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a 10.0mL vinegar sample was completely neutralized by 22.5mL 0.2M NaOH solution. calculate molarity and % of acetic acid in vinegar.
please help

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confused

  1. I assume we call the density of the vinegar 1.00 g/mL so the 10 mL has a mass of 10.0 grams.

    moles NaOH used = M x L = ??
    moles of acetic acid in the 10.0 mL = ?? (same as moles NaOH).
    M acetic acid = moles/L = ??/0.1 = xx M.

    Is that percent w/w or w/v. I assume w/v.
    You have M and that is moles/L.
    moles x molar mass = grams so that is g/L
    You want g/100 mL = ??

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    DrBob222

  2. i'm still lost can you show me step by step please

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    confused

  4. I worked the problem for you. All you need to do is substitute a few numbers. Post your work and tell me what you don't understand at the next step and I'll try to help you through BUT I am not going to do the problem for you.

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    DrBob222

  5. is the answer M= .08M
    and is the % of mass= 44%

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    Shelly

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