A concentrated acetic acid solution has a density of 1.05 g/mL at 25°C and is 17.4 M.What is the percent by mass of CH₃COOH in the solution? A)99.5% CH₃COOH by mass B)1.7% CH₃COOH by mass C)0.6% CH₃COOH by mass D)2.8% CH₃COOH by mass E)57.1% CH₃COOH by mass

Question

Quizplus · Accepted Answer

To calculate the percent by mass of acetic acid (CH₃COOH) in the solution, first find the mass of acetic acid in 1 liter (1000 mL) of solution. The molarity (M) is 17.4 moles/L, and the molar mass of acetic acid is approximately 60.05 g/mol. So, the mass of acetic acid in 1 L is 17.4 moles * 60.05 g/mol = 1044.87 g. The mass of the solution is 1.05 g/mL * 1000 mL = 1050 g. The percent by mass is (1044.87 g / 1050 g) * 100% ≈ 99.5%.

A Concentrated Acetic Acid Solution Has a Density of 1