Question

calculate the the no. of aluminium ions present in 0.051 g of aluminium oxide . (Al = 27 u and O= 16 u )

Answer 1

Aluminum oxide has the molecular formula of Al ₂ O 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 ₂ O ₃ = 0.051/101.9648

= 0.0005 mol

As 1 mole Al ₂ O ₃ contains 2 mole Al.

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

= 0.001mol Al

As 1 mole = 6.022 ×10 ²³ ions

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

= 6.022 × 10 ²⁰ ions

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