0.28 g an organic compound in Dumas method liberated 22.4 ml of nitrogen at STP. The percentage of nitrogen in the compound is

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answer is B.

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Detailed Solution

$\begin{array}{r} % of N = \frac{V_{N_{2}} (m L) at STP}{8 \times wt.of org compound} \\ ω_{organic compound} = 0.289 \\ % of N = \frac{22.4}{8 \times 0.28} = 10 \end{array}$

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