For the estimation of nitrogen, 1.4 g of an organic compound was digested by Kjeldahl method and the evolved ammonia was absorbed in 60 mL of $\frac{M}{10}$ sodium acid. The unreacted acid required 20 mL of $\frac{M}{10}$ sodium hydroxide for complete neutralization. The percentage of nitrogen in the compound is :

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10%

answer is B.

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Detailed Solution

$Meq. of acid (H_{2} {SO}_{4}) taken = \frac{1}{10} \times 2 \times 60 = 12 Meq. of acid used by NaOH = Meq. of NaOH = \frac{1}{10} \times 20 = 2 ∴ Meq. of acid used by {NH}_{3} = 12 - 2 = 10 = Meq. of {NH}_{3} % of N = \frac{Meq. of {NH}_{3} \times 1 .4}{Mass of organic compound} = \frac{10 \times 1 .4}{1 .4} = 10$