(A) Weight of organic compound $= 0.2 \ g$. Volume of $N_2$ at $STP$ $= 22.400 \ mL = 22.4 \times 10^{-3} \ L$. Moles of $N_2 = \frac{22.4 \times 10^{

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

(A) Weight of organic compound $= 0.2 \ g$. Volume of $N_2$ at $STP$ $= 22.400 \ mL = 22.4 \times 10^{-3} \ L$. Moles of $N_2 = \frac{22.4 \times 10^{-3} \ L}{22.4 \ L \ mol^{-1}} = 10^{-3} \ mol$. Mass of $N_2$ evolved $= 10^{-3} \ mol \times 28 \ g \ mol^{-1} = 28 \times 10^{-3} \ g$. Percentage of nitrogen $= \frac{\text{Mass of } N_2}{\text{Mass of compound}} \times 100$. Percentage of nitrogen $= \frac{28 \times 10^{-3} \ g}{0.2 \ g} \times 100 = \frac{0.028}{0.2} \times 100 = 14 \%$.

Class 11 Chemistry 3 Organic Chemistry: Purification and Characterisation Quantitative Analysis

Medium

$0.2 \ g$ of an organic compound was subjected to estimation of nitrogen by Dumas method in which volume of $N_2$ evolved (at $STP$) was found to be $22.400 \ mL$. The percentage of nitrogen in the compound is $.......$. [nearest integer] (Given: Molar mass of $N_2$ is $28 \ g \ mol^{-1}$. Molar volume of $N_2$ at $STP$ : $22.4 \ L \ mol^{-1}$ )

JEE MAIN 2022 All JEE MAIN

A
$14$
B
$21$
C
$18$
D
$56$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Chemistry Questions

Chapter

3 Organic Chemistry: Purification and Characterisation

Subtopic

Quantitative Analysis

Previous Year

JEE MAIN 2022 Paper All JEE MAIN years →

$0.5 \ g$ of an organic compound containing phosphorus was heated with conc. $HNO_3$ in a Carius tube. The phosphoric acid thus formed was precipitated with magnesia mixture $[MgNH_4PO_4]$,which on ignition gave a residue of $1.0 \ g$ of magnesium pyrophosphate $(Mg_2P_2O_7)$. The percentage of phosphorus in the organic compound is:

Difficult

View Solution

In the Kjeldahl's method for estimation of nitrogen present in a soil sample,ammonia evolved from $0.75 \ g$ of sample neutralized $10 \ mL$ of $1 \ M \ H_2SO_4$. The percentage of nitrogen in the soil is (in $.33$)

DifficultNEET 2014

View Solution

$0.3780 \, g$ of an organic chloro compound gave $0.5740 \, g$ of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound. (in $\%$)

Medium

View Solution

Which compound is formed by heating with nitric acid in the estimation of phosphorus in the Carius method?

Easy

View Solution

For the estimation of nitrogen,$1.4 \ g$ of an organic compound was digested by Kjeldahl method and the evolved ammonia was absorbed in $60 \ mL$ of $\frac{M}{10}$ sulphuric acid. The unreacted acid required $20 \ mL$ of $\frac{M}{10}$ sodium hydroxide for complete neutralization. The percentage of nitrogen in the compound is .....$\%$

DifficultJEE MAIN 2014

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Similar Questions

In the Kjeldahl's method for estimation of nitrogen present in a soil sample,ammonia evolved from $0.75 \ g$ of sample neutralized $10 \ mL$ of $1 \ M \ H_2SO_4$. The percentage of nitrogen in the soil is (in $.33$)

$0.3780 \, g$ of an organic chloro compound gave $0.5740 \, g$ of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound. (in $\%$)

Which compound is formed by heating with nitric acid in the estimation of phosphorus in the Carius method?

Vedclass Test Series

Exam Paper Generator

Online Exam Module