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I'm suppose to calculate the weight percent per volume (grams pre 100cm3) of ethanoic acid in a commercial sample of vineger from the following: (a) 0.4321g of pure momoprotic acid( formula weight 204.2 amu) required 23.45cm3 of naoh for neutralizalion.

  • chem - ,

    You need to know the molarity of the NaOH.
    The 204.2 sounds like you standardized the NaOH by using 0.4321 g of pure potassium hydrogen phthalate.
    moles KHP = 0.4321/204.22 = ?
    moles NaOH = moles KHP
    M NaOH = moles NaOH/0.02435

    You can work out the M NaOH.
    Now all we need to know is the volume of the vinegar sample you used.

    M NaOH x L NaOH = moles NaOH used for the sample.
    moles NaOH for the sample = moles acid in the vinegar.
    grams acid in the vinegar = moles x molar mass (60).
    All of that gives you grams acetic acid/volume used. Convert that to g/100 cc and you have it.
    Post your work if you get stuck and PLEASE type ALL of the problem.

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