Problem

A laboratory procedure calls for making 410.0 mL of a 1.2 M NaNO_{3} solution.

What mass of NaNO_{3} (in g) is needed?

Solution

We’re being asked to **calculate the mass of NaNO _{3}**

**To do so, we shall do these steps:**

*Step 1*: Calculate the moles of NaNO_{3}.

*Step 2*: Calculate the mass of NaNO_{3}.

Recall that ** molarity** is the ratio of the moles of solute and the volume of solution (in liters).

In other words:

$\overline{){\mathbf{Molarity}}{\mathbf{\left(}}{\mathbf{M}}{\mathbf{\right)}}{\mathbf{=}}\frac{\mathbf{moles}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solute}}{\mathbf{Liters}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solution}}}$

Molarity

→ Molarity