Question

Please help me :(

How many atoms are present in 179.0 g of iridium?

A. 6.464x10
B. 1.157x10
C. 5696x10
D. 1078 x 10

Answer 1

Answer: There are
`5.696 * 10^(23)` atoms present in 179.0 g of iridium.

Step-by-step 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.

`Moles = (mass)/(molar mass)\\= (179.0 g)/(192.217 g/mol)\\= 0.931 mol`

According to the mole concept, 1 mole of every substance contains
`6.022 * 10^(23)` atoms.

Therefore, atoms present in 0.931 moles are calculated as follows.

`0.931 mol * 6.022 * 10^(23) atoms/mol\\= 5.696 * 10^(23) atoms`

Thus, we can conclude that there are
`5.696 * 10^(23)` atoms present in 179.0 g of iridium.