Question

Calculate the number of Aluminum ions present in 1368g of Aluminum sulphate. (Given Atomic mass Al = 27u, S = 32u, O = 16u). Formulae of Aluminum sulphate (Al2(SO4)3)

Answer 1

1368 g of aluminum sulphate contains 5.1168 × 10²⁴ ions of aluminium.

1) Molar weight of Al2(SO4)3 = (2×27u) + (3×32u) + (4×3×16u) = 34u + 96u + 192u = 322u = 322g

2) Now, 322g of aluminum sulphate contains Avogadro's number of molecules, i.e., 6.022 × 10²³ molecules.

3) Therefore, 1368 g aluminum sulphate contains (1368 g / 322 g) × 6.022 × 10²³ molecules = 25.584 × 10²³ molecules.

4) From the formula, it is evident that one molecule of Al2(SO4)3 contains 2 ions of Aluminum. Therefore, 25.584 × 10²³ molecules of aluminum sulphate will contain 2 × 25.584 × 10²³ ions = 51.168 × 10²³ ions = 5.1168 × 10²⁴ ions.