1. Write the Kjeldahl relation

In Kjeldahl’s method, nitrogen in the organic compound is converted to ammonia, and the evolved $\mathrm{NH_3}$ is absorbed by excess hydrochloric acid:

$\mathrm{NH_3 + HCl \to NH_4Cl}$

So, moles of ammonia evolved = moles of HCl consumed.

2. Initial moles of HCl taken

Given:

• Volume of HCl $= 20\,\mathrm{mL} = 0.020\,\mathrm{L}$
• Molarity of HCl $= 0.1\,\mathrm{M}$

$n(\mathrm{HCl,initial}) = M \times V = 0.1 \times 0.020 = 0.002\,\mathrm{mol}$

3. Moles of excess HCl left after absorbing ammonia

This excess HCl required $15\,\mathrm{mL}$ of $0.1\,\mathrm{M}$ NaOH for neutralization.

Reaction:

$\mathrm{HCl + NaOH \to NaCl + H_2O}$

Thus, moles of excess HCl = moles of NaOH used.

• Volume of NaOH $= 15\,\mathrm{mL} = 0.015\,\mathrm{L}$
• Molarity of NaOH $= 0.1\,\mathrm{M}$

$n(\mathrm{NaOH}) = 0.1 \times 0.015 = 0.0015\,\mathrm{mol}$

So,

$n(\mathrm{HCl,excess}) = 0.0015\,\mathrm{mol}$

4. Moles of HCl consumed by ammonia

$n(\mathrm{HCl\ consumed}) = n(\mathrm{HCl,initial}) - n(\mathrm{HCl,excess})$

$= 0.002 - 0.0015 = 0.0005\,\mathrm{mol}$

Therefore,

$n(\mathrm{NH_3}) = 0.0005\,\mathrm{mol}$

Since each mole of $\mathrm{NH_3}$ contains one mole of nitrogen,

$n(\mathrm{N}) = 0.0005\,\mathrm{mol}$

5. Mass of nitrogen in the sample

$m(\mathrm{N}) = 0.0005 \times 14 = 0.007\,\mathrm{g} = 7.0\,\mathrm{mg}$

6. Calculate percentage of nitrogen

Mass of organic compound = $29.5\,\mathrm{mg}$

$\%\,\mathrm{N} = \frac{7.0}{29.5} \times 100$

$\%\,\mathrm{N} = 23.73\% \approx 23.7\%$

7. Match with options

The correct option is:

$\boxed{\text{C: }23.7}$