The chemical formula for calcium fluoride is [tex]CaF_2[/tex].
The [tex]CaF_2[/tex] dissociates into ions as:
[tex]CaF_2\rightarrow Ca^{2+} + 2F^{-}[/tex]
From the above reaction it is clear that 1 mole of [tex]CaF_2[/tex] gives 1 mole of [tex]Ca^{2+}[/tex] and 2 moles of [tex]F^{-}[/tex] ions.
Molar mass of [tex]CaF_2[/tex] is [tex]40.078 + 2\times 18.998 = 78.074 g/mol[/tex]
1 mole of [tex][tex]CaF_2[/tex][/tex] contains [tex]78.074 g[/tex] of [tex]CaF_2[/tex].
1 mole of [tex]CaF_2[/tex] contains 2 moles of [tex]F^{-}[/tex] ions.
So, the number of fluoride ions in 175 g of [tex]CaF_2[/tex] is:
[tex]175 g \times \frac{1 mole CaF_2}{78.074 g/mol CaF_2}\times \frac{2 mole F^{-}}{1 mole CaF_2}\times 6.022\times 10^{23} F^{-} ions[/tex]
= [tex]26.999\times 10^{23}[/tex]
Hence, the number of fluoride ions present in 175 g of calcium fluoride is [tex]26.999\times 10^{23}[/tex].
