Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
Vapour pressure of solution = 23.39 mm Hg; Relative lowering = 0.0173
Explanation
This problem applies Raoult's law for a non-volatile solute (urea) dissolved in a volatile solvent (water). The relative lowering of vapour pressure equals the mole fraction of the solute. The molar mass of urea (NH₂CONH₂) is calculated as 60 g/mol, and water as 18 g/mol.
Solution Steps
Step 1: Calculate molar masses Molar mass of urea (NH₂CONH₂) = 2(14) + 4(1) + 12 + 16 = 60 g mol⁻¹ Molar mass of water (H₂O) = 2(1) + 16 = 18 g mol⁻¹
Step 2: Calculate moles of solute and solvent Moles of urea (n₂) = 50/60 = 0.833 mol Moles of water (n₁) = 850/18 = 47.22 mol
Step 3: Calculate mole fraction of solvent Mole fraction of water (X₁) = n₁/(n₁ + n₂) = 47.22/(47.22 + 0.833) = 47.22/48.053 = 0.9827
Step 4: Calculate vapour pressure of solution Using P = P₀ × X₁ P = = 23.39 mm Hg
Step 5: Calculate relative lowering Relative lowering = (P₀ - P)/P₀ = X₂ = n₂/(n₁ + n₂) = 0.833/48.053 = 0.0173