NaOH + CH3COOH ==> H2O + CH3COONa
moles NaOH = M x L = 0.20 x 0.030 L = 0.006
You see 1 mol CH3COOH to 1 mol NaOH; therefore, 0.006 mols NaOH is due to 0.006 mols CH3COOH (vinegar--;i.e., acetic acid).
M acetic acid = moles/L = 0.006/0.010 = 0.60 M.
To do mass percent you need the mass of the 10 mL (you don't have that and no way to get it that I see). You also need the mass of the acetic acid in the sample. That is grams = mols x molar mass = ?
The titration of 10ml sample of vinegar requires 30ml of 0.20M of NaOH solution. Find molarity and mass percent concentration of the acetic acid.
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