JEE MAIN - Chemistry (2022 - 29th June Morning Shift - No. 17)
Kjeldahl's method was used for the estimation of nitrogen in an organic compound. The ammonia evolved from 0.55 g of the compound neutralised 12.5 mL of 1 M H2SO4 solution. The percentage of nitrogen in the compound is _____________. (Nearest integer)
Answer
64
Explanation
Meq. of $\mathrm{H}_2 \mathrm{SO}_4=$ Meq. of $\mathrm{NH}_3$
12. $5 \times 1 \times 2=25$ meq. of $\mathrm{NH}_3$ $=25$ millimoles of $\mathrm{NH}_3$
So Millimoles of ${ }^{\prime} \mathrm{N}^{\prime}=25$
Moles of $\mathrm{N}^{\prime}=25 \times 10^{-3}$
wt. of $N=14 \times 25 \times 10^{-3}$
$$ \begin{aligned} & \% \mathrm{~N}=\frac{14 \times 25 \times 10^{-3}}{0.55} \times 100 \\\\ & =63.66 \\\\ & \approx 64 \% \end{aligned} $$
12. $5 \times 1 \times 2=25$ meq. of $\mathrm{NH}_3$ $=25$ millimoles of $\mathrm{NH}_3$
So Millimoles of ${ }^{\prime} \mathrm{N}^{\prime}=25$
Moles of $\mathrm{N}^{\prime}=25 \times 10^{-3}$
wt. of $N=14 \times 25 \times 10^{-3}$
$$ \begin{aligned} & \% \mathrm{~N}=\frac{14 \times 25 \times 10^{-3}}{0.55} \times 100 \\\\ & =63.66 \\\\ & \approx 64 \% \end{aligned} $$
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