An aqueous solution of acetic acid ( CH ₃ COOH ) has a molarity of 2.0 M and a density of 1.02 g mL ⁻¹ . The mole fraction of acetic acid in the solution is closest to: [Given: Molar mass of C, H and O are 12 , 1 and 16 g mol ⁻¹ respectively.]

Step-by-step solution

Consider 1 L ( 1000 mL) of the solution. Moles of acetic acid in 1 L solution = 2.0 mol Molar mass of acetic acid ( CH ₃ COOH ) = 12 + 3 + 12 + 16 + 16 + 1 = 60 g mol ⁻¹ Mass of acetic acid = 2.0 60 = 120 g Mass of 1 L solution = Volume Density = 1000 1.02 = 1020 g Mass of water (solvent) = Mass of solution - Mass of solute Mass of water = 1020 - 120 = 900 g Moles of water = 900 18 = 50 mol Total moles in solution = n_ acetic acid + n_ water = 2.0 + 50 = 52 mol Mole fraction of acetic acid = n_ acetic acid Total mo