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Question
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)
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Solution
1 mole of aluminium oxide (Al2O3) = 2 × 27 + 3 × 16 = 102 g
i.e., 102 g of Al2O3 = 6.022 × 1023 molecules of Al2O3
Then, 0.051 g of Al2O3 contains`=(6.022xx10^(23))/102xx0.051" molecules"`
= 3.011 × 1020 molecules of 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
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