Question

C calculate the the number of Aluminium ions present in 0.051 gram of Aluminium oxide

Answer 1

Answer:

6.022 × 1020

Explanation:

1 mole of Aluminium Oxide (Al2O3) = 2 × 27 + 3 × 16  = 102 g

102 g of Aluminium Oxide (Al2O3) = 6.022 × 1023 molecules of Aluminium Oxide (Al2O3)

= (0.051 ÷ 102 ) × 6.022 × 1023 molecules of Aluminium Oxide (Al2O3)  

= 3.011 × 1020 molecules of Aluminium Oxide (Al2O3)

= 2 × 3.011 × 1020

= 6.022 × 10 raised to 20