Answer
The molar solubility of $BaF_2$ in pure water is equal to $0.0183M$.
Work Step by Step
1. Write the $K_{sp}$ expression:
$ BaF_2(s) \lt -- \gt 1Ba^{2+}(aq) + 2F^-(aq)$
$2.45 \times 10^{-5} = [Ba^{2+}]^ 1[F^-]^ 2$
2. Considering a pure solution: $[Ba^{2+}] = 1S$ and $[F^-] = 2S$
$2.45 \times 10^{-5}= ( 1S)^ 1 \times ( 2S)^ 2$
$2.45 \times 10^{-5} = 4S^ 3$
$6.13 \times 10^{-6} = S^ 3$
$ \sqrt [ 3] {6.13 \times 10^{-6}} = S$
$0.0183M = S$
