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  • A sample of an organic compound was digested with <img alt="\mathrm{conc.H_{2}S

A \mathrm{0.388\, g} sample of an organic compound was digested with \mathrm{conc.H_{2}SO_{4}} and then distilled with \mathrm{NaOH}. The ammonia gas evolved was passed through \mathrm{50.0\, mL\, of\: 0.5\, N\, H_{2}SO_{4}}. The used acid required \mathrm{30.0\, mL\, of\: 0.25\, N\, NaOH} for complete neutralization. The mass percentage of nitrogen in the organic ccompound is __________.

Option: 1

63


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Let \mathrm{m} moles of \mathrm{NH_{3}} obtained be \mathrm{x}

\mathrm{\therefore }  From law of Equivalence,

\mathrm{meq.\, of \, \mathrm{H}_{2} \mathrm{SO}_{4}= meq\, \, of \, \, \mathrm{NH}_{3}+ meq \: of\: \mathrm{NaOH}}
\mathrm{\Rightarrow \: 50 \times \frac{1}{2}\: =\: x+30 \times \frac{1}{4}}
\mathrm{\Rightarrow x=17.5}

\therefore  \mathrm{ moles \: of\: \mathrm{N} \: in\: organic \: compound =17. 5 \times 10^{-3}}
\therefore   \mathrm{ mass \, of \, N=17.5 \times 10^{-3} \times 14=0. 245}
\mathrm{ \therefore \, \% \text { of } N=\frac{0. 245}{0. 388} \times 100=63.14 \%}

Hence, the answer is 63

Posted by

avinash.dongre

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