Answer: There are [tex]5.696 \times 10^{23}[/tex] atoms present in 179.0 g of iridium.
Explanation:
Given: Mass = 179.0 g
Moles is the mass of a substance divided by its molar mass. So, moles of iridium (molar mass = 192.217 g/mol) is as follows.
[tex]Moles = \frac{mass}{molar mass}\\= \frac{179.0 g}{192.217 g/mol}\\= 0.931 mol[/tex]
According to the mole concept, 1 mole of every substance contains [tex]6.022 \times 10^{23}[/tex] atoms.
Therefore, atoms present in 0.931 moles are calculated as follows.
[tex]0.931 mol \times 6.022 \times 10^{23} atoms/mol\\= 5.696 \times 10^{23} atoms[/tex]
Thus, we can conclude that there are [tex]5.696 \times 10^{23}[/tex] atoms present in 179.0 g of iridium.
