At 310K, the solubility of $Ca{F}_{2 }$in water is $2.34\times {10}^{-3}$ g/100ml. The solubility product of CaF₂ is  _________ $\times {10}^{-8}$mol/lit)³   (Given molar mass of $Ca{F}_2 = 78 g$ mol^‑)

In 100ml solubility is $2.34\times {10}^{-3}gm$

In 1000ml solubility is  $\frac{2.34\times {10}^{-3}\times 1000}{100}$

No of moles $=\frac{2.34\times {10}^{-3}\times 1000}{100\times 78}=3\times {10}^{-4}$

${\mathrm{K}}_{\mathrm{sp}}\text{ of }{\mathrm{CaF}}_2=4{\mathrm{S}}^3$

$\begin{array}{l}=4\times {\left(3\times {10}^{-4}\right)}^3=108\times {10}^{-12}\\ =0.0108\times {10}^{-8}\end{array}$