Vinegar (0.9862 g) was titrated withj 0.120 M NaOH. 26.28 cm3 were required. What is the (mass) % CH3COOH in the vinegar?

I got an answer of 19.2% but I'm not sure can anyone help

Here's my working so you know I'm not just randomly agreeing with you. I haven't explained every step this time because if your right you must know what you're doing.


0.120 x (26.28 /1000) = 0.003M

MW of CH3COOH = 60.05

mass (g) = amount (moles) x molecular weight (g/mole)


g of CH3COOH = 0.189

(0.189 g of acetic acid/0.9862g of vinegar)/100= 19.2%

= YOUR RIGHT YAY!!