Question

how much NaNO3 must be weighed out to make 50 cm3 of an aqueous solution containing 70 mgNa+ per cm3

Answer 1

Task:

how much NaNO3 must be weighed out to make 50 cm3 of an aqueous solution containing 70 mgNa+ per cm3

Solution:

There must be 70 mg Na in 1 cm3. In 50 cm3 there will be

$$\begin{array}{l} \mathrm{m}(\mathrm{Na}) = 70 \cdot 50 = 3500 \mathrm{mg} = 3.5 \mathrm{g} \\ \text{The number of moles of Na in } 3.5 \mathrm{g} \text{ is} \\ \mathrm{n}(\mathrm{Na}) = \mathrm{m}(\mathrm{Na}) / \mathrm{MW}(\mathrm{Na}) = 3.5 / 23 = 0.15 \mathrm{mol} \\ \text{The number of moles of NaNO3 is equal to the number of moles of Na} \\ \text{The mass of } \mathrm{NaNO}_{3} \text{ is} \\ \mathrm{m}(\mathrm{NaNO}_{3}) = \mathrm{n}(\mathrm{NaNO}_{3}) \cdot \mathrm{MW}(\mathrm{NaNO}_{3}) = 0.15 \cdot 85 = 12.8 \mathrm{g} \\ \end{array}$$

Answer: $\mathrm{m}(\mathrm{NaNO}_{3}) = 12.8 \mathrm{g}$