25 cm3 of 0 02 M KOH neutralized 0 03 g of a mo... - JAMB Chemistry 1997 Question

Conc. of KOH = 39 x 0.02 = 0.78 g/dm³

25cm³ require 0.03 g of C_nH_{2n + 1} COOH 1000CM³ → x x = (1000) / (25) x 0.03 = 1.2g

C_nH_{2n + 1}COOH + KOH → C_nH_{2n + 1} COOK + H₂O

from (M_AV_A) / (M_B x V_B) = (1) / (1) (M_A x 1.2) / (0.02 x 0.78) = (1) / (1)

M_A = 0.013

Con. = Molarity x Molecular Mass

Molecular Mass = (0.78) / (0.013) x 60

C_nH_{2n + 1} COOH = 60

12n + 1 (2n + 1) + 12 + 32 + 1 = 6

14n = 60 - 46; n = (14) / (14) = 1

∴ CH₃COOH