Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.
The vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water is 17.44 mm Hg.
Explanation
This problem applies Raoult's law for solutions containing non-volatile solutes. When a non-volatile solute (glucose) is dissolved in a volatile solvent (water), the vapour pressure of the solution decreases proportionally to the mole fraction of the solvent. The formula used is P₁ = P₁⁰ × X₁, where P₁ is the vapour pressure of the solution, P₁⁰ is the vapour pressure of the pure solvent, and X₁ is the mole fraction of the solvent.
Solution Steps
Step 1: Calculate moles of glucose (solute). Molar mass of glucose (C₆H₁₂O₆) = 180 g mol⁻¹ Moles of glucose = 25 g / 180 g mol⁻¹ = 0.1389 mol
Step 2: Calculate moles of water (solvent). Molar mass of water (H₂O) = 18 g mol⁻¹ Moles of water = 450 g / 18 g mol⁻¹ = 25 mol
Step 3: Calculate mole fraction of water. Total moles = 0.1389 + 25 = 25.1389 mol Mole fraction of water (X₁) = 25 / 25.1389 = 0.9945
Step 4: Apply Raoult's law. P₁ = P₁⁰ × X₁ P₁ = 17.535 mm Hg × 0.9945 = 17.44 mm Hg