Question

calculate the number of aluminum ions present in 0.5g of aluminum oxide.​

Answer 1

Answer:

n = N / N O.

Given mass of Aluminium ion (Al 2 O 3 )= m = 0.051g.

N = 6.022 x 10 23 x (0.051 / 102)

= 6.022 x 10 23 x (51 x 10 -3 ) / 102.

Thus, 6.022 x 10 20 Al 3+ ions are present in 0.051 molecules of Al 2 O 3

Answer 2

Mole of aluminium oxide (Al2O3) is

⇒ 2 x 27 + 3 x 16

Mole of aluminium oxide = 102 g

i.e., 102 g of Al2O3= 6.022 x 1023 molecules of Al2O3

Then, 0.05 g of Al2O3 contains = 6.022 x 1023 / (102 x 0.05 molecules)

= 3.011 x 1020 molecules of Al2O3

The number of aluminium ions (Al3+) present in one molecule of aluminium oxide is 2.

Therefore, the number of aluminium ions (Al3+) present in 3.11 × 1020 molecules (0.05g) of aluminium oxide (Al2O3)

= 2 × 3.011 × 1020

= 6.022 × 1020