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)

To calculate the number of aluminum ions present in 0.051 grams of aluminum oxide (Al₂O₃), follow these steps:

1. Determine the molar mass of aluminum oxide (Al₂O₃):
   • The atomic mass of aluminum (Al) is 27 g/mol.
   • The atomic mass of oxygen (O) is 16 g/mol.
   • Molar mass of Al₂O₃ =2×27 g/mol+3×16 g/mol=54 g/mol+48 g/mol=102 g/mol
2. Calculate the number of moles of Al₂O₃ in 0.051 grams:
   • The number of moles (n) is given by the formula: n=mass/molar mass​
   • For 0.051 grams of Al₂O₃: n=0.051 g/102 g/mol≈0.0005 moles
3. Determine the number of aluminum ions in Al₂O₃:
   • Each formula unit of Al₂O₃ contains 2 aluminum ions (Al³⁺).
   • Therefore, the number of moles of aluminum ions is twice the number of moles of Al₂O₃: Moles of Al ions=2×0.0005 moles=0.001 moles
4. Calculate the number of aluminum ions using Avogadro’s number:
   • Avogadro’s number (NAN​) is 6.022 × 10²³ entities/mol
   • The number of aluminum ions (N) is given by: N=moles of Al ions× NA
   • Substituting the values: N=0.001 moles × 6.022× 10²³ions/mol=6.022 × 10²⁰ ions

Therefore, the number of aluminum ions present in 0.051 grams of aluminum oxide is 6.022 × 10²⁰ ions

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