In the estimation of nitrogen,0.3 g of an or | vedclass

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(C) $1$. Calculate the total millimoles of $H_2SO_4$ taken: $100 \ mL 	imes 0.1 \ M = 10 \ mmol$. $2$. Calculate the millimoles of $NaOH$ used for ne

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Urea

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(C) $1$. Calculate the total millimoles of $H_2SO_4$ taken: $100 \ mL 	imes 0.1 \ M = 10 \ mmol$. $2$. Calculate the millimoles of $NaOH$ used for neutralization: $20 \ mL 	imes 0.5 \ M = 10 \ mmol$. $3$. Since $2 \ mol$ of $NaOH$ reacts with $1 \ mol$ of $H_2SO_4$,the millimoles of $H_2SO_4$ neutralized by $NaOH$ is $10 / 2 = 5 \ mmol$. $4$. Millimoles of $H_2SO_4$ reacted with $NH_3$ = Total $H_2SO_4$ - $H_2SO_4$ neutralized by $NaOH$ = $10 - 5 = 5 \ mmol$. $5$. Since $2 \ mol$ of $NH_3$ reacts with $1 \ mol$ of $H_2SO_4$,millimoles of $NH_3$ = $5 	imes 2 = 10 \ mmol$. $6$. Mass of nitrogen = $(14 	imes 10 	imes 10^{-3}) \ g = 0.14 \ g$. $7$. Percentage of nitrogen = $(0.14 / 0.3) 	imes 100 = 46.67 \%$. $8$. Calculating nitrogen percentage for options: Urea $(NH_2CONH_2)$ has $28 \ g$ nitrogen in $60 \ g$ molar mass,so $\%N = (28/60) 	imes 100 = 46.67 \%$. Thus,the compound is Urea.

In the estimation of nitrogen,$0.3 \ g$ of an organic compound was digested,and the evolved ammonia was passed into $100 \ mL$ of $0.1 \ M \ H_2SO_4$. The excess acid required $20 \ mL$ of $0.5 \ M \ NaOH$ for complete neutralization. What is the organic compound?

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