Calculate the molar concentration of acetic acid (CH3COOH) in a 5.00-mL sample of vinegar (density is 1.00 g/mL) if it is titrated with 25.00 mL of NaOH.

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  1. Determine moles acetic acid. moles NaOH = M NaOH x L NaOH.
    Moles acetic acid = moles NaOH.

    M = # moles/L of solution.

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  2. Thank you very much.

    The molarity of the NaOH solution is 0.162.

    So I did the following:

    25.00 mL NaOH*(1 L / 1000 mL)*(0.160 mol NaOH / 1 L NaOH)*(1 mol CH3COOH / 1 mol NaOH) = 0.004 mol CH3COOH

    (0.004 mol CH3COOH / 5.00 mL) * (1000 mL / 1 L) = 0.8 M CH3COOH

    Now I'm supposed to determine the mass percent of the acetic acid in the vinegar sample. I assume this is why I was given the density...

    how do I use the concentration of acetic acid and the density of vinegar to determine the mass percent?

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  3. typo...
    "The molarity of the NaOH solution is 0.160." The number in the calculations is correct.

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  4. moles CH3COOH x molar mass acetic acid = grams CH3COOH.
    (g CH3COOH/100 g soln)*100 = mass percent. which you can do one of two ways.
    I have 0.24 g/5 g so how much is in 100?
    That will be 0.24 x 100/5 = 0.24 x 20 = 4.8 which is 4.8%.
    The other way, I think, is simpler:
    (0.24/5)*100 =

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