A certain volume of pure acetic acid ( CH ₃ COOH ) is dissolved in 1 kg of water. The degree of dissociation of acetic acid in the solution is 10 % . If the observed freezing point of the solution is -0.2046^ C , the volume of acetic acid added is: (Given: Density of pure acetic acid = 1.2 g mL ⁻¹ , K_f for water = 1.86 K kg mol ⁻¹ , Molar mass of acetic acid = 60 g mol ⁻¹ )

Step-by-step solution

The observed freezing point is -0.2046^ C , so the depression in freezing point T_f = 0.2046 K . For acetic acid, the degree of dissociation = 0.1 . The Van't Hoff factor i = 1 + = 1 + 0.1 = 1.1 . Using the formula for depression in freezing point: T_f = i K_f m 0.2046 = 1.1 1.86 m m = 0.2046 2.046 = 0.1 mol kg ⁻¹ Since the mass of the solvent (water) is 1 kg , the number of moles of acetic acid is 0.1 mol . Mass of acetic acid = moles molar mass = 0.1 mol 60 g mol ⁻¹ = 6.0 g . Volume of acetic acid = Mass Density