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$29.5\, mg$ 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 ......... $\%$
$59.0$
$47.4$
$23.7$
$29.5$
Solution
Moles of $H C l$ taken $=20 \times 0.1 \times 10^{-3}=2 \times 10^{-3}$
Moles of $H C l$ neutralised by $N a O H$ solution
$=15 \times 0.1 \times 10^{-3}=1.5 \times 10^{-3}$
Moles of $H C l$ neutralised by ammonia
${=2 \times 10^{-3}-1.5 \times 10^{-3}}$
${=0.5 \times 10^{-3}}$
$\%$ of nitrogen $=\frac{1.4 \times N \times V}{w . t . \text { of } S u b s \tan c e} \times 100$
$=\frac{1.4 \times 0.5 \times 10^{-3}}{29.5 \times 10^{-3}} \times 100=23.7 \%$
