25 cm3 of 0.02 M KOH neutralized 0.03 g of a monobasic organic acid having the general formula CnH2n + 1 COOH. The molecular formula of the acid is? (C = 12, H = 1, O = 16)
A. HCOOH B. C2H5COOH C. CH3COOH D. C3H7COOH
Correct Answer: C
Explanation
Conc. of KOH = 39 x 0.02 = 0.78 g/dm3 25cm3 require 0.03 g of CnH2n + 1 COOH 1000CM3 → x x = (1000) / (25) x 0.03 = 1.2g CnH2n + 1COOH + KOH → CnH2n + 1 COOK + H2O from (MAVA) / (MB x VB) = (1) / (1) (MA x 1.2) / (0.02 x 0.78) = (1) / (1) MA = 0.013 Con. = Molarity x Molecular Mass Molecular Mass = (0.78) / (0.013) x 60 CnH2n + 1 COOH = 60 12n + 1 (2n + 1) + 12 + 32 + 1 = 6 14n = 60 - 46; n = (14) / (14) = 1 ∴ CH3COOH