An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
The molar mass of the solute is 41.35 g mol⁻¹.
Explanation
This question applies Raoult's law for solutions containing non-volatile solutes. At the normal boiling point of water, the pure solvent has a vapour pressure of 1.013 bar. The solution's vapour pressure is lower (1.004 bar) due to the presence of the non-volatile solute. Using the relative lowering of vapour pressure formula and knowing that a 2% solution contains 2 g solute in 98 g of solvent, we can calculate the molar mass.
Solution Steps
Step 1: Identify given values: 2% solution means 2 g solute in 100 g solution, so mass of solute (w₂) = 2 g and mass of solvent (w₁) = 98 g. Vapour pressure of pure solvent (p₁⁰) = 1.013 bar (at normal boiling point). Vapour pressure of solution (p) = 1.004 bar. Molar mass of water (M₁) = 18 g mol⁻¹.
Step 2: Apply Raoult's law formula for relative lowering of vapour pressure: (p₁⁰ - p)/p₁⁰ = (w₂ × M₁)/(M₂ × w₁)
Step 3: Substitute values: (1.013 - 1.004)/1.013 = ()/(M₂ × 98)
Step 4: Solve: 0.009/1.013 = 36/(98 × M₂)
Step 5: Calculate: 0.00889 = 36/(98 × M₂), therefore M₂ = 36/() = 41.35 g mol⁻¹