When nitrogen gas is passed through 100 mL o | vedclass

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

Question

(C) $1$. Total millimoles of $H_2SO_4$ taken = $100 \ mL 	imes 0.1 \ M = 10 \ mmol$. $2$. Millimoles of $H_2SO_4$ neutralized by $NaOH$ = $\frac{1}{2

Vedclass · Accepted Answer

Urea

Vedclass · Answer

(C) $1$. Total millimoles of $H_2SO_4$ taken = $100 \ mL 	imes 0.1 \ M = 10 \ mmol$. $2$. Millimoles of $H_2SO_4$ neutralized by $NaOH$ = $\frac{1}{2} 	imes (20 \ mL 	imes 0.5 \ M) = 5 \ mmol$ (since $H_2SO_4 + 2NaOH ightarrow Na_2SO_4 + 2H_2O$). $3$. Millimoles of $H_2SO_4$ consumed by $NH_3$ = $10 - 5 = 5 \ mmol$. $4$. Since $2NH_3 + H_2SO_4 ightarrow (NH_4)_2SO_4$,millimoles of $NH_3$ = $2 	imes 5 = 10 \ mmol$. $5$. Mass of nitrogen = $\frac{10 	imes 10^{-3} 	imes 14}{0.30} 	imes 100 = 46.67\%$. $6$. For Urea $(NH_2CONH_2)$,molar mass = $60 \ g/mol$. Nitrogen content = $\frac{28}{60} 	imes 100 = 46.67\%$. $7$. Thus,the compound is Urea.

When nitrogen gas is passed through $100 \ mL$ of $0.1 \ M$ $H_2SO_4$ containing $0.30 \ g$ of an organic compound,the ammonia evolved is neutralized. To neutralize the excess acid,$20 \ mL$ of $0.5 \ M$ $NaOH$ is required. Identify the organic compound.

Similar Questions

In the standardization of $Na_2S_2O_3$ using $K_2Cr_2O_7$ by iodometry,the equivalent weight of $K_2Cr_2O_7$ is

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

Which of the following statements is not true about Mohr's salt?

$10 \, mL$ of concentrated $HCl$ were diluted to $1 \, L$. $20 \, mL$ of this diluted solution required $25 \, mL$ of $0.1 \, N$ sodium hydroxide solution for complete neutralization. The normality of the concentrated hydrochloric acid will be:

In the iodometric estimation in the laboratory,which process is involved?

Vedclass Test Series

Exam Paper Generator

Online Exam Module