Question

Calculate no of alluminium ions present in 0.065 g of alluminium oxide.

Answer 1

Molecular mass of Al₂O₃ = 27 x 2 + 16 x 3 = 102
%age composition of Al in Al₂O₃ =54/102 x 100 = 52.94 %
So, 0.065 g of aluminium oxide will have 52.94 % Aluminium ion = 0.065 x 0.5294 = 0.0344 g
Hence, 27 gm of Al⁺³ ion contains = 1 mole Al⁺³ ions = 6.022 x 1023 ions
Hence, 0.0344 g Al ⁺³ ion will have = (6.022 x 10²³ / 27 ) x 0.0344 = 7.67 x 10²⁰ ions.