 \text{ mol}
   \]

2. **Calculate the moles of glucose:**

   The molar mass of glucose (C₆H₁₂O₆) is:
   \[
   M_{\text{glucose}} = 180 \text{ g/mol}
   \]

   Given mass of glucose is 25 g, so the number of moles of glucose is:
   \[
   n_{\text{glucose}} = \frac{25 \text{ g}}{180 \text{ g/mol}} = 0.1389 \text{ mol}
   \]

3. **Calculate the mole fraction of water:**

   The mole fraction of water is calculated as:
   \[
   X_{\text{water}} = \frac{n_{\text{water}}}{n_{\text{water}} + n_{\text{glucose}}}
   \]

   Substituting the values:
   \[
   X_{\text{water}} = \frac{25}{25 + 0.1389} \approx \frac{25}{25.1389} \approx 0.9945
   \]

4. **Calculate the vapor pressure of the solution:**

   Using Raoult’s Law:
   \[
   P_{\text{solution}} = 17.535 \times 0.9945 \approx 17.439 \text{ mm Hg}
   \]

### Final Answer:
The vapor pressure of the solution at 293K is approximately **17.44 mm Hg**.

---

Would you like more details on any of these steps, or do you have any questions? 

Here are 5 related questions to explore this topic further:
1. How does the number of solute particles affect vapor pressure?
2. What would happen to the vapor pressure if more glucose is added to the solution?
3. Can Raoult's Law be applied to volatile solutes?
4. How does temperature affect the vapor pressure of a solvent?
5. What are the assumptions made when applying Raoult’s Law?

### Tip: 
For non-volatile solutes like glucose, the vapor pressure of the solution is always lower than that of the pure solvent due to the reduction in mole fraction of the solvent.

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

Learn how to calculate the vapour pressure of a solution when 25g of glucose is dissolved in 450g of water at 293K. The solution covers Raoult's Law, mole fractions, and the reduction in vapour pressure due to the solute. Detailed for advanced chemistry students.

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