Calculate the number of aluminum ions present in 0.051 g of aluminum 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)

1 mole of aluminium oxide (Al₂O₃) = 2 × 27 + 3 × 16 = 102g
i.e., 102g of Al₂O₃ = 6.022 × 10²³ molecules of Al₂O₃
Then, 0.051 g of Al₂O₃ contains = 6.022×10²³/102×0.051 molecules
= 3.011 × 10²⁰ molecules of Al₂O₃
The number of aluminium ions (Al³⁺) present in one molecules of aluminium oxide is 2. Therefore, The number of aluminium ions (Al³⁺) present in
3.11 × 10²⁰ molecules (0.051g) of aluminium oxide (Al2O3) = 2 × 3.011 × 10²⁰
= 6.022 × 10²⁰

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