The label on a bottle of vinegar indicates that it is 3.5% acetic acid (ch3cooh). if the density of the solution is 1.90 g/ml, what is the molarity of the solution?

[CH₃COOH] = 0.61 M

Explanation:

3.5% acetic acid indicates that 3.5mL are contained in 100 mL of solution.

So we need acetic acid, density to solve this.

1.05 g/mL

Acetic acid density = acetic acid mass / acetic acid volume

1.05 g/mL = acetic acid mass / 3.5 mL

Acetic acid mass = 1.05 g/mL .  3.5mL = 3.675 grams

Let's convert this mass in moles ( mass / molar mass)

3.675 g / 60 g/m = 0.061 moles

As this moles, are contained in 100 mL of solution, molarity is (mol /L)

100 mL = 0.1L

Molarity (mol/L) = 0.061 m/ 0.1L = 0.61 M

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