Pure acetic acid (HC₂H₃O₂)is a liquid and is known as glacial acetic acid.Calculate the molarity of a solution prepared by dissolving 20.00 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. A)2.52 × 10³ B)42.0 C)0.0420 D)0.699 E)6.99 × 10^-4

Question

Quizplus · Accepted Answer

First, we need to calculate the number of moles of glacial acetic acid present in 20.00 mL of the solution.

Mass of glacial acetic acid = volume x density = 20.00 mL x 1.05 g/mL = 21.00 g

Molar mass of acetic acid = 60.05 g/mol

Number of moles of acetic acid = mass/molar mass = 21.00 g/60.05 g/mol = 0.3496 mol

Now, we can calculate the molarity of the solution:

Molarity = number of moles of solute/volume of solution in liters

Volume of solution = 500.0 mL = 0.5000 L

Molarity = 0.3496 mol / 0.5000 L = 0.699 M

Therefore, the correct answer is (D) 0.699.

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

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