Pure acetic acid (HC₂H₃O₂)is a liquid and is known as glacial acetic acid. Calculate the molarity of a solution prepared by dissolving 10.00 mL of glacial acetic acid at 25 °C in sufficient water to give 500.0 mL of solution. The density of glacial acetic acid at 25 °C is 1.05 g mL^-1.

Question

Quizplus · Accepted Answer

First, calculate the mass of acetic acid using its volume and density: 10.00 mL × 1.05 g/mL = 10.50 g. Then, convert the mass to moles using the molar mass of acetic acid (60.05 g/mol): 10.50 g ÷ 60.05 g/mol = 0.1748 mol. Finally, calculate the molarity by dividing the moles by the volume of the solution in liters: 0.1748 mol ÷ 0.500 L = 0.350 mol/L.

Pure Acetic Acid (HC2H3O2)is a Liquid and Is Known as Glacial

Pure Acetic Acid (HC₂H₃O₂)is a Liquid and Is Known as Glacial