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$29.5 \mathrm{mg}$ of an organic compound containing nitrogen was digested according to Kjeldahl's method and the evolved ammonia was absorbed in $20 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{HCl}$ solution. The excess of the acid required $15 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{NaOH}$ solution for complete neutralization. The percentage of nitrogen in the compound is
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Verified Answer
The correct answer is:
$23.7$
$23.7$
Moles of $\mathrm{HCl}$ reacting with ammonia
$$
\begin{aligned}
& =(\text { moles of } \mathrm{HCl} \text { absorbed })-(\text { moles of } \mathrm{NaOH} \text { solution required) } \\
& =\left(20 \times 0.1 \times 10^{-3}\right)-\left(15 \times 0.1 \times 10^{-3}\right) \\
& =\text { moles of } \mathrm{NH}_3 \text { evolved. } \\
& =\text { moles of nitrogen in organic compound }
\end{aligned}
$$
$$
\begin{aligned}
& \therefore \text { wt. of nitrogen in org. comp }=0.5 \times 10^{-3} \times 14 \\
& =7 \times 10^{-3} \mathrm{~g} \\
& \% w t=\frac{7 \times 10^{-3}}{29.5 \times 10^{-3}}=23.7 \% \\
&
\end{aligned}
$$
$$
\begin{aligned}
& =(\text { moles of } \mathrm{HCl} \text { absorbed })-(\text { moles of } \mathrm{NaOH} \text { solution required) } \\
& =\left(20 \times 0.1 \times 10^{-3}\right)-\left(15 \times 0.1 \times 10^{-3}\right) \\
& =\text { moles of } \mathrm{NH}_3 \text { evolved. } \\
& =\text { moles of nitrogen in organic compound }
\end{aligned}
$$
$$
\begin{aligned}
& \therefore \text { wt. of nitrogen in org. comp }=0.5 \times 10^{-3} \times 14 \\
& =7 \times 10^{-3} \mathrm{~g} \\
& \% w t=\frac{7 \times 10^{-3}}{29.5 \times 10^{-3}}=23.7 \% \\
&
\end{aligned}
$$
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