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 $x \times 10^{-8} \, (mol / L)^{3}$. Find the value of $x$. (Given molar mass: $CaF_{2} = 78 \, g \, mol^{-1}$)

  • A
    $0.0108$
  • B
    $0.108$
  • C
    $1.08$
  • D
    $10.8$

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