Question

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

Answer 1

Answer:
`5.2* 10^(22) atoms` of iodine are present in 12.75 grams of
`CaI_2`.

Step-by-step explanation:

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

Moles of Calcium iodide :

Moles of
`CaI_2=(12.75 g)/(293.8 g/mol)=0.0433 moles`

number of molecules of
`CaI_2` in 0.0433 moles =
`N_A* \text{number of moles}=6.022* 10^(23)* 0.0433=2.60* 10^(22) molecules`

In one molecule of calcium iodide there are two iodine atoms, then number of iodine atoms in
`2.60* 10^(22) molecules` of
`CaI_2`

`2* 2.60* 10^(22) atoms=5.2* 10^(22) atoms`

Hence, there are
`5.2* 10^(22) atoms` of iodine are present in 12.75 grams of
`CaI_2`.

Answer 2

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.