Answer:
7.03×10²³ atoms of Al
Explanation:
Corundum density = 3.97g/cm³
Corundum = Al₂O₃. Then, you have 2 moles of Al and 3 moles of O in 1 mol of the compound.
Let's determine the mass of corundum, by density
Corundum density = Corundum mass / Corundum volume
3.97 g/cm³ = Corundum mass / 15cm³
3.97 g/cm³ . 15cm³ = Corundum mass →59.55 g
Now, let's find out how many moles are in that mass of corundum
1 mol of Al₂O₃ weighs 101.96 g/mol
Then, 101.96 grams of oxide have 2 moles of Al
59.55 g of oxide, will have (59.55 .2)/101.96 = 1.17 moles
Now, we can know the quantity of atoms, by this rule of three
1 mol has NA atoms (6.02×10²³)
1.17 moles will have (1.17 . NA) = 7.03×10²³ atoms of Al
