JEE MAIN - Chemistry (2022 - 25th July Morning Shift - No. 18)
While estimating the nitrogen present in an organic compound by Kjeldahl's method, the ammonia evolved from $$0.25 \mathrm{~g}$$ of the compound neutralized $$2.5 \mathrm{~mL}$$ of $$2 \,\mathrm{M} \,\mathrm{H}_{2} \mathrm{SO}_{4}$$. The percentage of nitrogen present in organic compound is ______________.
Answer
56
Explanation
$$\mathrm{NH}_{3}$$ gas is neutralized by $$2.5 \mathrm{~mL}$$ of $$2 \mathrm{M} \,\,\mathrm{H}_{2} \mathrm{SO}_{4}$$
$$\therefore$$ Moles of $$\mathrm{NH}_{3}$$ neutralized $$=2.5 \times 2 \times 2$$ millimole
$$ =10 \times 10^{-3} \text { moles } $$
$$\therefore$$ Weight of $$\mathrm{N}$$ present in the compound will be
$$ \begin{aligned} &=10 \times 10^{-3} \times 14 \\ &=0.14 \mathrm{~g} \end{aligned} $$
$$\therefore \%$$ of ' $$\mathrm{N}$$ ' in compound
$$ \begin{aligned} &=\frac{0.14}{0.25} \times 100 \\ &=56 \% \end{aligned} $$
$$\therefore$$ Moles of $$\mathrm{NH}_{3}$$ neutralized $$=2.5 \times 2 \times 2$$ millimole
$$ =10 \times 10^{-3} \text { moles } $$
$$\therefore$$ Weight of $$\mathrm{N}$$ present in the compound will be
$$ \begin{aligned} &=10 \times 10^{-3} \times 14 \\ &=0.14 \mathrm{~g} \end{aligned} $$
$$\therefore \%$$ of ' $$\mathrm{N}$$ ' in compound
$$ \begin{aligned} &=\frac{0.14}{0.25} \times 100 \\ &=56 \% \end{aligned} $$
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