What is the molarity of a solution that contains 3.25 moles of "KCl" in 2.25 L of the solution?

Mar 31, 2018

${\text{1.44 mol L}}^{- 1}$

Explanation:

The idea here is that in order to find the molarity of the solution, you need to figure out how many moles of solute are present for every $\text{1.00 L}$ of the solution.

In your case, you know that the solution contains $3.25$ moles of potassium chloride, the solute, in a total volume of $\text{2.25 L}$.

You can calculate the number of moles of potassium chloride present in $\text{1.00 L}$ of the solution by using the solution's known composition as a conversion factor.

1.00 color(red)(cancel(color(black)("L solution"))) * overbrace("3.25 moles KCl"/(2.25 color(red)(cancel(color(black)("L solution")))))^(color(blue)("the solution's composition")) = "1.44 moles KCl"

So if $\text{1.00 L}$ of this solution contains $1.44$ moles of solute, you can say that the molarity of the solution is equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molarity = 1.44 mol L}}^{- 1}}}}$

The answer is rounded to three sig figs.

Notice that you can get the molarity of the solution by diving the number of moles of solute by the total volume of the solution.

This will get you

["KCl"] = "3.25 moles"/"2.25 L" = 3.25/2.25 quad "mol"/"L" = "1.44 mol L"^(-1)