An aqueous solution of acetic acid has a molarity of 2 M and a density of 1.02 g/mL . The mole fraction of acetic acid in the solution is [Given: Molar mass of acetic acid = 60 g mol ⁻¹ , Molar mass of water = 18 g mol ⁻¹ ]

Step-by-step solution

Given, molarity = 2 M , which means 2 moles of acetic acid are present in 1 L ( 1000 mL ) of solution. Mass of 1000 mL of solution = Volume Density = 1000 mL 1.02 g/mL = 1020 g Mass of acetic acid (solute) = moles molar mass = 2 60 = 120 g Mass of solvent (water) = Mass of solution - Mass of solute Mass of water = 1020 g - 120 g = 900 g Number of moles of water = 900 g 18 g mol ⁻¹ = 50 mol Mole fraction of acetic acid = Moles of acetic acid Moles of acetic acid + Moles of water Mole fraction = 2 2 + 50 = 2 52 = 1 2