What is the molarity of a sodium chloride solution that contains "3.8 moles" of sodium chloride in "2.5 L" of the solution?

Aug 12, 2016

${\text{1.5 mol L}}^{- 1}$

Explanation:

The problem provides you with the composition of a sodium chloride solution in terms of moles of solute and total volume of solution.

In order to find the solution's molarity, all you have to do is figure out how many moles of solute would be present in $\text{1 L}$ of solution. This is the definition of molarity -- the number of moles of solute, which in your case is sodium chloride, present in One liter of solution.

You already know that $3.8$ moles are present in $\text{2.5 L}$, so use this known composition to "scale down" the solution to a volume of $\text{1 L}$. You can do that because, by definition, the particles of solute are evenly mixed with the particles of solvent.

You can thus say that you have

1 color(red)(cancel(color(black)("L solution"))) * overbrace("3.8 moles NaCl"/(2.5color(red)(cancel(color(black)("L solution")))))^(color(blue)("known composition")) = "1.52 moles NaCl"

Since one liter of solution contains $1.52$ moles of solute, you can say that the moalrity of the solution, $c$, will be

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {\text{1.5 mol L}}^{- 1}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to two sig figs.