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In Dumas method, $0.3 \mathrm{~g}$ of an organic compound gave $45 \mathrm{~mL}$ of nitrogen at STP. The percentage of nitrogen is
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The correct answer is:
$18.7$
In Dumes method,
$\%$ of nitrogen $=\frac{28 \mathrm{~V} \times 100}{22400 \times W}=\frac{28 \times 45 \times 100}{22400 \times 0.3}=18.75$
$\%$ of nitrogen $=\frac{28 \mathrm{~V} \times 100}{22400 \times W}=\frac{28 \times 45 \times 100}{22400 \times 0.3}=18.75$
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