Explanation

(c)
Let unreacted 0.1M$( = 0.2N)H_{2}SO_{4} = V^{'}mL$
∴ 20 mL of 0.5M NaOH
=V’mL of 02 $N\ H_{2}SO_{4}$
$\therefore 20 \times 0.5 = V^{'} \times 0.2$
$\therefore{\ \ \ \ \ \ \ \ \ \ \ \ V}^{'} = 50mL$
Used $H_{2}SO_{4} = 100 - 50 = 50mL$
$\ \ \% Nitrogen = \frac{1.4\ NV}{w}$
where, N=normality of $H_{2}SO_{4}$
V=volume of $H_{2}SO_{4}$ used
$\therefore\%\ nitrogen = \frac{1.4 \times 0.2 \times 50}{0.30}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 46.67\%$
% of nitrogen in
(a) $CH_{3}CONH_{2} = \frac{14 \times 100}{59} = 23.73\%$
(b) $C_{6}H_{5}CONH_{2} = \frac{14 \times 100}{122} = 11.48\%$
(c)$NH_{2}CONH_{2} = \frac{28 \times 100}{60} = 46.67\%$
(d) $NH_{2}CSNH_{2} = \frac{28 \times 100}{76} = 36.84\%$
Therefore, the organic compound is urea.