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21 February, 22:16

Pure acetic acid (hc2h3o2) is a liquid and is known as glacial acetic acid. calculate the molarity of a solution prepared by dissolving 50.00 ml of glacial acetic acid at 25 °c in sufficient water to give 500.0 ml of solution. the density of glacial acetic acid at 25 °c is 1.05 g/ml

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  1. 21 February, 22:29
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    In order to find the molarity of the solution, we first require the moles of acetic acid added. For this, we need the mass which is:

    Mass = volume * density

    Mass = 50 * 1.05

    Mass = 52.5 grams

    Moles = mass / molecular weight

    Moles = 52.5 / 60.05

    Moles = 0.874 mol

    Next, we know that the molarity of a solution is:

    Molarity = moles / liter

    Molarity = 0.874 / 0.5

    Molarity = 1.75 M
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