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    I did a vinegar titration & I'm struggling with the calculations for mass, density & percentage by mass of ethanoic acid in vinegar. I used 25cm-3 0.1M NaOH, phenolphthalein indicator & 25 cm-3 diluted with distilled water to 250 cm-3.

    The mean titre was 21.85 cm-3

    The calculations I have so far are:

    The chemical equation for this reaction is: CH3COOH(aq) + NaOH(aq) -> CH3COONa(aq) + H2O(l)
    This equation shows that 1 mol of NaOH reacts with 1 mol of CH3COOH so the molar ratio of CH3COOH to NaOH is 1:1

    Moles NaOH = (0.1 x 25)/1000 = 2.5x10-3
    Moles CH3COOH diluted =(0.1 x 250)/1000 = 2.5x10-4
    Moles CH3COOH undiluted = (0.1 x (250/10))/1000 = 2.5x10-3
    Concentration undiluted CH3COOH = 2.5x10-3 / (21.28/1000) = 0.117 moldm-3

    Mr CH3COOH = 60 g mol-1

    This is where I get stuck!

    I think I should use mass = mol x Mr but I don't know if I should use the figures from the diluted or undiluted figures, then which figures to use to calculate density. No matter which figures I use I can't come up with a figure that's even close to what google says is the density of vinegar. All this is leading to me not being to even begin to calculate percentage by mass. Help!!
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    *resolved*
 
 
 
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