Since the concentration is given in milliequivalents per liter (meq/L), we need to convert this to moles. The valence of $\mathrm{Mg}^{2+}$ is 2, so:
\[
\text{moles of } \mathrm{Mg}^{2+} = \frac{\text{meq/L}}{\text{valence}} = \frac{45.1 \, \mathrm{meq/L}}{2} =
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