In 1 mole of [tex]CaCl_{2}[/tex], there are 3 moles of ions, 1 mole of Ca^{2+} and 2 mole of [tex]Cl^{-1}[/tex].
[tex]CaCl_{2}\rightarrow Ca^{2+}+2Cl^{-}[/tex]
Molar mass of [tex]CaCl_{2}[/tex] is 110.98 g/mol. Calculating number of moles from given mass as follows:
[tex]n=\frac{m}{M}=\frac{347 g}{110.98 g/mol}=3.12 mol[/tex]
Thus, number of moles of ions will be [tex]3\times 3.12 mol=9.38 mol[/tex].
Since, 1 mole of any substance has [tex]6.023\times 10^{23}[/tex] units of that substance where [tex]6.023\times 10^{23}[/tex] is Avogadro's number.
Thus, 9.38 mol of ions will have [tex]9.38\times 6.023\times 10^{23}=5.65\times 10^{24}[/tex] number of ions.
Therefore, total number of ions in 347 g of [tex]CaCl_{2}[/tex] is [tex]5.65\times 10^{24}[/tex].
