simple volumetric work

On diluting a sample of ethanoic acid five times, it was found that 25cm3 of the diluted solution reqired 30cm3 of 0.1M sodium hydroxide solution for complete neutralisation. what was the conc. of original ethanoic acid in
1. mol dm-3 ?
2. g dm^-3 ?


1.
Consider the neutralization of the diluted acid with sodium hydroxide solution.
CH3COOH + NaOH → CH3COONa + H2O
Mole ratio CH3COOH : NaOH = 1 : 1

No. of moles of NaOH used = 0.1 x (30/1000) = 0.003 mol
No. of moles of CH3COOH used = 0.003 mol
Molarity of the diluted solution = 0.003/(25/1000) = 0.12 mol dm-3
Molarity of the original solution = 0.12 x 5 = 0.6 mol dm-3


2.
Molar mass of CH3COOH = 12x2 + 1x4 + 16x2 = 60 g mol-1
Concentration of the original solution = 0.6 x 60 = 36 g dm-3