From 0.50 g of an organic compound,ammonia i | vedclass

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(B) $1$. Total millimoles of $H_2SO_4$ taken = $50 \ mL 	imes 0.5 \ M = 25 \ mmol$. $2$. Millimoles of $NaOH$ used for titration of excess $H_2SO_4$

Vedclass · Accepted Answer

$56$

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(B) $1$. Total millimoles of $H_2SO_4$ taken = $50 \ mL 	imes 0.5 \ M = 25 \ mmol$. $2$. Millimoles of $NaOH$ used for titration of excess $H_2SO_4$ = $60 \ mL 	imes 0.5 \ M = 30 \ mmol$. $3$. Since $2 \ mol$ of $NaOH$ react with $1 \ mol$ of $H_2SO_4$,millimoles of excess $H_2SO_4$ = $30 / 2 = 15 \ mmol$. $4$. Millimoles of $H_2SO_4$ reacted with $NH_3$ = $25 - 15 = 10 \ mmol$. $5$. Since $2 \ mol$ of $NH_3$ react with $1 \ mol$ of $H_2SO_4$,millimoles of $NH_3$ = $10 	imes 2 = 20 \ mmol$. $6$. Mass of nitrogen = $20 	imes 10^{-3} \ mol 	imes 14 \ g/mol = 0.28 \ g$. $7$. Percentage of nitrogen = $(0.28 / 0.50) 	imes 100 = 56 \%$.

From $0.50 \ g$ of an organic compound,ammonia is evolved and absorbed in $50 \ mL$ of $0.5 \ M \ H_2SO_4$. In the titration of the excess $H_2SO_4$,$60 \ mL$ of $0.5 \ M \ NaOH$ is used. Calculate the percentage of nitrogen in the compound. (in $\%$)

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