Question

How many fluoride ions are present in 175 g of calcium fluoride?

Answer 1

The chemical formula for calcium fluoride is $CaF_2$.

The $CaF_2$ dissociates into ions as:

$CaF_2\rightarrow Ca^{2+} + 2F^{-}$

From the above reaction it is clear that 1 mole of $CaF_2$ gives 1 mole of $Ca^{2+}$ and 2 moles of $F^{-}$ ions.

Molar mass of $CaF_2$ is $40.078 + 2\times 18.998 = 78.074 g/mol$

1 mole of $[tex]CaF_2$[/tex] contains $78.074 g$ of $CaF_2$.

1 mole of $CaF_2$ contains 2 moles of $F^{-}$ ions.

So, the number of fluoride ions in 175 g of $CaF_2$ is:

$175 g \times \frac{1 mole CaF_2}{78.074 g/mol CaF_2}\times \frac{2 mole F^{-}}{1 mole CaF_2}\times 6.022\times 10^{23} F^{-} ions$

= $26.999\times 10^{23}$

Hence, the number of fluoride ions present in 175 g of calcium fluoride is $26.999\times 10^{23}$.