A volume of 31.7 cm3 of a 0.1-normal NaOH is required for the neutralization of 0.19 g of an organic — Organic Chemistry Chemistry Question

a) The supposed acid may be: HA, H₂A, H₃A, etc.
n(NaOH) = c V = 0.1 mol dm^-3 × 0.0317 dm₃ = 3.17 × 10-3 mol
v × n(acid) = 3.17 × 10-3 mol
where v = 1, 2, 3,......
n(acid) = m(acid) / M(acid)
3.17 × 10-3 mol = v × 0.19 g / M(acid)
M(acid) = v × 60 g mol^-1 (1)

b) From the ideal gas law we can obtain:
ρ1 / ρ2 = M₁ / M₂
M(H₂) = 2 g mol^-1
M(acid) = 30 × 2 = 60 g mol^-1
By comparing with (1): v = 1
The acid concerned is a monoprotic acid and its molar mass is 60 g mol^-1.
The acid is acetic acid: CH3−COOH