$$29.5$$ 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

The Kjeldahl method estimates nitrogen by measuring the $$NH_3$$ that neutralises a standard acid.

Step 1 : Moles of $$HCl$$ taken
Volume $$=20\ \text{mL}=0.020\ \text{L}$$, Molarity $$=0.1\ \text{M}$$.
$$n_{\,\text{HCl,\,initial}} = M \times V = 0.1 \times 0.020 = 0.002\ \text{mol}$$

Step 2 : Moles of $$HCl$$ left after absorbing $$NH_3$$
Excess acid needed $$15\ \text{mL}=0.015\ \text{L}$$ of $$0.1\ \text{M}$$ $$NaOH$$.
Because $$NaOH + HCl \rightarrow NaCl + H_2O$$ is 1 : 1,
$$n_{\,\text{HCl,\,excess}} = 0.1 \times 0.015 = 0.0015\ \text{mol}$$

Step 3 : Moles of $$HCl$$ that reacted with $$NH_3$$
$$n_{\,\text{HCl,\,reacted}} = 0.002 - 0.0015 = 0.0005\ \text{mol}$$

Step 4 : Moles of $$NH_3$$ (and hence of nitrogen)
Since $$NH_3$$ reacts with $$HCl$$ in 1 : 1 ratio,
$$n_{\,NH_3}= n_{\,N} = 0.0005\ \text{mol}$$

Step 5 : Mass of nitrogen
Molar mass of N $$=14\ \text{g mol}^{-1}$$.
$$m_{\,N}= 0.0005 \times 14 = 0.007\ \text{g}=7\ \text{mg}$$

Step 6 : Percentage of nitrogen in the sample
Mass of sample $$=29.5\ \text{mg}$$.
$$\%N = \frac{7}{29.5}\times100 = 23.7\%$$ (to one decimal place)

Option C which is: 23.7