Question

Calculate the no. of aluminium atoms [ions]. present in 0.051 gm aluminium chloride. [atomic mass of aluminium =27.

Answer 1

Aluminum chloride has the molecular formula of AlCl₃ .
Atomic mass of aluminum = 26.98 g/mol
Atomic mass of chlorine = 35.5 g/mol
The molecular mass of aluminium chloride is = 26.98 + (3 × 35.5)
= 133.48 g/mol
Number of moles = given mass/ molar mass
Number of moles in 0.051g of AlCl₃ = 0.051/133.48
= 0.00038 mol
As 1 mole AlCl₃ contains 1 mole Al.So, 0.00038 mole of AlCl₃ contains = 0.00038×1 = =0.00038 mol Al
As 1 mole = 6.022 ×10²³ ions
Therefore, 0.00038 mol of AlCl₃ = 6.022 × 10²³ × 0.00038
= 2.29 × 10²⁰ ions
Hence 0.051 g of AlCl₃ contains 2.29 ×10²⁰ ions.