How many Al atoms are there in 40.5 g of Al foil? Al:27 A) 6.02×10²³
B) 3.01×10²³
C) 9.03×10²³
D) 1.20×10²⁴

c) 9.03 x 10^23

Explanation:

find the molar mass of Al

Al is 27.0 grams

Then use that, to find the number of moles in Aluminum.

Then use Avogadro's number which is 6.02 * 10^23

After that, write all of that down with dimensional analysis.

40.5 g * 1 mol/ 27.0 g of Al  * 6.02 x 10^23 / 1mol

As your final answer, you will get 9.03 * 10^23 atoms with sig figs.

Hope it helped!

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