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Answer the following 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)
(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 aluminium oxide (Al_2O_3).
= (2 x 27) + (3 x 16).
= 54 + 48 = .
= (2 x 27) + (3 x 16).
= 54 + 48 = .
Therefore,
102 g of Al_2O_3 contains = 2 x 6.022 x 10^2{^3} aluminium ions
0.051 g of Al_2O_3 contains = 2 x 6.022 x 10^2{^3} / 102 x 0.051
= x 10^2{^3} x 0.051 / 102
= 0.614 x 10^2{^3} /
= 0.006022 x 10^2{^3}
= x 10^2{^0} Al^3{^+} ions.
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