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

Answer:

#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.