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In Duma's method for estimation of nitrogen, 0.25 g of an organic compound gave 40 mL of nitrogen collected at 300 K temperature and 725 mm pressure. If the aqueous tension at 300 K is 25 mm, the percentage of nitrogen in the compound is:
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The correct answer is:
16.76
\(\begin{aligned}
& n_{N_2}=\frac{P V}{R T} \\
& =\frac{(725-25) 40 \times 10^{-3}}{760 \times 0.082 \times 300} \\
& n_{N_2}=\frac{28}{760 \times 24.6} \\
& w_{N_2}=\frac{28 \times 28}{760 \times 24.6} \mathrm{~g} \text { of } N_2 \\
& \text {Weight of } \mathrm{N}_2=\frac{28 \times 28}{760 \times 24.6} \mathrm{gm} \\
& \% N_2=\frac{28 \times 28}{\frac{1}{660 \times 24.6}} \times \frac{1}{0.25} \times 100 \\
& =16.77 \%
\end{aligned}\)
& n_{N_2}=\frac{P V}{R T} \\
& =\frac{(725-25) 40 \times 10^{-3}}{760 \times 0.082 \times 300} \\
& n_{N_2}=\frac{28}{760 \times 24.6} \\
& w_{N_2}=\frac{28 \times 28}{760 \times 24.6} \mathrm{~g} \text { of } N_2 \\
& \text {Weight of } \mathrm{N}_2=\frac{28 \times 28}{760 \times 24.6} \mathrm{gm} \\
& \% N_2=\frac{28 \times 28}{\frac{1}{660 \times 24.6}} \times \frac{1}{0.25} \times 100 \\
& =16.77 \%
\end{aligned}\)
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