At $298 \ K$ temperature,the $K_{sp}$ of $CaF_2$ is $1.7 \times 10^{-10}$. If a person drinks $2.5 \ L$ of $CaF_2$ saturated water daily,how many grams of $CaF_2$ are present in the water consumed? (Molecular mass of $CaF_2$ is $78 \ g \ mol^{-1}$)

(A) The dissociation of $CaF_2$ is: $CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq)$.
Let the solubility be $S \ mol \ L^{-1}$. Then $K_{sp} = [Ca^{2+}][F^-]^2 = (S)(2S)^2 = 4S^3$.
Given $K_{sp} = 1.7 \times 10^{-10}$,so $4S^3 = 1.7 \times 10^{-10}$.
$S^3 = 0.425 \times 10^{-10} = 42.5 \times 10^{-12}$.
$S = \sqrt[3]{42.5} \times 10^{-4} \approx 3.49 \times 10^{-4} \ mol \ L^{-1}$.
Amount in $2.5 \ L$ in moles $= S \times V = 3.49 \times 10^{-4} \ mol \ L^{-1} \times 2.5 \ L = 8.725 \times 10^{-4} \ mol$.
Mass of $CaF_2 = \text{moles} \times \text{molar mass} = 8.725 \times 10^{-4} \ mol \times 78 \ g \ mol^{-1} \approx 0.068 \ g$.