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)

Answer :

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

This is an answered question from Chapter 3. Atoms And Molecules of CBSE Class 9th Science

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