Problem: A sodium nitrate solution is 12.5 % NaNO3 by mass and has a density of 1.02 g/mL. Calculate the molarity of the solution.

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FREE Expert Solution

Recall:

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

Given:

mass percent = 12.5% NaNO₃ solution

density = 1.02 g/mL

Recall:

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

This means: 12.5 g NaNO₃ g in 100 g solution