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JAMB Chemistry 1997 Question 47

Question 47

Alkanoic Acids and Esters

25 cm3 of 0.02 M KOH neutralized 0.03 g of a monobasic organic acid having the general formula CnH2n + 1COOH. The molecular formula of the acid is [C = 12, H = 1, O = 16].

  1. A.HCOOH
  2. B.C2H5COOH
  3. C.CH3COOH
  4. D.C3H7COOH
Show the answer and explanation

Answer: C (CH3COOH)

Explanation

Moles of KOH = 0.025 × 0.02 = 0.0005 = moles of acid. Molar mass = 0.03/0.0005 = 60 g mol−1. 14n + 46 = 60 gives n = 1: CH3COOH.