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Q.
The solubility of $CaF _{2}$ in water at $298 \,K$ is $1.7 \times 10^{-3}$ $gm$ per $100 \,cm ^{3}$. The solubility product of $CaF _{2}$ at $298\, K$ is
Equilibrium
Solution:
Solubility of $CaF _{2}=1.7 \times 10^{-3} gm$ per $100\ ,cm ^{3}$
$=1.7 \times 10^{-3} \times \frac{1000}{100} g L ^{-1}$
$=\frac{1.7 \times 10^{-2}}{78}$
$=2.18 \times 10^{-4} \,mol\,L ^{-1}$
$CaF _{2} \rightleftharpoons Ca ^{2+}+2 F ^{-}$
$K _{ sp } \cdot\left[ Ca ^{2+}\right]\left[ F ^{-}\right]^{2}$
$=\left(2.18 \times 10^{-4}\right)\left(2 \times 2.18 \times 10^{-4}\right)^{2}$
$=4.14 \times 10^{-11}$