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

Md Ariful Islam

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Md Ariful Islam

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14.5g of $NaCl$ is equal to 0.248 mol according to the equation

$n=m/M$

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- $Na^+$ and $Cl^-$ 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.

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