# How many moles of each element are present in 1.000L of a 0.295 M solution of sodium chloride?

Feb 25, 2018

$0.295$.

#### Explanation:

You know that molarity is defined as the number of moles of solute present in exactly $\text{1 L}$ of a solution.

Basically, what this means is that if you know the molarity of a solution, you know how many moles of solute are present for every $\text{1 L}$ of that solution.

In your case, you're dealing with a $\text{0.295-M}$ solution of sodium chloride, so right from the start, you can say that this solution will contain $0.295$ moles of sodium chloride, the solute, for every $\text{1 L}$ of the solution.

So in your $\text{1.000-L}$ sample of this solution, you will have $0.295$ moles of sodium chloride.

Now, sodium chloride is soluble in water, which implies that it exists as ions in aqueous solution.

${\text{NaCl"_ ((aq)) -> "Na"_ ((aq))^(+) + "Cl}}_{\left(a q\right)}^{-}$

So for every mole of sodium chloride that you dissolve in water to make your solution, you will end up with $1$ mole of sodium cations, ${\text{Na}}^{+}$, and $1$ mole of chloride anions, ${\text{Cl}}^{-}$.

This means that your $\text{1.000-L}$ sample contains $0.295$ moles of sodium cations and $0.295$ moles of chloride anions.

If you want the number of moles of each element, you can say that the $0.295$ moles of sodium cations correspond to $0.295$ moles of sodium, the parent element of the cation and that the $0.295$ moles of chloride anions correspond to $0.295$ moles of chlorine, the parent element of the anion.