Solubility of NaCl in water
Solubility of solid in a liquid depends on the nature of the solid.

It depends on whether the dissolution process is exothermic or endothermic. Using Le Chatelier's principle, if the temperature is increased for an exothermic process then dissolution decreases and if the temperature is increased for an endothermic process then dissolution increases.

I am going to stick to the case of $NaCl$, if you want to know more about Le Chatelier's principle, click on the link below and look for the 2nd answer https://physics.stackexchange.com/questions/311409/why-do-most-gases-dissolve-easier-at-colder-temperatures/311429#311429



There are some solids for whom change in temperature doesn't affect the degree of dissolution.

There are some substance which on heating, change on a particular temperature from one polymorphic form to another (like $\alpha$ to $\beta$ as in the case of ammonium nitrate) or from one hydrated form to another ($CaCl_2.6H_2 O -&gt; CaCl_2.4H_2 O$) or from hydrated to anhydrous form ($Na_2SO_4.10H_2O -&gt; Na_2SO_4$).

Generally, dissolution process in a solvent is a disintegration process. This process needs energy. In such cases, energy is absorbed. But in some cases, besides the process of breaking or ionization, there is hydrate formation. During hydration heat is evolved. The net result is that heat is either evolved or absorbed. 

There are also cases in which heat of separation of ions in just equal to the heat of hydration and there is very little heat effect as in the case of $NaCl$. The heat of solution of $NaCl$ is very small as the heat of ionization is nearly equal to the heat of hydration. Therefore, temperature doesn't affect the dissolution of $NaCl$ as the heat of hydration almost equalizes heat of dissolution.