# Calculate the number of sodium ions and chlorine ions and total number of ions in 14.5g of NaCl ?

Mar 26, 2018

Here is what I get:

#### Explanation:

14.5g of $N a C l$ is equal to 0.248 mol according to the equation

$n = \frac{m}{M}$

Now, this equates to $1.495 \times {10}^{23}$ Sodium chloride molecules.

We get this if we multiply 0.248 mol with Avogadro's number, $6.022 \times {10}^{23}$

Each molecule of Sodium chroride holds two atomic ions bonded in an ionic bond- $N {a}^{+}$ and $C {l}^{-}$ in a 1:1 ratio. This means that the amount of molecules correspond with each of the individual ions.

So :
Number of Sodium ions=Number of Chloride ions= $1.495 \times {10}^{23}$ ions

So the total amount of ions is twice this number, $2.990 \times {10}^{23}$ ions in total.