Question

11. Csuculate the number of aluminium ions present in 0.051 g of
alu minium 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:

1 mole of Aluminium Oxide (Al2O3) = 2 × 27 + 3 × 16 = 102 g. ...

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)

Explanation:

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