calculate the number of aluminum ions present in 0.051 g of aluminum oxide (al2o3) (atomic mass : all =27 u and o =16 u )

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Solution

Aluminum oxide has the molecular formula of Al ₂O ₃.

Atomic mass of aluminum = 26.98 g/mol

Atomic mass of oxygen = 16 g/mol

The molecular mass of aluminium oxide is = (2 × 26.98) + (3 ×16)

= 101.9648 g/mol

Number of moles = given mass/ molar mass

Number of moles in 0.051g of Al ₂O ₃= 0.051/101.9648

= 0.0005 mol

As 1 mole Al ₂O ₃contains 2 mole Al.

So, 0.0005 mole of Al ₂O ₃contains = 0.0005×2

= 0.001mol Al

As 1 mole = 6.022 ×10 ²³ions

Therefore, 0.001mol of Al ₂O ₃= 6.022 × 10 ²³× 0.001

= 6.022 × 10 ²⁰ions

Hence 0.051 g of Al ₂O ₃contains 6.022×10 ²⁰ions.

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