$C a F_{2}$ has a solubility of $1.56 \times 10^{- 3} g$ per 100 mL of solution at $25 º C .$ Its solubility product is (Atomic mass Ca = 40, F = 19)

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$2 \times 10^{- 1} M^{3}$

$3.2 \times 10^{- 11} M^{3}$

$8 \times 10^{- 12} M^{3}$

$6.24 \times 10^{- 9} M^{3}$

answer is B.

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Detailed Solution

$S = 1.56 \times 10^{- 3} g / 100 m L$ $= 1.56 \times 10^{- 3} \times 10$

$g / L = 1.56 \times 10^{- 2} g / L$

$S = \frac{1.56 \times 10^{- 2}}{78} = 2 \times 10^{- 4} m o l / L$

$K_{s p} = 4 s^{3} = 4 \times {(2 \times 10^{- 4})}^{3} = 4 \times 8 \times 10^{- 12} = 3.2 \times 10^{- 11}$

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