Question

How many total ions are present in 347g of cacl2?

Answer 1

In 1 mole of
`CaCl_(2)`, there are 3 moles of ions, 1 mole of Ca^{2+} and 2 mole of
`Cl^(-1)`.

`CaCl_(2)\rightarrow Ca^(2+)+2Cl^(-)`

Molar mass of
`CaCl_(2)` is 110.98 g/mol. Calculating number of moles from given mass as follows:

`n=(m)/(M)=(347 g)/(110.98 g/mol)=3.12 mol`

Thus, number of moles of ions will be
`3* 3.12 mol=9.38 mol`.

Since, 1 mole of any substance has
`6.023* 10^(23)` units of that substance where
`6.023* 10^(23)` is Avogadro's number.

Thus, 9.38 mol of ions will have
`9.38* 6.023* 10^(23)=5.65* 10^(24)` number of ions.

Therefore, total number of ions in 347 g of
`CaCl_(2)` is
`5.65* 10^(24)`.