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)
Molar Mass of Al2O3 = 2 x Atomic mass of Al + 3 x Atomic mass of O
= 2 x 27 + 3 x 16
= 54 + 48 = 102 g
102 g of Al2O3 contain = 2 x 6.022 x 1023 aluminium ion
= 0.051 g of Al2O3 = 2 x 6.022 x 1023 / 102 x 0.051
= 6.022 x 1023 x 10-3 = 6.022 x 1020
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Fig. 8.11
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Fig. 8.12
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