At $310 \mathrm{~K}$, the solubility of $\mathrm{CaF}_{2}$ in water is $2.34 \times 10^{-3} \mathrm{~g} / 100 \mathrm{~mL}$. The solubility product of $\mathrm{CaF}_{2}$ is ____________ $\times 10^{-8}(\mathrm{~mol} / \mathrm{L})^{3}$. (Give molar mass : $\mathrm{CaF}_{2}=78 \mathrm{~g} \mathrm{~mol}^{-1}$)

$\mathrm{CaF}_{2} \stackrel{\mathrm{s}}{\rightleftharpoons} \underset{ \mathrm{s}}{\mathrm{Ca}^{2+}}+\underset{2 \mathrm{~s}}{2 \mathrm{~F}^{-}}$ <br/><br/> $$ \begin{aligned} \mathrm{K}_{\mathrm{sp}} &=\mathrm{s}(2 \mathrm{~s})^{2} \\\\ &=4 \mathrm{~s}^{3} \end{aligned} $$ <br/><br/> Solubility $(\mathrm{s})=2.34 \times 10^{-3} \mathrm{~g} / 100 \mathrm{~mL}$<br/><br/> $=\frac{2 \cdot 34 \times 10^{-3} \times 10}{78}$ mole $/$ lit<br/><br/> $=3 \times 10^{-4} \mathrm{~mole} / \mathrm{lit}$ <br/><br/> $\therefore \mathrm{K}_{\mathrm{sp}}=4 \times\left(3 \times 10^{-4}\right)^{3}$ <br/><br/> $$ \begin{aligned} &=108 \times 10^{-12} \\\\ &=0.0108 \times 10^{-8}(\mathrm{~mole} / \mathrm{lit})^{3} \end{aligned} $$<br/><br/> $$ \begin{aligned} & \therefore x \approx 0 \end{aligned} $$ <br/><br/>