(A) $n$-Octane is a non-polar solvent. According to the principle 'like dissolves like',non-polar solutes are more soluble in non-polar solvents.The p

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(A) $n$-Octane is a non-polar solvent. According to the principle 'like dissolves like',non-polar solutes are more soluble in non-polar solvents.The polarity order of the given compounds is: $KCl$ (ionic) > $CH_3OH$ (polar) > $CH_3CN$ (polar) > Cyclohexane (non-polar).Since solubility in a non-polar solvent ($n$-octane) increases as the polarity of the solute decreases, the order of solubility is:$KCl Thus, the correct order is $II < III < IV < I$.

Class 12 Chemistry Solutions Solubility

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Which is the correct order for the solubility of the following compounds in $n$-octane (at identical conditions)? $I$) Cyclohexane $II$) $KCl$ $III$) $CH_3OH$ $IV$) $CH_3CN$

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A
$II < III < IV < I$
B
$II < III < IV < I$ (Note: Corrected sequence based on polarity: $II < III < IV < I$)
C
$I < IV < III < II$
D
$I < III < IV < II$