Calculate the number of Aluminium ions present in 0.051 gram of alumin

1.	Calculate the number of Aluminium ions present in 0.051 gram of aluminium oxide the mass of iron is same as that of an account of the same element atomic mass of a l is equal to 27 uAnswer this question with full of explanation
Answer» Answer: Explanation: 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 10^23 molecules of Al2O3 Then, 0.051 g of Al2O3 CONTAINS = 6.022 x 10^23 / (10^2 x 0.051 molecules) = 3.011 x 10^20 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 × 10^20 molecules (0.051g) of aluminium oxide (Al2O3) = 2 × 3.011 × 10^20 = 6.022 × 10^20

Calculate the number of Aluminium ions present in 0.051 gram of aluminium oxide the mass of iron is same as that of an account of the same element atomic mass of a l is equal to 27 uAnswer this question with full of explanation

Answer»

Answer:

Explanation:

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 10^23 molecules of Al2O3

Then, 0.051 g of Al2O3 CONTAINS = 6.022 x 10^23 / (10^2 x 0.051 molecules)

= 3.011 x 10^20 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 × 10^20 molecules (0.051g) of aluminium oxide (Al2O3)

= 2 × 3.011 × 10^20 = 6.022 × 10^20