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7 January, 18:34

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?

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  1. 7 January, 20:07
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    [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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