एक अज्ञात क्लोरोहाइड्रोकार्बन में \(3.55 \%\) क्लोरीन है। यदि हाइड्रोकार्बन के प्रत्येक अणु में केवल एक क्लोरीन परमाणु है, तो \(1\; g\) क्लोरोहाइड्रोकार्बन में उपस्थित क्लोरीन परमाणु हैं। ( \(Cl\) का परमाणु भार \(=35.5 \;u\); अवोगाद्रो नियतांक \(=6.023 \times 10^{23}\; mol ^{-1}\) )

A \(6.023 \times {10^9}\)
B \(6.023 \times {10^{23}}\)
C \(6.023 \times {10^{21}}\)
D \(6.023 \times {10^{20}}\)

Verified Solution

Answer & Solution

Correct Answer

(D) \(6.023 \times {10^{20}}\)

Step-by-step Solution

Detailed explanation

Given percentage of chlorine in an hydrocarbon \(= 3.55\%\) i,e., \(100\,g\) of chlorohydrocarbon has \(3.55\,g\) of chlorine. \(1\,g\) of chlorohydrocarbon will have \(\frac{{3.55}}{{100}} = 0.0355\,g\) of chlorine. Atomic wt. of \(Cl = 35.5\,g/mol\) Number of moles of \(Cl\)…

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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