You must make 1 L of 0.2 M acetic acid(CH3COOH). All you have available is concentrated glacial acetic acid (assay value, 98%; specific gravity, 1.05 g/mL). It will take ______ milliliters of acetic acid to make this solution. Assume a gram molecular weight of 60.05 grams.

You must make 1 L of 0.2 M acetic acid(CH3COOH). All you have available is concentrated glacial acetic acid (assay value, 98%; specific gravity, 1.05 g/mL). It will take ______ milliliters of acetic acid to make this solution. Assume a gram molecular weight of 60.05 grams.

• A. 11.7
• B. 1.029
• C. 3.42
• D. 12.01

Answer: (A)

To make a 1 L solution of 0.2 M acetic acid from concentrated glacial acetic acid, you convert the desired amount of solute into mass and then relate that mass to the mass and volume of the stock solution using its percent by weight and density. First, you need 0.2 moles in 1 liter, which is 0.2 mol. The mass of pure acetic acid required is 0.2 mol × 60.05 g/mol = 12.01 g. Since the concentrated acid is 98% by weight, you must supply that 12.01 g of solute with a solution that has 0.98 g of acetic acid per 1 g of solution, so the mass of stock solution needed is 12.01 g / 0.98 ≈ 12.26 g. The density of the stock is 1.05 g/mL, so the volume of stock solution required is 12.26 g / 1.05 g/mL ≈ 11.69 mL, which rounds to 11.7 mL. So, about 11.7 mL of glacial acetic acid is needed to prepare the 1 L of 0.2 M solution.

To make a 1 L solution of 0.2 M acetic acid from concentrated glacial acetic acid, you convert the desired amount of solute into mass and then relate that mass to the mass and volume of the stock solution using its percent by weight and density.

First, you need 0.2 moles in 1 liter, which is 0.2 mol. The mass of pure acetic acid required is 0.2 mol × 60.05 g/mol = 12.01 g. Since the concentrated acid is 98% by weight, you must supply that 12.01 g of solute with a solution that has 0.98 g of acetic acid per 1 g of solution, so the mass of stock solution needed is 12.01 g / 0.98 ≈ 12.26 g. The density of the stock is 1.05 g/mL, so the volume of stock solution required is 12.26 g / 1.05 g/mL ≈ 11.69 mL, which rounds to 11.7 mL.

So, about 11.7 mL of glacial acetic acid is needed to prepare the 1 L of 0.2 M solution.