$29.5 \mathrm{mg}$ of an organic compound containing nitrogen was digested according to Kjeldahl's method…

$29.5 \mathrm{mg}$ 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
  1. $59.0$
  2. $47.4$
  3. $23.7$
  4. $29.5$

Solution

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} $

Asked in: JEE Main 2010

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