The ammonia evolved from the treatment of 0.3 | vedclass

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

Question

(A) $1$. Calculate the milliequivalents of $H_2SO_4$ taken: $100 \, mL 	imes 0.1 \, M 	imes 2 \, (	ext{basicity}) = 20 \, meq$. $2$. Calculate the

Vedclass · Accepted Answer

Urea

Vedclass · Answer

(A) $1$. Calculate the milliequivalents of $H_2SO_4$ taken: $100 \, mL 	imes 0.1 \, M 	imes 2 \, (	ext{basicity}) = 20 \, meq$. $2$. Calculate the milliequivalents of $NaOH$ used for back titration: $20 \, mL 	imes 0.5 \, M 	imes 1 \, (	ext{acidity}) = 10 \, meq$. $3$. Milliequivalents of $NH_3$ evolved = $20 - 10 = 10 \, meq = 0.01 \, eq$. $4$. Percentage of nitrogen = $\frac{1.4 	imes 	ext{meq of } NH_3}{	ext{mass of compound}} = \frac{1.4 	imes 10}{0.30} = 46.66 \%$. $5$. Nitrogen percentage in Urea $(NH_2)_2CO$: $\frac{28}{60} 	imes 100 = 46.66 \%$. $6$. Since the percentage matches,the compound is Urea.

The ammonia evolved from the treatment of $0.30 \, g$ of an organic compound for the estimation of nitrogen was passed in $100 \, mL$ of $0.1 \, M$ sulphuric acid. The excess of acid required $20 \, mL$ of $0.5 \, M$ sodium hydroxide solution for complete neutralization. The organic compound is:

Similar Questions

$2.52 \ g$ of oxalic acid dihydrate was dissolved in $100 \ mL$ of water. $10 \ mL$ of this solution was diluted to $500 \ mL$. The normality of the final solution (in $N$) and the amount of oxalic acid (in $mg/mL$) in the solution are respectively

$K_2Cr_2O_7$ is preferred to $Na_2Cr_2O_7$ for use in volumetric analysis as a primary standard because

Assertion : $K_2Cr_2O_7$ is used as a primary standard in volumetric analysis.
Reason : It has a good solubility in water.

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

Which of the following is used as an oxidizing agent in volumetric analysis?

Vedclass Test Series

Exam Paper Generator

Online Exam Module