Question

The ${\mathrm{NaNO}}_3$ weighed out to make $50\mathrm{mL}$ of an aqueous solution containing$70.0\mathrm{mg}{\mathrm{Na}}^{+}$ per mL is $\mathrm{\_\_\_\_\_\_\_\_}$g.

Answer 1

Step 1:${\mathrm{W}}_{\mathrm{Na}}$ in $50\mathrm{mL}$ solution :

• $$\begin{gathered} 1\mathrm{mL}\mathrm{of}\mathrm{solution}=70\mathrm{mg}\mathrm{of}{\mathrm{Na}}^{+} \\ 50\mathrm{mL}\mathrm{of}\mathrm{solution}=70\mathrm{x}50\mathrm{mg}\mathrm{of}{\mathrm{Na}}^{+} \end{gathered}$$
• ${\mathrm{w}}_{{\mathrm{Na}}^{+}}=70\times 50\mathrm{mg}=3500\mathrm{mg}=3.5\mathrm{g}$

Step 2:Moles of ${\mathrm{Na}}^{+}$in $50\mathrm{mL}$ solution :

• $\frac{\mathrm{given}\mathrm{mass}}{\mathrm{molar}\mathrm{mass}}=\frac{3.5}{23}$

Step 3:Finding mass of ${\mathrm{NaNO}}_3$:

• Moles of ${\mathrm{NaNO}}_3$ = moles of ${\mathrm{Na}}^{+}$ = $\frac{3.5}{23}$ mol
• Mass of ${\mathrm{NaNO}}_3$ =$\frac{3.5}{23}\times 85=12.94\simeq 13\mathrm{g}$