Question

State Raoult's law related to lowering of vapour pressure. Calculate the vapour pressure of pure water at 293 K temperature when 25 g of glucose is dissolved in 450 g of water. Vapour pressure of pure water at 293 K temperature is 1535 mmHg.

Hint:

Raoult's law applies to ideal solutions and relates vapour pressure lowering to solute concentration.

Answer 1

Raoult's Law: The relative lowering of vapour pressure is equal to the mole fraction of the solute: \[ P_1 = P_1^0 \cdot (1 - x_2), \] where \( P_1^0 \) is the vapour pressure of pure solvent, \( P_1 \) is the vapour pressure of solution, and \( x_2 \) is the mole fraction of solute. 

Calculation: - Moles of glucose: \[ \text{Moles of glucose} = \frac{25}{180} = 0.1389 \, \text{mol}. \] - Moles of water: \[ \text{Moles of water} = \frac{450}{18} = 25 \, \text{mol}. \] - Mole fraction of glucose: \[ x_2 = \frac{\text{Moles of glucose}}{\text{Total moles}} = \frac{0.1389}{25 + 0.1389} = 0.00554. \] - Vapour pressure of solution: \[ P_1 = 17.535 \cdot (1 - 0.00554) = 17.435 \, \text{mmHg}. \]