The solubility product

The amount of fluoride ions $(F^{-})$ at $ 25^oC $ in a saturated solution of pure strontium fluoride is $0.0220$ g in $100$ mL of solution. The molarity of the fluoride in the saturated solution is therefore .......... $\frac{mol}{L}$ The equation for the dissolution of pure strontium fluoride $SrF_2(s)$ $Sr^{2+}(aq)$ $+$ $2F^{-}(aq)$ shows that then the concentration of $Sr^{2+}$ is equal to .......... $\frac{mol}{L}$ So, the solubility product at $15^oC$ of $[SrF_2]$ is $K_s=[Sr^{2+}][F^{-}]^2$ = .......... $\frac{mol^3}{L^3}$