In Dumas method, $0.3 \mathrm{~g}$ of an organic compound gave $45 \mathrm{~mL}$ of nitrogen at STP. The…
In Dumas method, $0.3 \mathrm{~g}$ of an organic compound gave $45 \mathrm{~mL}$ of nitrogen at STP. The percentage of nitrogen is
- $16.9$
- $18.7$
- $23.2$
- $29.6$
Solution
In Dumes method,
$\%$ of nitrogen $=\frac{28 \mathrm{~V} \times 100}{22400 \times W}=\frac{28 \times 45 \times 100}{22400 \times 0.3}=18.75$
Asked in: JEE-TOPICTESTS-CHEMISTRY
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