A quantity of 1.16 g CH₃(CH₂)ₙCOOH was burnt in excess of air and the resultant gases (CO₂ and H₂O) were passed through excess NaOH solution. The resulting solution was divided into two equal parts. O

Question

A quantity of 1.16 g CH₃(CH₂)ₙCOOH was burnt in excess of air and the resultant gases (CO₂ and H₂O) were passed through excess NaOH solution. The resulting solution was divided into two equal parts. One part requires 50 ml of N - HCl for neutralization using phenolphthalein indicator. Another part required 80 ml of N - HCl for neutralization using methyl orange indicator. The value of n is

TestHub · Accepted Answer

CO₂ ends up as Na₂CO₃ (excess NaOH). Per part, let b = mol Na₂CO₃ and c = mol excess NaOH. Phenolphthalein: c + b = 0.050; methyl orange: c + 2b = 0.080. So b = 0.03 mol per part, i.e. total CO₂ = 0.06 mol. Acid moles = 0.06/(n+2), and 1.16 = 0.06/n+2(14n + 60) gives 1.16n + 2.32 = 0.84n + 3.6 ⇒ 0.32n = 1.28 ⇒ n = 4. Ans: 4.

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