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0.2gm of organic compound is subjected to kjeldahl’s method, the ammonia evolved was absorbed in 60 ml of 0.1M $H_{2} {SO}_{4}$ , excess of acid (unused) required 40ml of 0.1 M NaOH for complete neutralisation . Percentage of Nitrogen in the compound is

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46 %

56%

28%

17%

answer is B.

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Detailed Solution

$N_{a} V_{a}$ = 12, $N_{b} V_{b} = 4$

No. of milli equivalents of acid neutralized by $N H_{3} = 12 - 4 = 8$ % of nitrogen = $\frac{1.4 \times 8}{0.2} = 56$

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