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(C) $1$. Total millimoles of $H_2SO_4$ taken = $100 \ mL 	imes 0.1 \ M = 10 \ mmol$. $2$. Millimoles of $H_2SO_4$ neutralized by $NaOH$ = $\frac{1}{2

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Urea

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(C) $1$. Total millimoles of $H_2SO_4$ taken = $100 \ mL 	imes 0.1 \ M = 10 \ mmol$. $2$. Millimoles of $H_2SO_4$ neutralized by $NaOH$ = $\frac{1}{2} 	imes (20 \ mL 	imes 0.5 \ M) = 5 \ mmol$ (since $H_2SO_4 + 2NaOH ightarrow Na_2SO_4 + 2H_2O$). $3$. Millimoles of $H_2SO_4$ consumed by $NH_3$ = $10 - 5 = 5 \ mmol$. $4$. Since $2NH_3 + H_2SO_4 ightarrow (NH_4)_2SO_4$,millimoles of $NH_3$ = $2 	imes 5 = 10 \ mmol$. $5$. Mass of nitrogen = $\frac{10 	imes 10^{-3} 	imes 14}{0.30} 	imes 100 = 46.67\%$. $6$. For Urea $(NH_2CONH_2)$,molar mass = $60 \ g/mol$. Nitrogen content = $\frac{28}{60} 	imes 100 = 46.67\%$. $7$. Thus,the compound is Urea.

When nitrogen gas is passed through $100 \ mL$ of $0.1 \ M$ $H_2SO_4$ containing $0.30 \ g$ of an organic compound,the ammonia evolved is neutralized. To neutralize the excess acid,$20 \ mL$ of $0.5 \ M$ $NaOH$ is required. Identify the organic compound.

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