The ammonia evolved from the treatment of 0.30 g of an organic compound for the estimation of nitrogen was passed in 100 mL of 0.1 M sulphuric acid. The excess of acid required 20 mL of 0.5 M sodium hydroxide solution for complete neutralization. The percentage nitrogen in the organic compound is(Report the answer in the nearest integer)

Question

The ammonia evolved from the treatment of 0.30 g of an organic compound for the estimation of nitrogen was passed in 100 mL of 0.1 M sulphuric acid. The excess of acid required 20 mL of 0.5 M sodium hydroxide solution for complete neutralization. The percentage nitrogen in the organic compound is(Report the answer in the nearest integer) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Equivalents of NH3 evolved = (1 0 0 × 0 ⋅ 1 × 2/1 0 0 0) - (2 0 × 0 ⋅ 5/1 0 0 0) = (1/1 0 0) Percent of nitrogen in the unknown organic compound = (1/1 0 0) × (1 4/0. 3) × 1 0 0 = 4 6 . 67 ≅ 47

Equivalents of NH3​ evolved =1000100×0⋅1×2​−100020×0⋅5​=1001​ Percent of nitrogen in the unknown organic compound =1001​×0.314​×100=46.67 ≅ 47

Equivalents of $N H_{3}$ evolved
$= \frac{100 \times 0 \cdot 1 \times 2}{1000} - \frac{20 \times 0 \cdot 5}{1000} = \frac{1}{100}$
Percent of nitrogen in the unknown organic compound
$= \frac{1}{100} \times \frac{14}{0.3} \times 100 = 46.67$ $≅$ $47$