In the estimation of nitrogen by Duma’s method 1.18g of an organic compound gave 224ml of N2 at STP. The percentage of nitrogen in the compound is

% of N= Vol.of N2 gas wt.of org.compound ⋅2822400⋅100=2241.18⋅2822400⋅100=23.7%

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In the estimation of nitrogen by Duma’s method 1.18g of an organic compound gave 224ml of N₂at STP. The percentage of nitrogen in the compound is

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answer is 23.7.

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Detailed Solution

$\begin{array}{l} % of N = \frac{Vol . of N_{2} gas}{wt . of org . compound} \cdot \frac{28}{22400} \cdot 100 \\ = \frac{224}{1.18} \cdot \frac{28}{22400} \cdot 100 = 23.7 % \end{array}$

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