Consentration on vineger?

Question

vineger is a solution of ethanoic acid. a 10cm^3 portion of a certain vinerger needs 55cm^3 of 0.2mol/dm^3.... wot is the consentration o f enthnoic acid in vineger?

Answer 1

You have missed out vital information, but because no one else has had a go I will try to help.

I will assume that what was needed was 55cm^3 of 0.2 molar NaOH.

Moles NaOH = conc NaOH x vol NaOH = 0.2 mol/L x 55 mL x 1L/1,000 mL

1 dm^3 is just a confusing name for a L, and 1 cm^3 = 1 mL = 1/1,000 L

So you have 0.011 mol NaOH

CH3COOH (ethanoic acid) + NaOH >>> CH3COONa + H2O

So mols NaOH = mols ethanoic acid

concentration = amount/volume, so

Concentration of acid in vinegar = (0.011 mol ethanoic acid)/(0.010 L vinegar) = 1.1 molar or 1.1mol/dm^3

If you want the concentration in g/dm^3, multiply by the molar mass of ethanoic acid, CH3COOH

Best wishes, and good luck!