How much $𝑁𝑎𝑁{𝑂}_3$ must be weighed out to make 50 𝑚𝐿 of an aqueous solution containing 70 𝑚𝑔 of $𝑁{𝑎}^{+}per𝑚𝐿$?

Calculating the moles of $𝑵{𝒂}^{+}$ in 50 𝒎𝑳 of the solution

In 1 𝑚𝐿 of the solution = 70 𝑚𝑔 of $𝑁{𝑎}^{+}$ is present.

So, in 50 𝑚𝐿 of the solution
$={\displaystyle \frac{70mg}{1mL}}\times 50mL=3500mgofN{a}^{+}$ is present.

$\because 3500mg=3.5g(\because 1mg={10}^{-3}g)$

Therefore, in 50 𝑚𝐿 of the solution, $3.5𝑔of𝑁{𝑎}^{+}$ is present.

Thus, number of moles of $𝑁{𝑎}^{+}$
$={\displaystyle \frac{Massofsodiumions}{Molarmassofsodium}}$
$={\displaystyle \frac{3.5}{23}}=0.152173mol$

Calculating the weight of $𝑵𝒂𝑵{𝑶}_{\mathrm{𝟑}}$ required to make the 𝟓𝟎 𝒎𝑳 solution containing 𝟕𝟎 𝒎𝒈 of $𝑵{𝒂}^{+}$ per 𝒎𝑳

0.152173 𝑚𝑜𝑙 of $𝑁{𝑎}^{+}$ ions are present in 50 𝑚𝐿 of the solution.
1 𝑚𝑜𝑙 of $𝑁{𝑎}^{+}$ ions is obtained from 1 𝑚𝑜𝑙 of $𝑁𝑎𝑁{𝑂}_3$.

Thus, 0.152173 𝑚𝑜𝑙 of $𝑁{𝑎}^{+}$ ions will be obtained from 0.152173 𝑚𝑜𝑙 of $𝑁𝑎𝑁{𝑂}_3$. Mass of $𝑁𝑎𝑁{𝑂}_3$ = 𝑚𝑜𝑙𝑒𝑠 × 𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠s
$=0.152173𝑚𝑜𝑙\times 85𝑔𝑚𝑜{𝑙}^{-1}$
$=12.934$ g

Thus, 12.934 𝑔 of $𝑁𝑎𝑁{𝑂}_3$ is required to make a 50 𝑚𝐿 solution containing 70 𝑚𝑔 of $𝑁{𝑎}^{+}$ per 𝑚𝐿.

So, option (D) is the correct answer.