**Recall:**

$\overline{){\mathbf{Molarity}}{\mathbf{\hspace{0.17em}}}{\mathbf{\left(}}{\mathbf{M}}{\mathbf{\right)}}{\mathbf{=}}\frac{\mathbf{moles}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solute}}{\mathbf{Liters}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solution}}}$

**Given:**

mass percent = **12.5% NaNO _{3} solution**

density = **1.02 g/mL**

**Recall:**

$\overline{){\mathbf{\%}}{\mathbf{\hspace{0.17em}}}{\mathbf{Mass}}{\mathbf{=}}\frac{\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{NaNO}}_{\mathbf{3}}}{\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solution}}{\mathbf{\times}}{\mathbf{100}}}$

This means: **12.5 g NaNO _{3} g** in

A sodium nitrate solution is 12.5 % NaNO_{3} by mass and has a density of 1.02 g/mL. Calculate the molarity of the solution.

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