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:

$n=\frac{m}{M}=\frac{347 g}{110.98 g/mol}=3.12 mol$

Thus, number of moles of ions will be $3\times 3.12 mol=9.38 mol$.

Since, 1 mole of any substance has $6.023\times 10^{23}$ units of that substance where $6.023\times 10^{23}$  is Avogadro's number.

Thus, 9.38 mol of ions will have $9.38\times 6.023\times 10^{23}=5.65\times 10^{24}$ number of ions.

Therefore, total number of ions in 347 g of $CaCl_{2}$ is  $5.65\times 10^{24}$.