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

1 Answer
Feb 25, 2018

0.295.

Explanation:

You know that molarity is defined as the number of moles of solute present in exactly "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 "1 L" of that solution.

In your case, you're dealing with a "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 "1 L" of the solution.

So in your "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.

"NaCl"_ ((aq)) -> "Na"_ ((aq))^(+) + "Cl"_ ((aq))^(-)

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, "Na"^(+), and 1 mole of chloride anions, "Cl"^(-).

This means that your "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.