Question

calculate the number of aluminium ions present in 0.102 gram of Aluminium oxide​

Answer 1

Answer:

Explanation:1 mole of Aluminium Oxide (Al2O3) = 2 × 27 + 3 × 16  = 102 g

We know, 102 g of Aluminium Oxide (Al2O3) = 6.022 × 1023 molecules of Aluminium Oxide (Al2O3)

We know, The number of atoms present in given mass = (Given mass ÷ molar mass) × Avogadro number

Then, 0.051 g of Aluminium Oxide (Al2O3) contains

= (0.051 ÷ 102 ) × 6.022 × 1023 molecules of Aluminium Oxide (Al2O3)  

= 3.011 × 1020 molecules of Aluminium Oxide (Al2O3)

The number of Aluminium ions (Al3+) present in one molecule of aluminium oxide is 2.

Therefore, the number of Aluminium ions (Al3+) present in 3.011 × 1020 molecules (0.051 g ) of Aluminium Oxide (Al2O3)

= 2 × 3.011 × 1020

= 6.022 × 1020

Answer 2

Answer:-

1.2044×10²¹ ions

Explanation:-

In this problem, mass of an aluminium ion is same as that of an aluminium atom.

1 mole of Al₂O₃ = Molar mass of Al₂O₃

= 27×2 + 16×3

= 54 + 48

= 102g

1 mole of Al₂O₃ contains 2 moles of Al.

Mass of Al in 1 mole of Al₂O₃ :-

= 27×2 = 54g

Now, we know that 102g of Al₂O₃ contains 54g of Aluminium.

Thus, 0.102g Aluminum oxide will contain:-

= 54/102 × 0.102

= 0.054g Al

• Avogadro number = 6.022×10²³

∵ 27g of Aluminium has 6.022×10²³ ions.

∴ 0.054g of Aluminium will have:-

= [6.022×10²³/27] × 0.054

= 0.012044×10²³

= 1.2044×10²¹ ions

Thus, 1.2044×10²¹ ions are present in 0.102g of Aluminium oxide.