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0.2 gm of an organic compound was analysed by Kjeldahl method. Ammonia evolved was absorbed in 60 ml of N/5 $H_{2} {SO}_{4}$ . Unused acid required 40 ml of N/10 $NaOH$ for complete neutralisation. The percentage of nitrogen in the compound is

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answer is B.

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Detailed Solution

$% N = \frac{1 .4 (\frac{1}{5} \times 60 - \frac{1}{10} \times 40)}{0 .2} = 56$

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