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Based on solute-solvent interactions, arrange the following in order of increasing solubility in \(n\)-octane and explain.
Cyclohexane, \(\mathrm{KCl}, \mathrm{CH}_3 \mathrm{OH}, \mathrm{CH}_3 \mathrm{CN}\).
Cyclohexane, \(\mathrm{KCl}, \mathrm{CH}_3 \mathrm{OH}, \mathrm{CH}_3 \mathrm{CN}\).
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(a) Cyclohexane and \(n\)-octane both are non-polar. They mix completely in all proportions.
(b) \(\mathrm{KCl}\) is an ionic compound, \(\mathrm{KCl}\) will not dissolve in \(n\)-octane.
(c) \(\mathrm{CH}_3 \mathrm{OH}\) is polar. \(\mathrm{CH}_3 \mathrm{OH}\) will dissolve in \(n\)-octane.
(d) \(\mathrm{CH}_3 \mathrm{CN}\) is polar but lesser than \(\mathrm{CH}_3 \mathrm{OH}\). Therefore, it will dissolve in \(n\)-octane but to a greater extent as compared to \(\mathrm{CH}_3 \mathrm{OH}\).
Hence, the order is \(\mathrm{KCl} < \mathrm{CH}_3 \mathrm{OH} < \mathrm{CH}_3 \mathrm{CN} < \) Cyclohexane.
(b) \(\mathrm{KCl}\) is an ionic compound, \(\mathrm{KCl}\) will not dissolve in \(n\)-octane.
(c) \(\mathrm{CH}_3 \mathrm{OH}\) is polar. \(\mathrm{CH}_3 \mathrm{OH}\) will dissolve in \(n\)-octane.
(d) \(\mathrm{CH}_3 \mathrm{CN}\) is polar but lesser than \(\mathrm{CH}_3 \mathrm{OH}\). Therefore, it will dissolve in \(n\)-octane but to a greater extent as compared to \(\mathrm{CH}_3 \mathrm{OH}\).
Hence, the order is \(\mathrm{KCl} < \mathrm{CH}_3 \mathrm{OH} < \mathrm{CH}_3 \mathrm{CN} < \) Cyclohexane.
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