Question

How many atoms are present in 179.0 g of iridium?

Answer 1

Avogadro's number represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units/mole. This number can be used to convert the number of atoms or molecules into a number of moles. We do as follows:
 179.0 g of iridium (1 mol / 192.217 g) ( 6.022 x 10^23 atoms /  1 mol ) = 5.61 x 10^23 atoms of iridium 
Hope this answers the question.

Answer 2

Based on how to calculate number of atoms, the number of atoms of Iridium in 179.0 g sample is 5.59 × 10^23 atoms.

How many atoms are present in 179.0 g of iridium?

The number of atoms is determined using the formula below:

• Number of atoms = number of moles × 6.02 × 10^23
• Number of moles = mass / molar mass

molar mass of Iridium = 192.22

mass of Iridium = 179.0

number of moles = 179/192.22

number of moles = 0.93 moles of Iridium

Then:

Number of atoms = 0.93 × 6 × 10^23 atoms

Number of atoms = 5.59 × 10^23 atoms

Therefore, the number of atoms of Iridium in 179.0 g sample is 5.59 × 10^23 atoms.

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