The molar solubility of an ionic solid is the number of moles of that solid that dissolve per liter of saturated solution under the stated conditions. It links the macroscopic amount dissolved to the equilibrium ion concentrations that appear in $K_{sp}$.

Consider

$$\mathrm{CaF_2(s)\rightleftharpoons Ca^{2+}(aq)+2F^-(aq)}.$$

Let the molar solubility in pure water be $s$. Dissolving $s$ moles of $\mathrm{CaF_2}$ per liter produces

$$[\mathrm{Ca^{2+}}]=s$$

and

$$[\mathrm{F^-}]=2s.$$

Therefore

$$K_{sp}=[\mathrm{Ca^{2+}}][\mathrm{F^-}]^2=s(2s)^2=4s^3.$$

If

$$K_{sp}=3.2\times10^{-11},$$

then

$$s=\left(\frac{K_{sp}}{4}\right)^{1/3} =\left(8.0\times10^{-12}\right)^{1/3} \approx2.0\times10^{-4},\mathrm{mol/L}.$$

The stoichiometry matters. For a 1:1 salt such as

$$\mathrm{AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)},$$

pure-water dissolution gives

$$K_{sp}=s^2,$$

whereas a 1:2 salt such as $\mathrm{CaF_2}$ gives $K_{sp}=4s^3$. Two solids therefore cannot generally be ranked by comparing their $K_{sp}$ values alone unless their dissolution stoichiometries are the same.

The calculation can also be reversed. If the molar solubility is measured, the equilibrium ion concentrations follow from the balanced dissolution equation and can be substituted into the $K_{sp}$ expression.

Molar solubility is condition-dependent. The simple relations above assume that the dissolving solid is the only important source of its ions and that no additional equilibria remove or bind those ions. A solution containing a common ion, an acid-base reaction, or complex formation requires those effects to be included explicitly.