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* 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 * (1 mole CaF_2)/(78.074 g/mol CaF_2)* (2 mole F^(-))/(1 mole CaF_2)* 6.022* 10^(23) F^(-) ions`

=
`26.999* 10^(23)`

Hence, the number of fluoride ions present in 175 g of calcium fluoride is
`26.999* 10^(23)`.