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
  1. $16.9$
  2. $18.7$
  3. $23.2$
  4. $29.6$

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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