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A sample of 0.125 \mathrm{~g} of an organic compound when analyzed by Duma's method yields 22.78 \mathrm{~mL} of nitrogen gas collected over \mathrm{KOH} solution at 280 \mathrm{~K}$ and $759 \mathrm{~mm} \mathrm{Hg}. The percentage of nitrogen in the given organic compound is _____________
(Nearest integer)
Given :
(a) The vapour pressure of water of 280 \mathrm{~K}$ is $14.2 \mathrm{~mm} \mathrm{Hg}.
(b) \mathrm{R}=0.082 \mathrm{~L}$ atm $\mathrm{K}^{-1} \mathrm{~mol}^{-1}

Option: 1

22


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Given -

\mathrm{V= 22.78\, ml,T= 280K}
\mathrm{P_{Total}= 759\, mm\, Hg}
\mathrm{R= 0.082\, L\, atm\, K^{-1}mol^{-1}}

\mathrm{P_{N2}= P_{Total}-P_{Water}}
          \mathrm{= 759-14.2= 744.8\, mm\, Hg}


 \mathrm{PV= nRT\Rightarrow \eta _{N2}= \frac{744.8\times 22.78}{0.082\times 1000\times 760\times 280}= 0.00097}

\mathrm{W_{Nitrogen}= \eta _{N2}\times M_{N2}= \eta _{N2}\times 14}
                    = 0.00097\times 14= 0.02716

\mathrm{%N= \frac{0.02716}{0.125}\times 100= 21.728}

Hence answer is 22
                     


       

Posted by

vinayak

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