Question

Solubility order of compounds in water is:
\[A)BeC{L_2} < NaCl < LiCl\]
$B)NaCl < LiCl < BeC{l_2}$
$C)BeC{l_2} < LiCl < NaCl$
$D)LiCl < NaCl < BeC{l_2}$

Answer 1

Hint: Solubility depends on the charge density and the hydration energy of the compounds. Charge density is the amount of electric charge per unit length, surface area, or volume and Hydration energy increases as the size of the ion decreases and charges on it increases.

Complete answer:
Charge density is the amount of electric charge per unit length, surface area, or volume. Volume charge density is the quantity of charge per unit volume, measured in the $SI$system in coulombs per cubic meter, at any point in a volume.
According to the charge density of the given compounds the solubility order is $NaCl < LiCl < BeC{l_2}$
Among $NaCl,LiCl$ and $BeC{l_2}$.$BeC{l_2}$ is the most polar compound, thus its solubility is higher than others.
Also, this is the same order in which the hydration energy increases. Enthalpy of hydration is described as the amount of energy released on dilution of one mole of gaseous ions. It can be considered as enthalpy of solvation with the solvent being water. Hydration enthalpy is also called hydration energy and its values are always negative.
So, the correct option is $B)NaCl < LiCl < BeC{l_2}$

Note:
Most of the ionic compounds are insoluble in non-aqueous solutions but they show high solubility in water. The factor that determines the solubility of a salt is the interactions of the ions with the solvent. For instance the water is a polar molecule with a partial positive charge on hydrogen and partial negative charge on oxygen, interacts with the ions and forms a strong bond releasing energy.