91Ó°ÊÓ

One mole (abbreviated mol) is equal to $6.022×{10}^{23}$tiny particles of the material in question. As a result, $6.022×{10}^{23}$atoms are referred to as 1 mol l. The $7×{10}^{17}$atoms of iodine in the RDA we've been talking about would be $(7×{10}^{17})/(6.022×{10}^{23})$mol Br or 0.0000012 mol iodine.

To see how many electrons are used we need to write the equation for the reduction of iodine :

$I_2+2e^{-}↤⇒2I^{-}$

To calculate mols of iodine used we can use the following relation:

$\mathrm{n}\left({\mathrm{I}}_2\right)=\frac{\mathrm{I}.\mathrm{t}}{\mathrm{n}.\mathrm{F}}$

whereis the Faraday constant, F= 96485C/mol

By inserting the known data, we get:

$\mathrm{n}\left({\mathrm{l}}_2\right)=\frac{52.6×{10}^{-3}\mathrm{A}.812\mathrm{s}}{2.96485\mathrm{C}/\mathrm{mol}}=2.213×{10}^{-4}\mathrm{mol}$

We can see from the equation in the task question that the mols of iodine are equal to mols of ${\mathrm{H}}_2\mathrm{S}$so the mass concentration ${\mathrm{H}}_2\mathrm{S}$is as follows:

$$\begin{gathered} \mathrm{γ}\left({\mathrm{H}}_2\mathrm{S}\right)=\frac{\mathrm{m}\left({\mathrm{H}}_2\mathrm{S}\right)}{\mathrm{V}\left(\mathrm{solution}\right)}=\frac{\mathrm{n}\left({\mathrm{H}}_2\mathrm{S}\right).\mathrm{M}\left({\mathrm{H}}_2\mathrm{S}\right)}{\mathrm{V}\left(\mathrm{solution}\right)} \\ \mathrm{γ}\left({\mathrm{H}}_2\mathrm{S}\right)=\frac{2.213×{10}^{-10}\mathrm{mol}.34.08\mathrm{g}/\mathrm{mol}}{50\mathrm{mL}}=1.51×{10}^{-4}\mathrm{g}/\mathrm{ml}=151\mathrm{μ²\mu }/\mathrm{ml} \end{gathered}$$