# What is the molarity of the solution produced when "350.0 mL" of water and "75.0 mL" of "0.125 M" "NaNO"_3 are mixed an boiled down to "325.0 mL" ?

Feb 12, 2018

$\text{0.0288 M}$

#### Explanation:

The trick here is to realize that the number of moles of sodium nitrate present in the solution does not change.

When you dilute the solution by adding the $\text{350.0 mL}$ of water, the volume of the solution increases, but the number of moles of sodium nitrate remains unchanged.

Similarly, when you boil the solution, you decrease its volume to $\text{325.0 mL}$, but once again, the number of moles of sodium nitrate remains unchanged.

Basically, by diluting and boiling the solution, you're changing its volume without affecting the number of moles of solute it contains.

So, you know that the initial solution contains

75.0 color(red)(cancel(color(black)("mL solution"))) * "0.125 moles NaNO"_3/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.009375 moles NaNO"_3

This means that the final solution will contain $0.009375$ moles of sodium nitrate in a total volume of $\text{325.0 mL}$, which implies that the molarity of the solution will be--do not forget to convert the volume of the solution to liters!

["NaNO"_3] = "0.009375 moles"/(325.0 * 10^(-3) quad "L") = color(darkgreen)(ul(color(black)("0.0288 M")))

The answer is rounded to three sig figs.