Calculate acetic acid needed to make a 35 mM solution in 1 L. Note: 35 mM = 0.035 M = 0.035 mol/L.
1. Calculate how many moles you need for 1L.
2. Calculate how many grams of acetic acid that is.
3. Use the density of the glacial acetic acid, 1.05 g/mL, to calculate how many mL you
will need to use. Show your instructor or TA this calculation in your notebook to make
sure you have the correct answer before going forward.

1. 0.035 mole

2. 2.1 g

3. 2 mL

Explanation:

From the question given above, the following data were obtained:

Molarity of acetic acid = 0.035 mol/L.

1. Determination of the number of mole of acetic acid needed.

Molarity of acetic acid = 0.035 mol/L.

Volume of solution = 1 L

Mole of acetic acid =?

Molarity = mole /Volume

0.035 = mole / 1

Mole of acetic acid = 0.035 mole

2. Determination of the mass of acetic acid, CH₃COOH.

Mole of CH₃COOH = 0.035 mole

Molar mass of CH₃COOH = 12 + (3×1) + 12 + 16 + 16 + 1

= 12 + 3 + 12 + 16 + 16 + 1

= 60 g/mol

Mass of CH₃COOH =?

Mole = mass / Molar mass

0.035 = Mass of CH₃COOH / 60

Cross multiply

Mass of CH₃COOH = 0.035 × 60

Mass of CH₃COOH = 2.1 g

3. Determination of the volume.

Mass = 2.1 g

Density = 1.05 g/mL

Volume =?

Density = mass / volume

1.05 = 2.1 / volume

Cross multiply

1.05 × volume = 2.1

Divide both side by 1.05

Volume = 2.1 / 1.05

Volume = 2 mL

law of conservation of matter