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0.66g of a compound gave 112ml of nitrogen measured at STP in the Dumas method. The percentage of nitrogen in the compound is

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41.5

42.4

21.2

answer is D.

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Detailed Solution

$% of N = \frac{28}{22400} \times \frac{V of N_{2} in ml}{W_{O . C}} \times 100$

$% of N = \frac{28}{22400} \times \frac{112}{0.66} \times 100$

$% of N = \frac{3136}{14784} \times 100$

% of N = 21.21

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