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 \(K ^{-1} mol ^{-1}\)
- A \(22\)
- B \(23\)
- C \(21\)
- D \(20\)
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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