Question

18 g of glucose was dissolved in 360 g of water. The vapor pressure of the solution at 100°C will be

• A. 756.2 mm
• B. 755.2 mm
• C. 723.8 mm
• D. 76.0 mm

Hint:

Vapor pressure lowering depends on the mole fraction of the solute in the solution.

Answer 1

The Correct Option is B

Step 1: Vapor pressure lowering.
The vapor pressure lowering is given by: \[ \Delta P = P_0 \times \frac{n_{solute}}{n_{solvent} + n_{solute}} \] where: - \( P_0 \) is the vapor pressure of pure solvent (water at 100°C is 760 mmHg), - \( n_{solute} \) is the moles of solute (glucose), - \( n_{solvent} \) is the moles of solvent (water). Step 2: Substituting values.
- Moles of glucose: \( \frac{18}{180} = 0.1 \, \text{mol} \), - Moles of water: \( \frac{360}{18} = 20 \, \text{mol} \). Substitute into the vapor pressure lowering equation: \[ \Delta P = 760 \times \frac{0.1}{20 + 0.1} = 0.38 \, \text{mmHg} \] Step 3: Final vapor pressure.
The final vapor pressure is: \[ P = P_0 - \Delta P = 760 - 0.38 = 755.2 \, \text{mmHg} \] Step 4: Conclusion.
The vapor pressure of the solution is \( \boxed{755.2} \, \text{mmHg} \). The correct answer is (2) 755.2 mm.