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The molar solubility of $CaF_2$ ($K_{sp}=5.3\times10^{-11}$) in 0.1 M solution of NaF will be
Detailed Solution
$CaF_2\rightleftharpoons Ca^{2+}+2F^-$. Due to the common ion, $[F^-]\approx0.1$ M. $K_{sp}=s(0.1)^2 \Rightarrow s=\frac{5.3\times10^{-11}}{10^{-2}}=5.3\times10^{-9}$ mol L$^{-1}$.