At 310,K , solubility of CaF _{2} in water is 2.34 × 10^{-3},g / 100,mL . solubility product of CaF

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6-2.Equilibrium-II (Ionic Equilibrium)

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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 :$CaF _{2}=78\,g\,mol ^{-1} \text { ) }$

A

$1$

B

$0$

C

$2$

D

$3$

(JEE MAIN-2022)

Solution

Solubility of $CaF _{2}= S$ mole/ $L$

$S =\frac{2.34 \times 10^{-3}}{0.1 \times 78}=\frac{2.34}{78} \times 10^{-2}=3 \times 10^{-4}\,mol / L$

$K _{ sp }\left( CaF _{2}\right)=4 S ^{3}=4\left(3 \times 10^{-4}\right)^{3}$

$=108 \times 10^{-12}$

$=0.0108 \times 10^{-8}( mol / L )^{3}$

Standard 11

Chemistry

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