Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • 0.2 g of an organic compound was subjected to estimation of nitrogen by Dumas method in which volume of <img alt="\mathrm{N_{2}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BN_

0.2 g of an organic compound was subjected to estimation of nitrogen by Dumas method in which volume of \mathrm{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 \mathrm{N_{2}}, is \mathrm{28 \;g\; mol^{-1}} Molar volume of \mathrm{N_{2}} at STP : 22.4L )

Option: 1

14


Option: 2

----------


Option: 3

---------


Option: 4

--------


Answers (1)

best_answer

The estimation by Duma's method involves the reaction:

\mathrm{C_{x} H_{y} N_{z}+(2 x+\frac{y}{2}) C u O \rightarrow x C O_{2}+\frac{y}{2} H_{2} O+\frac{z}{2} N_{2}+(2 x+\frac{y}{2}) C u}

\mathrm{\text { mole of } \mathrm{N}_{2} \text { Volume }=22.400 \mathrm{~mL}=22.4 \mathrm{\times 10^{-3}L}

\begin{aligned} \text { mole of } \mathrm{N_{2} }&=\frac{22.4 \times 10^{-3}}{22.4} \\ \text { mole of }\mathrm{N_{2} } &=10^{-3} \text { mole } \\ \text { mass of }\mathrm{N_{2} } &=28 \times 10^{-3} \text { gram } \end{aligned}

% of nitrogen in the compound

\begin{aligned} &=\frac{28 \times 10^{-3}}{0.2} \times 100 \\ &=14\% \end{aligned}

Hence, the answer is 14

Posted by

Devendra Khairwa

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JCECEB 2025 registration for BTech admission through JEE Main rank begins jceceb.jharkhand.gov.in
anam-khan

JEE Main 2025: BTech CSE cut-offs for PwD candidates at NITs make entry easy; top institutes still selective
anam-khan

73 or 99 percentile? JEE Main result questioned, J-K candidate says, ‘Don’t create hype’

Latest Question