The solubility of `CaF_(2) " in water at " 298 K is 1.7 xx 10^(-3)` grams per 100 mL of the solution. Calculate solubility product of `CaF_(2).`

Explanation

To calculate the solubility product (Ksp) of calcium fluoride (CaF₂) based on its solubility in water at 298 K, we can follow these steps:

### Step 1: Convert the solubility from grams to moles
The given solubility of CaF₂ is 1.7 x 10^(-3) grams per 100 mL of solution. First, we need to convert this to grams per liter (g/L):

\[ \text{Solubility in g/L} = 1.7 \times 10^{-3} \text{ g/100 mL} \times 1000 \text{ mL/L} = 17 \times 10^{-3} \text{ g/L} \]
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