What should be the nomality of sodium hydroxide solution if exactly 25.00 mL are to be used to titrate?

25.00 mL of a vinegar sample containing 4.80% acetic acid?

Answer:
4.80 % by mass means
4.80 g acetic acid in 100 g of solution
If d = 1
we have 4.80 g of acetic acid ( MM = 60 g/mol) ==> 0.08 moles CH3COOH in 100 mL = 0.1L
[CH3COOH ] = 0.08/0.1 = 0.8 M
Moles CH3COOH in 25 mL = 25 x 0.8 /1000 = 0.02
Moles NaOH = moles CH3COOH at the equivalent point
[NaOH ] = 0.02 / 0.025 L = 0.8 M or 0.8 N
For NaOH M = N

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