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$\overline{)\mathbf{mass}\mathbf{}\mathbf{percent}\mathbf{=}\frac{\mathbf{mass}\mathbf{}\mathbf{element}}{\mathbf{total}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{compound}}\mathbf{\times}\mathbf{100}}$

iodine RDA = 150 µg/day

mass percent iodine (I) = 76.45% I

Calculate mass KI:

$\mathbf{mass}\mathbf{}\mathbf{KI}\mathbf{=}\frac{\mathbf{mass}\mathbf{}\mathbf{element}}{\mathbf{mass}\mathbf{}\mathbf{percent}\mathbf{}\mathbf{I}}\mathbf{\times}\mathbf{100}\phantom{\rule{0ex}{0ex}}\mathbf{mass}\mathbf{}\mathbf{KI}\mathbf{=}\frac{\mathbf{150}\mathbf{}{\displaystyle \raisebox{1ex}{$\mathbf{\mu g}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{day}$}\right.}}{\mathbf{76}\mathbf{.}\mathbf{45}\mathbf{\%}}\mathbf{\times}\mathbf{100}$

The iodide ion is a dietary mineral essential to good nutrition. In countries where potassium iodide is added to salt, iodine deficiency or goiter has been almost completely eliminated. The recommended daily allowance (RDA) for iodine is 150 µg/day. How much potassium iodide (76.45% I) should be consumed to meet the RDA?

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Based on our data, we think this problem is relevant for Professor Person's class at CSU.