If an iodized salt contains \(1 \%\) of \(\mathrm{KI}\) and a person takes \(2 \mathrm{~g}\) of the salt every day, the iodine ions going into his body everyday would be approximately \((\mathrm{K}=39\), \(\mathrm{I}=127\) ) (a) \(7.2 \times 10^{21}\) (b) \(7.2 \times 10^{19}\) (c) \(3.6 \times 10^{21}\) (d) \(9.5 \times 10^{19}\)

Short Answer

Expert verified
Answer: (b) 7.2 x 10^19 ions

Step by step solution

01

Calculate the amount of KI

Given that 1% of the iodized salt is KI, we will multiply the daily consumption (2 g) by the percentage (1%) to find the mass of KI: \(\text{Mass of KI} = 2\, \text{g} \times \frac{1}{100} = 0.02\, \text{g}\)
02

Molar masses of K and I

We are provided with the molar mass of potassium (K) as 39 g/mol and iodine (I) as 127 g/mol.
03

Molar mass of KI

To find the molar mass of potassium iodide (KI), we simply add the molar masses of potassium and iodine: \(\text{Molar mass of KI} = (\text{Molar mass of K}) + (\text{Molar mass of I}) = 39 + 127 = 166\, \text{g/mol}\)
04

Calculate the moles of KI

To find the moles of KI, we divide the mass of KI by its molar mass: \(\text{Moles of KI} = \frac{\text{Mass of KI}}{\text{Molar mass of KI}} = \frac{0.02\, \text{g}}{166\, \text{g/mol}} = 1.2 \times 10^{-4}\, \text{mol}\)
05

Calculate the number of iodine ions

Since each KI molecule has one iodine ion, the number of iodine ions is equal to the number of KI molecules. To find the number, we multiply the moles of KI by Avogadro's constant: $\text{Number of iodine ions} = (\text{Moles of KI}) \times (\text{Avogadro's constant}) = (1.2 \times 10^{-4}\, \text{mol}) \times (6.02 \times 10^{23}\, \text{ions/mol}) = 7.2 \times 10^{19}\, \text{ions}$ The number of iodine ions going into the person's body every day is approximately 7.2 x 10^19 ions, which corresponds to answer choice (b).

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