\( = 60 \times \frac{{M \times 2}}{{10}} = 12\)
Mili equivalents of \(NaOH = 20 \times \frac{M}{{10}} = 2\)
Mili equivalents of \(N{H_3} = 12 - 2 = 10\)
\(\% \,\) of nitrogen \( = \frac{{1.4 \times (N \times V)N{H_3}}}{{(Wt.\,of\,organic\,compound)}}\)
\(\frac{{1.4 \times 10}}{{1.4}} = 10\)