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 $=\frac{1.4 \times N \times V}{w . t . \text { of } S u b s \tan c e} \times 100$

$=\frac{1.4 \times 0.5 \times 10^{-3}}{29.5 \times 10^{-3}} \times 100=23.7 \%$