JEE MAIN - Chemistry (2022 - 28th July Evening Shift - No. 18)
A sample of $$0.125 \mathrm{~g}$$ of an organic compound when analyzed by Duma's method yields $$22.78 \mathrm{~mL}$$ of nitrogen gas collected over $$\mathrm{KOH}$$ solution at $$280 \mathrm{~K}$$ and $$759 \mathrm{~mm}\, \mathrm{Hg}$$. The percentage of nitrogen in the given organic compound is __________. (Nearest integer)
Given :
(a) The vapour pressure of water of $$280 \mathrm{~K}$$ is $$14.2 \mathrm{~mm} \,\mathrm{Hg}$$.
(b) $$\mathrm{R}=0.082 \mathrm{~L}$$ atm $$\mathrm{K}^{-1} \mathrm{~mol}^{-1}$$
Answer
22
Explanation
$P_{\text {actual }}=759-14.2=744.8 \,\mathrm{mmHg}$
$$ \begin{aligned} \mathrm{n}_{\mathrm{N}_{2}} &=\frac{744.8 \times 22.78}{760 \times 0.0821 \times 280 \times 1000} \\\\ &=0.000971 \mathrm{~mol} \end{aligned} $$
Mass of $\mathrm{N}_{2}=0.02719 \,\mathrm{gm}$
Percentage of nitrogen
$$ =\frac{0.0271}{0.125} \times 100=21.75 \simeq 22 $$
$$ \begin{aligned} \mathrm{n}_{\mathrm{N}_{2}} &=\frac{744.8 \times 22.78}{760 \times 0.0821 \times 280 \times 1000} \\\\ &=0.000971 \mathrm{~mol} \end{aligned} $$
Mass of $\mathrm{N}_{2}=0.02719 \,\mathrm{gm}$
Percentage of nitrogen
$$ =\frac{0.0271}{0.125} \times 100=21.75 \simeq 22 $$
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