Question

11. 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

Answer:

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Explanation:

Mole of aluminium oxide (Al2O3) = 2 x 27 + 3 x 16

= 102 g

i.e., 102 g of Al2O3= 6.022 x 1023 molecules of Al2O3

Then, 0.051 g of Al2O3 contains = 6.022 x 1023 / 102 x 0.051 molecules

= 3.011 x 1020 molecules of Al2O3

The number of aluminium ions (Al3+) present in one molecule of aluminium oxide is 2.