(c)
Weight of organic compound =29.5mg
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ NH_{3} + HCl \longrightarrow Na_{4}Cl$
$HCl\ \ \ \ \ \ \ \ \ + \ \ \ \ NaOH\ \ \ \ \ \ \ \ \ \ \longrightarrow NaCl + H_{2}O$
(remaining) $15 \times 0.1\ M$
=1.5 mmol
Total millimole of HCl=2
Millimole used by $NH_{3} = 2 - 1.5 = 0.5$
Weight of $NH_{3}$=0.5 ×17mg=8.5mg
Weight of nitrogen $= \frac{14}{17} \times 8.5mg = 7mg$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \%\ Nitrogen = \frac{7}{29.5} \times 1100 = 23.7\%$