\(\{\) ચિત્ર \(\}\) ઉપરોક્ત પ્રક્રિયામાં, \(3.9\, g\) બેન્ઝીન નાઈટ્રેશન પર \(4.92\, g\) નાઇટ્રોબેન્ઝિન આપે છે, ઉપરોક્ત પ્રક્રિયામાં નાઇટ્રોબેન્ઝિનની ટકાવારી નીપજ ............. \(\%\). (આણ્વિય દળ આપેલ છે: \(C : 12.0\, u , H : 1.0\, u\)\(O : 16.0\, u , N : 14.0\, u )\)

A \(20\)
B \(80\)
C \(75\)
D \(50\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(80\)

Step-by-step Solution

Detailed explanation

\(1 mole \quad1 mole\) \(78 gm \quad 123 gm\) \(3.9 gm \quad \frac{123}{78} \times 3.9=6.15 gm\) But actual amount of nitrobenzene formed is \(4.92 gm\) and hence. Percentage yield \(=\frac{4.92}{6.15} \times 100=80 \%\)

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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