A solution is prepared by dissolving 12.0 g of a weak monoprotic acid HA in water to make 1000 mL of solution. At 300 K , the osmotic pressure of this solution is measured to be 5.478 10⁵ Pa . The percentage dissociation of the weak acid is ________. (Given: Molar mass of HA = 60 g mol ⁻¹ , R = 0.083 L bar K ⁻¹ mol ⁻¹ , 1 bar = 10⁵ Pa )

Step-by-step solution

Moles of HA = 12.0 60 = 0.2 mol Molarity (C) = 0.2 mol 1 L = 0.2 M Osmotic pressure ( ) in bar = 5.478 10⁵ Pa 10⁵ Pa bar ⁻¹ = 5.478 bar Using the formula for osmotic pressure, = iCRT : 5.478 = i 0.2 0.083 300 5.478 = i 4.98 i = 5.478 4.98 = 1.1 For a weak monoprotic acid, the van't Hoff factor i = 1 + , where is the degree of dissociation. 1.1 = 1 + = 0.1 Percentage dissociation = 100 = 0.1 100 = 10 Answer: 10