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Why is solubility of \[NaCl\] in water \[311g{l^{ - 1}}N\] but is zero in benzene?

Last updated date: 13th Jun 2024
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Hint:In the given question firstly we have to define why exactly does the \[NaCl\] dissolve in the water. Now after that we have the reason that the solubility is due to the ionic nature of the water and the polarity of the water. As we know being a covalent compound the ionic nature is missing from the Benzene and that it has no polarity in its structure. Now that makes the case for insolubility.

Complete step by step answer:
In the given question we have to first find out the process of dissolution, as basically the question asked by the statement is regarding the insolubility of the \[NaCl\] in Benzene.

Now firstly we need to know the nature of all the components involved in the question. In the question there is only one solute that is \[NaCl\] which is highly ionic in nature. Now apart from that the water is also the ionic compound.

The process of dissolution of the \[NaCl\] involves the breakage of the ionic bond in order to form the ions and that is fulfilled by water due to two factors. First that it is ionic in nature which provides an excellent solvent for the ionic \[NaCl\] . Now the other factor is the polarity which leads to the dissolution.

In the case of Benzene both these factors are missing as it is non-ionic in nature and not polarising in nature too.

And according to the principle of "like dissolve like '', we can say that the ionic solvent needs the ionic solute and the nonionic needs non-ionic.

Note:The reason why all the non-polar molecules have capability to dissolve in nonpolar substances, is because of the fact that there is no need to compete with the attraction between any of the molecules because there essentially isn’t any. Now apart from that the other reason why polar solutes typically do not dissolve in non-polar solvents is because there is nothing to keep the polar and nonpolar molecules near each due to the absence of the attraction forces.