Which is the correct order for solubility of following compound in n-octane (at Identical Condition)
I) Cyclohexane
II) KCl
III) \(CH _3 OH\)
IV) \(CH _3 CN\)

A II \(<\) III \(<\) IV \(<\) I
B II \(<\) IV \(<\) III \(<\) I
C I \(<\) IV \(<\) III \(<\) II
D I \(<\) III \(<\) IV \(<\) II

Verified Solution

Answer & Solution

Correct Answer

(B) II \(<\) IV \(<\) III \(<\) I

Step-by-step Solution

Detailed explanation

\(KCl\) is an ionic compound, virtually insoluble in nonpolar n-octane. \(CH_3CN\) is very polar (dipole moment \( \approx 3.9 D \)), having strong dipole-dipole interactions.

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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Which is the correct order for solubility of following compound in n-octane (at Identical Condition)I) CyclohexaneII) KClIII) \(CH _3 OH\)IV) \(CH _3 CN\)

Step-by-step Solution

Which is the correct order for solubility of following compound in n-octane (at Identical Condition)
I) Cyclohexane
II) KCl
III) \(CH _3 OH\)
IV) \(CH _3 CN\)