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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 0.5M sodium hydroxide solution for complete neutralization. The percentage nitrogen in the organic compound is

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answer is 47.

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Detailed Solution

Equivalents of $N H_{3}$ evolved
$= \frac{100 \times 0.1 \times 2}{1000} - \frac{20 \times 0.5}{1000} = \frac{1}{10}$
Percent of nitrogen in the unknown organic compound
$= \frac{1}{100} \times \frac{14}{0.3} \times 100 = 46.6$

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