JEE MAIN - Chemistry (2010 - No. 17)
29.5 mg of an organic compound containing nitrogen was digested according to Kjeldahl’s method
and the evolved ammonia was absorbed in 20 mL of 0.1 M HCl solution. The excess of the acid
required 15 mL of 0.1 M NaOH solution for complete neutralization. The percentage of nitrogen in
the compound is
59.0
47.4
23.7
29.5
설명
Moles of $$HCl$$ taken $$ = 20 \times 0.1 \times {10^{ - 3}} = 2 \times {10^{ - 3}}$$
Moles of $$HCl$$ neutralised by $$NaOH$$ solution
$$ = 15 \times 0.1 \times {10^{ - 3}} = 1.5 \times {10^{ - 3}}$$
Moles of $$HCl$$ neutralised by ammonia
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2 \times {10^{ - 3}} - 1.5 \times {10^{ - 3}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.5 \times {10^{ - 3}}$$
$$\% $$ of nitrogen $$ = {{1.4 \times N \times V} \over {w.t.\,\,of\,\,Subs\tan ce}} \times 100$$
$$ = {{1.4 \times 0.5 \times {{10}^{ - 3}}} \over {29.5 \times {{10}^{ - 3}}}} \times 100 = 23.7\% $$
Moles of $$HCl$$ neutralised by $$NaOH$$ solution
$$ = 15 \times 0.1 \times {10^{ - 3}} = 1.5 \times {10^{ - 3}}$$
Moles of $$HCl$$ neutralised by ammonia
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2 \times {10^{ - 3}} - 1.5 \times {10^{ - 3}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.5 \times {10^{ - 3}}$$
$$\% $$ of nitrogen $$ = {{1.4 \times N \times V} \over {w.t.\,\,of\,\,Subs\tan ce}} \times 100$$
$$ = {{1.4 \times 0.5 \times {{10}^{ - 3}}} \over {29.5 \times {{10}^{ - 3}}}} \times 100 = 23.7\% $$