Could someone explain q.11 plz with working and solution
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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 × 102
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