If left[ {{F^ - }} right] = 2.0 × {10^{ - 5}} M in water. Then, how many gram of CaCl_{2} wi

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6-2.Equilibrium-II (Ionic Equilibrium)

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If the $\left[ {{F^ - }} \right] = 2.0 \times {10^{ - 5}}$ $M$ in water. Then, how many gram of $CaCl_{2}$ will be added for precipitation of ${{F^ - }}$ ? ${K_{sp}}$ for $Ca{F_2}$ $ = 1.7 \times {10^{ - 10}}$ . ( Molecular mass of $CaCl_{2}$ $= 111$ $g$ $mol^{-1}$ )

Option A

Option B

Option C

Option D

Solution

More then $47.175 \mathrm{gm}$ of $\mathrm{CaCl}_{2}$ is to be added

Standard 11

Chemistry

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