Question

Calculate the number of aluminium ions present in 0.051g of aluminum oxide by brainly

Answer 1

The number of aluminum ions present in 0.051 g of aluminum oxide is 6 × 10²⁰ .

• The formula for aluminum oxide is Al₂O₃.

• The molar weight of Al₂O₃ is (2 × 27 g) + (3 × 16 g) = 54 g + 48 g = 102 g

• We know that, according to Avogadro's law, one mole of Al₂O₃ contains 6.022 × 10²³ molecules of Al₂O₃.

=> 102 g of Al₂O₃ conatins 6.022 × 10²³ molecules of Al₂O₃

• Therefore, 0.051 g of Al₂O₃ will contain (6.022 × 10²³ / 102 ) × 0.051 number of molecules.

Or, 0.051 g of Al₂O₃ contains (6.022 × 10²³ × 0.051 ) / 102 number of molecules

Or, 0.051 g of Al₂O₃ contains 0.003 × 10²³  number of molecules

• From the formula, we know that one molecule of Al₂O₃ contains 2 ions of aluminum.

• Therefore, 0.003 × 10²³  molecules of Al₂O₃ will contain 2 × 0.003 × 10²³ ions of aluminum.

Or, 0.051 g of Al₂O₃ contains 0.006 × 10²³ ions of aluminum.

Or, 0.051 g of Al₂O₃ contains 6 × 10²⁰  ions of aluminum.