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)

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

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