MCQ
A sample of $0.125\,g$ of an organic compound when analysed by Duma's method yields $22.78\,mL$ of nitrogen gas collected over $KOH$ solution at $280\,K$ and $759\,mm\,Hg$. The percentage of nitrogen in the given organic compound is.(Nearest integer).

$(a)$ The vapour pressure of water at $280 K$ is $14.2$ $mm Hg$

$(b)$ $R =0.082 L$ atm $K ^{-1} mol ^{-1}$

  • $22$
  • B
    $23$
  • C
    $21$
  • D
    $20$

Answer

Correct option: A.
$22$
a
$V =22.78\,ml , T =280\,K$

$P _{\text {total }}=759\,mmHg$

$P _{ N _{2}}=759-14.2=744.8\,mmHg$

$n _{ N _{2}}=\frac{744.8 \times 22.78}{760 \times 1000 \times 0.082 \times 280}=0.00097$

$W _{\text {Nitrogen }}=0.02716$

$\% N =\frac{0.02716}{0.125} \times 1000=21.728$

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