Question

How many atoms of Iodine are in a 12.75g sample of CaI2?

Answer 1

Answer is: 5.22·10²² atoms of Iodine.

m(CaI₂) = 12.75 g; mass of calcium iodide.

M(CaI₂) = 293.9 g/mol; molar mass of calcium iodide.

n(CaI₂) = m(CaI₂) ÷ M(CaI₂).

n(CaI₂) = 12.75 g ÷ 293.9 g/mol.

n(CaI₂) = 0.043 mol; amount of calcium iodide.

In one molecule of calcium iodide, there are two iodine atoms

n(I) = 2 · n(CaI₂).

n(I) = 0.086 mol; amount of iodine atoms.

Na = 6.022·10²³ 1/mol; Avogadro number.

N(I) = n(I) · Na.

N(I) = 0.086 mol · 6.022·10²³ 1/mol.

N(I) = 5.22·10²²; number of iodine atoms.

Answer 2

Answer: $5.2\times 10^{22} atoms$ of iodine are present in 12.75 grams of $CaI_2$.

Explanation:

$moles=\frac{\text{given mass of the compound}}{\text{molar mass of the compound}}$

Moles of Calcium iodide :

Moles of $CaI_2=\frac{12.75 g}{293.8 g/mol}=0.0433 moles$

number of molecules of $CaI_2$ in 0.0433 moles = $N_A\times \text{number of moles}=6.022\times 10^{23}\times 0.0433=2.60\times 10^{22} molecules$

In one molecule of calcium iodide there are two iodine atoms, then number of iodine atoms in $2.60\times 10^{22} molecules$ of $CaI_2$

$2\times 2.60\times 10^{22} atoms=5.2\times 10^{22} atoms$

Hence, there are $5.2\times 10^{22} atoms$ of iodine are present in 12.75 grams of $CaI_2$.