Explanation

(c)
Lead unreached 0.1 M (=0.2 N)$H_{2}SO_{4} = V$ mL
20 Ml of 0.5 NaOH=$V$ mL of 0.2 N $H_{2}SO_{4}$
$20 \times 0.5 = V \times 0.2$
$V = \frac{20 \times 0.5}{0.2} = 50$ mL
Used $H_{2}SO_{4} = 100 - 50 = 50$ mL
$\%\ of\ N = \frac{1.4\ NV}{w} = \frac{1.4 \times 0.2 \times 50}{0.30}$
$= 46.67\ \%$
% of nitrogen in
(a) $CH_{3}CONH_{2} = \frac{14 \times 100}{59} = 23.73\ \%$
(b) $C_{6}H_{5}CONH_{2} = \frac{14 \times 100}{121} = 11.57\ \%$
(c) $NH_{2}CONH_{2} = \frac{28 \times 100}{60} = 46.67\ \%$
(d) $NH_{2}CSNH_{2} = \frac{28 \times 100}{76} = 36.84\ \%$