Titration + dilution calculation~HELP

Consider the titration between 50 mL diluted vinegar and NaOH.
CH3COOH + NaOH → CH3COONa + H2O
Mole ratio CH3COOH : NaOH = 1 : 1
No. of moles of NaOH = MV = 0.096 x (34.88/1000) = 0.003348 mol
No. of moles of CH3COOH in 50 mL of diluted vinegar = 0.003348 mol

No. of moles of CH3COOH in 250 mL of diluted vinegar
= 0.003348 x (250/50)
= 0.01674 mol

No. of moles of CH3COOH in 25 mL of original vinegar = 0.01674 mol
Molar mass of CH3COOH = 12x2 + 1x4 + 16x2 = 60 g mol-1

Mass of CH3COOH in 25 mL of original vinegar
= mol x (molar mass)
= 0.01674 x 60
= 1.004 g

Acidity of the original vinegar
= (1.004/25) x 100%
= 4.016%

2008-06-15 11:09:44 補充：
The 「acidity of the original vinegar」should be the「concentration of the original vinegar」instead.

H+ + NaOH --> Na+ + H2O (1)


NO. OF MOLE OF NaOH: 34.88 /1000 X 0.096=0.00334848


BY (1) No.of mole of vinegar = no .of mole of NaOH =0.00334848


No. of mole = Volume X acidity

Acidity : 0.00334848/ (50/1000) =0.0669696


THE ORIGIN (NOT YEY DILUTE) VINEGAE ACIDIRY IS"


0.0669696/ (25/250) = 0.669696
=0.67 M