## Friday 5 January 2018

### Calculate the number of aluminium ions present in 0.051 g of aluminium oxide. (Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u)

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

### Check for Complete Exercise Solution for Chapter 3. Atoms And Molecules | CBSE Class 9th Science

1. Why the no of aluminium ions more than total

1. Because avagardo number : 6.022*10²³. The number of alluminium molecule is 6.022*10²⁰. Subtract from avagrado number the left over would be the number of oxygen atoms and it will be same as 3.011*10²⁰. Ok

2. it's about numbers of atom if there Al2O3 consist of 3.011×10^20( it is only about molecukes) if this is molecules then number of atom will be more that is why it is ×2

2. Thanks sir

3. thanks a lot sir

4. Yes how is it more than whole oxide

5. the solution is totally wrong

6. Thank you be