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.

Question

Quizplus · Accepted Answer

First, we need to calculate the moles of acetic acid present in 10.00 mL of glacial acetic acid using its density:

mass of glacial acetic acid = 10.00 mL x 1.05 g/mL = 10.5 g
moles of glacial acetic acid = mass/molar mass = 10.5 g/60.05 g/mol = 0.175 mol

Now we can use this value to calculate the molarity of the solution:

Molarity = moles of solute/volume of solution in liters
Volume of solution = 500.0 mL = 0.5 L

Molarity = 0.175 mol/0.5 L = 0.350 M

Therefore, the correct answer is D.

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

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