At 310 ~K, the solubility of CaF_2 in water is 2.34 10^-3 ~g / 100 ~mL. The solubility product of CaF_2 is ____________ 10^-8(~mol / L)^3. (Give molar mass : CaF_2=78 ~g ~mol^-1)

CaF_2 s sCa^2++2 ~s2 ~F^- aligned K_sp &=s(2 ~s)^2 \\\\ &=4 ~s^3 aligned Solubility (s)=2.34 10^-3 ~g / 100 ~mL =2 34 10^-3 1078 mole / lit<

At 310 K the solubility of CaF2 in water is 234 x 103 g 100 : (Chemistry)

$\mathrm{CaF}_{2} \stackrel{\mathrm{s}}{\rightleftharpoons} \underset{ \mathrm{s}}{\mathrm{Ca}^{2+}}+\underset{2 \mathrm{~s}}{2 \mathrm{~F}^{-}}$

$\begin{aligned} \mathrm{K}_{\mathrm{sp}} &=\mathrm{s}(2 \mathrm{~s})^{2} \\\\ &=4 \mathrm{~s}^{3} \end{aligned}$

Solubility $(\mathrm{s})=2.34 \times 10^{-3} \mathrm{~g} / 100 \mathrm{~mL}$

$=\frac{2 \cdot 34 \times 10^{-3} \times 10}{78}$ mole $/$ lit

$=3 \times 10^{-4} \mathrm{~mole} / \mathrm{lit}$

$\therefore \mathrm{K}_{\mathrm{sp}}=4 \times\left(3 \times 10^{-4}\right)^{3}$

$\begin{aligned} &=108 \times 10^{-12} \\\\ &=0.0108 \times 10^{-8}(\mathrm{~mole} / \mathrm{lit})^{3} \end{aligned}$

$\begin{aligned} & \therefore x \approx 0 \end{aligned}$