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Using Duma's method for estimating nitrogen $\mathrm{l} \mathrm{g}$ of an organic compound $X$ gave $82 \mathrm{~mL}$ of nitrogen, collected at $27^{\circ} \mathrm{C}$ and $750 \mathrm{~mm} \mathrm{Hg}$ pressure. If the aqueous tension at $27^{\circ} \mathrm{C}$ is $30 \mathrm{~mm} \mathrm{Hg}$ pressure, then the percentage of nitrogen in the given compound $X$ is
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The correct answer is:
88.36 %
For 750 mm pressure, subtract aqueous tension 30 mm to obtain pressure of nitrogen.
p = 750 − 30 = 720mm of Hg
Now, volume of nitrogen at STP $=\frac{V \times p \times 273}{T \times 760}$
$$
\begin{aligned}
& =\frac{82 \times 720 \times 273}{300 \times 760} \\
& =\frac{1.6117 \times 10^7}{228000}=\frac{1.6117 \times 10^4}{228} \\
& =70.68 \mathrm{~mL}
\end{aligned}
$$
$\therefore$ Percentage of nitrogen
$$
\begin{aligned}
& =\frac{\text { vol. of } \mathrm{N}_2 \text { at STP }}{\text { wt. of organic compound }} \times \frac{28}{22400} \times 100 \\
& =\frac{70.68}{1} \times \frac{28}{22400} \times 100=88.36 \%
\end{aligned}
$$
The percentage composition of nitrogen in the compound would be $88.36 \%$.
p = 750 − 30 = 720mm of Hg
Now, volume of nitrogen at STP $=\frac{V \times p \times 273}{T \times 760}$
$$
\begin{aligned}
& =\frac{82 \times 720 \times 273}{300 \times 760} \\
& =\frac{1.6117 \times 10^7}{228000}=\frac{1.6117 \times 10^4}{228} \\
& =70.68 \mathrm{~mL}
\end{aligned}
$$
$\therefore$ Percentage of nitrogen
$$
\begin{aligned}
& =\frac{\text { vol. of } \mathrm{N}_2 \text { at STP }}{\text { wt. of organic compound }} \times \frac{28}{22400} \times 100 \\
& =\frac{70.68}{1} \times \frac{28}{22400} \times 100=88.36 \%
\end{aligned}
$$
The percentage composition of nitrogen in the compound would be $88.36 \%$.
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