Question

How many atoms of iodine are in 12.75g of CaI2? Hint. How many Iodines are there in one CaI2 particle?

(Answer) atoms of I (4 SF)

Answer 1

Answer:

$5.225x10^{22}atoms\ I$

Explanation:

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In this case, since 12.75 g of calcium iodide has the following number of moles (molar mass = 293.89 g/mol):

$n_{CaI_2}=12.75gCaI_2*\frac{1molCaI_2}{293.89gCaI_2}=0.0434molCaI_2$

In such a way, since 1 mole of calcium iodide contains 2 moles of atoms of iodine, and one mole of atoms of iodine contains 6.022x10²³ atoms (Avogadro's number), we compute the resulting atoms as shown below:

$atoms\ I=0.0434molCaI_2*\frac{2molI}{1molCaI_2} *\frac{6.022x10^{23}atoms\ I}{1molI} \\\\atoms\ I = 5.225x10^{22}atoms\ I$

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