How much \(𝑁𝑎𝑁𝑂_{3}\) must be weighed out to make 50 𝑚𝐿 of an aqueous solution containing 70 𝑚𝑔 of \(𝑁𝑎^{ +} per ~𝑚𝐿\)?

Calculating the moles of \(𝑵𝒂^{+}\) in 50 𝒎𝑳 of the solution

In 1 𝑚𝐿 of the solution = 70 𝑚𝑔 of \(𝑁𝑎^{ +}\) is present.

So, in 50 𝑚𝐿 of the solution
\(=\dfrac{70~mg}{1~mL}\times50~mL=3500~mg~of~Na^{+}\) is present.

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

Therefore, in 50 𝑚𝐿 of the solution, \(3.5~ 𝑔 ~of~ 𝑁𝑎 ^{+}\) is present.

Thus, number of moles of \(𝑁𝑎^{ +}\)
\(=\dfrac{Mass ~of~ sodium~ions}{Molar~mass~of~sodium}\)
\(=\dfrac{3.5}{23}=0.152173~mol\)

Calculating the weight of \(𝑵𝒂𝑵𝑶_{𝟑}\) required to make the 𝟓𝟎 𝒎𝑳 solution containing 𝟕𝟎 𝒎𝒈 of \(𝑵𝒂^{ +}\) per 𝒎𝑳

0.152173 𝑚𝑜𝑙 of \(𝑁𝑎^{ +}\) ions are present in 50 𝑚𝐿 of the solution.
1 𝑚𝑜𝑙 of \(𝑁𝑎^{ +}\) ions is obtained from 1 𝑚𝑜𝑙 of \(𝑁𝑎𝑁𝑂_{3}\).

Thus, 0.152173 𝑚𝑜𝑙 of \(𝑁𝑎^{ +}\) ions will be obtained from 0.152173 𝑚𝑜𝑙 of \(𝑁𝑎𝑁𝑂_{3}\). Mass of \(𝑁𝑎𝑁𝑂_{3}\) = 𝑚𝑜𝑙𝑒𝑠 × 𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠s
\(= 0.152173~ 𝑚𝑜𝑙 \times 85~ 𝑔~ 𝑚𝑜𝑙 ^{−1}\)
\(= 12.934\) g

Thus, 12.934 𝑔 of \(𝑁𝑎𝑁𝑂_{3}\) is required to make a 50 𝑚𝐿 solution containing 70 𝑚𝑔 of \(𝑁𝑎^{ +}\) per 𝑚𝐿.

So, option (D) is the correct answer.