Answer is : (18)
\(V_{N_2} \ \text{at STP} = \frac{273 \times (715- 15)\times 50}{300\times 760}\)
\(= 41.9 \ \text{ml}\)
\(m _{n_2} = \frac{41.9}{22400} \times 28 = 0.052 \ g\)
\(\%\ \text{of N} = \frac{0.052 \ g}{0.292} \times 100 = 17.94 \%
\)
\(\simeq 18 \%
\)