Pure acetic acid (HC2H3O2) is a liquid and is known as glacial acetic acid. Calculate the molarity of a solution prepared by dissolving 7.50 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.

Answer 1

First, we need to calculate the moles of acetic acid in 7.50 mL of glacial acetic acid.

Mass of acetic acid = volume x density = 7.50 mL x 1.05 g/mL = 7.88 g
Molar mass of acetic acid (HC2H3O2) = 60.05 g/mol

Number of moles = mass / molar mass = 7.88 g / 60.05 g/mol = 0.131 mol

Now, we can calculate the molarity of the solution.

Molarity (M) = moles of solute / liters of solution
Liters of solution = 500.0 mL / 1000 mL/L = 0.500 L

Molarity = 0.131 mol / 0.500 L = 0.262 M

Therefore, the molarity of the solution prepared by dissolving 7.50 mL of glacial acetic acid in sufficient water to give 500.0 mL of solution is 0.262 M.