Question
A prescription order calls for a $500 \mathrm{~mL}$ solution of $\mathrm{NaCl}$ containing $10 \mathrm{mEq}$ of $\mathrm{Na}^{+}$. How many milligrams of $\mathrm{NaCl}$ are required?
Step 1
The molecular weight of NaCl is calculated by adding the atomic weights of sodium (Na) and chlorine (Cl). The atomic weight of Na is 23 g/mol and the atomic weight of Cl is 35.5 g/mol. Therefore, the molecular weight of NaCl is 23 + 35.5 = 58.5 g/mol. Show more…
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