Question

How many atoms of iron (Fe) are in a sample of 7.38 mol Fe? 1.23 x 1023 atoms Fe 5.58 x 1023 atoms Fe 3.37 x 1024 atoms Fe 4.44 x 1024 atoms Fe

Answer 1

Answer: $4.44\times 10^{24}$ atoms of Fe

Explanation: According to Avogadro's law, 1 mole of every substance contains Avogadro's number $6.023\times 10^{23}$ of particles. 1 mole of iron atom contains $6.023\times 10^{23}$ atoms of Fe.

If 1 mole of Fe contains= $6.023\times 10^{23}$ atoms of Fe.

then 7.38 moles of Fe contains= $\frac{6.023\times 10^{23}}{1}\times 7.38=44.4\times 10^{23}or 4.44\times 10^{24}$ atoms of Fe.

Answer 2

The amount of iron (Fe) atoms in a sample of 7.38 moles of iron (Fe) is $4.44$ × $10^{24}$ atoms.

Given the following data:

• Number of moles = 7.38 moles.

To find the amount of iron (Fe) atoms in a sample of 7.38 moles of iron (Fe), we would apply Avogadro's law:

Avogadro's law states that at the same temperature and pressure (STP), equal volumes of all gases have the same number of molecules.

Avogadro's constant = $6.02$ × $10^{23}$

1 mole of iron (Fe) = $6.02$ × $10^{23}$ atoms.

7.38 moles of iron (Fe) = X atoms.

Cross-multiplying, we have:

$X = 7.38$ × $6.02$ × $10^{23}$

X = $4.44$ × $10^{24}$ atoms

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