Question

If 1 mole of aluminum contains 6.02 ×10^23 atoms aluminum. how many atoms are contained in 0.9g of aluminum (Al=27)​

Answer 1

Step-by-step explanation:

Hi there!

We have,

1 mole of Al = 6.02*
`10^(23)` atoms

1 mole of Al = 27 g

Now, from the above relation;

27 g = 6.02*
`10^(23)` atoms

1 g =
`(6.02*10^(23) )/(27)` atoms

or, 0.9 g of Al =
`(6.02*10^(23) )/(27)`*0.9 atoms

= 2.0066*
`10^(22)` atoms

Therefore, 0.9 g of Al contains 2.0066*
`10^(22)` atoms.

Hope it helps!