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JEE Mains · Chemistry · STD 11 - 8.3. Organic chemistry purification and characterization

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 \(K ^{-1} mol ^{-1}\)

  1. A \(22\)
  2. B \(23\)
  3. C \(21\)
  4. D \(20\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(22\)

Step-by-step Solution

Detailed explanation

\(V =22.78\,ml , T =280\,K\) \(P _{\text {total }}=759\,mmHg\) \(P _{ N _{2}}=759-14.2=744.8\,mmHg\) \(n _{ N _{2}}=\frac{744.8 \times 22.78}{760 \times 1000 \times 0.082 \times 280}=0.00097\) \(W _{\text {Nitrogen }}=0.02716\) \(\% N =\frac{0.02716}{0.125} \times 1000=21.728\)
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