Question

Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, \[\text{KCl , C}{{\text{H}}_{\text{3}}}\text{OH , C}{{\text{H}}_{\text{3}}}\text{CN}\]

Answer 1

Hint: The amount of solute dissolved in the given amount of solvent depends on the nature of the solute and the solvent. The solute-solvent interaction generally depends on the “Like dissolves like “rule. This is a general rule for the solubility of solute in solvent. The n-octane is a non-polar solvent and thus non-polar solutes are easily soluble in it.

Complete answer:
The polar solvents more readily dissolve the polar solutes but cannot dissolve the non-polar solutes. Thus the sodium chloride salt which is an ionic compound readily dissolves in the water which is a polar solvent. However, it is not readily soluble in ether, benzene, or hexane which are non-polar solvents.
When the molecule does not own a polar bond it does not possess the permanent charge difference between its molecules and such molecules are called non-polar. None of the bonds in the hydrocarbon exhibit the permanent charge difference or do not form the dipole thus they are considered as the non-polar.
Thus here the solvent n-octane $\text{n-}{{\text{C}}_{\text{8}}}{{\text{H}}_{\text{18}}}$ is a non-polar solvent. Now according to the solute-solvent interaction rule, the non-polar solute dissolves in the non-polar solvent. It cannot dissolve the polar or ionic solutes.
Let us have a look at the given solutes:
The cyclohexane is a cyclic molecule. It has six carbon atoms forming a ring. Even though the $\text{C-H}$ is slightly polar, the six dipoles which are formed due to six $\text{C-H}$ bonds cancel out each other. So the overall molecule does not have a dipole moment and it is nonpolar. Thus it is soluble in n-octane.
Methanol is an asymmetrical molecule so it is not a nonpolar molecule. Since the dipole produced cannot cancel out each other. Oxygen is more electronegative than carbon and hydrogen and the net dipole moment is non zero. Thus methanol is the polar solute.
The nitrile group in the acetonitrile is highly polarised. Since the triple bonds bind together the carbon and nitrogen. It has a strong dipole moment of about 3.2 debyes.it is more polar than methanol. Thus the solubility of acetonitrile is less than that of the methanol in n-octane.
The $\text{KCl}$ is an ionic compound. It dissociates into ${{\text{K}}^{\text{+}}}$ and $\text{C}{{\text{l}}^{\text{-}}}$ ions. Thus it is insoluble in n-octane.
Therefore the increasing strength of solubility of compounds in n-octane as follows:
\[\begin{align}
  & \text{ } \\
 & \begin{matrix}
   \text{KCl} & \langle & \text{C}{{\text{H}}_{\text{3}}}\text{OH} & \langle & \text{C}{{\text{H}}_{\text{3}}}\text{CN} & \langle & \text{Cyclohexane} \\
   \text{(Ionic)} & {} & {} & \text{(Polar)} & {} & {} & \text{(Non-polar)} \\
\end{matrix} \\
 & \begin{matrix}
   \text{Solubility increases in n-octane } & \to & {} & {} & {} & {} & {} \\
\end{matrix} \\
\end{align}\]

Note:
Non polar solvents do not form the hydrogen bonding with the solvent. But they dissolve the solute by weak van der Waals forces among the solute and solvent. In solving such questions remember the rule “like dissolves like”.