$ = 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)}}$
$\frac{{1.4 \times 10}}{{1.4}} = 10$
[આણ્વિય દળ: ${Ag}=108, {Br}=80$ ]
(મોલર દળ $N _{2}=28 \,g\, mol ^{-1}$, $STP$ એ $N _{2}$નું મોલર કદ $: 22.4\,L$)