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

\(0.125\,g\) કાર્બનિક સંયોજનના એક નમૂનાનું ડ્યુમા પદ્ધતિ વડે પૃથ્થકરણ કરતાં પ્રાપ્ત થતા \(22.78\, mL\) નાઈટ્રોજન વાયુ ને \(280 \,K\) અને \(759\, mm \,Hg\) પર \(KOH\) ના દ્વાવણ ઉપર ભેગો કરવામાં આવ્યો. આપેલ કાર્બાનિક સંયોજનમાં નાઈટ્રોજનની ટકાવારી \(......\) છે. (નજીકનો પૂર્ણાંક ) આપેલું છે : \((a)\,280\, K\) પર પાણીનું બાષ્પદબાણ= \(14.2\, mm\, Hg\). \((b)\,R =0.082 \,L \operatorname{atm~} \,K ^{-1} \,mol ^{-1}\)

A \(22\)
B \(23\)
C \(21\)
D \(20\)

Verified Solution

Answer & Solution

Correct Answer

(A) \(22\)

Step-by-step Solution

Detailed explanation

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

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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