Calculate the mole fraction of ethanoic acid ($CH_3COOH$) in an aqueous solution containing 20% of $CH_3COOH$ by mass.

0.0697

The correct answer is Option (1) → 0.0697

Given: 20% CH₃COOH by mass ⇒ In 100 g of solution:

• Mass of CH₃COOH = 20 g
• Mass of water = 80 g

Step 1: Calculate moles

• Molar mass of CH₃COOH = 60 g/mol Moles of CH₃COOH = 20 / 60 = 0.3333 mol
• Molar mass of H₂O = 18 g/mol Moles of H₂O = 80 / 18 = 4.444 mol

Step 2: Total moles

Total moles = 0.3333 + 4.444 = 4.7773 mol

Step 3: Mole fraction of CH₃COOH

X(CH₃COOH) = moles of CH₃COOH / total moles = 0.3333 / 4.7773 ≈ 0.0698 ≈ 0.0697 (on rounding)