A 0.166 g sample of an organic compound was | vedclass

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(D) Step $1$: Calculate the milliequivalents $(m_{eq})$ of $NaOH$ used for back titration: $m_{eq} = 30.0 \ mL 	imes 0.25 \ N = 7.5 \ m_{eq}$.Step $2

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$148$

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(D) Step $1$: Calculate the milliequivalents $(m_{eq})$ of $NaOH$ used for back titration: $m_{eq} = 30.0 \ mL 	imes 0.25 \ N = 7.5 \ m_{eq}$.Step $2$: Calculate the total $m_{eq}$ of $H_2SO_4$ taken: $m_{eq} = 50.0 \ mL 	imes 0.5 \ N = 25.0 \ m_{eq}$.Step $3$: Calculate the $m_{eq}$ of $H_2SO_4$ reacted with $NH_3$: $m_{eq} = 25.0 - 7.5 = 17.5 \ m_{eq}$.Step $4$: Since $m_{eq}$ of $NH_3 = m_{eq}$ of $H_2SO_4$ reacted,$m_{eq}$ of $NH_3 = 17.5 \ m_{eq}$.Step $5$: Calculate the mass of nitrogen: $	ext{Mass of } N = 17.5 	imes 10^{-3} 	imes 14 \ g = 0.245 \ g$.Step $6$: Calculate the percentage of nitrogen: $\% N = (0.245 / 0.166) 	imes 100 \approx 147.59 \%$. Rounding to the nearest integer,we get $148 \%$.

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