Question

# The reason for the extremely low solubility of carbon dioxide in benzene (${C_6}{H_6}$) at room temperature is due to which of the following?A.The increased disorder due to the mixing of the solute and solvent.B.The relatively low strength of the intermolecular forces between carbon dioxide and benzene.C.The strong hydrogen bonding in benzene.D.The weak solvation of carbon and oxygen ions by benzene.

Hint:
Dissolution usually takes place between compounds that have similar polarities in their molecules. If we place a solid which is nonpolar into a nonpolar liquid, then "like dissolves like" implies that the solid will dissolve in the given liquid. However, the only forces that will cause the liquid to be attracted to the solid are weak London dispersion forces.

Due to the linear structure of the carbon dioxide molecule, the dipoles of the molecule cancel each other out and hence the $C{O_2}$ is nonpolar.
Similarly, the benzene molecule is nonpolar due to its symmetrical molecule, which cancels out all the dipoles within the molecule.
Dissolution usually takes place between compounds that have similar polarities in their molecules. Since both the carbon dioxide molecule and the benzene molecule are nonpolar in nature, only dispersive forces are present between these two molecules. Also, the intermolecular forces are also too weak for the solute to be strongly attracted to the solvent.
Hence, at room temperature, the extremely low solubility of carbon dioxide in benzene is due to the relatively low strength of the intermolecular forces between carbon dioxide and benzene.

Hence, Option B is the correct option.

Note:
A molecule is called to be a nonpolar molecule if there is an equivalent sharing of electrons between two atoms of a molecule which is diatomic in nature, or if there is symmetry in the polar bonds of a significantly complex molecule.