JEE MAIN - Chemistry (2025 - 3rd April Evening Shift - No. 20)
In Dumas' method for estimation of nitrogen 0.4 g of an organic compound gave 60 mL of nitrogen collected at 300 K temperature and 715 mm Hg pressure. The percentage composition of nitrogen in the compound is
(Given : Aqueous tension at $300 \mathrm{~K}=15 \mathrm{~mm} \mathrm{Hg}$ )
$15.71 \%$
$17.46 \%$
$7.85 \%$
$20.95 \%$
Explanation
$$\begin{aligned} \text { Pressure of } \mathrm{N}_2 \text { gas evolved } & =715-15 \\ & =700 \mathrm{~mm} \mathrm{Hg} \\ & =\frac{700}{760} \mathrm{~atm} . \end{aligned}$$
$\therefore$ Mole of $\mathrm{N}_2$ evolved $=\frac{\mathrm{PV}}{\mathrm{RT}}$
$$\begin{aligned} & =\frac{700 \times 60 \times 10^{-3}}{760 \times 0.0821 \times 300} \\ & =0.0022 \mathrm{~mole} \end{aligned}$$
$\therefore$ wt. of $\mathrm{N}_2$ evolved $=0.0022 \times 28=0.063 \mathrm{gm}$
$\therefore \mathrm{wt} . \%$ of nitrogen in compound
$$\begin{aligned} & =\frac{\text { wt. of nitrogen }}{\text { wt. of compound }} \times 100 \\ & =\frac{0.063}{0.4} \times 100 \\ & =15.71 \% \end{aligned}$$
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