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  4. 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. (Given molar mass: .CaF 2=78 g mol -1)

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Q. At $310 \,K$, the solubility of $CaF _2$ in water is $2.34 \times 10^{-3} g / 100 \,mL$. The solubility product of $CaF _2$ is __ $\times 10^{-8}( mol / L )^3$. (Given molar mass : $\left.CaF _2=78\, g\, mol ^{-1}\right)$

JEE MainJEE Main 2022Equilibrium

Solution:

Solubility of $CaF _2= S$ mole $/ L$
$S =\frac{2.34 \times 10^{-3}}{0.1 \times 78}=\frac{2.34}{78} \times 10^{-2}=3 \times 10^{-4} mol / L$
$ K _{\text {sp }}\left( CaF _2\right)=4 S ^3=4\left(3 \times 10^{-4}\right)^3$
$=108 \times 10^{-12} $
$ =0.0108 \times 10^{-8}( mol / L )^3$