Question

Calculate the no. of aluminium ions present in 0.051 gm of Aluminium oxide.

Answer 1

Aluminum oxide has the molecular formula of Al₂O₃ .

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 Al₂O₃ = 0.051/101.9648

= 0.0005 mol

As 1 mole Al₂O₃contains 2 mole Al.

So, 0.0005 mole of Al₂O₃ contains = 0.0005×2

= 0.001mol Al

As 1 mole = 6.022 ×10²³ ions

Therefore, 0.001mol of Al₂O₃ = 6.022 × 10²³ × 0.001

= 6.022 × 10²⁰ ions

Hence 0.051 g of Al₂O₃ contains 6.022×10²⁰ ions.