Question

calculate the number of ions of aluminium in 0.051 gram of aluminium oxide.
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Answer 1

Answer:

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, 0.051 g of Al2O3 contains = 6.022 x 1023 / (102 x 0.051 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 (0.051g) of aluminium oxide (Al2O3)

= 2 × 3.011 × 1020 = 6.022 × 1020

Answer 2

Answer:

Mole of aluminium oxide (A|2 ) is

Rightarrow2*27+3*16

Mole of aluminium oxide e = 102g

i.e., 102 g of AI2O3=6.022*102 molecules of Al2O3

Then, 0.051 g of Al2O3 ins = 6.022x 1023/(102*0.051 molecules)

=3.011*102 molecules of A1203

The number of aluminium ions (A|3 +) present in one molecule of aluminium oxide is 2.

Therefore, the number of aluminium ions (A|3+) present in 3.11*1( molecules (0.051g) of aluminium oxide (A1203)

=2*3.011*1020=6.022*10^ prime

Explanation:

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