# If given 0.090 moles of sodium sulfate in 12 mL of solution, what is the concentration?

Jul 7, 2016

$\text{7.5 M}$

#### Explanation:

In order to find a solution's molar concentration, or molarity, you need to determine how many moles of solute, which in your case is sodium sulfate, ${\text{Na"_2"SO}}_{4}$, you get in one liter of solution.

That is how molarity was defined -- the number of moles of solute in one liter of solution.

So, you know that you have $0.090$ moles of solute in $\text{12 mL}$ of solution. Your goal here will be to scale up this solution by using this information as a conversion factor to help you determine the number of moles of solute present in

$\text{1 L" = 10^3"mL}$

of solution. Convert the volume from milliliters to liters first

12 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = "0.012 L"

If $\text{0.012 L}$ of solution contain $0.090$ moles of solute, it follows that $\text{1 L}$ will contain

1 color(red)(cancel(color(black)("L solution"))) * ("0.09 moles Na"_2"SO"_4)/(0.012color(red)(cancel(color(black)("L solution")))) = "7.5 moles Na"_2"SO"_4

So, if $\text{1 L}$ of solution contains $7.5$ moles of solute, it follows that the solution's molarity is

"molarity" = color(green)(|bar(ul(color(white)(a/a)color(black)("7.5 mol L"^(-1) = "7.5 molar" = "7.5 M")color(white)(a/a)|)))

The answer is rounded to two sig figs.