A 0.166 g sample of an organic compound was | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$148$

Vedclass · Answer

(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 $148 \%$.

Similar Questions

Why is it essential to wash the precipitate with water before estimating it quantitatively?

$KMnO_4$ is titrated with ferrous ammonium sulphate hexahydrate in the presence of dilute $H_2SO_4$. The number of water molecules produced for $2$ molecules of $KMnO_4$ is $..........$.

The volume of $\frac{N}{10}$ $NaOH$ required to neutralize $100 \ mL$ of $\frac{N}{25}$ $HCl$ is......$mL$

Which of the following cannot give iodometric titrations?

Which substance quickly absorbs oxygen?

Vedclass Test Series

Exam Paper Generator

Online Exam Module