Question

A 25.0 mL aliquot of vinegar was diluted to 250 mL in a volumetric
flask . Titration of 50 mL aliquot of the diluted solution required an
average of 34.88 mLof 0.096 M NaOH. Express the acidity of vinegar
in terms of the percentage w/v of acetic acid?

Answer 1

CH3COOH + NaOH ==> CH3COONa + H2O
millimoles NaOH = mL x M = 34.88 mL x 0.096 M = approximately 3.35 but you should confirm that.
The equation tells you that 1 mol CH3COOH = 1 mol NaOH; therefore,
millimoles CH3COOH in the 50 mL aliquot = approx 3.35. That's the millimoles vinegar in the 50 mL; convert that to millimols acid in the 250 mL flask. That's 3.35 millimoles x (250/50) = approximately 16.8 millimoles acid in the initial 25.0 mL sample. Again, confirm that. How many grams is that. grams = mols x molar mass = 0.0168 mols x 60 g/mol = approx 1 gram acetic acid. You had that in 25.0 mL so
g/mL = 1/25 = about 0.04 g/mL or 4% w/v, You should recalculate each step. You haven't handled the number of significant figures very well; for example you give the volume to 4 places (34.88 but only give 2 for the M of NaOH as 0.096 so I've just estimated the numbers above.