Pure acetic acid,CH₃CO₂H(l),has a density of 1.049 g/mL.To prepare 1.00 L of 6.00 M CH₃CO₂H(aq),one may dilute:

Question

Quizplus · Accepted Answer

To prepare a 6.00 M solution of acetic acid in 1.00 L, you need 6.00 moles of acetic acid. The molar mass of acetic acid (CH3COOH) is approximately 60.05 g/mol. Therefore, 6.00 moles of acetic acid is equivalent to 360.3 g (6.00 moles * 60.05 g/mol). Given the density of acetic acid is 1.049 g/mL, the volume of acetic acid needed is 360.3 g / 1.049 g/mL = 343.47 mL, which is closest to option B.

Pure Acetic Acid,CH3CO2H(l),has a Density of 1

Pure Acetic Acid,CH₃CO₂H(l),has a Density of 1