A sample of 0.50 , g of an organic compound w | vedclass

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(A) Given that,total mass of organic compound $= 0.50 \, g$. $60 \, mL$ of $0.5 \, M \, NaOH$ was required by residual acid for neutralisation. Since

Vedclass · Accepted Answer

$56$

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(A) Given that,total mass of organic compound $= 0.50 \, g$. $60 \, mL$ of $0.5 \, M \, NaOH$ was required by residual acid for neutralisation. Since $2 \, mol$ of $NaOH$ neutralises $1 \, mol$ of $H_{2}SO_{4}$,$60 \, mL$ of $0.5 \, M \, NaOH$ is equivalent to $30 \, mL$ of $0.5 \, M \, H_{2}SO_{4}$. Acid consumed by ammonia $= (50 - 30) \, mL = 20 \, mL$ of $0.5 \, M \, H_{2}SO_{4}$. Since $1 \, mol$ of $H_{2}SO_{4}$ reacts with $2 \, mol$ of $NH_{3}$,$20 \, mL$ of $0.5 \, M \, H_{2}SO_{4}$ is equivalent to $40 \, mL$ of $0.5 \, M \, NH_{3}$. Mass of nitrogen $= \frac{14 	imes M 	imes V}{1000} = \frac{14 	imes 0.5 	imes 40}{1000} = 0.28 \, g$. Percentage of nitrogen $= \frac{0.28}{0.50} 	imes 100 = 56 \%$.

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