Question

The vinegar sold in the grocery stores is described as 5.0%(V/V) acetic acid?

What is the molarity of this solution (density of 100% acetic acid is 1.05 g/mL)?

What is the mass % of acetic acid in this solution (density of vinegar is 1.01 g/mL)?

What is the mole fraction of acetic acid in this solution?

What is the molality of this solution?

Please explain as you solve

Answer 1

Let volume of solution be 1 L.

Then volume of acetic acid = 5% of 1 L = 0.05 L=50 mL

Mass of acetic acid = density * volume = 1.05*50 = 52.5 gm

molar mass of acetic acid (CH3COOH) = 60 g

number of moles of acetic acid = mass / molar mass =52.5/60 = 0.875 mol

Molairty= number of moles/ volume = 0.875 mol / 1L = 0.875 M

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Mass of acetic acid has been calculated above.It is 52.5 gm

Volume of solution is 1L

Volume of Vinegar = 95% of 1 L = 0.95 L = 950 ml

density of Vinegar = 1.01 g/mL

Mass of vinegar = density * volume = 1.01*950 = 959.5 gm

mass % of acetic acid = mass of acetic acid *100/ total mass

                                        =52.5 * 100 / (52.5 + 959.5)

                                        = 5.19 %

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mass of vinegar = 959.5 gm

molar mass of vinegar = 60 gm/mol

number of moles of vinegar = mass / molar mass= 959.5/60 = 16 mol

number of moles of acetic acid = 0.875 mol

mole fraction of acetic acid = mole of acetic acid / total moles

                                               = (0.875) / (0.875+ 16)

                                                = 0.052

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Molality = number of moles of acetic acid / mass of vinegar in Kg

              =0.875 mol/ 0.9595 Kg

             = 0.91 molal